STATIC AND DYNAMIC ELECTRICITY
ELECTROMAGNETICS LIBRARY C. E. BAUM, Editor
Baum and Kritikos Electromagnetic Symmetry Baum and Stone Transient Lens Synthesis: Differential Geometry in Electromagnetic Theory Bowman, Senior, and Uslenghi Electromagnetic and Acoustic Scattering by Simple Shapes
Gardner Lightning Electromagnetics Hoppe and Rahmat Samii Impedance Boundary Conditions in Electromagnetics -
Lee EMP Interaction: Principles, Techniques, and Reference Data
Mittra Computer Techniques for Electromagnetics Smythe Static and Dynamic Electricity
Taylor and Giri High-Power Microwave Systems and Effects Van Bladel Electromagnetic Fields
STATIC AND DYNAMIC ELECTRICITY Third Edition, Revised Printing
WILLIAM R. SMYTHE Professor Emeritus of Physics California Institute of Technology
A SUMMA BOOK
Taylor &Francis Publishers since 1798
STATIC AND DYNAMIC ELECTRICITY: Third Edition, Revised Printing Copyright © 1989 by Hemisphere Publishing Corporation. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 234567890 BRBR 898765 Library of Congress Cataloging in Publication Data -
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Smythe, William Ralph, date. Static and dynamic electricity. (A SUMMA book) Includes bibliographies and index. 1. Electrostatics. 2. Electrodynamics. I. Title. QC571.S59 1989 88-24722 537'.2 ISBN 0-89116-916-4 (hard) ISBN 0-89116-917-2 (soft)
CONTENTS
TABLE OF SYMBOLS
xi
PREFACE ID THE THIRD EDITION, REVISED PRINTING
xvii
PREFACE TO THE THIRD EDITION
xix
PREFACE TO THE SECOND EDITION
xxi
PREFACE ID THE FIRST EDITION
xxiii
CHAPTER I BASIC IDEAS OF ELECTROSTATICS
1
Electrification, conductors, and insulators—Positive and negative electricity — Coulomb's law, unit charge, dielectrics—Limitations of the inverse-square law—Electrical induction—The elementary electric charges—Electric field intensity—Electrostatic potential—Electric dipoles and multipoles—Interaction of dipoles—Lines of force—Equipotential surfaces—Gauss's electric flux theorem — Lines of force from collinear charges—Lines of force at infinity—Potential maxima and minima. Earnshaw's theorem—Potential of electric double layer — Electric displacement and tubes of force—Stresses in an electric field— Gauss's electric flux theorem for nonhomogeneous mediums—Boundary conditions and stresses on the surface of conductors—Boundary conditions and stresses on the surface of a dielectric—Displacement and intensity in solid dielectric—Crystalline dielectrics—Problems—References.
CHAPTER II CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
25
Uniqueness theorem—Capacitance—Capacitors in series and parallel— Spherical capacitors—Cylindrical capacitors—Parallel-plate capacitors— Guard rings—Energy of a charged capacitor—Energy in an electric fieldParallel-plate capacitor with crystalline dielectric—Stresses when the capacitivity is a function of density—Electrostriction in liquid dielectrics—Force on conductor in dielectric—Green's reciprocation theorem—Superposition of fields—Induced charges on earthed conductors—Self- and mutual elastanceSelf- and mutual capacitance—Electric screening—Elastances and capacitances for two distance conductors—Energy of a charged system—Forces and torques on charged conductors—Problems—References.
CHAPTER III GENERAL THEOREMS
Gauss's theorem—Stokes' s theorem—Equations of Poisson and Laplace— Orthogonal curvilinear coordinates—Curl in orthogonal curvilinear coordinates v
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CONTENTS — 4 (eVV) in other coordinate systems—Green's theorems—Green's reciprocation theorem for dielectrics—Green's function—Solution of Poisson's equation —Uniqueness theorem with dielectrics present—Introduction of new conductor —Green's equivalent stratum—Energy of a dielectric body in an electric field— Effect of an increase of capacitivity—Potential of axially symmetrical fieldProblems—References.
CHAPTER IV TWO—DIMENSIONAL POTENTIAL DISTRIBUTIONS
63
Field and potential in two dimensions—Circular harmonics—Harmonic expansion of line charge potential—Conducting or dielectric cylinder in uniform field—Dielectric cylinder. Method of images—Image in conducting cylinder—Image in plane face of dielectric or conductor. Intersecting conducting planes—Dielectric wedge—Complex quantities—Conjugate functions — The stream function—Electric field intensity. Electric flux—Functions for a line charge—Capacitance between two circular cylinders—Capacitance between cylinder and plane and between two similar cylinders—Conformal transformations—Given equations of boundary in parametric form—Determination of required conjugate functions—The Schwarz transformation—Polygons with one positive angle—Polygon with angle zero—Polygon with one negative angle. Doublet. Inversion—Images by two-dimensional inversion—Polygon with two angles—Slotted plane—Riemann surfaces—Circular cylinder into elliptic cylinder—Dielectric boundary conditions—Elliptic dielectric cylinder—Torque on dielectric cylinder—Polygon with rounded corner—Plane grating of large cylindrical wires—Elliptic function transformations. Two coplanar strips— Unequal coplanar strips by inversion—Charged circular cylinder between plates —Problems—References.
CHAPTER
V
THREE—DIMENSIONAL POTENTIAL DISTRIBUTIONS
When can a set of surfaces be equipotentials2—Potentials for confocal conicoids —Charged conducting ellipsoid—Elliptic and circular disks—Method of images. Conducting planes—Plane boundary between dielectrics—Image in spherical conductor—Example of images of point charge—Infinite set of images. Two spheres—Difference equations. Two spheres—Sphere and plane and two equal spheres—Inversion in three dimensions. Geometrical properties—Inverse of potential and image systems—Example of inversion of images—Inversion of charged conducting surface—Capacitance by inversion— Three-dimensional harmonics—Surface of revolution and orthogonal wedge-Spherical harmonics—General property of surface harmonics—Potential of harmonic charge distribution—Differential equations for surface harmonics- Surface zonal harmonics. Legendre's equation—Series solution of Legendre's equation—Legendre polynomidls. Rodrigues's formula—Legendre coefficients. Inverse distance—Recurrence formulas for Legendre polynomials— Integral of product of Legendre polynomials—Expansion of function in Legendre polynomials—Table of Legendre polynomials—Legendre polynomial with imaginary variable—Potential of charged ring—Charged ring in conducting sphere—Dielectric shell in uniform field—Off-center spherical capacitor-
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CONTENTS
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Simple conical boundary—Zonal harmonics of the second kind—Recurrence formulas for Legendre functions of the second kind—Legendre functions of the second kind in terms of Legendre polynomials—Special values of Legendre functions of the second kind—Legendre function of the second kind with imaginary variable—Use of Legendre function of the second kind in potential problems—Nonintegral zonal harmonics—Associated Legendre functions—Integrals of products of associated functions—Associated functions with imaginary argument—Recurrence formulas for Legendre associated functions—Special values of associated Legendre functions—Neutral points and lines—Biaxial harmonics—Conical boundaries—Nonintegral associated Legendre functions-Green's function for a cone—Green's function for a conical box—Oblate spheroidal coordinates—Oblate spheroidal harmonics—Conducting sheet with circular hole—Torque on disk in uniform field—Potential of charge distribution on spheroid—Potential of point charge in oblate spheroidal harmonics— Prolate spheroidal harmonics—Prolate spheroid in uniform field—Laplace's equation in cylindrical coordinates—Bessel's equation and Bessel functions —Modified Bessel equation and functions—Solution of Bessel's equation— Recurrence formulas for Bessel functions—Values of Bessel functions at infinity—Integrals of Bessel functions—Expansion in series of Bessel functions—Green's function for cylinder. Inverse distance—Green's function for cylindrical box—Bessel functions of zero order—Roots and numerical values of Bessel functions of zero order—Derivatives and integrals of Bessel functions of zero order—Point charge and dielectric plate—Potential inside hollow cylindrical ring—Nonintegral order and spherical Bessel functions—Modified Bessel functions—Recurrence formulas for modified Bessel functions—Values of modified Bessel functions at infinity—Integral of a product of complex modified Bessel functions—Green's function for a hollow cylindrical ring—Modified Bessel functions of zero order—Definite integrals for the modified Bessel function of the second kind. Value at infinity—Definite integrals for Bessel functions of zero order—Inverse distance in terms of modified Bessel functionsCylinarical dielectric boundaries—Potential inside hollow cylindrical ring— Modified Bessel functions of nonintegral order—Wedge functions—Mixed boundaries. Charged right circular cylinder—Mixed spherical, spheroidal, and cylindrical boundaries—Sphere in concentric, coaxial, earthed spheroid — Charged sphere in cylinder—Cylinder in hole in sheet—Off-axis charged sphere in cylinder—Problems—References. ,
CHAPTER VI ELECTRIC CURRENT
Electric current density. Equation of continuity—Electromotance—Ohm's law. Resistivity—Heating effect of electric current—Steady currents in extended mediums—General theorems—Current flow in two dimensions—Long strip with abrupt change in width—Current flow in three dimensions—Systems of electrodes. Two spheres. Distant electrodes—Source and sink in solid sphere. Spherical bubble—Solid conducting cylinder—Earth resistance — Currents in thin curved sheets—Current distribution on spherical shell— Surface of revolution—Limits of resistance—Currents in nonisotropic mediums. Earth strata—Vector potential. Flow around sphere in tube—Spacecharge current. Child's equation—Problems—References.
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CONTENTS
CHAPTER VII MAGNETIC INTERACTION OF CURRENTS
280
Definition of the ampere in terms of the magnetic moment—Magnetic induction and permeability—Magnetic vector potential. Uniform field—Uniqueness theorems for magnetostatics—Orthogonal expansions for vector potential— Vector potential in cylindrical coordinates—Vector potential in spherical coordinates—Vector potential in terms of magnetic induction on axis—Equation of axially symmetrical tubes of induction—Vector potential and field of bifilar circuit—Vector potential and field of circular loop—Field of currents in spherical shell—Zonal currents in spherical shell—Field of circular loop in spherical harmonics—Biot and Savart's law. Field of straight wire—Field of helical solenoid—Field in cylindrical hole in conducting rod—Field of rectilinear currents in cylindrical conducting shell—Force on electric circuit in magnetic field—Examples of forces between electric circuits—Vector potential and magnetization—Magnetic boundary conditions—Example of the use of A and A— Current images in plane face—Magnetic induction and permeability in crystals —Two-dimensional magnetic fields—Magnetic shielding of bifilar circuit— Current images in two dimensions—Magnetomotance and magnetic intensity— The magnetic circuit. Anchor ring—Air gaps in magnetic circuits—Field in shell-type transformer—Slotted pole piece. Effective air gap—ProblemsReferences.
CHAPTER VIII ELECTROMAGNETIC INDUCTION
329
Faraday's law of induction—Mutual energy of two circuits—Energy in a magnetic field—Mutual inductance—Boundary conditions on A—Mutual inductance of simple circuits—Mutual inductance of circular loops—Variable mutual inductance—Self-inductance—Computation of self-inductance. Thin wire—Self-inductance of circular loop—Self-inductance of solenoid—Selfinductance of bifilar lead—Energy of n circuits—Stresses in a magnetic field— Energy of a permeable body in a magnetostatic field—Problems—References.
CHAPTER IX
349
MAGNETISM
Paramagnetism and diamagnetism—Magnetic susceptibility—Magnetic properties of crystals—Crystalline sphere in uniform magnetic field—Ferromagnetism—Hysteresis. Permanent magnetism—The nature of permanent magnetism—Uniform magnetization. Equivalent current shell—Magnetized sphere and cylinder. Magnetic poles—Boundary conditions on permanent magnets—Spherical permanent magnet in uniform field—Lifting power of horseshoe magnet—Field of cylindrical magnet—Magnetic needles—Problems —References.
CHAPTER X EDDY CURRENTS
Induced currents in extended conductors—Solution for vector potential of eddy currents—Steady-state skin e f fect—Skin effect on tubular conductor-
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CONTENTS
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Skin ef fect on solid cylindrical conductor—Solution in spherical coordinates for axial symmetry—Conducting sphere in alternating field—Power absorbed by sphere in alternating magnetic field—Transients in conducting sphere— Eddy currents in plane sheets—Eddy currents in infinite plane sheet by image method—Torque on small rotating current loop or magnetic dipole—Eddy currents from rotating dipole—Shielding of circular coil by thin conducting sheet— Rotating sheet in magnetic gap—Rotating disk in magnetic gap—Zonal eddy currents in spherical shell—Spherical shell in alternating field solenoid gap— General eddy currents in spherical shell—Torque on spinning spherical shell between magnet poles—Eddy currents in thin cylindrical shell—Eddy currents in rotating finite cylindrical shell—Eddy current damping—Transient shielding by a thick cylindrical shell—Problems—References.
CHAPTER XI PLANE ELECTROMAGNETIC WAVES
415
Maxwell's field equations—Propagation equation. Dynamic potentials. Gauges. Hertz vector—Poynting's vector—Plane waves in homogeneous uncharged dielectric insulator—Plane wave velocity in anisotropic mediums— Ray surface and polarization in anisotropic mediums—Energy, pressure, and momentum of a plane wave—Refraction and reflection of a plane wave— Intensity of reflected and refracted waves—Frequency, wave length, elliptic polarization—Total reflection—Electromagnetic waves in homogeneous conductors—Plane waves in homogeneous isotropic conductors—Reflection from conducting surface—Plane waves on cylindrical perfect conductors—Intrinsic impedance of a medium—Reflection at a discontinuity. Matching section— Complex Poynting vector—Nearly plane waves on imperfect conductors. Lecher wires—Group velocity—Problems—References.
CHAPTER XII ELECTROMAGNETIC RADIATION
448
The radiation problem—Spherical electromagnetic waves. Dipole and quadrupole radiation—Retarded Potentials—Radiation from linear antenna— Distant radiation from linear antenna—Radiation from progressive waves— Conical transmission lines—The biconical antenna—Antenna arrays—Earth effects—Uniqueness of solution—Solutions of the wave equation in spherical coordinates—Polynomial expansion for a plane wave—Radiation from uniform current loop. Magnetic dipole—Free oscillations of a conducting sphere —Forced oscillations of dielectric or conducting sphere—Solution of propagation equation in cylindrical coordinates—Expansion in cylindrical harmonics for a plane wave—Radiation from apertures in plane conducting screens— Diffraction from rectangular aperture in conducting plane—Orthogonal functions in diffraction problems. Coaxial line—Problems—References.
CHAPTER XIII WAVE GUIDES AND CAVITY RESONATORS
Waves in hollow cylindrical tubes—Attenuation in hollow wave guides—The rectangular wave guide—Green's function for rectangular guide—Aperture excitation of wave guides. Coaxial opening—Rectangular guide filled by two
493
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CONTENTS mediums—Thin iris in rectangular guide—A variational method for improving approximate solutions—Inductive iris susceptance by variational method— The circular wave guide=-Green's function for a circular guide—Loop coupling with circular guide—Orifice coupling with circular guide—The coaxial wave guide—Plane discontinuities in coaxial lines—Guides of arbitrary section. Elliptic guides—Cavity resonators. Normal modes—Independent oscillation modes of a cavity—Energy and damping of normal modes. Quality—Normal modes of a cylindrical cavity—Properties of a rectangular cavity—Properties of a right circular cylindrical cavity—Multiply connected cylindrical cavities— Coaxial cable resonators—Normal modes of a spherical cavity—Normal modes of an imperfect cavity—Effect of small foreign body on cavity mode frequency— Wall deformation effect on mode frequencies—Stationary formulas for cavity resonance frequency—Complex cavities—Excitation of cavities. Inductive coupling—Inductive coupling to a circular cylindrical cavity—Excitation of cavity by internal electrode—Cavity excitation through orifice—ProblemsReferences. CHAPTER XIV
SPECIAL RELATIVITY AND THE MOTION OF CHARGED PARTICLES
560
The postulates of special relativity—The Lorentz transformation equations— Transformation equations for velocity and acceleration—Variation of mass with velocity—The transformation equations for force—Force on charge moving in magnetic field—Motion of charge in uniform magnetic field—Energy of a charged moving particle—Magnetic cutoff of thermionic rectifier—Path of cosmic particle in uniform field—Magnetic field of moving charge—Retarded fields and potentials of moving charge—Radiation from an accelerated charge— Charge radiation when acceleration parallels velocity—Charge radiation with acceleration at right angles to velocity—Cherenkov radiation—Transformation of Maxwell's equations—Ground speed of an airplane—Motion of charged particle in crossed electric and magnetic fields—Aberration and Doppler effectProblems—References.
APPENDIX SYSTEMS OF ELECTRICAL UNITS
591
TABLE I. RELATIONS BETWEEN CGS AND MKS MECHANICAL UNITS
592
TABLE II. TRANSPOSITION OF MKS FORMULAS INTO CGS ESU
592
TABLE III. TRANSPOSITION OF MKS FORMULAS INTO CGS EMU
593
TABLE IV. TRANSPOSITION OF CGS ESU OR CGS EMU FORMULAS INTO MKS
594
TABLE V. DIMENSIONS OF ELECTRIC AND MAGNETIC QUANTITIES
596
TABLE VI. NUMERICAL VALUES
597
INDEX
599
TABLE OF SYMBOLS
Note: In this table bold-face symbols (v, u, 4), . . .) are space vectors. Phasors (I, E, . . .), phasor space vectors (E, B, II, . . .), conjugate phasors (I, E, . . .) and conjugate phasor space vectors (E, B, II, . . .) are shown by an erect ( or an inverted (^) flat vee above the symbol. Magnitudes of vectors and scalars, whether time-dependent or not, are written without designation.
A, Ao, Az, etc. Vector potential. A° Normalized vector potential. A, A., etc. Quasi-vector potential B, Bz, etc. Magnetic induction B Susceptance. B° Normalized or relative susceptance, BZk . C Capacitance. A constant. C° Normalized or relative capacitance, CZk. Velocity of light. A length. Self-capacitance. Cnn Cmn Mutual capacitance. D, Dz, etc. Electric displacement. Dw Dwight integral tables. ds Differential element along s. dr Differential change in r. E, E, E, E etc. Electric field intensity. E(k) Complete elliptic integral. e Electronic charge. 2.71828. e, 6, E, E, etc. Electromotance. se Effective or rms electromotance. F, Fz Force G Conductance, Y = G jB. Acceleration of gravity. H, H, H, H, etc. Magnetic field intensity. H(1), 11(:) (v), H.;,!)(v) Hankel functions. h Planck's constant. 4112, h3 In orthogonal curvilinear coordinates. Length elements are hidui, h2du2, hadu3. :), !IT )(v Spherical Hankel functions. o , h;," (v), 11( xi ,
xii
TABLE OF SYMBOLS 1,1,1,1, etc. Electric current.
/e Effective or rms current. ie Effective or rms current density. 1, i, ix, etc. Current density. Current. k Unit x, y, z vectors. Jn, J, (v) Bessel functions.
j (-1)'. Spherical Bessel functions. K Relative capacitivity, a/ e,,. K(k) Complete elliptic integral. Km Relative permeability, K., K.(v) Modified Bessel functions. k„, k„(v) Modified spherical Bessel functions. k Boltzman constant. L, Lnn, Ln Inductance. Lm. Mutual inductance. L° Normalized or relative inductance, L/Zk. Direction cosines with x, y, z axis. M, M Magnetization. M Mutual inductance. M, M, etc. Dipole or loop moment. M', M' Classical magnetic dipole moment (IX). m Mass. A number (usually integer). N Electric or magnetic flux. n Unit normal vector. n Index of refraction. A number. nn(y) Spherical Bessel function. nn, 2n 2n!! 2 .-4 . 6 (2n + 1)!! 1 3 . 5 • • • (2n + 1) P Polarization P, P Power. P Average power. P7(1) Associated Legendre function. Pc Peirce integral tables. p, p Momentum. p A number. wiry. w. Q Electric charge. Quality of cavity. Q Quadrupole moment. (27, Q7(.1.) Associated Legendre function. q Point or variable charge. R, R, R.., R.,. Resistance. R, R(r) Function of r only. R., R.(v) Solution of Bessel's equation. j n, jn(v)
TABLE OF SYMBOLS
R;)„ nu) Solution of Bessel's modified equation. R, R Distance between two points. r, r Distance from origin. S Area or surface. So, So Cavity cross-section areas. S, S',7 Surface harmonic. S, Sn, Smn, Snn Elastance. Self-elastance. s71n s„,,, Mutual elastance. s Distance along curve. An integer. T, T Torque. T Absolute temperature. Period. t Time. TE Transverse electric. TM Transverse magnetic. to Subscript for TE wave quantities. tm Subscript for TM wave quantities. U Stream or potential function. [U] dU around constant V curve. u, u Velocity. u cos O. u2, 2/3 Orthogonal curvilinear coordinates. V Potential or steam function. dV around constant U curve. [V] v Volume. v, v Velocity. W, W, W, W Solutions of scalar wave equation. W Energy. U jV. Wte Solutions yielding TE waves. Wt„, Solutions yielding TM waves. X Reactance. x, y, z Rectangular coordinates. Yn, Y.,„(v) Bessel functions. Y, Y Admittance, G jB. Y° Normalized or relative admittance, YZk . Z, Z Hertz vector. 2, Z, Z, Z.., Z. Impedance. 2k Characteristic impedance. Z° Normalized or relative impedance, 2/2k. Z, Z(z) Function of z only. z Complex variable, x jy. a, 0, y, 8, el
Often used for angles.
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TABLE OF SYMBOLS
p„). 0 Ratio v/c. Ratio (p. — p,)/(12 0 Free space wave number, (0(1.40 1. „ 0:,„ Wave-guide wave number, (02 — 02.7,)1. Cutoff wave number. 0..p Cavity resonance wave number. F, P Phasor propagation constant. y Electrical conductivity. A, Ara Determinant. A small part of. (5 Skin depth. Phase difference. 5 A small quantity. A small part of. 67,7 Kronecker delta, zero if m n, one if m = n. • Capacitivity. A small quantity. E„ Free space capacitivity. E, En Phase angle. n Intrinsic impedance. 0, 0(0) Function of 0 only. Colatitude angle. Unit vector in 0 direction. 0, 0', 0" Angles of incidence, reflection, and refraction. • Magnetic susceptibility. (1 — /32)-1. X Wave length. Cutoff wave length. Xg, Wave-guide wave length. X.„„ Cavity resonance wave length. Permeability. cos 0. 11„ Free space permeability, 47 X 10-7 . • Frequency in cycles per second. limn Cutoff frequency. P mn p Cavity resonance frequency. :EZ(E) Function of E only. E, 0 Oblate spheroidal coordinates. n, 4 Prolate spheroidal coordinates. H, H, lI Poynting vector. lI Rms Poynting vector. p Distance from z- or 0-axis. Charge density. Ni Unit vector in p-direction. a Surface electric charge density. s Area or surface resistivity. • Volume resistivity. Density. Time. 4' Function of 0 only. + Unit vector in 0-direction. 0 Longitude angle. Phase angle. 1' Scalar potential.
TABLE OF SYMBOLS
St Magnetomotance. Solid angle. 0.1 Frequency in radians per second. V Vector operator, is/ax is/ay ka/az. V2 Two-dimensional vector operator. a x b Vector product of a and b. a • b Scalar product of a and b. V2 Laplace operator. [v] Retarded v.
xv
PREFACE TO THE THIRD EDITION, REVISED PRINTING
It is a source of great satisfaction to my father on his 95th birthday that a revision of this book is being published. In the past 50 years it has been used as a textbook by countless students who have, sometimes painfully, learned the value of a rigorous problem course. Others have found it valuable as a reference that contains solutions to hundreds of difficult electromagnetic problems with widespread applicability. WILLIAM RODMAN SMYTHE
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PREFACE TO THE THIRD EDITION
The arrival of the digital computer since the appearance of the second edition has considerably altered the electromagnetic computation problem and made methods feasible that were probably discarded years ago as impractical. These devices iterate so efficiently that the summation of a simple, slowly convergent series is often faster than closed-form evaluations requiring interpolation in an inadequate function table. Solution of linear simultaneous algebraic equations is now easy and very useful for obtaining coefficients in the series solutions of mixed boundary and mixed coordinate problems such as those added to the text and problems of Chap. V. Many of these appear here for the first time. About 25 useful elliptic function conformal transformations are added to text and problems of Chap. IV as well as minor alterations and additions to Chaps. VI, VII, and XI (former Chap. XIII). Old Chaps. IX and X have been deleted because for about 10 years all physics and electrical engineering graduate students at the California Institute of Technology, and probably elsewhere, have seemed to be already familiar with linear circuits. The coverage of eddy currents in Chap. X (former Chap. XI) is greatly extended. Little or none of this added material appears elsewhere. Chapter XII (former Chap. XIV) on wave guides and cavities is entirely rewritten, with the addition of new matter and use of the variational method. The treatment of radiation from accelerated particles in Chap. XIII (former Chap. XV) has been expanded. The new material herein, unlike that in former editions, is unchecked by students. As much as possible of the almost error-free last edition has been left undisturbed. The author has checked the new material as well as he can but knows from past experience that some errors may survive. It is hoped that those finding them will inform him so that they may be corrected in future printings.
WILLIAM R. SMYTHE
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PREFACE TO THE SECOND EDITION
The wide use of rationalized mks units and the increased importance of microwaves made this radical revision of the first edition imperative. The units are changed throughout. The resultant extensive resetting of the text permits a modernization of nomenclature through such changes as "capacitor" for "condenser" and "electromotance" for "electromotive force," The original wording has been preserved only in the Cambridge problems. In static-field chapters, 40 problems of aboveaverage difficulty have been added, usually covering boundary conditions omitted in the first edition. The expanded treatment of electromagnetic waves made necessary the rewriting of the parts of Chap. V dealing with Bessel functions and led to the introduction of vector surface harmonics, which greatly simplify some calculations. Much of Chap. XI on eddy currents has been rewritten, and two of the three electromagnetic-wave chapters are entirely new. Both the text and the 150 problems include methods and results not found in the literature. Two groups of advanced Ph.D. students worked over this material to get practice in attacking every type of wave-field problem. Many are too difficult for first-year graduate students, but every problem was solved by at least one of the advanced students. They can be worked either directly from the text or by fairly obvious extensions of it. Some useful results appear in the problems and are listed in the Index, which should be consulted by engineers with boundary value problems to solve. Chapter XV of the first edition is omitted because none of the remaining theory is based on it and because to bring it up to date would require an excessive amount of space. None of the new topics appears to lie outside the scope of the mathematical preparation assumed for readers of the first edition. That the successful solution of electrical problems depends on physical rather than mathematical insight is borne out by the author's experience with the first edition, which shows that graduate students in electrical engineering and physics greatly excel those in mathematics. It is believed that very few of the errors and obscure or ambiguous statements in the first edition escaped the scrutiny of the 375 students at the California Institute of Technology who worked it through. No infallible system for locating errors caused by the transposition of units has been found, and the author will appreciate letters from readers pointing them out.
WILLIAM R. SMYTHE July, 1950
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PREFACE TO THE FIRST EDITION
It has been found that, in most cases, the average graduate student, even though he seems to be thoroughly familiar with advanced electrical theory, is unable to solve the electrical problems he encounters in research if they fall outside the routine types and so must be worked out from first principles. The present book is the result of having taught, for the last twelve years, a course designed to train first-year graduate students in physics, electrical engineering, geophysics, and mathematics to apply the principles of electricity and magnetism. Students in these fields must show a proficiency equivalent to an average grade in this course to be admitted to candidacy for the Ph.D. degree at this institute. It is hoped that this book will also provide a reference where workers in these fields may find the methods of attack on those problems, common in research, for which the formulas in the handbooks are inadequate. It is assumed that the reader has the mathematical preparation usually acquired in a good undergraduate course in introduction to mathematical physics. This implies a reasonable familiarity with vector analysis, the calculus, and elementary differential equations. All the mathematical development beyond this point is done herein by methods that a reader with such training can follow. The author has succeeded, with some difficulty, in avoiding the use of contour integration but believes that anyone going further into the subject should acquire a working knowledge of this powerful mathematical tool. As has already been implied, this book is written for the experimental research physicist and engineer rather than for the theoretical man. For this reason, only that theory which has applications is included, and this is developed by the most concise method compatible with the assumed preparation of the reader. No subject is included for its historical interest alone. More than the usual number of problems have been worked out in the text, and these have been selected for one or more of the following reasons: the result has important applications, it clarifies the theory, it illustrates some useful mathematical device, it proves the utility of some concept in the theory. At the end of each chapter is an extensive collection of problems involving nearly all the theory in the text. Many of these are taken from the Cambridge University examination questions as printed in Jeans's "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925, and are included with the permission of the Cambridge University Press to whom we express our thanks. Although the best students can work all the problems, the average student cannot. They provide an opporxxiii
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PREFACE TO THE FIRST EDITION
tunity for the reader who is working up the subject by himself to test his proficiency. In many cases important results, which lack of -space prevented working out in the text, are included as problems and are listed in the index. The treatment and arrangement of the subject matter depart from the most common practice in several respects. In the first place, it will be noticed that all developments are based directly on the macroscopic experimental facts rather than on the hypothetical microscopic structure of conductors and dielectrics. There are two reasons for this. The first is that, although the microscopic theory meets the crucial test, namely, gives the observed macroscopic laws within the precision of observation, this does not imply that the theory is unique or that other deductions from it are correct. The second is that the development of the most satisfactory theories, those based on wave mechanics, require mathematical technique with which we have assumed the reader to be unfamiliar when starting this book, but with which he will be quite familiar after finishing two-thirds of it. The development of these theories, necessarily brief, is therefore postponed until the last chapter. The second departure from the most common practice consists in the complete omission of the concept of the isolated magnetic pole. All magnetic theory is therefore based on the interactions of electric currents or moving charges. This logically leads to the use of the magnetic vector potential rather than the scalar potential, so that the former is used extensively, although not exclusively, in all magnetic and electromagnetic theory. It will surprise many to find that this leads to considerable simplification in certain places, notably in the calculation of inductance and the treatment of eddy currents and electromagnetic radiation. Other minor departures from the usual practice include the more extensive use of Bessel functions and conformal transformations and the treatment of forces between moving charges exclusively by the methods of special relativity. The latter procedure, besides having a very firm experimental basis, makes hypotheses as to the shape or size of electric charges unnecessary and gives a clearer idea of the limits within which the standard formulas can be applied without using such hypotheses. Several subjects usually included in books on electricity and magnetism are omitted here. The theories of electrolytic conduction, thermionic emission, photoelectric phenomena, thermoelectric effects, etc., of which the reader is assumed to have an elementary knowledge, are entirely omitted because it is believed that a treatment of them, on the same level as the remainder of the book, would require a background of physical chemistry, thermodynamics, and quantum theory not possessed by the average reader. The theory of electrical machines and instruments, including vacuum tubes, is also omitted because it is
PREFACE TO THE FIRST EDITION
xxv
believed that such subjects are best treated in connection with laboratory courses. Certain alternative methods such, for example, as the Heaviside operational method and the dynamic method of circuit analysis have been omitted for lack of space. Before starting this book the average reader will have become familiar with all the common systems of electrical units and will have his own preference. The actual system used will make little difference provided it is clearly indicated. For each section of the subject, the author has used that system which seemed simplest to work with. Thus the c.g.s. electrostatic units are used in Chaps. I to V, the c.g.s. electromagnetic units in Chaps. VII to XII, and the Gaussian system in Chaps. XIII, XIV, and XV. To avoid confusion, the units used are noted at the bottom of each page and a very complete system of conversion tables is given in the Appendix, enabling the results of any calculation to be expressed in any units. To see whether the use of rational units would have simplified calculations, all the numbered formulas occurring in the preliminary lithoprinted edition were scrutinized. It was found that the complexity of 1196 formulas would be unchanged, that of 169 would be decreased, and that of 123 would be increased. Thus in the theory there is little to choose between rationalized and unrationalized units. The results in problems were not investigated. It has been the practice throughout, whenever an integration is performed or a mathematical transformation made, to refer, by number, to the appropriate formula in both Peirce's "A Short Table of Integrals," Ginn, 1929, and Dwight's "Tables of Integrals and Other Mathematical Data," Macmillan, 1934 [4th ed., 1961]. It is therefore advisable for the prospective reader to procure a copy of one of these inexpensive tables. The bibliographies at the ends of the chapters are not by any means complete but include merely those books that have come to the author's attention which contain useful additional material or instructive alternative treatments. The author has used every device he can think of to eliminate mistakes but is perfectly certain that they still persist and will be grateful to anyone pointing them out. The author wishes to express his gratitude to the hundred graduate students, especially Dr. Pasternack, who worked through the preliminary lithoprinted edition, zealously detecting errors and obscure spots, and checking the answers to problems. He also wishes to thank Professors Ira S. Bowen and William V. Houston for reading sections of the manuscript and for valuable discussions. In particular, he wishes to thank Dr. Charles H. Townes for checking all the derivations in the final manuscript and the answers to those problems not appearing in the preliminary edition. WILLIAM R. SMYTHE August, 1939
CHAPTER I BASIC IDEAS OF ELECTROSTATICS 1.00. Electrification, Conductors, and Insulators.—The word "electricity" is derived from the Greek word for amber. Apparently Thales of Miletus first discovered, about 600 B.C. that amber, when rubbed, attracts light objects to itself. It is now known that most substances possess this property to some extent. If we rub with a piece of silk a glass rod or a metal rod supported by a glass handle, we find that both will attract small bits of paper, and we say they are electrified. We see that only in the case of the metal rod can we destroy the electrification by touching with the finger. Furthermore, we find when we touch the electrified metal rod with pieces of various materials held in the hand that metals and damp objects destroy the electrification whereas substances like glass and silk do not. We call substances that remove the electrification conductors and those which do not insulators. We can find materials that remove the electrification very slowly and that we might call poor conductors or poor insulators. Thus there is no definite division between the two classes. 1.01. Positive and Negative Electricity.—If we rub a glass rod with silk and touch some light conducting body, such as a gilded pith ball suspended by a silk string, with either the rod or the silk, we find it is electrified. If two balls are electrified by the glass or by the silk, they repel each other; but if one is electrified by the rod and one by the silk, they attract. Thus we conclude that there are two kinds of electrification and that bodies similarly electrified repel each other and those oppositely electrified attract each other. By experimenting with many substances, we conclude that there are only two kinds of electricity. Arbitrarily we define the electricity on the glass rod to be positive and that on the silk negative. 1.02. Coulomb's Law, Unit Charge, Dielectrics.—We next observe that the force between the balls decreases rapidly as they are separated. Coulomb investigated these forces with a torsion balance and found that the force between two small electrified bodies is directed along the line joining them, is proportional to the product of the charges, and is inversely proportional to the square of the distance between them. This is known as Coulomb's or Priestley's law. The statement of this law implies the definition of a quantity of electrification or of electric 1
2
BASIC IDEAS OF ELECTROSTATICS
§1.03
charge. We formulate this as follows: An electrostatic unit of charge is that charge which, when placed one centimeter away from a like charge in a vacuum, repels it with a force of one dyne. The practical unit of charge is the coulomb which is 3.0 X 109 electrostatic units. In many homogeneous, isotropic, nonconducting mediums, the inversesquare law is found to hold, but the force between the same charges is reduced, so that for such a medium we may write this law, in mks units, 4
F =
21
4/rer2
(1)
7.1
where F is the force on the charge of q' coulombs due to the charge q, r the vector from q to q' whose magnitude is r meters, r1a unit vector along r, and e the capacitivity which is a constant characteristic of the medium. In vacuo the numerical value of e is 8.85 X 10-12 farad per meter and is written Ev. The specific inductive capacity or relative capacitivity K is the ratio E/E,, which is dimensionless and has the same numerical value for all systems of units. 1.021. Limitations of the Inverse square Law. The precision of Coulomb's measurement has been greatly exceeded by modern methods, and we may now state that the exponent of r in 1.02 (1) has been verified, recently, to about 1 part in 109 for measurable values of r. It should be borne in mind that we can apply Coulomb's law with certainty only to dimensions for which it has been verified. We shall endeavor throughout this book to avoid basing any macroscopic theory on the assumption of the validity of this law at atomic distances. The law applies strictly only to charged bodies whose dimensions are small compared with the distances between them. Their shape and composition are immaterial. 1.03. Electrical Induction. The fact that the electrical charge on a conductor is mobile indicates that when an electric charge is brought near an uncharged conductor electricity of the opposite sign will move to the parts nearer the charge and that of the same sign to the parts more remote from it. From Coulomb's law, the force between nearer charges will be greater so that the charge will attract the conductor. These charges on the conductor are known as induced charges. Unless the more remote charge is removed, for example, by touching with the finger, the conductor will return to its neutral state when the inducing charge is removed. If, with the inducing charge in place, we separate the near and far parts of the conductor, being careful to keep them insulated, we find that the two parts are oppositely charged, as we would expect. Many "static machines" have been devised that repeat this operation automatically and accumulate the separated charges. When sufficiently sensitive methods are used, it is found that a charge -
—
—
§1.05
ELECTRIC FIELD INTENSITY
3
also exerts a small attractive force on uncharged insulators. This seems to indicate that even in insulators charges are present and are not absolutely fixed but may suffer some displacement. The hypotheses concerning the actual behavior of the electric charges in conductors and insulators will not be discussed here. The theories are still imperfect but have greatly improved since 1930. 1.04. The Elementary Electric Charges.—It has been found that an electric charge cannot be subdivided indefinitely. The smallest known negative charge is that of the negative electron and mesotron, first determined with considerable precision by Millikan. The accepted value is now 1.60 X 10-" coulomb. The smallest known positive charge is that on the positron, or positive electron, on the mesotron, and on the proton. To a high order of precision, all these elementary charges are the same in magnitude. The mass of the negative electron and also, probably, that of the positron is 9.1 x 10 -31kg. The mass of the proton is about 1837 times that of the electron. In handling electrical problems, we shall treat electrical charges as infinitely divisible, using charge density as if this were so. Clearly such a procedure is justifiable only when we are dealing with quantities much greater than 1.60 X 10-'9 coulomb and certainly is useless if we go to atomic dimensions. We have seen that the electricity in conductors is free to move, so that if it possess inertia we might expect it to lag behind when the body is accelerated, thus producing an electric current which could be detected by its magnetic field. Maxwell predicted such phenomena but their magnitude is so small that they were detected and measured by Tolman, Barnett, and others not until long after his death. The results show that the mobile electricity in conductors is negative and that the ratio of its charge to its mass equals, within experimental error, the ratio for the negative electron. It appears that almost all the phenomena with which we deal in this book involve the distribution or motion of such electrons, a positive charge appearing when there is a dearth of electrons. As far as the mathematics is concerned, it is immaterial whether the positive, the negative, or both take part in the transfer of electric charge. 1.05. Electric Field Intensity.—When an infinitesimal electric charge, placed in a region, experiences a force, we say an electric field exists there. We define the electric intensity to be a vector equal to the force per unit charge acting on a positive charge placed at that point, the charge being so minute that no redistribution of charge takes place due to its presence. The last qualification is necessary because of the phenomenon of electric induction. Just as a number of mechanical forces acting on a body can be resolved into a single resultant force by taking their vector sum, so the resultant
§1.06
BASIC IDEAS OF ELECTROSTATICS
4
electric intensity due to any charge distribution can be obtained by taking the vector sum of the electric intensities due to each element of the distribution. Thus the electric intensity at point P due to n charges is
E=
1
qi
4re
r2
(1)
i=i
where E„ is the electric intensity in volts per meter, ri is the vector of magnitude ri directed from P to qi, and e is the capacitivity of the infinite homogeneous medium in which the system is immersed. 1.06. Electrostatic Potential.—Work is done against electric forces when a charge is moved in an electric field. The potential in volts at a point P in an electrostatic field is the work in joules per coulomb to bring a charge from the point of zero potential to the point 1‘,P, Vie charge being so 0 0 small that there is no redistribution FIG. 1.06. of electricity due to its presence. The choice of the point of zero potential is a matter of convenience. It is frequently, but not always, chosen at infinity. To avoid other than electrostatic effects, it will be necessary to move the charge very slowly. Let us compute the potential due to a point charge q. The work dV required to move a unit charge a distance ds in a field of intensity E is —E • ds or —E ds cos 0 where 0 is the angle between E and ds. In the case of the point charge, this becomes
dV = _q cos 0 ds 4irer2
where r is the vector from the charge q to the element of path ds and 0 is the angle between r and ds as shown in the figure. Evidently dr = ds cos 0, so that we have for the potential in volts
1.
VP
q dV = — —
'
dr
4ire fr 2
Or
VP =
471.6 \r,,
—
)
ro /
(1)
If r0is chosen at infinity, this becomes
V, =
(2)
q
47i- Erp
the electrostatic potential is a scalar point function and is independent of the path by which we bring our charge to the point. The potential at any point in an electrostatic field can be obtained by adding up the potentials due to the individual charges producing the field; thus qi
Vp = 1
LlireLJri 1=1
where ri is the distance from P to qi in meters.
(3)
ELECTRIC DIPOLES AND MULTIPOLES
§1.07
5
Since it is usually much easier to take a scalar sum than a vector sum, it is evident why (3) is to be preferred to 1.05 (1) for purposes of computation. The field intensity at P can be obtained from (3) by taking the gradient; thus E = —grad V = —17V (4) In rectangular coordinates, the components are
av Ex = ax
ay E, = --, ay
E, = —
aV az
(5)
The components of the gradient in any other fixed coordinate system can be obtained by expressing V and the components of the gradient in terms of that system. A method of doing this is given in 3.03 and 3.05. If the elementary charges are very close together compared with the dimensions involved, as is always the case in ordinary practice, we may treat the distribution as continuous and speak of the charge per unit volume as volume charge density p and the charge per unit area as the surface charge density o. The summation in (3) now becomes an integral V, =
1 fpdv zire v r
1 radS zirEJ s r
(6)
where dv is the element of volume and dS is the element of surface. It should be noted that this formula applies only when all space, including the interior of any material bodies present, has the capacitivity e. When this is not the case, we must use the methods of Chaps. IV and V. 1.07. Electric Dipoles and Multipoles. Consider the potential at a point P at x, y, z due to charges — q at xo, Ye, zo and +q at xo h,yo, zo for small h. If the distance r_ = f(xo, yo, zo, x, y, z) from — q to the field point P is [(x — xo) 2 (y — yo)2 (z — zo)91, then from +q to P it will be r + = f(xo h, yo, zo, x, y, z). Write down this potential from 1.06 (2) and expand r;1in a Taylor series by Dw 39, which gives —
V
4ire\r+
4reLr_
a2 1 \ axAr_) 2! axo2V_)
If h 0 and q 00 in such a way that hq = M remains a finite constant, the result is an electric dipole of moment M in the x-direction. Only the h term in the expression above survives, so that, if 0 is the angle between 11 and r, which replaces r_ as h 0, there results V—
M a (1) — m a (1) m cos o m • r 4re axo r 4ire ax r 4rer2 Lirer3
(1)
Evidently this can be extended so that if the potential V, at the point P due to a set of n charges, the radius vector from q, to P being r„ is given by 1.06 (3), then the potential P, due to a set of n dipoles of the same strength and sign located at the same points with axes parallel to x is
BASIC IDEAS OF ELECTROSTATICS
6
vt
§1.071
n
p aV
ax,
(xp —
(2)
41-Er2 := 1
By differentiating the expression for the potential of a unit electric dipole with respect to any of the rectangular coordinates, we get the potential due to a unit quadrupole whose dimensions are QL2; thus 1 a2(1/r)
1 a2(1/r) 47E axe
etc.
ax ay
represent the potentials due to a linear quadrupole (Fig. 1.07a) and a square quadrupole (Fig. 1.07b), etc. Further differentiations give us the potentials due to more complicated multipole arrangements but always ones in which the total charge is zero. For other cases see 1.12. y .dxrdx 4-
•
4 -2q +4
x
dy +q • dx •-q
(b)
(a) FIG. 1.07.
The translational force on a dipole M in a field E will be the vector sum of the forces on each charge. Since the charges are equal and opposite, this means the vector difference in the field strength (ds • V )E, at the two ends, multiplied by q; thus F = q(ds • V)E = (m • V)E
(3) In a uniform field, this force is zero. In a uniform field, the charges are subject to forces -FqE and — qE applied at a distance ds sin 0 apart where 0 is the angle between ds and E. Thus there is a torque acting on the dipole of amount (4) T = tEq ds sin 0 = tME sin 0 = M x E where t is a unit vector normal to M and E. 1.071. Interaction of Dipoles. The potential energy of a dipole, in any field whose potential is V, equals the total work done, against this field, in bringing each charge separately into place. If the potential of the field at P1where +q is placed is Vi and at P2 where — q is placed it is V2, then av iv = q(V — V 2) = qPir- 2av (1) as = m as where s is in the direction of the dipole axis and M is its moment. In vector notation, this is IV = (m • V) V (2) —
LINES OF FORCE
*1.08
7
If miand M2 are the vector moments of two dipoles A and B and if r is the vector from A to B, then the potential at B due to A is, from 1.07 (1), i cos 0 mm• i
47re V -
r2
r = -mi • V( 1)
(3)
Substituting this value for V in (2) gives 471-eW = ±m2 • V(mi • r7-3) = m2 • (7-3Vmi • r Ml •T or-3) M2 • (m1r-3 - M1 • r3rr-5) = M1 • m27-3- 3M1 • TM2 • rr-5 If M1 and M2 make angles 0 and 0' with r and ¢ with each other, we have W=
mi m
2 (cos
4rer 3
47 - 3 cos 0 cos 0')
(4)
If ik is the angle between the planes intersecting in r which contain M1 and M2, then, taking r in the direction of x and M1in the xy-plane, the direction cosines are i1 = cos 0,12 = cos 0', m l =sin 0, m2 = sin 0' cos and n1 = 0, so that cos 4,and W become, respectively, cos 47 = 1112 + mim2 W=
mim2
4rer 3
n1n2 = cos 0 cos 0' + sin 0 sin 0' cos
(si n0 sin 0' cos ¢ - 2 cos
cos 0')
(5)
This gives the radial force between two dipoles, by differentiation, to be
F=
aw = 3mim2(sin 0 sin 0' cos ¢- 2 cos 0 cos 0') car
4rfr4
(6)
This has its maximum value when IP = 0, 0 = 0' = 0. The torque trying to rotate the dipole in the direction a is obtained similarly, giving
aw
T = -as
(7)
1.08. Lines of Force.—A most useful method of visualizing an electric field is by drawing the "lines of force" and " equipotentials." A line of force is a directed curve in an electric field such that the forward drawn tangent at any point has the direction of the electric intensity there. It follows that if ds is an element of this curve, ds = AE (1) where X is a scalar factor. Writing out the components in rectangular coordinates and equating values of X, we have the differential equation of the lines of force
dx _ dy dz (2) E. E„ E L Analogous equations can be written in terms of other coordinate systems by using 3.03 and 3.05. Usually much easier methods of getting the equations of the lines of force exist than by integrating these equations.
8
BASIC IDEAS OF ELECTROSTATICS
§1.08
We shall, however, give one example of the integration of this equation. Consider the field due to two charges +q at x = a and ±q at x = —a. Since, from symmetry, any section of the field by a plane including the x-axis will look the same, we may take the section made by the xy-plane. The sum of the x-components of the intensity at any point due to the two charges is E, where
ex(x——a)a)2]i — [y2 ex(x+ a)a)21i 4reEl — [y2 + If we substitute, = x + a and v — x — a y y this becomes qv 47EL y2(1 + v2)1 y2(1 + u2)1 Similarly, 47rEE'r —
y2(1 + v2)1
(3)
+
y2(1 + u2)1 Equation (2) becomes dy _ E, _ (1+ v2)i ± (1 + u2)1 u(1 v2)1 ± v(1 + u2)1 dx Solving (3) for y and x and taking the ratio of the differentials give
dv—du dx u dv — v du
dy
Comparing these two expressions for dy/dx, we see that
du_ (1 ± WV' + ) dv = Separating variables and integrating, we have u(1 + u2)-1 ± v(1 + v2)-1 = C In terms of x and y, this becomes
(x
a)[(x
a)2 + y2]-1 ± (x — a)[(x — a)2 + y2]-1 = C
(4)
This is the equation giving the lines of force shown in Figs. 1.08a and 1.08b in which the value of C for each line is marked. An easier method for getting this equation, by Gauss's electric flux theorem, is given in 1.101. We may write the left side of (4) in the form
(x
a)r-1(1 + 2axr-2 + a2r-2)-1— (x — a)r-1(1 — 2axr-2 + a2r-2)-1
where r2 = x2 + y2.If we let a —> 0, expand the radicals by Pc 750 or DID 9.03, neglect the square and higher powers of a, and write C' for C/(2a), we obtain
EQUIPOTENTIAL SURFACES
§1.09
Y2
— 3
= CI =
sin2
9 (5)
which is the equation of the lines of force for an electric dipole, shown in Fig. 1.08c.
FIG. 1.08a.—Field about equal charges of opposite sign. Lines of force and equipotential lines are shown by solid and dotted lines, respectively.
1.09. Equipotential Surfaces.—An equipotential surface in an electric field is one at all points of which the potential is the same. The equation of an equipotential surface is therefore
V= C
(1)
where C is a constant. Numerous maps of electric fields showing equipotentials and lines of force will be found in subsequent chapters. It should be noted that since no work is done in moving charges over an equipotential surface, the lines of force and equipotentials must intersect orthogonally. As an example of the use of this equation, we shall take the case just considered. The potential at any point P will be C, where q[(x — a)2
Y2]-1-T q[(x
a)2
y2]-1= 4rreC
(2)
10
BASIC IDEAS OF ELECTROSTATICS
§1.10
This is the equation of the equipotential surfaces shown in section in Figs. 1.08a and 1.08b by broken lines. C values are for q = 4re. There are often points or lines in an electrostatic field where an equipotential surface crosses itself at least twice so that V V vanishes there.
1.08b.—Field about equal charges of the same sign. Lines of force and equipotential lines are shown by solid and dotted lines, respectively.
These are called neutral or equilibrium or singular points or lines. The origin in Fig. 1.08b is such a point. Some of the properties of such points are considered in 5.235. 1.10. Gauss's Electric Flux Theorem.—We shall derive this flux theorem from the inverse-square law on the assumption, to be modified later, that all space is filled with a uniform dielectric.
§1.10
GAUSS'S ELECTRIC FLUX THEOREM
11
Consider a small element dS of a closed surface (Fig. 1.10), whose outward normal makes an angle a with the radius vector from a point charge q at P. Now draw a line from every point of the boundary of dS to the point P so that the small cone so formed cuts out an area dI from the spherical surface, having P as a center and passing through Q. Then
= dS cos a. The normal component of the intensity at Q due to a charge q at P is —
qr • n _ q cos a 471-Er3 4T-Er2
The normal component of the flux through dS is defined as
dAr= eEn dS =
q cos a dS _ q dZ 471-r2 tirr 2
But the solid angle subtended by dS at P is dIr-2= d2, so that 4r dN = q d12 We notice that if the point is inside the surface the cone cuts the surface n times where n is an odd integer, and the angle a will be acute *(n + 1) times and obtuse i(n — 1) times so that the net value of the flux through the cone is (q/47) da But if the point is outside the surface, n is even and we have the same number of positive and negative values of (12 so that the net contribution is zero. To obtain the total flux through the surface surrounding the charge, we integrate the normal component and obtain or f dN = qJo N= q Js By adding up the flux due to all the charges inside, we arrive at Gauss's electric flux theorem which is: If any closed surface is taken in an electric field and if E is the electric intensity at any point on the surface and n the 4r
12
BASIC IDEAS OF ELECTROSTATICS
§1101
unit outward normal vector to the surface, then
efsE • n dS = q
(1)
where the integration extends over the whole surface which includes the charge q. If the space outside this surface is not homogeneous but contains various dielectric and conducting bodies, it is necessary to make certain hypotheses as to the electrical behavior of matter in electrostatic fields. It is assumed therefore that, as regards such fields, matter is entirely electrical in nature consisting of positive and negative charges whose fields obey the inverse-square law. With this hypothesis the electrostatic effect of any material body is obtained by superimposing the fields of its constituent charges as calculated from this law. Equation (1) is valid, therefore, no matter what the nature of the dielectric or conducting material outside the surface considered may be because the fields of external charges were considered in its derivation. The above hypothesis is used, explicitly or implicitly, in most treatments of electrostatics. 1.101. Lines of Force from Collinear Charges. To illustrate the application of this theorem, we shall use it to find the equation of the —
lines of force about any set of collinear electric charges qi, q2, q3 . . . situated at x,, x2, x3. . . along the x-axis. From symmetry, no lines of force will pass through the surfaces of revolution generated by rotating the lines of force lying in the xy-plane about the x-axis. Gauss's electric flux theorem, applied to the charge free space inside such a surface between the planes x = A and x = B (Fig. 1.101), tells us that the total normal flux N entering through section A equals that leaving through section B since none passes through the surface. The equation of the surface is therefore obtained by setting N equal to a constant. From 1.05 (1), N equals the sum of the normal fluxes due to each charge and, as we have just found in proving Gauss's electric flux theorem, this is 47N =
q202 q30 3 ± • ' •
13
POTENTIAL MAXIMA AND MINIMA
§1.11
where 03, 22, 23 . . are the solid angles which section A subtends at Xi, x2, x8 . . . In terms of the angles al, a2, a3 . . . in Fig. 1.101 this is
N = Ziq,(1 — cos «,) = C' — -}Zqi cos ai i=i i-1 Collecting the constants on one side of the equation and substituting for the cosines in terms of the coordinates x, y of a line of force in the xy-plane, we have C = Zq,(x — xi)[(x — x,.)2 +y2]-1
(1)
for the equation of a line of force. This agrees with 1.08 (4). 1.102. Lines of Force at Infinity.—If f = [( x — X)2 y9 1we neglect 2 and write f and n [(2 — /f]n when x —
c
=x
211qi
f-
Fi Lf
(
x — X)2]qi(2 f3
i=i
=
z —
qi (1) iml
where 2 is the coordinate of the "center of gravity" of the charges. Thus at infinity, the field is the same as if the algebraic sum of the charges were concentrated at their center of gravity. We may extend this rule to noncollinear charges by pairing off members of a group of charges and taking the centers of gravity of each pair by (1), then pairing off these centers of gravity, etc., until we arrive at the center of gravity of the group.
1.11. Potential Maxima and Minima. Earnshaw's Theorem.—Consider a small spherical surface enclosing a point P in an electric field. The average value of the potential over this sphere is
rr
1 2' V dS = — f irf V sin 0 d0 47r2s 47 o Taking derivatives and applying Gauss's electric flux theorem gives V =
1
cif _ 1 far 21. v sin n 0 de 45 dr 47rjo Jo ar f dV — dS = — q 4irr2 47r-r2 s dr where q is the charge inside the sphere, giving, on integration, the result
=4irer + c If q = 0, the average value of the potential over the small sphere enclosing P is the same as at P. Hence the theorem that the potential cannot have a maximum or a minimum value at any point in space not occupied
14
BASIC IDEAS OF ELECTROSTATICS
§1.13
by an electric charge. It follows from the definition of potential that to be in stable equilibrium a positive point charge must be at a point of minimum potential and a negative point charge must be at a point of maximum potential, the potential due to the charge itself being excluded. From the theorem just proved, this is impossible, giving us Earnshaw's theorem which states that a charge, acted on by electric forces only, cannot rest in stable equilibrium in an electric field. This shows that if we are to consider matter as purely electrical in nature and as made up of positive and negative charges controlled by electrical forces only, these forces must be different from those operating on a laboratory scale in electrostatics. 1.12. Potential of Electric Double Layer.—We saw in 1.07 that the potential due to a dipole may be obtained from that due to a single charge by differentiation in the direction of the dipole axis. From this, we see that if the potential at P due to an element dS of a surface having a charge density a is
dV =
dS 4rer where r is the distance from dS to P, then
dS
4re is the potential at P due to a dipole of strength a- dS which has the direction of n. Therefore, if we have an electric double layer of moment (I3 per unit area, the potential at P due to it is
v=
14,
I 4,n • r dS = dS (1) 4rej s r3 an\r But we saw in 1.10 that n • r r-3dS = 6/0 where dfl is the solid angle subtended at P by the element dS. Thus we have (
V = f c3 d (2) 4re If the double layer is of uniform strength NI!, this becomes qf V= (3) 4ire where 12 is the total solid angle subtended by it on the positive side at P. 1.13. Electric Displacement and Tubes of Force. The product of the capacitivity by the electric intensity occurs frequently and, in an isotropic dielectric, this product is designated as the electric displacement D; thus (1) D = eE In the mks system of units, D is measured in coulombs per square meter and E in volts per meter. Lines of electric displacement, which are analogous to lines of electric —
§1.14
STRESSES IN AN ELECTRIC FIELD
15
intensity, will have the same direction as the latter in an isotropic dielectric, but if e is greater than E,, they will be more closely spaced. Taking a small element of area normal to the displacement and drawing lines of displacement through its boundary, we cut out a tubular region in space known as a tube of force. Applying Gauss's electric flux theorem to the charge free space bounded by two normal cross sections of such a tube, we see that, since there is no contribution Si to the surface integral by the side walls of the tube, the flux entering one end must equal that leaving the other, so that if the areas of the sections are Si and 82 the flux through the tube is
N = SID]. = 82D2 Thus we define a unit tube of force to be one in which the flux through any section is unity. Many diagrams showing tubes of force will be found in subsequent chapters. Since there are 41- square meters surface on the sphere of unit radius surrounding a point charge q and since the displacement on this surface is q/(47r), it follows that there are q unit tubes of force emerging from a charge q. Thus the charge on the end of a unit tube is 1 coulomb. 1.14. Stresses in an Electric Field.—We have introduced the conception of lines and tubes of force merely as an aid toward visualizing an electric field. It is possible to carry this idea considerably further, as Faraday did, and actually look upon these tubes as transmitting electrical forces. Since this is sometimes a -±q most useful method of attacking a problem, let us see if a system of FIG 1.14. stresses can be postulated to account for observed electrical forces. Let us determine what function of the electric intensity the tension along the tubes of force must be in order to account for the observed Coulomb force between two equal charges of opposite sign at a distance 2a apart. Call this function 43(E). From 1.05 (1), the intensity in the plane of symmetry is q cos 3 0 2aq E= 27rea2 4re(a2+ y 2)1 The geometry is shown in Fig. 1.14. The ring element of area is 2ira2 sin 0 dS — dO cos' 0
BASIC IDEAS OF ELECTROSTATICS
16
§1.14
Writing the Coulomb force on the left and the tension across the yz-plane on the right and dividing both sides by 27T-a2 give n2
3272ea4
—
f2 ( q cos' Oyin 0 de 1 f CE) dS = 2n-Ea2 cos3 27r C2
(1)
Let x = q/(27rea2), and expand cp in powers of E; then
e 2 8
1 2 7ra2
2 f cos3(n-1) 0sin 0 dB
C„E' dS = n =0
n =0
This equation must hold for all values of q and a and hence for all values of x. Thus C. = 0 except when n = 2. For this we have, canceling out x, r
2
e g = C2
- -
cos3 0 sin 0 d0 =
C2
4
Thus we find that
4,(E) = e
2
(2)
This is the tension along the lines of force required to account for the Coulomb law of attraction between charges of opposite sign. It is evident in the preceding case that if there were only a tension along the tubes of force they would shorten as much as possible and all would lie in the line joining the charges. Since we know that in equilibrium they fill the entire space surrounding the charges, there must be some repulsion between them to prevent this. To determine this pressure NI,(E), let us consider the force between two equal charges of the same sign. This differs from the case just considered because lines of force terminate at infinity. The tension per unit area across a spherical surface of large radius falls off as the inverse fourth power of its radius because of (2) and the inverse-square law. The area of the sphere increases only with the square of its radius so that no force is transmitted this way. Therefore the entire force may be considered as due to the repulsion of the lines of force across the plane of symmetry. From 1.05 (1), the intensity in this plane is
E—
= q cos2 0sin 0
2qy
2irea2 Proceeding as before, we obtain instead of (1) q2
471-6(a2
1 f
y2)i
2
(3)
cos2 0 sin 0)sin d0 24) 271-ta COS3 cos For the same reasons as before, 4,(E) must be of the form C2E2, and we evaluate C2 in the same way which gives 32,2,a4
271-a2
NP(E) dS
=
Jo
§1.15
GAUSS'S ELECTRIC FLUX THEOREM
17
r
2
-- = 8
C2f
Sins 0 cos 0 dO =
C2
Thus we find that
(E) = — eE
2
(5
)
This is the repulsive force per unit area between adjacent lines of force necessary to account for the Coulomb law of repulsion between charges of the same sign. These results may be put in the equivalent forms
eE2 E •D=D2 2 2 2€
(6)
Since (1) and NI/ are functions of E and E only, the origin or shape of the field is immaterial and they must have the same form in all fields.
1.15. Gauss's Electric Flux Theorem for Nonhomogeneous Mediums. We are now prepared to extend Gauss's electric flux theorem to isotropic mediums within which the capacidS tivity varies continuously from place to place. Let us suppose that a point charge q is situated at a point P somewhere inside a closed surface S in such a medium. We shall draw about P a sphere S' so small that e' is constant throughout it. Now take an element dS of the surface S, so small that E is constant over dS, and consider the tube of force, of which dS is a section, that cuts out an element dS' on S' and ends on q. Now apply Gauss's theorem to the uncharged dielectric inside the tube between dS and dS'. Since the normal component of D over the walls is zero, the only contribution to the surface integral is from dS and dS', so that
e'E' • n' dS' = eE • n dS Integrating over the two surfaces, we have
e' fs,E' • n' dS' = fseE • n dS since e' is the same for all elements dS'. But we have already proved in 1.10 that the left-hand integral is q, so that
fseE • n dS = q
(1)
where e and E are both functions of position. This may be extended as before, so that q includes all the charges inside S. A complicated field may result from simple sources. The application
18
BASIC IDEAS OF ELECTROSTATICS
§1.17
of (1) in such a case is often greatly simplified by calculating the flux from each source separately and adding the result. Thus j(Ei E2
+ • • • + En) • n
dS = jEl •n dS 5E2 • n dS + • • • + 5E. • n dS
(2)
This permits many problem answers to be written down by inspection.
1.16. Boundary Conditions and Stresses on the Surface of Conductors. When the electric charge on a conductor is in static equilibrium, there can be no fields in the interior or along the surface. Otherwise, since by definition charges are free to move on a conductor, there would be a movement of the charges, contradicting the postulated equilibrium. It follows that the whole conductor is at one potential and that lines of force meet its surface normally and end there. The charge density on the surface is a- coulombs per square meter, and there is one unit tube of force, pointing outward if 0- is positive, from each charge; so we have (1) D = eE = Since lines of force leave the conducting surface normally, they can intersect each other only at infinitely sharp points or edges. We have seen that this occurs at a mathematical point or edge. The converse also holds. At the bottom of a sharp V-shaped groove or conical depression, D and a are zero. We have seen in 1.14 that there is a tension along these lines of force of the amount 2 F = D2 a (2) 2e 2€ and so this is the outward force per square meter on a charged conducting surface. This force is independent of the sign of the charge. It should be noted that we have neglected the hydrostatic forces that may be present in the dielectric owing to its tendency to expand or contract in an electric field. An expression including such forces will be derived later in 2.09.
1.17. Boundary Conditions and Stresses on the Surface of a Dielectric.—Let us apply Gauss's electric flux theorem to the small disk-shaped volume whose flat surfaces of area dS lie on opposite sides of the plane boundary between two dielectrics e' and e" (Fig. 1.17a). The disk is so thin that the area of the edges is negligible compared with that of the faces. If there is no free charge on the surface and the normal components of displacement are D„ ", we find from 1.15 that ' and Da D'„ dS = D',,' dS
or
D' = D„" (1)
The stress on this surface due to the normal components of the displacement must be the difference between the stresses on the two sides, so
BOUNDARY CONDITIONS AND STRESSES
§1.17
19
that, from 1.14 (6), we have — e") K' — K" T =D'n2 D'.' 2 (2) 2e'e" 2E,, 2e' 2e" = KW" Let us consider the work required to take a unit charge around the path shown in Fig. 1.17b in which the length of path normal to the boundary is vanishingly small. Starting at any point, the work done in taking
7
/ / / 7 dS D
.,s' , ,\\
1,.,,
Et
a's / = , ====
ex if
X ds
\ ( \ \ , ,
Et
FIG. 1.17b.
FIG. 1.17a.
a unit charge around the path is, if energy is to be conserved, zero, so that E; ds = g' ds or E; = E;' (3) The pressure on the interface will be the difference in the pressure on the two sides; thus, from 1.14 (5), we have
— -€"g'2 = ig2(e — e") = riE;2e,,(K' — K")
P,. =
(4)
We may state then that at the uncharged interface between two dielectrics, the normal component of the displacement and the tangential component of the intensity are continuous. In terms of the potential these boundary conditions may be written
,av' an
E
„av an
"
K' a' v
or
n
Knav" an
(5)
V' = V"
(6)
where V' and V" are the potentials in e' and e". Equation (6) implies that the zero of potential is chosen in both mediums, so that at some point on the boundary V' = V". It then follows, by integration of (3), that (6) holds for all points on the boundary. The normal stresses directed from e" to e' on the interface between two dielectrics may also be written, from (2) and (4),
F = T. — P. =
K' — K"{D;2 2E ,K'
K'
e' — e"(D'2 t D'2) n
K" =
2e'
E'
e"
(7)
The foregoing formulas do not take account of the fact that some dielectrics tend to contract or expand in the presence of an electric field. In such a medium, there will be an additional force of a hydrostatic nature acting on the surface. An expression including this force will be derived later, in 2.09. At the boundary between two isotropic dielectrics, the lines of force and the lines of displacement will be refracted in the same way. Let
20
BASIC IDEAS OF ELECTROSTATICS
§1.18
the angle between E1 or D1, in the dielectric E1 and the normal to the boundary be al, and let the corresponding angle in €2 be a2 (Fig. 1.17c). Then from (1) and (3), we have D1cos al= 61E1 cos al = D2 cos a2 = €2E2 cos a2 DI ET' sin al = E1 sin al = D2€21 sin a2 = E2 sin a2 Dividing the first equation by the second gives E l cot al = €2 cot a2 (8) ,
This gives the law of refraction for both D and E at the boundary between two isotropic mediums with different FIG. 1.17c. capacitivities. 1.18. Displacement and Intensity in Solid Dielectric.—The capacitivity was first introduced in the statement of Coulomb's law [(1.02 (1)] as a factor that depends on the medium in which the electric force is measured. This hypothetical measurement is difficult to conceive in a solid dielectric, but by using the boundary conditions just derived we may devise a method of determining the displacement and intensity and hence the capacitivity in such a medium. To determine the displacement and intensity in a solid dielectric, let us excavate a small evacuated disk-shaped cavity whose thickness is very small compared with its radius. The intensity inside the cavity E,
FIG. 1.18.
far from the edges will be determined entirely by the boundary conditions over the flat surfaces, as shown in Fig. 1.18a. To determine the displacement, we orient the cavity so that the intensity inside is normal to the flat faces in Fig. 1.18a. From 1.17 (1), we know that the displacement in the dielectric equals that in the cavity. Thus by measuring the intensity in the cavity and multiplying by ev, we determine the displacement in the dielectric. To determine the intensity in the solid dielectric, we orient a long thin cylindrical cavity so that the intensity inside is parallel to the axis as in Fig. 1.18b. Then from 1.17 (3), we know that the intensity inside is the same as that in the solid dielectric. The ratio of the displacement to the intensity, the cavities being small compared with all dimensions of the dielectric mass and the external field remaining constant, gives the capacitivity of the medium.
CRYSTALLINE DIELECTRICS
§1.19
21
The values of displacement and intensity found in this way certainly do not represent the actual fields on a molecular scale found inside a dielectric but are some type of average. Other types of average might give results contradicting macroscopic observations. 1.19. Crystalline Dielectrics.—We shall now use the experimental tests of the last article to find the relation between displacement D and intensity E in a homogeneous crystalline dielectric. Let us cut three plane parallel plates of thickness d from the positive x, y, and z faces of a large cube of this material. We now deposit a conducting layer on each face of each plate and apply a difference of potential V. If we consider areas of the plate sufficiently far from its edges we see that the potential boundary conditions on all such areas of all plates are identical so that the potential distribution in the central areas of all plates is the same. Thus the equipotentials near the center of each plate are parallel to its faces and from 1.06 (4) the intensity E equals V/d. By experimenting with our disk-shaped cavity which is taken small compared with d so that it does not disturb the charge distribution on the conducting faces we find that D is proportional to E but not in the same direction. Thus we have in the x, y and z plates, respectively, (D.). = eilE., (D.). = e13E. (Dy). = 612E., (1) (D0y = E22Ey, (DA, = €21Ey, (D.)y = 623E/, (D.). = € 31E.,
(Dr). = E324
(D.). = € 33E.
Even when E is the same in all plates the normal components of D will in general differ. However in all cases it is found experimentally that (D.),E. = (Dy).Ey,
(D.).E. = (D.).E.,
(Dy).Ey = (D.)„E.
(2)
Superimposing the results in (1) gives
Dx = €11E. + €21Ey -I- €31E. €32E. D. = €13E. + €23Ey + €33E.
Dy = e12Ex + €22Ey +
(3)
Comparing (2) and (1), we see that €12 = €21,
€13 =
€31,
€23 = e32
(4)
Thus the single simple factor e connecting the quantities D and E in the case of an isotropic medium becomes, for a crystal, a quantity known as a symmetrical tensor, having nine components of which six are different. Let us see if it is possible to orient our axes so as to simplify (3). The product E • D, being a scalar quantity, must be independent of the choice of axes. Writing out in terms of the components of E, we have, from (3), using (4),
E • D = €11E1 + €22E,2, + €33E1 + 2€12E.Ei, + 2e13E.E. + 2€23EyEz (5) This is the equation of a quadric surface in E, Ey, and E. We can vary
22
BASIC IDEAS OF ELECTROSTATICS
§1.19
these coordinates in any way we please, keeping E! + E2y + Ei constant, by rotating our axes. We may therefore orient our axes so that the cross products EzE,,, EzEz, and E„Ez disappear. The equation of the quadric referred to the new axes is E3E! (6) E•D= E2E: The components of the displacement referred to these axes are Dz =€1Ez Dy = €2E,, Dz = €3E2 ( 7) The directions of the coordinate axes in (7) are called the electrical axes of the crystal. If €1, €2, €3are all the same, we have an isotropic medium; if only two are the same, we have a uniaxial crystal; and if all three are different, we have a biaxial crystal. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. The work required to bring a point charge q up to the centers, which are a distance b apart, of two thin parallel conducting coaxial rings, each of radius a, is W1 and W2, respectively. Show that the charges on the rings are
Crea
(a2 + b2)1[(a2 + b2)1W1.2 — aW2.1]
Q1.2 = b2q
2. Four equal parallel line charges lie at the corners of a square prism, those at the ends of one diagonal being positive and the others negative. Find the fraction of the total flux that enters the prism. 3. There is a charge q at x = +a, y = 0, z = 0. Find how large a charge must be placed at x = —a, y = 0, z = 0 so that a flux N passes in the positive direction through the circle x = 0, y2 + z2 = a2. 4. Two thin concentric coplanar rings of radii 1 and 2 carry charges —Q and +(27)1(2, respectively. Show that the only neutral points in the field are at x = 0 and ± 2-1. 5. Show that the equation of the lines of force between two parallel line charges, of strength q and —q per unit length, at x = a and x = —a, in terms of the flux per unit length N, between the line of force and the positive x-axis is 277-N)] 2
[y — a cot (— + x2 = a2 csc2 (— 4
6C. Charges +4q, —q are placcd at the points A, B, and C is the point of equilibrium. Prove that the line of force that passes through C meets AB at an angle of 60° at A and at right angles at C. Find the angle at A between AB and the line of force that leaves B at right angles to AB. (Write potential for small region about C in polar coordinates with origin at C.) 7G. Two positive charges qt and q2 are placed at the points A and B, respectively. Show that the tangent at infinity to the line of force, which starts from q1making an angle a with BA produced, makes an angle 2 thn-' (Ng' q2)-1sin with BA and passes through the point C in AB such that AC: CB = q2: 41.
PROBLEMS
23
8C. Point charges +q, -q are placed at the points A, B. The line of force that leaves A making an angle a with AB meets the plane that bisects AB at right angles in P. Show that a PAB sin - = 21sin 2 2 9C. If any closed surface is drawn not enclosing a charged body or any part of one, show that at every point of a certain closed line on the surface it intersects the equipotential surface through the point at right angles. 10C. Charges 3q, -q, -q are placed at A, B, C, respectively, where B is the middle point of AC. Draw a rough diagram of the lines of force, show that a line of force that starts from A making an angle a with AB > cos-' ( i) will not reach B or C, and show that the asymptote of the line of force for which a = cos-1 ( -I) is at right angles to AC. 11C. If there are three electrified points A, B, C in a straight line, such that AC = f, BC = a2/f, and the charges are q, -qa/f, and 471-eVa, respectively, show that there is always a spherical equipotential surface, and discuss the position of the equilibrium points on line ABC if 47r€V = q(f + a)/(f - a) 2and if 4ireV = q(f - a)/(f + a)2. 12C. A and C are spherical conductors with charges q + q' and -q, respectively. Show that there is either a point or a line of equilibrium depending on the relative size and positions of the spheres and on q' /q. Draw a diagram for each case, giving the lines of force and the sections of the equipotentials by a plane through the centers. 13C. An electrified body is placed in the vicinity of a conductor in the form of a surface of anticlastic curvature. Show that at that point of any line of force passing from the body to the conductor, at which the force is a minimum, the principal curvatures of the equipotential surface are equal and opposite. 14C. If two charged concentric shells are connected by a wire, the inner one is wholly discharged. If the force law were r-(2+P), prove that there would be a charge B on the inner shell such that if A were the charge on the outer shell and f, g the sum and difference of the radii 2gB = -Ap[(f - g) In (f + g) - f In f + g In g], approximately. 15C. Three infinite parallel wires cut a plane perpendicular to them in the angular points A, B, C of an equilateral triangle and have charges q, q, -q' per unit length, respectively. Prove that the extreme lines of force which pass from A to C make at starting angles (2q - 5q')7(6q)-' and - (2q q')7(6q)-1with AC, provided that q' > 2q. 16C. A negative point charge -q2 lies between two positive point charges qi and qa on the line joining them and at distances a, $ from them, respectively. Show that if the magnitudes of the charges are given by quErt = q3a-1 = q2X3(a + 15)-1and if 1 < X2 < (a + 0)2(a - ft)-2, there is a circle at every point of which the force vanishes. Determine the general form of the equipotential surface on which this circle lies. 17C. Charges of electricity qi, -q2, q3, (q3 > a.-e placed in a straight line, the negative charge being midway between the other two. Show that, if 4q2 lie between (q31 q11)3 and (q31 OA the number of unit tubes of force that pass from qi to q2 is q2 - q3) 3(214)-1(q31 - q13)(q11 - 210 + 18C. An infinite plane is charged to surface density f, and P is a point distant in. from the plane. Show that of the total intensity icr/e at P half is due to the charges at points that are within 1 in. of P and half to the charges beyond. 19C. A disk of vulcanite (nonconducting), of radius 5 in., is charged to a uniform surface density by friction. Find the electric intensities at points on the axis of the disk distant, respectively, 1, 3, 5, 7 in. from the surface.
24
BASIC IDEAS OF ELECTROSTATICS
20. Two parallel coaxial circular rings of radii a and b carry uniformly distributed charges Q1and Qz. The distance between their planes is c. Show that the force between them is
F—
ck 3(2,(22
E
1672e(ab)11 — k2)
where k2 —
4ab c2 -I- (a
b)2
and E is a complete elliptic integral of modulus k. 21. Show that at a great distance the field of a ring charge —Q of radius b concentric and coplanar with a ring charge Q of radius c is identical with that of a linear quadrupole in which the end charges —Q are at a distance a from the center charge 2Q, provided that b2 — c2 = 4a2. References Useful treatments of the material of this chapter occur in the following books: and R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Uses vector notation and treats simple cases. CORSON, D. R., and P. LORRAIN: "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Well-illustrated dielectric discussion. DuitAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Complete and well-illustrated treatment. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Berlin, 1927. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapters I and IV are especially pertinent. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Uses long notation but gives comprehensive treatment. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives an excellent treatment using vectors. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation with many good field drawings. OWEN, G. E.: "Electromagnetic Theory," Allyn and Bacon, 1963. Has clear and well-illustrated discussion of multipoles. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Chapter I discusses clearly the material in this chapter. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives extensive mathematical treatment based on Maxwell's equations. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Gives clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. ABRAHAM, M.,
CHAPTER II CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS 2.00. Uniqueness Theorem.—Before trying to find the potentials in a system of conductors when the charges are given or vice versa, it is well to be sure that there is only one correct solution. First let us assume that a surface density a on the conductors produces the same potentials as a different surface density a'. The potential at a point P on the surface of one of the conductors due to the surface density a — a' will be, from 1.06 (6),
VP
cr — =
s 4rer
ak)
1
1 f
„
= — — — — — aO ilire S sr 47re ,s r
where r is the distance from P to the surface element dS and the integration covers the surface of all conductors. The two integrals on the right were initially assumed equal, so that VP is zero. The surface density — a therefore makes all conductors at zero potential, so that there is no electric field and a — a is zero. Therefore a = a, and the distribution is unique. We have proved then that only one distribution of electric charge will give a specified potential to every conductor in an electric field. Now let us assume that a surface density a on the conductors gives the same total charge Q in each conductor as a different surface density a. The surface density a — a' will then give a total charge zero on each conductor. The density a — of may be zero over the whole conductor or negative on some areas and positive on others. If the latter is the case, the tubes of force ending on negative areas must originate at points of higher potential and those ending on positive areas at points of lower potential. This reasoning applies to every conductor in the field so that, with the density a — a', no conductor can be at the maximum or the minimum of potential. Therefore, all of them must be at the same potential, and the field vanishes, which gives a = a'. We have now proved that there is only one surface charge distribution which will give a specified total charge to each conductor in an electric field. 2.01. Capacitance.—One consequence of the fact (1.14) that all electrical stresses in a medium in an electric field depend on the intensity in the same way is that the equilibrium is undisturbed if the intensity everywhere in the field is changed by the same factor. Thus if we double the surface density at every point in a system of charged conductors, the 25
26
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.02
configuration of the field is unchanged but the intensity is doubled, and hence the work required to take a small charge from one conductor to another is also doubled. This constant charge to potential ratio of an isolated conductor is its capacitance, and the reciprocal ratio is its -21astance. These terms are not precise when other conductors are in the field unless all are both earthed and uncharged. If Q is the charge in coulombs, C the capacitance in farads, S the elastance in darafs, and V the potential in volts, then the definitive equations are Q = CV V = SQ
(1)
Two insulated conductors near together constitute a simple capacitor. If these two conductors are given equal and opposite charges, the capacitance of the capacitor is the ratio of the charge on one to the difference of potential between them. The ratio is always taken so as to make the capacitance a positive quantity. Thus, for a capacitor we have V1— V2 = SQ (2) Q = C(Vi — V2) 2.02. Capacitors in Series and Parallel.—If we take n simple uncharged capacitors, connect one member of each capacitor to a terminal A and the other member to a terminal B, as shown in Fig. 2.02a, and then apply a potential V between A and B, we have
Q = C1V,
Q2
Q. = C.V
= C2V,
where Q1, Qz, . . . , Q. are the charges on the capacitors whose capacitances are C1, Cz, . . . , C,,, respectively. The total charge is then
Q = Q1 + Q2 + • • + Q. = V(Ci + C2
+ • • • + Cn)
Thus these capacitors, which are said to be connected in parallel or "multiple arc," behave like a single capacitor whose capacitance is C
CI ± C2 + • • • + C.
(1)
Let us take n simple uncharged capacitors, connecting the first member of capacitor 1 to A and the second member to the first member of capacitor 2, then connecting the second member of 2 to the first member of 3, and so forth, finally connecting the second member of the nth capacitor to B as shown in Fig. 2.02b. These capacitors are now said to be connected in series or "cascade." The application of a potential between A and B gives V = V1 + V2 + • • + V. = S1Q1 + S2Q2 + • • • + S.Q. where Vi is the potential across the ith capacitor. Since each pair of connected conductors has remained insulated, its net charge is zero. Hence, if all the tubes of force leaving one member of a capacitor end on the other member of the same capacitor, we have Qi =
Q2 = • • • =
Q.
§2.03
SPHERICAL CAPACITORS
27
so that the charge factors out of the right side of the equation. Thus these capacitors, connected in series, behave like a single capacitor of capacitance C or elastance S where S
= SI + 82
1
+ • • • +5,,,
=1+
± • • • + c.
(2)
We observe that, in general, it will not be possible to confine the fields in the manner assumed so that this formula is an approximation. The additional "distributed" capacitance due to stray fields is usually negligible in cases where the members of a capacitor are close together, and when the capacitivity of the region between is high compared with that outside the capacitor. In the case of air capacitors with widely separated members, this formula may be worthless A
°11 141 -0240
3
A. 4
Ca
C2
C'
B FIG. 2.02a.
C4
2.02b.
2.03. Spherical Capacitors.—Consider a pair of concentric spherical conducting shells, the inner, of radius a, carrying a charge +Q and the outer, of radius b, carrying a charge —Q, the space between being filled by a homogeneous isotropic dielectric of capacitivity E. From symmetry, the electric displacement must be directed radially and its magnitude can depend only on r. Hence, applying Gauss's electric flux theorem to a concentric spherical surface of radius r where b > r > a, we have
fseE • n dS = 47rr2eE = Q
so that
aV car
Q
47rer2
The potential difference between the spheres is therefore Va
Q radr — — Lirejb r2
Q (1 47re \a
b
_ b — aQ 47reab
This gives the capacitance of a spherical capacitor to be
C= 4ireab b—a
(1)
The capacitance of a single sphere of radius a immersed in a dielectric of capacitivity e can be obtained from (1) by letting b —) Go giving (2) C = 47rea
28
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.05
It should be noted that in deriving (1) it is assumed that there is no charge on the outside of b, which requires that b be at zero potential. If this is not the case, the additional capacitance between the outside of b and infinity, computed from (2), must be considered. 2.04. Cylindrical Capacitors.—Consider a pair of concentric circular conducting cylinders of infinite length, the inner, of radius a, carrying a charge Q per unit length and the outer, of radius b, carrying a charge —Q per unit length, the space between being filled with a homogeneous isotropic medium of capacitivity e. From symmetry, the electric displacement must be directed radially outward from the axis and lie in a plane normal to the axis, and its magnitude must depend only on r. Apply Gauss's electric flux theorem to the volume enclosed by two planes, set normal to the axis and one meter apart, and the concentric circular cylinder of radius r when b > r > a. The plane walls contribute nothing to the surface integral; we have therefore
so that
fseE • n dS = 2rreE = Q Q
E_ aV ar
(1)
2irer
The potential difference between the cylinders is therefore
Q fadr 27rejb r
Q a 2ire b
Thus the capacitance per unit length of a long cylindrical capacitor is C—
2re In
(b/a)
(2)
If we let b—> co, we see that C —› 0. Therefore, if there is a finite charge per unit length on a circular cylinder of finite radius and infinite length, the potential difference between its surface and infinity is infinite. Since physically we deal only with cylinders of finite length, this difficulty does not arise, but it indicates that the results of this article apply only where the distance to the surface of the cylinder is small compared with the distance to the ends. 2.05. Parallel-plate Capacitors.—When two infinite parallel conducting planes, carrying charges ±Q and — Q, are a distance a apart, the space between being filled with a homogeneous isotropic dielectric, we see from symmetry that the field between them must be uniform and normal to the plates. If cr is the charge per square meter, then there must be, from 1.13, a unit tubes leaving every square meter of the plates. Thus we have for the displacement and intensity between the plates the relation
GUARD RINGS
*2.06
29
av D = eE = —e— =o
ax
The difference of potential between the plates V2
—
V1
is
= —Crefoa dx = va
Therefore the capacitance per unit area is e/a, and the capacitance of an area A is ry
eA
=— a
(1)
In practice, the field will be uniform only far from the edges of the plate, so that this formula is an approximation which is good if a is small compared with all surface dimensions of the plate and still better if, in addition, the capacitivity of the region between the plates is higher than that of the region beyond the edges. 2.06. Guard Rings.—The derivation of formula 2.04 (2) for the capacitance per unit length of a cylindrical capacitor and that of 2.05 (1) for the capacitance of a parallelplate capacitor both involve the / ( ( y ) hypothesis of conductors of infi11, (172 nite dimensions. To permit
FIG.
2.06a.
FIG.
2.06b.
the application of these formulas to actual capacitors, a device known as a guard ring is used. For the cylindrical capacitors, shown in Fig. 2.06a, the end sections of one of the members are separated from the center section by narrow cracks but are maintained at the same potential. Thus the distorted field near the edges does not affect the center section, except for a very small effect produced by the cracks, so that the charge on this section is related to the potential difference by 2.04 (1.1). A similar arrangement is used for the parallel-plate capacitor by leaving a narrow gap between the central section of one plate and the area surrounding it, maintaining both at the same potential as shown in Fig. 2.06b. The field between the central areas is now uniform except for the negligible effect of the small gap, and the charge on this section is related to the potential difference by 2.05 (1.1).
30
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.08
2.07. Energy of a Charged Capacitor.—We can compute the mutual energy of any system of charges directly from the definition of potential. The work in joules to put the jth charge in place will be, from 1.06 (3), n
IV; =
=
4re
where i
rii
j
The total work to put all charges in place is n
n
W= 1 ~~ q 1 871-E
where i
j
(1)
1=1 j=1
The factor 1 is necessary because the summation includes not only the work done in bringing the ith charge into its position in the field of the jth charge but also that done in bringing the jth charge into the field of the ith charge, which is the same. If Viis the potential at the spot where the ith charge is situated, this may be written, from 1.06 (3), (2) i=i When all charges lie on the same conductor a, they are at the same potential and if their sum is Qa, we may write W =
1
= Wa 2 q:Va = over a
Va •
q1 =
Va
(3)
over a
If the capacitance of a conductor is C, we have, from 2.01 (1), the following equivalent expressions for the energy of a charged conductor: W = IQ V = IQ2/C = iCV2
(4)
For a capacitor, the two members of which carry charges Q and —Q at potentials V1 and V2, respectively, the energy becomes TV = 1071 — rlQV2 = 1Q(T71 — T72) (5) From 2.01 (2), (5) has the same equivalent forms as (4). 2.08. Energy in an Electric Field.—We have seen that visualizing electric forces as transmitted by stresses in the region occupied by an electric field gives results consistent with the observed laws of electrostatics. Where stresses exist, potential energy must be stored. We shall now compute this energy density. Consider an infinitesimal diskshaped element of volume oriented in such a way that the two faces are equipotential surfaces. If this element is taken sufficiently small, the faces of the disk will be flat and parallel and the field inside uniform and identical with that in an infinitesimal parallel-plate capacitor. Let the thickness of the disk be ds; then the difference of potential between the
§2.081
PARALLEL-PLATE CAPACITOR
31
faces is (av /as) ds = —E ds. Let n be the unit vector normal to the faces so that E = En. The charge on an area dS of the face is
D • n dS — D • E dS and the capacitor volume is dv = ds dS, so that 2.07 (4) gives
dW — D • E dv This gives, for the energy density in the field,
dW _ D • E dv 2
(1)
In an isotropic dielectric, D • E = DE, giving
dW _ €E2 _ DE _ D2 dv — 2 — 2 — 2€
(2)
In a crystalline dielectric, we have, from 1.19 (5), dW
= 1(€11E1 + €22E2 + €33Ei 2612E1E2 + 2E13E1E3 + 2E23E2E3) (3) dv or if the coordinate axes are chosen to coincide with the electric axes of the crystal, we have, from 1.19 (6), dW = dv
2 + €2E1; + €3ED
(4)
2.081. Parallel-plate Capacitor with Crystalline Dielectric.—Let us now calculate the capacitance per square meter of a parallel-plate capacitor where the dielectric consists of a slab of crystal of thickness d. Let the capacitivities along the crystal axes x, y, and z be El, €2, and E3, respectively, and let the direction cosines of the angle made by the normal to the capacitor plates with these axes be /,., and n. Since, electrically, one square meter section is like any other, the equipotentials must be parallel to the plates and equally spaced and the electric intensity must lie along the normal. Thus we have
V E = (X2 ± Y2 4- Z2)-1 = 71where V is the potential across the capacitor. Thus
nv nt V Z= X = IV — Y= (1) d Substituting in 2.08 (4), multiplying by the volume d of a square meter section, and using 2.07 (4), we have d)21 2C1V2 2 1[6112G)2 €2m2G)2 e 2(V
32
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
so that the capacitance per square meter is 0€3 „,2,2 12 61
CI =
§2.09
(2)
2.09. Stresses When the Capacitivity Is a Function of Density.—In considering stresses in a dielectric heretofore in 1.16 and 1.17, we have ignored the possibility that the capacitivity may actually change with density T so that there may be a hydrostatic stress tending to expand or contract the dielectric. By working with a volume element of the shape and orientation used in 2.08, we can simplify the investigation to that of a small parallel-plate capacitor of area AS and spacing 3 in which the charge on the plates is considered as embedded in the dielectric at the boundary. Combining 2.07 (4) with 2.05 (1) and assuming an isotropic dielectric of capacitivity E, we have for the energy of our capacitor
mD2
Q2 _ Q2 _ AS 2e AS — 2er AS 2er where we have let m be the mass of the dielectric per unit area between the plates so that m = ra. If m is assumed constant and E is taken as a function of T, the force on an area AS of the plate is a(AW) a (A W) a, = D2 a(er) As AF = — AW=
as
as
Or
— 2E2 a,
Thus the stress or force per unit area pulling on the surface of the conductor is _AF _ D2a(ET) (1) AS — 2E2 a, Carrying out the differentiation and comparing with 1.16 (2), we see that the additional hydrostatic stress is €,E2 arc D2a€ E2 (2) — 2E2Tor = — 2 Tar = — 2 r a, At the surface between two dielectrics, we shall now have to add to those stresses already considered the difference in this hydrostatic pressure giving, in place of 1.17 (7), for the total stress directed from K' to K" the value 1 1-1(' — K"(g2 13'2) D'2r' aK' D"2," 0K" ] F = (3) K"2 n K' \K' K" K' 2 a,' a," At the surface between a dielectric and a vacuum set K" = 1 and alc"/ar" = 0, we have
F
lc - 1 N2 D ,2)
[
24 e„E' 2[
2
(K'
1)
D/2T'OKI
K' 2 a,' j
K' \K' ,aK'j T
a
eE;,2
1 (K' (K — 1))2
2
(4)
§2.11
FORCE ON CONDUCTOR IN DIELECTRIC
where En = g2
en2
=
33
ID;A2
The sign of aK'/ar' may be either positive or negative, and if this term is large enough to predominate in (4) the dielectric may either expand or contract in the field. The phenomenon is known as electrostriction and has been observed by Quincke and others. 2.10. Electrostriction in Liquid Dielectrics.—There is a relation known as the Clausius-Mossotti formula which connects the relative capacitivity with the density for a given liquid.
K — 1 = CT K-{- 2
(1)
C is a constant characteristic of the liquid. This formula, although theoretically derived on incomplete assumptions, has been extremely well confirmed in many cases experimentally (see Geiger-Scheel, "Handbuch der Physik," Band XII, p. 518). Differentiating, it gives 2)(K — 1) a K _ (K 2) 2 (K C — tar 3r 3 Substituting in 2.09 (2), we find that the hydrostatic pressure tending to contract the liquid is e E2(K -1- 2) (K — 1) P —' (2) 2 3 We have assumed throughout that the liquid is nearly incompressible so that r is nearly constant. Thus we see that if we had a charged sphere immersed in a large body of liquid dielectric the pressure given by (2) would vary inversely as the fourth power of the distance from the center of the sphere, and if the liquid were slightly compressible it would be most dense at the sphere's surface, when other effects such as gravitation are neglected. 2.11. Force on Conductor in Dielectric.—So far we have considered the force at the charged boundary between a dielectric and a conductor as if the charge were on the dielectric side of the boundary so that the entire field lies in the dielectric. This gives, from 1.16 (2), the force per unit area to be D2/(2€), neglecting electrostriction. We know from considering the energy of a charged capacitor that this result is correct. Now let us investigate the force if the charge is taken to be on the conducting side of the boundary, the capacitivity of the conductor being e'. The force then becomes D2/(269. The surface of the dielectric is now in the field so that there will be a tension on it pressing against the conductor whose amount is given by 2.09 (3). Neglecting electrostriction and remembering that the field is normal to the surface we obtain for the sum.
34
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.12
of these forces F2 —
K — Kr D2— 2€,,KK'
— e' D2 2c€'
The total force acting on the conductor will now be F = F1—
D2 K — K K' D2 = = 2E„KK' 2€
F2
(1)
Thus either point of view as to the location of the charge gives the same resultant force. If, instead of being discontinuous, the relative capacitivity is considered to change rapidly but continuously in crossing the boundary, the stresses must be determined by integrating. The total resultant stresses come out the same as before although their distribution near the boundary is different. 2.12. Green's Reciprocation Theorem.—We shall now prove that if charges Q 1, Q2, . . . Qn on the conductors of a system give rise to potentials V1, V2, . . . , V. and if charges Qi, Q3 . . . , Q, give rise to potentials VI, V2i . . . , V,„ then ,
ZQJT'i =
(1)
Let us first consider a system of point charges and write down in the form of a matrix n2 terms, each of which comprises the product of one point charge by the potential of a second point charge. Using 1.06 (3), we shall write the sum of the columns in the bottom row and the sum of the horizontal rows in the right-hand column. Thus we obtain
0 d- qq1
47rEr2i
gli g2 _F
-r
0
47r r12 qt q;q3 A ±
airEri 3
Viqn
4rer23
43rer3i
-F
-r
T Aqq2 A-ire 32
m
-F 0 + •
_F
+
q;q.
4rEr1n
47rEr2n
4rer3n
VIVI
V2 V2
g3 V3
— qi
'IV ggri En1
V42 — 47rErn 2
• . •
d_
q2-17
V43 . q3r3
ilrern
0 = qnn qfn Vn
The order of summation being immaterial, the sum may be obtained by adding the terms of the last row or last column. Equating these gives
1q;Va = Zgara e =1
13.°1
(2)
§2.14
INDUCED CHARGES ON EARTHED CONDUCTORS
35
It should be noted that for these point charges V, is the potential at qa due to all unprimed charges except q, itself. All charges located on the same conductor are multiplied by the same potential and so may be collected; thus IVY = = Q'V giving (1). One important application of this theorem is obtained by putting Q1, V3, • • • , Qn = 0, Q2, Q3, • . , Q. = 0, and Q1 = Qs in (1). This gives V', = V2. Thus the potential to which an uncharged conductor A is raised by putting a charge Q on B is the same as that to which B, when uncharged, will be raised by putting the charge Q on A. This theorem is still valid if dielectric boundaries are present as is proved in 3.07. 2.13. Superposition of Fields.—By adding MQ,,Vi or EQ',W, to both sides of 2.12 (1) and comparing with 2.12 (1), we see that if charges Qi, Q2, . . . , Q. produce potentials VI, V2, . . . , V. etc., then charges Vn Qi + VI, Q2 + (4 • • • , Q. + Q'n give V1 + VII, V2 ± A great many new problems can be solved by superimposing the solutions of known problems. As an example of this, suppose we are given the charges Qi, Q2, . . . , Qn on n concentric conducting spheres of radii r1, r2, . . . , r. and wish to find the potential of any sphere, say the sth. We may superimpose the potential produced by each of these charged shells separately, giving 42r€V. = (Q1 +
Q2 + • •
+ Qs)rTi
(23-1-177.4!.14- • • • + Qnr-n-1 (1)
since the potential outside a charged shell at a distance r from its center is independent of its radius and equals Q(471-Er)-1, and the potential inside a conducting spherical shell of radius a is equal to Q(4rEa)-1. 2.14. Induced Charges on Earthed Conductors.—If we place a point charge q at P in the neighborhood of a group of earthed conductors, induced charges will appear on them. We can find the induced charge Q on any one of these conductors if we know the potential V; to which the point P would be raised, the charge q being absent, by raising that conductor to a potential V'. For, from 2.12 (1), QV' ±
qc 0 ± q2 • 0 + • • • = Q' • 0 ± 0 • V ,
0 ± 0-F. • -
so that Q=—
V' 1
(1)
If a conducting sphere is the only conductor and P is at a distance r from its center, we have, from 2.03 (2), V' = qa4irEa)-1and V; = qa42rer)-', so that the charge induced on it by a charge q at P is Q=
aq r
(2)
36
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.15
If a point P lies between two conductors, one of which encloses the other, and if the potential at P is known when the conductors are at potentials V'1 and -172, respectively, then the charge induced on either, when both are earthed, by a charge q at P can be found by 2.12 (1). Let the induced charges be Q1 and Q2; then qV; = 0 Also since all tubes of force from q end on these conductors, we have the relation Q1 ± Q2 = —q. Solving gives
Q1 =
2
—
V1 -
V' -17;q
and
Q2 =
V' — V' 1 V i4 TP2 —
(3)
Thus from 2.03 if the point P lies between two earthed spheres, the charge induced on the inner and outer spheres are, respectively, Q1 = —
ri(r2 — r) r(r2 — r1)'2'
and
Q2
—
r2(r — r1) r(r2 — ri)q
(4)
where r is the distance of P from the center and r1 < r < r2. From 2.04 if P lies between two earthed cylinders, the charge induced on the inner and outer are, respectively, -
In (r2/r) In (r2/r1)
and
Q2 =
In (ri/r) In (ri/r2)q
(5)
From 2.05 if P lies between two earthed plates at a distance a from the first plate and b from the second, we have aq bq (2, = and (6) Q2 a ±b a±b 2.15. Self- and Mutual Elastance.—Consider n initially uncharged conductors, fixed in position and shape. We have seen that putting a charge on any conductor of the group will affect the potential of all other conductors in a definite way which depends only on the geometrical configuration of the system and the capacitivity. The ratio of the rise in potential V,. of conductor r to the charge Q, placed on conductor s to produce this rise is called the coefficient of potential or mutual elastance s.„.. The first application of Green's reciprocation theorem (2.12) shows that 8, = Sr,. A superposition of solutions for charges Q„ Q„ Qt, etc., on conductors r, s, t gives V1 = siiQi
821Q2
V2 = S12Q1
S22Q2
••+ • • •
Sn1Qn Sn2Qn
(1)
V = si.Qi T 82.Q2 - • + snnQ,, Thus ssr is the potential to which the rth conductor is raised when a unit charge is placed on the sth conductor, all the other conductors being „
ELECTRIC SCREENING
§2.17
37
present but uncharged. Putting a positive charge on a conductor always raises the potential of neighboring insulated conductors so s„ is always positive. The coefficient s„ is called self-elastance. 2.16. Self- and Mutual Capacitance.—We may solve the set of equations 2.15 (1) and obtain the charge on a conductor in terms of the potentials of neighboring conductors. The solutions are Q1 = C11V 1 + C21V2 + • • • + Cn1V n Q2 = Cl2V1
Qn
= cinVi
• • • 4" Cn2Vn
C22V 2
(1)
c2nV2 + • • • + c..Vn
where S22832 •
S21831 •
• • Sn2
1 S23S33 • • • Sn3 cll = — A
C12
=
C21
=
• • Snl
1 S23S33 • • • Sn3
S2nS3n • • • Snn
82nS3n • • • Snn
and S11S21 • • • Snl
A =
812S22 • • • Sn2
(2)
S1nS2n • • • Snn
so that c, is the cofactor of s„ in A divided by A. The quantity Cr, is called the coefficient of capacitance or the selfcapacitance and is defined as the charge to potential ratio on the rth conductor when the other conductors are present but earthed. Since the potential has the same sign as the charge, c, is always positive. The quantity c„ is called the coefficient of induction or the mutual capacitance and may be defined as the ratio of the induced charge on the rth conductor to the potential of the sth conductor when all conductors, except the sth, are grounded. The induced charge is always opposite in sign to the inducing charge so c„is always negative or zero. 2.17. Electric Screening.—Suppose Fm. 2.17. conductor 2 surrounds conductor 1 as in Fig. 2.17. All tubes of force from 1 end on 2 so that, if V2 = 0, the charge on 1 depends only on its own potential, which means that, in 2.16 (1), Cal = C41 = • • • = Cnl = 0
(1)
Thus there is no inductive effect between 1 and any of the conductors outside of 2. In this case, we say that they are electrically screened from
38
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.19
1. We notice also that, since Q2 = —Qi when all but 1 are earthed, = —C12. Because of the complete similarity of 2.15 (1) and 2.16(1), we see that we can express the elastances in terms of the capacitances by an interchange of s and c in 2.16 (2). When 1 is screened by 2, we put in the values given by (1) and obtain (2)
1 V2, tubes of force pass from V1 to V2 and Qi is positive so that sii— s12 is positive, and since, from (2), s12 = 522, we have 511 >
Si2
and
sii >
(3)
522
2.18. Elastances and Capacitances for Two Distant Conductors.—Let two conductors 1 and 2 have capacitances C1 and C2 when alone. When 1 is uncharged, let us bring up 2, which has a charge Q2, to a mean distance r from 1, r being large compared with the linear dimensions, of magnitude a, of either conductor. The potential to which 1 is raised is Q2(47-er)-', if we neglect the variation of this potential over the region occupied by 1 which is of the order a(470-2)-1. Comparison of this with the first equation of 2.15 (1) shows that 821 = (471-er)-1. A charge of opposite sign to Q2 and of order of magnitude Q2a(4rEr)-' will be induced on the nearer parts of 1 and an equal charge of the same sign on the more remote parts. Here we have equal and opposite charges, separated by a distance smaller than a which is small compared with r, so that the field at the distance r is essentially a dipole field, and the potential [1.07 (1)] at 2 due to the presence of 1 uncharged is at most of order Q2a2(4rer3)-1. To this order of magnitude, therefore, 1 does not affect the potential at 2 so that, from the second equation in 2.15 (1), we have S22 = V2W1 = CV- and similarly sii = Cr'. Thus to a first approximation, 511 =
1 Ci
812 = S21 = 4E 1 r'
8 22 =
1
(1)
C2
Solving for the self- and mutual capacitances by 2.16 (2), we have, neglecting r--3terms, 167,2,2r2c1 161.2,2r2c2 C,C2
Cu —
1e7r 2E 2r 2
-
ir; ; v2
C12 = C21 =
4rer
C22 = 162.2,2r2
C1C v2
(2)
2.19. Energy of a Charged System.—If the electrical intensity and the displacement were known at all points outside a set of charged conductors, it would be possible to get the energy of the system by integration from
§2.20
FORCES AND TORQUES ON CHARGED CONDUCTORS
39
2.08 (1); thus W = ifvD • E dv
(1)
where the integration extends over the whole volume outside the conductors. Frequently we do not know the field at all points but do know the potentials and charges on conductors and the self- or mutual capacitances. To determine the energy, suppose we add to the charges Qice, Q2a, • • • , Qnce on the conductors, by infinitesimal steps, (21 da, Q2 da, . . . , Qn da, starting with the initial uncharged state where a = 0 and ending with the final state where a = 1. Since the potentials are Via, V2a, . . . , V ,,a when the charges are Qia, Q2a, . . . , Q.a, the work done in a single step is da
dW = V aia da V2Q2a da -I- • • • +
The energy in the final state is then
W=
ViQifola da =
(2)
V iQi
If we substitute for from 2.16 (1), we have Wv = i(ciiVi
2c12V1V2
c22-V2
••)
(3)
• )
(4)
If we substitute for Vifrom 2.15 (1), we have WQ = Esial
40 -4-. 281219,11,2
1,1 4-.
s2210
•
2.20. Forces and Torques on Charged Conductors.—If the electrical intensity and displacement are known at all points on the surface of a conductor, the total resultant force on the conductor in the direction of the unit vector p can be obtained from 2.11 (1) by integration over the surface of the conductor, giving FP
—
fs
D• E p • n cl3 2
(1)
where n is the unit vector normal to the surface element dS. When the charges and the elastances of a set of conductors are known, the potential energy of the system is given by 2.19 (4) in which si b 812, etc., are functions of the configuration of the system. As in mechanics, we determine the force or torque tending to produce a change in any element of this configuration by taking the negative derivative of the potential energy with respect to this element.
aw an
=
1(asn.Q1 — 2 an
+ 2— anQ1Q2 +
•••)
(2)
This gives either a force or a torque tending to increase n depending on whether n is a length or an angle. In this case where the charge is kept
40
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
§2.20
constant, the change in electrical energy equals the mechanical work done. If, on the other hand, we work with 2.19 (3), keeping the potentials constant by a battery or equivalent device, we may supply additional energy to the system. The force in both cases must be the same, since it depends only on the initial state of the system which may be expressed in terms of either charge or potential. To determine the force in terms of the potential, let us combine 2.19 (4), (2), and (3) in the equation TiTv— IV iQi = 0
=
(3)
Taking the total differential, we have
a* dQi aQi
= i=i
dn. = 0
dVi
(4)
i=i
But from (3), 2.19 (4), and 2.15 (1)
a* aQi
a WQ aQi
.177
Sii I —
Vi= 0
CiiVi
Qi
5=1
and from (3), 2.19 (3), and 2.16 (1) n
aq, a wv a vi avi
Vz
=
Substituting in (4) gives dn. -= 0 This sum is zero for any values of dn. we choose; so each term must be separately equal to zero. Substituting for NI,from (3) gives
axp , 7 aW V an. = 0 aW an. an. = (5) on. on. on. But we know that —a WQ/an. is the force or torque tending to increase n., so that in terms of the potentials this is
_LaWv = — 1(acii —Vi2 an.
2 an.
2aci2 viv2
an.
(6)
When the charges are kept constant, the work done making a small displacement is given by (2), and when the potentials are kept constant it is given by (6). The difference must represent the work done by the device maintaining the potential giving
Cawv aw,) an.
=2
aw, dn, an.
(7)
41
PROBLEMS Problems
Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Three identical spheres of radius a are placed with their centers in line, the intervals being r1 and r2, and are numbered in order. At first, only the central one 2 carries a charge Q. It is now connected with 1. This connection is then broken, and it is connected with 3. Show that, if the intervals are large compared with a, the final charge on 3 is are Q + 1] Q3 = 4 [ri (ri ±r2)(r2 — a) 2. Four identical uncharged conducting spheres are p aced at the corners of a square and numbered in rotation. A charge is given to 1, which is then connected for an instant by a thin wire to 2, 3, and 4 in turn. Show that finally Q Q4= Q
511 - 524
Q
Sji - 25,4 + S24
8 0
511 - 514
Q1 -
0 Sll - :314
3. If, initially, spheres 1 and 2 in the last problem had charges +Q and —Q, respectively, and if, as described, 1 is connected for an instant to 3 and 4 in turn, and if the side of the square r is large compared with the radius a of the spheres, show that, finally, Q4 = 2-1Q[2ir — (21— 3)a] / (r — a) approximately, and find Q3 and Qi. 4. Each of three similar spheres, situated a considerable distance apart at the corners of an equilateral triangle, has initially a charge Q. Each in turn is now earthed for an instant and then insulated. Show that the final charge on 3 is a2r-2Q(3 — 2a/r) and find the final charges on 1 and 2. 5. Four similar conductors are arranged at the corners of a regular tetrahedron in such a way that each is perfectly symmetrical with respect to the other three. All are initially uncharged. One is now given a charge Q by means of a battery of voltage V and is then insulated. It is now connected for an instant to each of the other three in turn and then to earth. The charge is now —Q1. Show that all the mutual capacitances are 56Q2Q1/[V(24Q1 + 7Q)(8Q1 7Q)] and find the elastances. 6. Two conductors are mirror images of each other and are initially uncharged. A sphere with a charge Q is now brought in contact with a certain point on one conductor and then with the image point on the other. If, in each case, it shares its charge equally with the conductor, show that after a large number of alternate contacts the charge will be equally distributed between the three conductors. 7. Three similar insulated conductors are so arranged that each is perfectly symmetrical with respect to the other two. A wire from a battery of unknown voltage is touched to each in turn. The charges on the first two are found to be Q1 and Q2. What is the charge on the third? 8. The inner and outer of three concentric spherical conducting shells of radii a, b, and c are grounded. The intervening sphere is split into halves and charged. Find how large a must be in order just to prevent the halves from separating if a b > a. Show that the charge induced on the sphere is — Kabq I tr[b (K — 1)a]}. 15. A spherical capacitor with inner radius a and outer radius b is filled with two spherical layers of capacitivities el and ez, the boundary between being given by r = i(a b). If, when both shells are earthed, a point charge on the dielectric boundary induces equal charges on the inner and outer shells, find the ratio e1/c2. 16. An uncharged conducting spherical shell of mass M floats with one quarter of its volume submerged in a liquid of capacitivity e. To what potential must it be charged to float half submerged? 17C. If the algebraic sum of the charges on a system of conductors is positive, then on one at least the surface density is everywhere positive. 18C. There are a number of insulated conductors in given fixed positions. The capacities of any two of them in their given positions are C, and Cz, and their mutual coefficient of induction is B. Prove that if these conductors are joined by a thin wire, the capacity of the combined conductor is CI +C2 + 2B. 19C. In a system of insulated conductors, having been charged in any manner, charges are transferred from one conductor to another, till they are all brought to the same potential V. Show that V = E As' + 28z), where Si, sz are the algebraic sums of the coefficients of capacity and induction, respectively, and E is the sum of the charges. 20C. Prove that the effect of the operation described in the last question is a decrease of the electrostatic energy equal to what would be the energy of the system if each of the original potentials were diminished by V. 21C. Two equal similar condensers, each consisting of two spherical shells, radii a, b, are insulated and placed at a great distance r apart. Charges e, e' are given to the inner shells. If the outer surfaces are now joined by a wire, show that the loss of energy is, approximately (1670-1(e — e')2(b-1— r-1). 22C. A condenser is formed of two thin concentric shells, radii a, b. A small hole exists in the outer sheet through which an insulated wire passes connecting the inner sheet with a third conductor of capacity c at a great distance r from the condenser. The outer sheet of the condenser is put to earth, and the charge on the two connected conductors is E. Prove that the force on the third conductor is, approximately, ac2E2(47rer3)-,[47reab(a — b)-1
c]-2
23C. Two closed equipotentials VI, Vc) are such that V, contains Vo, and Vp is the potential at any point P between them. If now a charge E is put at P and both equipotentials are replaced by conducting shells and earth connected, then the charges
PROBLEMS
43
E1, E., induced on the two surfaces are given by E,(Vo — V8)-1 = Eo(Vp — V1)-1 = E(V1— Vo)-' 24C. A conductor is charged from an electrophorus by repeated contacts with a plate, which after each contact is recharged with a quantity E of electricity from the electrophorus. Prove that if e is the charge of the conductor after the first operation, the ultimate charge is Ee(E — e)-1. 25C. Four equal uncharged insulated conductors are placed symmetrically at the corners of a regular tetrahedron and are touched in turn by a moving spherical conductor at the points nearest to the center of the tetrahedron, receiving charges el, e2, es, e4. Show that the charges are in geometrical progression. 26C. In 25C, replace "tetrahedron" by "square" and prove that (ei— e2)(eie3— 4) = ei(e2e3 — e1e4) 27C. Two insulated fixed conductors are at given potentials when alone in the electric field and charged with quantities E1, E2 of electricity. Their coefficients of potential are s11, 812, 822. But if they are surrounded by a spherical conductor of very large radius R at zero potential with its center near them, the two conductors require charges E'„ E2fto produce the given potentials. Prove, neglecting R-2, that (.0; — E1)(E2' — E2)-1= (822 — 812) (811 — .312)-1 28C. Show that the locus of the positions, in which a unit charge will induce a given charge on a given uninsulated conductor, is an equipotential surface of that conductor supposed freely electrified. 29C. Prove (1) that if a conductor, insulated in free space and raised to unit potential, produces at any external point P a potential denoted by (P), then a unit charge placed at P in the presence of this conductor uninsulated will induce on it a charge — (P); (2) that if the potential at a point Q due to the induced charge is denoted by (PQ), then (PQ) is a symmetrical function of the positions of P and Q. 30C. Two small uninsulated spheres are placed near together between two large parallel planes, one of which is charged, and the other connected to earth. Show by figures the nature of the disturbance so produced in the uniform field, when the line of centers is (1) perpendicular, (2) parallel to the planes. 31C. A hollow conductor A is at zero potential and contains in its cavity two other insulated conductors B and C, which are mutually external. B has a positive charge, and C is uncharged. Analyze the different types of lines of force that are possible within the cavity, classifying with respect to the conductor from which the line starts and the conductor at which it ends, and proving the impossibility of the geometrically possible types that are rejected. Hence, prove that B and C are at positive potentials, the potential of C being less than that of B. 32C. A portion P of a conductor, the capacity of which is C, can be separated from the conductor. The capacity of this portion, when at a long distance from other bodies, is c. The conductor is insulated, and the part P when at a considerable distance from the remainder is charged with a quantity e and allowed to move under the mutual attraction up to it. Describe and explain the changes that take place in the electrical energy of the system. 33C. A conductor having a charge (21is surrounded by a second conductor with charge Q2. The inner is connected by a wire to a very distant uncharged conductor. It is then disconnected, and the outer conductor connected. Show that the charges V2 are now (m n)Q2-1- mnQ: (2; — nQ2 , (2; = n m m n mn
44
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
where C, C(1 m) are the coefficients of capacity of the near conductors, and Cn is the capacity of the distant one. 34C. If one conductor contains all the others, and there are n 1 in all, show that there are n relations between either the coefficients of potential or the coefficients of induction, and if the potential of the largest is Vo and that of the others V1, V2, . . . , V,,, then the most general expression for the energy is icv,1 increased by a quadratic function of V1— Vo, Vz — Vo, . . . , V. — Vo, where C is a definite constant for all positions of the inner conductors. 35C. The inner sphere of a spherical condenser (radii a, b) has a constant charge E, and the outer conductor is at potential zero. Under the internal forces, the outer conductor contracts from radius b to radius b1. Prove that the work done by the electric forces is E 2(b — bi) I (87 cobb 1) • 36C. If, in the last question, the inner conductor has a constant potential V, its charge being variable, show that the work done is 27r€V2a2(b — b1)/[(51 — a)(b — a)] and investigate the quantity of energy supplied by the battery. 37C. With the usual notation, prove that (sii 12 +SIRS + 8 13 ) , -11-23 > -12 S13. 823) > (, S 38C. Show that if srr, s”, s,, are three coefficients before the introduction of a new conductor and slc, 481 s,', the same coefficients afterward, then (srrs.. — srts:,) 4 (Sr, — s'.)2 39C. A system consists of p q + 2 conductors A1, A 2, . . . , Ap, B1, B2, • • • B5, C, D. Prove that when the charges on the A's and on C and the potentials of the B's and of C are known, there cannot be more than one possible distribution in equilibrium, unless C is electrically screened from D. 40C. A, B, C, D are four conductors, of which B surrounds A and D surrounds C. Given the coefficients of capacity and induction (1) of A and B when C and D are removed, (2) of C and D when A and B are removed, (3) of B and D when A and C are removed, determine those for the complete system of four conductors. 41C. Two equal and similar conductors A and B are charged and placed symmetrically with regard to each other; a third movable conductor C is carried so as to occupy successively two positions, one practically wholly within A, the other within B, the positions being similar and such that the coefficients of potential of C in either position are p, q, r in ascending order of magnitude. In each position, C is in turn connected with the conductor surrounding it, put to earth, and then insulated. Determine the charges on the conductors after any number of cycles of such operations, and show that they ultimately lead to the ratios 1: )3:#2 1, where /3 is the positive root of rx 2— qx p — r = 0. 42C. Two conductors are of capacities C1 and C2, when each is alone in the field. They are both in the field at potentials V1 and V2, respectively, at a great distance r apart. Prove that the repulsion between the conductors is —
—
C1C2(471-Er V l — C2 V2) (47er V2 — VI) (476)/[(1672E2r2 C1C2)9 As far as what power of r-1is this result accurate? 43C. Two equal and similar insulated conductors are symmetrically placed with regard to each other, one of them being uncharged. Another insulated conductor is made to touch them alternately in a symmetrical manner, beginning with the one that has a charge. If e1, e2 are their charges when it has touched each once, show that their charges, when it has touched each r times, are, respectively, —
e?
[1 ( 1 e1 e)1 [1 + (el e1 e2 and 2e1— ez ) 2e1— ez (For 43C and 44C see difference equations, 5.081.)
PROBLEMS
45
44C. Three conductors A1, A 2, and As are such that As is practically inside A2. A1 is alternately connected with Ay and A3 by means of a fine wire, the first contact being with As. A1 has a charge E, initially, A2 and As being uncharged. Prove that the charge on A lafter it has been connected n times with As is
ES [
a+
1+
a —
flyi]
fi(a ± 7) a ± 7
where a, 13, -y stand for sil — 81 , 822 — Si x,and 833— 812. 45C. A spherical condenser, radii a, b, has air in the space between the spheres. The inner sphere receives a coat of paint of uniform thickness t and of a material of which the inductive capacity is K. Find the change produced in the capacity of the condenser. 46C. A conductor has a charge e, and V1, V2 are the potentials of two equipotential surfaces completely surrounding it (V1 > V2). The space between these two surfaces is now filled with a dielectric of inductive capacity K. Show that the change in the energy of the system is 1e(Vi — V2)(K — 1)/K. 47C. The surfaces of an air condenser are concentric spheres. If half the space between the spheres is filled with solid dielectric of specific inductive capacity K, the dividing surface between the solid and the air being a plane through the center of the spheres, show that the capacity will be the same as though the whole dielectric were of K). uniform specific inductive capacity 1(1 48C. The radii of the inner and outer shells of two equal spherical condensers, remote from each other and immersed in an infinite dielectric of inductive capacity K, are, respectively, a and b, and the inductive capacities of the dielectric inside the condensers are K1, K2. Both surfaces of the first condenser are insulated and charged, the second being uncharged. The inner surface of the second condenser is now connected to earth, and the outer surface is connected to the outer surface of the first condenser by a wire of negligible capacity. Show that the loss of energy is Q2[2(b — a)K aK2] 8rb4[(b — a)K aK z]
where Q is the quantity of electricity that flows along the wire. 49C. The outer coating of a long cylindrical condenser is a thin shell of radius a, and the dielectric between the cylinders has inductive capacity K on one side of a plane through the axis and K' on the other side. Show that when the inner cylinder is connected to earth and the outer has charge q per unit length, the resultant force on the outer cylinder is q2(K — K') per unit length 72e„a(K K')2
50C. A heterogeneous dielectric is formed of n concentric spherical layers of specific inductive capacities K1, K2, . . . , K,,, starting from the innermost dielectric, which forms a solid sphere; also, the outermost dielectric extends to infinity. The radii of the spherical boundary surfaces are al, az, . . . , a,,_1, respectively. Prove that the potential due to a quantity Q of electricity at the center of the spheres at a point distant r from the center in the dielectric K. is 1 [Q (1 47re„ K„ r
1)
Q (1 + K,÷1 a ,,
Q 1 I 1 ) a+1 + • • • + K„ a,_1
MC. A condenser is formed by two rectangular parallel conduct ng plates of breadth b and area A at distance d from each other. Also, a parallel slab of a dieleo.
46
CAPACITORS, DIELECTRICS, SYSTEMS OF CONDUCTORS
trio of thickness t and of the same area is between the plates. This slab is pulled along its length from between the plates so that only a length x is between the plates. Prove that the electric force sucking the slab back to its original position is E2 dbt'(d — t') xbt']2 24[A (d — t')
where t' = t(K — 1)/K. K is the specific inductive capacity of the slab, E is the charge, and the disturbances produced.bythe edges are neglected. 52C. Three closed surfaces 1, 2, 3 are equipotentials in an electric field. If the space between 1 and 2 is filled with a dielectric e, and that between 2 and 3 is filled with a dielectric e', show that the capacity of a condenser having 1 and 3 for faces is C, given by 1/C = (GI) e„/(e'B) where A, B are the capacities of air condensers having as faces the surfaces 1, 2 and 2, 3, respectively. 53C. The surface separating two dielectrics (K,, Kz) has an actual charge a per unit area. The electric intensities on the two sides of the boundary are F,, F2 at angles al, cz with the common normal. Show how to determine F,, and prove that K2 cot C2 = K1 cot — a/(e,K,F, cos ci)]. 54C. The space between two concentric spheres, radii a, b, which are kept at potentials A, B is filled with a heterogeneous dielectric of which the inductive capacity varies as the nth power of the distance from their common center. Show that the potential at any point between the surfaces is [(Aan+1— Bbn+1)/(a.+1— bn+1)] — (ab/r)n-"(A — B)/(0+1— bn+1) 55C. A condenser is formed of two parallel plates, distant h apart, one of which is at zero potential. The space between the plates is filled with a dielectric whose inductive capacity increases uniformly from one plate to the other. Show that the capacity per unit area is e„(K, — K,)/[h In (K2/K1)] where K1and K2 are the values of the inductive capacity at the surfaces of the plate. The inequalities of distribution at the edges of the plates are neglected. 56C. A spherical conductor of radius a is surrounded by a concentric spherical conducting shell whose internal radius is b, and the intervening space is occupied by a r)/r. dielectric whose specific inductive capacity at a distance r from the center is (c If the inner sphere is insulated and has a charge E, the shell being connected with the earth, prove that the potential in the dielectric at a distance r from the center is [E/ (47rEc)] In {[b/r][(c r) / (c b)]]. 57C. A spherical conductor of radius a is surrounded by a concentric spherical shell of radius b, and the space between them is filled with a dielectric of which the inductive capacity at distance r from the center is jle—P'71-3where p = ra--1. Prove that the capacity of the condenser so formed is Rare„lia/(eb2/"' — e). 58C. Show that the capacity of a condenser consisting of the conducting spheres r = a, r = b, and a heterogeneous dielectric of inductive capacity K = AO, cb) is e,,ab(b — a)-if ff(0, 0) sin fl do dqS. 59C. In an imaginary crystalline medium, the molecules are disks placed so as to be all parallel to the plane of zy. Show that the components of intensity and polarization (displacement in our notation) are connected by equations of the form f = eiiX + exiY,
g = tia
e22Y,
h = e33Z
60C. A slab of dielectric of inductive capacity K and of thickness z is placed inside a parallel-plate condenser so as to be parallel to the plates. Show that the surface of the slab experiences a tension (1a2 /e,)[1 — K 1— x d(K-1)/dx].
47
PROBLEMS
61C. For a gas K = 1 + Op, where p is the density and B is small. A conductor is immersed in the gas. Show that if 82 is neglected the mechanical force on the conductor is laz/e, per unit area. Give a physical interpretation of this result. 62C. The curve a— x a+x ± 1 y213} — a y9 / a)2 [(x — a)2 when rotated round the axis of x generates a single closed surface, which is made the bounding surface of a conductor. Show that its capacity will be 471-€,,a and that the surface density at the end of the axis will be e/(3ira2), where e is the total charge. 63. The potential ratios, of a system of n conductors may be defined by 9a 16{[(x
1
(x2 +y2)4
V1 = cril(21 Pi2V2 + • • • + V2 = P21 V, c21Q2 + • •
P2.V.
Pn2I72 + • • • + c,T,;(2. Show that Pr, is given in terms of the capacitances, if s # r, by Pr.
— CreC,7,1
References The author has found helpful treatments of the subject matter of this chapter in the following books: ABRAHAM,
M., and R.
BECKER:
"Classical Electricity and Magnetism," Blackie,
D., and P. LORRAIN: "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Second chapter includes much of this material. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Includes a well-illustrated treatment of this material.
CORSON,
J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives comprehensive treatment using long notation.
JEANS,
J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation and contains many good scale drawings of fields.
MAXWELL,
M. K. E. L.: "Theory of Electricity and Magnetism," Macmillan, 1932. Uses vector notation. RAMSEY, A. S.: "Electricity and Magnetism," Cambridge, 1937. Gives clear elementary treatment with problems. Uses vectors. RUSSELL, A.: "Alternating Currents," Cambridge, 1914. Gives applications to multiple conductors. THOMSON, J. J.: "Mathematical Theory of Electricity and Magnetism," Cambridge, 1921. THOMSON, W.: "Papers on Electrostatics and Magnetism," Macmillan, 1884. Solves many problems and gives numerical results on spheres and spherical segments. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Gives a clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. PLANCK,
CHAPTER III GENERAL THEOREMS 3.00. Gauss's Theorem.—We now proceed to find a relation between the integral over a closed surface S and over m — 1 other closed surfaces, inside it, of the normal component of a vector A, which is a continuous function of position in space, and the integral of its divergence defined as V • A, throughout the volume v between these surfaces. Let the components of A in rectangular coordinates be A, Au, Az so that
v•A
= aAz + aA, +aA. ay
ax
(1)
az
Let us suppose v divided up into slender prisms of rectangular cross section dy dz. One of these (Fig. 3.00) cuts elements dgi, dS'l from Si,
the coordinates of which are x; and x7 and the outward drawn normal unit vectors to which are n;'. The contributions of the elements d8; and d8;' to the surface integral are Az,,i •
dS;
But from Fig. 3.00, i • n; dS; = d8;' = dy dz so that the total contribution to all sections of that prism, cut from v, which penetrates surfaces p to q is
dy dz(A„,— Azi,) = 7 °P
dy dz f
z,,
aA ox
dx
7 =P
We now sum over all the prisms into which v is divided, and we have
1
si
Azi • nid51
= f aA z dx dy dz jv ax
48
§3.02
EQUATIONS OF POISSON AND LAPLACE
49
Working out similar expressions for A„ and A, and adding, we obtain, after substituting V • A from (1) and writing dv = dx dy dz, the expression
ZfsiA • ni dS; = fvV • A dv
(2) =1. This formula, apparently first given by Green, states a law widely known as Gauss's theorem. 3.01. Stokes's Theorem.—We can derive another important theorem directly from Gauss's theorem. Let us apply 3.00 (2) to an infinitesimal right cylinder of height h, base area S, curved area S', and peripheral length s. Let n be the unit vector normal to the base and n1 that normal to the side. Let F be some vector function and replace A by n x F. Since n is a constant in the following expression F•Vxn vanishes. V •nxF =F•V:n—n•V:F But [n x F] • n is zero on the flat faces so that the surface integral over S vanishes and 3.00 (2) becomes
fen:F • nidS' = f F•nl xndS' = — f n-V:F dv But dv = h dS and dS' = h ds, so that the surface and volume integral become line and surface integrals, respectively, giving
f F • ni x n ds =
—fsn • Vx F dS
Now n: ni= — n1x n is a unit vector directed along the boundary, so that if we choose it for the positive direction of s we have n1: n ds = —ds, and summing up for all the infinitesimal cylinders composing a surface we have
ZfF • ds = Zisn • V xF dS In summing up the line integrals, every interior contour is covered twice in opposite directions, so that they cancel each other leaving only the integral around the outer edge of the whole area. The sum of the surface integrals is of course the integral over the whole surface, so that
f F • ds = f n•V xF dS
(1)
This is Stokes's theorem and may be stated as follows: The line integral of a vector function F around any closed contour equals the surface integral of its curl over any surface bounded by this contour. 3.02. Equations of Poisson and Laplace.—In Gauss's theorem, let the vector A be the electric displacement D = EE and apply Gauss's electric
flux theorem [1.15 (1)] to the surface integral, and we obtain fvV • D dv = Q = fvp dv
GENERAL THEOREMS
50
§3.03
where p is the electric charge density. Thus we have, if dv becomes vanishingly small, dQ (1) div D = V • D = — = p
dv
If we write D = EE = —e grad V = — eVV, this becomes div (e grad V) = V • (eVV) = — p
(2)
This is Poisson's equation for a nonhomogeneous dielectric. If the dielectric is homogeneous, e is a constant and may be factored out, giving div grad V = V VV = V2 V =
(3)
If p = 0, we obtain Laplace's equation which may then be written, in rectangular coordinates, for a nonhomogeneous but isotropic dielectric
a aX
E
ax
,a( -r 6 ay ay
,— a(6-- _— az az
(4)
If the dielectric is homogeneous but not isotropic and if the coordinate axes lie on the electric axes of the crystal as in 1.19 (7), this becomes
a2 v E1
ax2
a 2v
+
a2v ( + E 3---T
= 0
(5)
If the dielectric is both homogeneous and isotropic, the equation becomes
a 2v ax2
a2va2v =o az2 ay2
(6)
3.03. Orthogonal Curvilinear Coordinates.—In most electrostatic problems, the data given are the charges on, or the potentials of, all conductors involved, the size and location of all other charges, and the capacitivity at all field points. The problem is considered solved when the potential at all points has been determined. To do this, it is necessary to find a solution of Laplace's equation that can be made to fit the given boundary conditions. There is usually one system of coordinates in terms of which these conditions can be most simply expressed, and it is therefore desirable to solve Laplace's equation in this coordinate system. All commonly used coordinate systems can be classed as orthogonal curvilinear coordinates. Consider three families of mutually orthogonal surfaces, one member of each family passing through every point of the region under consideration. Any member of the first family can be specified by giving the proper numerical value to u1and similarly for the other families by u2 or u3. An infinitesimal rectangular parallelepiped will be cut out of space by the six surfaces 21,1, u14- dui, u2, u2 duz, u3, u3 dui. Since the quantities u1, u2, and u3will represent distances in a few cases only,
CURL IN ORTHOGONAL CURVILINEAR COORDINATES
§3.04
51
it will, in general, be necessary to multiply dui, du2, and du3by certain factors* h1, h2, and h3to get the actual length of its edges. These factors may be variable from point to point in the field, and so they are functions of u1, U2p and u3. The lengths of the edges of our parallelepiped are therefore F ds3 = h3 du3 (1) ds2 = h2 du2, dsi = hi dui, These are shown in Fig. 3.03. If V is a scalar, then, by definition, components of the gradient are
av 0.32
av = av as, h, au,
—
av
C
ay
.33
h 2 0142'
(2) — hL3 ., OU3
0
Fro. 3.03.
To calculate the divergence in this coordinate system apply Gauss's theorem [3.00 (2)] to the infinitesimal volume shown in Fig. 3.03. The outward normal component of the flux of the vector A through the faces OCGB and ADFE is
a a a —(A, ds2ds3) ds, = as, ou,(h2h3A dui du2 du3— h,h2h3 au,(h2h3A,) dv Adding the two corresponding expressions for the other four faces and comparing with 3.00 (2), we see that div A = v • A — Now letting A = we have 1
1 r a (h2h324 1) hih2h31_ aul
— (h3h1A 2)
au2
a — (h,h2A3)] au3
(3)
VV and substituting in Poisson's equation [3.02 (2)],
r a (h2h3e
av)
a 03hie av)
hih2h3L au,‘ hi au, + au2
au2
a eih2€ aV au3 h 3 au3)] = — P
v • (e VV) = —p
(4)
Putting p = 0 gives us Laplace's equation for a nonhomogeneous isotropic dielectric. If the dielectric is homogeneous and isotropic, e as well as (h1h2h3)-1may be factored out of the equation. 3.04. Curl in Orthogonal Curvilinear Coordinates.—Let us apply Stokes's theorem to the face 0 ADC of the elementary curvilinear cube of Fig. 3.03. Let F1, F2, and F3 be the components of a vector along u1, u2, and u3. Then, from 3.03 (2), the line integral along OA and DC is
u2)F
u2) —
u2
du2)Fi(ui, u2 du2)] dui (h ,F7 au2
duidu2
* Our notation is that used by Houston, "Principles of Mathematical Physics,"
and Abraham-Becker, "Classical Electricity and Magnetism." The Peirce and Smithsonian Tables and Jeans, "Electricity and Magnetism," use h1, h2, and ha for the reciprocal quantities.
§3.05
GENERAL THEOREMS
52
and along AD and CO it is [h2(ui -1-- dui, u2)F2(ui + dui,
u2) — h2(UI, U2)F2 (U1, U2)] dU2 —
(h2F2)dUl dU2
aui
Adding these gives the line integral around this face, and, by Stokes's theorem, this equals the integral of the normal component of the curl of F over the area hih2 dui du2 of the face. Canceling out the factor dui du2, this gives ra(h2F2) a(h1F1)] (1) (V x F)3 = 1 au2 hih21_ aul and similarly for the other faces, we have 1 [a (haF3)
x F = h2h3 (V x F)2 =
a (h2F2)1
(2)
au3
au2
1 [a (hiF i) h3h1 au3
a(haF3)]
3)
(
au,.
3.05. V • (e VV) in Other Coordinate Systems.—In spherical polar coordinates where r = u1is the distance from the origin, 0 = u2 is the
(a) FIG. 3.05.
colatitude angle, and q5 = u3 is the longitude angle, we have h1 = 1 dsi = hi dui = dr ds2 = h2 du2= r d0 h2 = r h3 = r sin 0 ds3 = h3 du3 =r sin 0 det. From 3.03 (4), Laplace's equation becomes
1 a 2av T *2 ar \Er ar
1 a
.
r2 sin ao tsin
0a_ u
Ae
1 + r2 sin2 e
ad,
=0
(1)
In cylindrical coordinates where p = u1 is the distance from the z-axis, 4) = u2 is the longitude angle, and z = u3is the distance from the xy-plane, we have h1 = 1 dsi = h1dui = dp ds2 = h2 du2= p d4) h2 = p h3 = 1 ds3 = h3 du3 = dz
GREEN'S THEOREMS
§3.06
53
This arrangement is shown in Fig. 3.05b. From 3.03 (4), Laplace's equation becomes V • (e V V)
_1 a cay) + 1. a ( av) + ean = 0 pap\ apl p2 ao\ ail az\ az/
(2)
We shall have occasion to use still other coordinate systems, such as the confocal system. These will be taken up later in connection with special problems. 3.06. Green's Theorems.—In 3.00 (2), let A = (e grad 43)* = *e V43 where if and (13 are scalar quantities which are finite and continuous in the region of integration and can be differentiated twice, and e is a scalar quantity which may be differentiated once and which may be discontinuous at certain boundaries in the region. We shall exclude these discontinuities by drawing surfaces around them which fit them closely on both sides. Let n; and np" be the unit normal vectors drawn into the pth boundary from the two sides, and let the values of A on the two ' and Then if there are q such surfaces enclosing q dissides be Ap continuities, the integral over them is
fs,(A; • n; Ap • n';) dS, p=i where dS, is an element of the original surface of discontinuity. Adding these terms to 3.00 (2) and substituting A = \Fe In, we have
e* &to Is, an' dS•
f (epq/pZ epV s ,
P=1
=f
1r
dS, )
V. (e Vt.) dv e V* • V43 dv ± f ‘1,17 v
(1)
If we write a similar equation with' and ' interchanged and subtract the two, we obtain q
p=1
ac; a& — (1)"°*;:)1 dS„ an II eil`P' Panp ' 1 + e;:(41 \ an" \ van;— V P an", P P i m
1••1
fsi
€(* `L — an;
)d = f [4/V • (e V') — (1)17 • (e VNP)] dv (2)
an;
When e is a constant, without discontinuities, (1) becomes C13 d S = f (vim • 7.13 + s, on;
j.•1
'724,) dv
(3)
GENERAL THEOREMS
54
§3.08
and (2) becomes m
fs,(1/wi aci) — ca ll dSi =fv NI V 24) — cl) V2 T) dv
(4)
i=1 A useful vector analogue of (3) and (4) has been proved by Stratton as follows. In 3.00 (2) let A = x (V x 41) where Ar and 43 are vector quantities that are finite and continuous in the region of integration and can be differentiated twice. Then E [117 x (V x OA • ni d$5 = fvv • ['ir x
(v x 4,)]
dv
si i
lr • [(V 41)x n] dS; = fv [(V x w) • (V x 4)) — w • v x (V x 0)] dv (5) 5-1 Subtraction of a similar equation with AP and interchanged gives m
fsi
[Ar x (V x 4)) — x (V x I•)] • n dS;
5-1
=
rt
Jv
cp • [V x (V x 11?)] — AP • IV x (V x
dv (6)
3.07. Green's Reciprocation Theorem for Dielectrics.—In 3.06 (2), let = V be the potential of one distribution of electricity 1' = V' that of another and let e be the capacitivity. If the contact surfaces of the different dielectrics are uncharged, we have, from 1.17 (5), , air €7) an" € Pan' = ev an; = and, from 1.17 (6), = 4,,," and cP',, = c13„'' and a similar relation for Alf, so that the integrals over the surfaces of discontinuity disappear; and if there are no charges throughout the volume, V • (e VT) = 0 and V • (e V (I)) = 0, so that the volume integral vanishes, leaving
Or
Or
i=i
-1
fv
av
si\" i6 an;
Vifsi cr'
*a7j dSi = 0
dS; — V; fe d8i) = 0
171
EQ; V; = EQ;Vi i -1
(1)
Thus we have proved Green's reciprocation theorem 2.12 (1) to hold when dielectrics are present. 3.08. Green's Function.—Let NI,be the potential due to a unit charge at the point P, and let c be the potential due to the induced surface
SOLUTION OF POISSON'S EQUATION
43.09
55
density o on some closed surface S at zero potential, which surrounds P. We may consider the unit charge at P as consisting of charge density p which is everywhere zero except in an infinitesimal volume dv enclosing P. This element is so small that has a constant value c13, throughout it. On the surface 43 = -if, at a distance r from P, = (471-er)-1; also, V2c13 is zero throughout v, whereas V24, = -p/e in dv and is zero elsewhere. Substituting in 3.06 (4) gives
(47er)-11 dS = 4,„ f p dv = O n where, as in 3.00 and 3.06, the positive direction of n is from v into S. (47rer)-1, this becomes Writing G for {ON)
f
aG = 1 f 1 T dS = — 1 f 47 sr S an n. sr dS 4re
(1)
We shall refer to G as Green's function although some authors designate e by this name. It is evident that G is a solution of Laplace's equation which is zero over a given boundary and has a simple pole at a point P inside. Electrically, it represents the potential inside an earthed conducting surface under the influence of a unit charge at a point P within it. Equation (1), which is identical with 1.06 (6), is usually worthless for determining G as we rarely know a on the conductor. Many methods of determining G for various types of boundaries will appear in subsequent chapters. The force F1 acting along u1on a charge q at ui, 4, u3 in the coordinate system of 3.03 can be found from G(ui, u2, u3) by the formulas
q iim frOG(ui,4 ILO
L
1,,±3 [
OG(ui,
u01
q OG(u'i, u'2,
Jue_a} = 2
-hiaz4
(2)
The first is valid because, when 8 is small, the fields of the induced charges at and uc - 8 are the same, but those of the charge itself are equal and opposite and cancel out. The second form follows from 2.07 (2) because the field energy of the charge itself is unaltered by a change in u'1 so only the induced charge energy is affected. 3.09. Solution of Poisson's Equation.—In a vacuum, the potential at the point P due to a charge density p in the element of volume dv is, from 1.06 (2), dV = p(47rer)-1dv, where r is the distance from the element dv to P. Thus the potential at P due to all charges in space is
v P
1 p dv
47ejlr r
(1)
But we know from 3.02 (3) that this potential satisfies Poisson's equation
= -12 Thus (1) is a solution of (2).
(2)
56
GENERAL THEOREMS
§3.10
We may also solve (2) directly by means of Green's theorem. Let us apply this theorem to the region v between a very small sphere of area a and a larger sphere of area and radius R, both centers being at P. Write x1,= r--1and (I) = V in 3.06 (4), and we have I pi:: an
dS .111. az —
7. 1
T l(r)]
dS
= f [1V 2V — VV 2(1)1 dv (3) vr Consider the first of these integrals. On the small sphere, a/an = — a/ar and the solid angle subtended by any area dS is dS2 = 7-2dS, so that, since a V/ar is finite, the integral is
av
fV —rf — dS2 —dC2
ar
— 47V,
(4)
If the total charge Q producing the potential V lies in the finite region, then V —* Q(471-cR)-1as R —> .0 so on the surface Z, both terms in the integrand approach Q(477-ER3)-1whereas the surface area is 471-R2. Thus the second integral becomes zero as Q(ER)-1. Since r-1is a solution of Laplace's equation, v2 (r— 1) = 0 and (3) becomes, if (2) is used,
- =
v2v
1 p
dv = — dv v r e vr
(5)
f
which gives us (1) again. 3.10. Uniqueness Theorem with Dielectrics Present.—If there were several solutions to Laplace's equation satisfying the same boundary conditions so that we had to resort to experiment to choose the correct one, there would be little use for potential theory. We shall now prove that if the location and size of all fixed charges inside a certain region are specified, as well as the value of the potential over all its boundaries, with the possible exception of certain closed conducting surfaces on which only the total charge is given; then the value of V is uniquely determined in this region. Suppose there are two values V and V', both of which satisfy the boundary conditions. Since both satisfy Poisson's equation, their difference is, from 3.02 (2), V • [eV(V — V')] = 0 everywhere, since fixed charges cancel. Put = NI/ = V — V' in 3.06 (1), and remember that, as in 3.07, the integrals over the boundaries between dielectrics vanish if these are uncharged. The substitution gives m
c(V —
— V') dS; =
f[V(V — Vi)]2 dv (1) j=1 The surface integrals vanish in those terms on the left representing boundaries on which the potentials are specified, since V = V' at every s,
an,
GREEN'S EQUIVALENT STRATUM
§3.12
57
point. All surfaces on which the total charge is specified, but not the actual potential, are conducting surfaces so that V and V' are constant and may be taken out of the integrals, giving, for the sth such surface, by 1.15 (1),
- V;) dS, = (V, - V;) (Q, - Q;)
(V, - V's ) f s.ur4
By hypothesis the charge on this surface is fixed so that Q, - Q; is zero. The whole left side of (1) is, therefore, zero and, since the integrand on the right is always positive, V(V - V') must vanish everywhere. Thus V - V' must be a constant and since it is zero on the boundaries it must be zero everywhere and the theorem is proved. 3.11. Introduction of New Conductor.—We shall now prove that if a new uncharged or earthed conductor is introduced into the electric field produced by a system of charged conductors, the charge on all conductors remaining unchanged, the energy of the system is decreased. Let the energy, electric field, and the space occupied by this field be designated by W, E, and v, respectively, in the original system and by W', E', and v' in the final system; then we have W - W' =2 vE2 dv - !-f E'2 dv 2v
(1)
= 2fv-v•E2dv + --2efv,[(E - E') 2- 2E' • (E' - E)] dv Now putting vIf = V' and I. = V - V' in 3.06 (3) and remembering that V243 = 0 throughout the volume give
fv•
•v(v —
dv = f E' • (E - E') dv
f
v•
i -1
= 111V; ej
s,
v,(aV s/
- a') dS, =
an;
av) '3' an; (1
V;(Q, - Q;) = 0 —1
since Q; = Q,, so that the last term in the last integral (1) vanishes, and W - W' is a positive quantity. Since we do no work in making an earth connection, any motion of electric charge caused thereby must be at the expense of the electric field and so cause a further decrease in energy. 3.12. Green's Equivalent Stratum.—Let Vp be the potential at some point P, outside a closed surface S, due to an electric charge density p(x, y, z), throughout the volume v, enclosed by S. In 3.06 (4), let = r-1and 4 = V, so that V24. = -p/6 by 3.02 (3). This gives
lay
- — dS f V
an
a(
s an r dS =
f
p dv = — -4TV,
e v r
(1)
58
GENERAL THEOREMS
§3.13
by 1.06 (6), where r is the distance from P to dv. Thus we see that the potential Vp at P can be expressed either as a volume integral of the charge density in v or by two surface integrals over the surface, S, enclosing v. Referring to 1.06 (6) and 1.12 (1), we see that this means that Vp will be unchanged if we remove the volume distribution p and replace it by a surface stratum which coincides with S and carries a charge — €(8V/an) per unit area and an outward pointing dipole distribution of moment eV per unit area. If S is an equipotential surface, the second integral becomes, by 3.00 (2), when V is taken out, 1) dS = V f V 2(-Vi n • V( 1)dv v r
(2)
This is zero since 1/r is the potential of a point charge 4ire at P, and therefore V2(1/r) = 0 inside v. In this case, no dipole layer is required. This shows that an electric field is unchanged when any area of an equipotential surface is replaced by a very thin uncharged conducting sheet, because such a sheet may be regarded as two equipotential surfaces infinitely close together enclosing this area. By (1), they must have equal and opposite densities on the exterior faces to leave Vp unchanged. 3.13. Energy of a Dielectric Body in an Electric Field.—From the last article the forces exerted on external fixed charges by a given volume distribution and by its equivalent stratum are the same. It follows from Newton's laws that the reverse is also true. This fact facilitates the calculation of the work done in placing an uncharged dielectric body in the field produced by fixed sources in a region of capacitivity 6.. If the stratum is brought into the field fully formed, 1.06 and 1.071 (2) give the energy of an area dS to be dW = [aV (m • V)V] dS
(1)
where o is the charge density, M the dipole moment density, and V the potential of the external fixed charges. The dielectric polarization and hence the strength of the equivalent stratum are proportional in this case to the external field and are built up by it from zero to its final value so that the total work done is one-half that given by (1). Substitution for a and M from the last article and integration over the surface of the
stratum give W
= -14( — Vn • VV. ± Von • VV) dS
(2)
where V. is the final potential just outside the equivalent stratum or the dielectric surface. By 1.17 (5) and (6) the potential just inside the dielectric surface is related to V„ by the equations V„ = V,
and
eon • VV. = en • VV;
(3)
§3.14
EFFECT OF AN INCREASE OF CAPACITIVITY
59
where e is the capacitivity of the body. Substitution of (3) in (2) and application of Gauss's theorem 3.00 (2) to the result give W = ifs (e.,Vin • VV — eVn • VVi) dS = ifv(e0 — e)VV • VVi dv (4) since V 2V = V • eV V, = 0. Thus if the electric intensity produced in the volume v of a uniform isotropic medium of capacitivity eo by a fixed charge distribution is E, and when v is occupied by a uniform isotropic body e it is Ei, then the energy difference is Es dv
W = ifv (E0 —
(5)
The torque or force acting on the body in the direction 0 is F=—
OW
(6)
ao
3.14. Effect of an Increase of Capacitivity.—If the capacitivity at any point in the electric field produced by a system of charged conductors is increased, the energy of the system is decreased provided the charges on the conductors are kept constant. To prove this, let W be the energy of the system, Q;the charge on the jth conductor, p the volume density of charge, V the potential at any point, and e the capacitivity. We assume that Q, and p remain constant when E is varied but that V and W vary. Thus, W = ifv€E2 dv = krvE(VV)2 dv SW = ifvSe(VV) 2 dv fveVV • V(W) dv
(1)
Substitute SV for and V for 4 in 3.06 (1), set V • (E VV) = —p, and assume uncharged dielectric surfaces over which the integral vanishes. LEV(SV) • VV dv =
f 3Ve— av dSan'I 7 si
SV p dv V
i=i 6-17,5 dS ;
SV p dv
j-i Q; SV; f SV p dv = 2(5W
= -1 since we also have
1 W = QiVi i -1
1
.fvpV dv
GENERAL THEOREMS
60
§3.15
Substituting in (1) and transferring 23W to the left side, we have
SW = --f Oe(VV) 2 dv (2) 2 v Thus STV is negative if Se is positive. 3.15. Potential of Axially Symmetrical Field.—It can be verified by substitution in 3.05 (2) and integration by parts that a solution of Laplace's equation, if e is constant and V does not depend on 4), is
f
=
V(z,
jp sin 0) dB
0 I(Z
(1)
where cl)(z) is a real function of z whose Taylor expansion is, by Dw 39, 4)(z, p) = 1'(z)
(1)'(z)jp sin
0 + (2!)-4"(z)(jp sin 0)2
• • • (2)
Substitution of (2) in (1) and integration from 0 to 271- give C0
p 2n
V(z, p) = n
=o
(-1)' a")(z) 2 (n !) 2 az2n
(3)
It is evident that V(z, 0) is identical with 4'(z). If no axial charges exist, (3) gives V uniquely at all points attainable from the axis without crossing a charged surface. The proof resembles that used for 7.07 (5). Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Show that the components of the curl in cylindrical coordinates are ,
Mid pA =
aA. 1 OA — — p acp az
aA, aA,
=az -
aP
aA p] acs
i[a(pAgs)
curl.A
aP
P
2. Show that the components of the curl in spherical polar coordinates are [ (sin OA.)
1
curl,A
r sin =
o
1 [OA, r sin 0
l[a(rA8) curloA = r ar
ao sin
i
aA0 act,
]
(I a( rA,
or
aAr ao
3. Show that for ellipsoidal coordinates, as defined in 5.01, we write in 3.03 (1) 47,1 = (.2 - .0(.2 - .3)D2,
= (u2 - .2)(.2 - .2)D2,
= (u3 - .2)(., - .2)D3,
PROBLEMS
61
where D1.2,3 = [(a' + u1.2.3)(b2+ u1,2.3)(c2+ u1.2.3)]-13 c > b > a, -b2 < u2 < -a2, and -a2 < ul< co. For special cases of oblate and prolate spheroids, see 5.27 and 5.28. 4. If three sets of orthogonal surfaces are defined by the concentric spheres ui = x2 +y2 + z2 and by the two cones eu-i2 + y2(4 - b2)-1 z 2(4 - 0)-1 = 0 and x214, 2 + y2(u: - b2)-1+ z2(4 - c2)-1= 0, show that we must choose h1 =1, hs = u?(L4 — 4)[(4 — b2)(0 — 41-1, and h: = ul(u: - u:)[(b 2 - ul)(c2 - 4]-1. 5. If the three sets of orthogonal surfaces are x2c-2ur2 y2c-2(ul - 1)-1= 1, x2c-2uT' - y2c-2(1 - 4)-1= 1, and z = u3, show that hl = c2 (ul - 4)(ul h: = c2(u: - 4)(1 - 4)-1, and h3 = 1. 6. If the orthogonal surfaces are y2 = 4cuix + 4c24 y2 = -4cu2x + 4c24, and 1L z = u3, show that hl = c2'(ul + u2), = (33 + u2), h3 = 1. 7. If the sets of orthogonal surfaces are z = c(u1 U2), x2 +y2 = 4c2u1u2, and y = x tan u3, show that h1 = cIaTi(u1 + h2 = c[ici(u1 + u2)]1, h3 = 2c(a1a2)*. 8. If the orthogonal surfaces are (x2 + y2)1 = c sinh u1(cosh al - cos u2)-1, y = x tan u3, and z = c sin u2(cosh ui - cos u2)-1,prove h1 =c(cosh ai - cos u2)-1, h2 = h1, and h3 = c sinh ui(cosh ul - cos u2)-1. These are known as toroidal coordinates and give orthogonal anchor rings, spheres, and planes. 9. If the orthogonal surfaces are (x2 + y2)1 = c sin u2(cosh ul - cos u2)-1, y = x tan u3, and z = c sinh u2 (cosh 14 2 - cos u2)-1, prove 112 = c(cosh u1- cos u2)-1, h2 = h1, and h3 = c sin u2(cosh u1- cos u2)-1. These are called bipolar coordinates, because if 7.2 and 7.2 are the vectors from a point P to the points z = +c and z = -c, respectively, we have ul = In (r2ri-1) and cos u2 = (7'1- r2)771rii. 10C. If the specific inductive capacity varies as e-il', where r is the distance from a fixed point in the medium, verify that the differential equation solution satisfied by the potential is a2r-2[eria - 1 - ra-1 - r2(2a2)-9 cos 0, and hence determine the potential at any point of a sphere, whose inductive capacity is the above function of the distance from the center, when placed in a uniform field of force. 11C. If the electricity in the field is confined to a given system of conductors at given potentials and the inductive capacity of the dielectric is slightly altered according to any law such that at no point is it diminished and such that the differential coefficients of the increment are also small at all points, prove that the energy of the field is increased. 12. Find the condition where a set of two dimensional equipotentials V2 = f(z, y) can generate a set of equipotentials when rotated about the z axis. Show that if this is possible the potential is V = A f e-IF(V2)dV2 dV 2 + B
where F(V2) -
1
3V2
y(vV2) , ay
References The subject matter of this chapter is treated in the following books: ABRAHAM, M., and R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Gives lucid derivations of general theorems using vector notation. BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Gives a very complete treatment. Cowx, R. E.: "Field Theory of Guided Waves," McGraw-Hill, 1960. Chapter 2 includes a fine detailed treatment of Green's functions. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Extensive treatment in first four chapters. GEIGER-SCHEEL: "Handbuch der Physik," Vols. III and XII, Berlin, 1928 and 1927.
62
GENERAL THEOREMS
JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapter 1 treats this material. Static Green's functions are applied neatly in Chap. 3. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives comprehensive treatment using long notation. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives an excellent treatment using vector notation. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment using long notation. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. First two chapters cover this material. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives a very extensive and rigorous treatment. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Chapter 1 contains clear and detailed treatment. WEBSTER, A. G.: "Electricity and Magnetism," Macmillan, 1897. Clear treatment using long notation. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930.
CHAPTER IV TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS 4.00. Field and Potential in Two Dimensions.—A potential problem is said to be two-dimensional when all equipotential surfaces in the field are cylindrical, which means that each can be generated by moving an infinite straight line parallel to some fixed straight line. The unit of charge is now a uniform line charge parallel to this axis and having a strength of 1 coulomb per meter length. We have already seen in 2.04 (1) that in a homogeneous isotropic dielectric the field intensity at a distance r from such a charge is radial and its magnitude is E
q 2rer
(1)
The potential obtained by integrating this is
V =
2re
In r
C
(2)
Clearly the zero of potential cannot be chosen, conveniently, at infinity as this would make C infinite. We usually give C a value that makes the computation as simple as possible. Theoretically, a two-dimensional electrostatic problem can never occur as all conductors are finite. However, there are a vast number of important cases in which the lengths of the parallel cylindrical conductors are so great compared with the intervening spaces that the end effects are negligible, and the problem then becomes two-dimensional. 4.01. Circular Harmonics.—In its most general sense, the term "harmonic" applies to any solution of Laplace's equation. In the more usual but restricted sense, it applies to a solution of Laplace's equation in a specified coordinate system, which has the form of a product of three terms, each of which is a function of one coordinate only. The solution required to fit a given set of boundary conditions is then constructed by adding up a number of such harmonics which have been multiplied by suitable coefficients. In the ordinary cylindrical coordinates described in 3.05, we should therefore have cylindrical harmonics of the form V = R(p).13(0)Z(z) (1) In the special case where Z(z) is a constant, this reduces to a two-dimensional problem and the harmonics are called circular harmonics. For 63
64
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.01
this case, we usually write r for p and 0 for so that for a uniform isotropic dielectric Laplace's equation, 3.05 (2), becomes, when we multiply through by r2,
a (al , ao2 avv —
rOr r ar
(2)
0
We now wish to find solutions of this equation of the form
V = R(r)O(0)
(3)
Substituting in (2) and dividing through by (3) give 1(
aR
R r
2a2R) _1_329 e or. + r07.2
0 —
This equation will evidently be satisfied by solutions of the equations c/20
,7 = — n20 , d 2R _L 1 dR 2R
dr2
2- dr =
(4) (5)
A solution of (4), the simple harmonic motion equation, is e = A cos nO + B sin nO
(6)
and it is easily verified that the solution of (5) is R = Crn where n
Dr-n
(7)
0. If n = 0, we have the solutions
0 = AO + B R = C ln r + D
(8) (9) The number n is called the degree of the harmonic. The circular harmonics then are
V = (AO + B)(C In r D) V = (A cos nO + B sin n0) (Cr" + Dr-n)
Degree zero Degree not zero
(10) (11)
A sum of such terms, with different constants for each n, or an integral with respect to n is also a solution of (2). Thus
V = Zenitm
or
V = f f(n)O.Rm do
(12)
n
It should be noticed that we have placed no restrictions on n. If the equation 1(An cos nO + Bm sin nO) = l(Cm cos nO + Dm sin nO) (13) holds for all values of 0 where n is an integer, then we have the relations
Am = Cm
and
B. = D.
(14)
65
CONDUCTING CYLINDER
§4.03
To prove this, it is necessary only to multiply both sides of (13) by cos m0 and integrate from 0 to 2r, giving
Z(A,1027 cos nO cos m0 dO + B„f021.sin nO cos m0 do) = VC4027 cos nO cos me do + D„f021- sin nO cos m0 dO) If m n, using Pc 361 and 360 or Dw 445 and 465, we see that all terms on both sides drop out. If m = n, we use Pc 489 and 364 or Dw 450.11 and 858.4 and obtain frA„ = irC„ A similar procedure using the sine proves the second part of (14). 4.02. Harmonic Expansion of Line Charge Potential.—In using circular harmonics, it is frequently necessary to have the expansion due to a line charge whose coordinates are ro, 00. If R is the distance from the line charge q to the field point P (Fig. 4.02), then, from 4.00 (2) when C is zero, the potential V at P is given by treV = - 2q In R =
-
q ln [r2 -I- ro — 2rro cos (0
-
00]
= -2q In r - q In [1 - /le l i(4-6)][1 - ne-i(" 01 Using Pc 768 or Dw 601 where r 0 is less than r, we obtain
ro 1 ro 2 -2q In r + 4-[ei(e--9) + e-i0-60] + (-) [020-80-I- e-120-00)] -I- • - - 1 r r 1 7:2 = -2q[In r - 7)cos (0 - 00) - 2 cos 2(0 — 00 ••• r 2r -
Writing as a summation and expanding cos n(0 Dw 401.04 give
-
Bo) by Pc 592 or
40
n
V = j–-
-I- -
nr
27re
(cos nOocos nO + sin nO0 sin nO) - In r
I
(1)
nml.
This holds when r > ro. When r < ro, the same procedure gives, factoring out In r0 instead of In r, [.
V = -C 1-1(I. . (cos nO0 cos nO } sin nO0 sin nO) - In ro 24r E n\r 0)
(2)
n ∎1
These are the required expansions. 4.03. Conducting or Dielectric Cylinder in Uniform Field.—As an example involving boundary conditions at both dielectric and conducting
66
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.03
surfaces, let us find the field at all points when an infinite conducting cylinder of radius a surrounded by a layer, of relative capacitivity K and radius b, is set with its axis perpendicular to a uniform field of strength E. The original potential outside the cylinder is of the form V=Ex=Er cos U At infinity, the potential superimposed on this, due to the induced charges on the cylinder, must vanish so that no terms of the form r" can occur in it. Since this potential must be symmetrical about the x-axis, no terms involving sin nO can occur. The final potential outside must therefore be of the form Anr-n cos nO
V. -= Er cos 0 ±
(1)
n =1
Since the values r = 0 and r = 00 are excluded from the dielectric, both r" and r-n can occur there but sin nO terms are thrown out as before. The potential in the dielectric must therefore be of the form (B"r" Car-n) cos nO
=
(2)
n =1
If we take the center of the cylinder at the origin, then V = 0 throughout the conductor. We have therefor• found solutions of Laplace's equation that satisfy the boundary conditions at infinity and the symmetry conditions. It is now necessary only to determine A, B, and Ch so that the boundary conditions at the dielectric and conducting surfaces are satisfied. From 1.17 (5) and (6), at r = b the boundary conditions are
ay. €v
ar =
avi ar
or
avo Kavi ar = ar
and
V. = Vi
(3)
Substituting (1) and (2) in (3) gives E cos 0 — XnAhb-n-1cos nO = Mn(Bhbn-1— C,,b-n-1) cos nO Eb cos 0 ± ZA.b-n cos nO = Z(B.bn C.b-n) cos nO From 4.01 (14), we have, if n
1,
— —Anb-n--' = Anb-n = B.bn +
(4) (5)
At the conducting surface, r = a, Vi = 0, so that 0 = &an T C.a-n (6) Adding (4) X b to (5) gives (K 1)Bn = (K — 1)C.b-2n (7 ) The only way (6) and (7) can be satisfied is to have B. = C. = 0 or (K 1)/(K — 1) = — (a/b)2n. The latter condition is impossible since
§4.03
CONDUCTING CYLINDER
67
the left side is positive and the right side negative so that the first holds, and substituting in (5) we have An = B„ = C. = 0 (8) When n = 1 we have, instead of (4), (5), and (6),
E — Alb-2= KB,. — KC1b-2 E + Alb-2= B1+ C ib-2 0 = Bla + Cia-' Solving these equations for A1, B1, and C1gives
(K +1)a2 + (K — 1)b2 (K + 1)b2± (K — 1)a2 2Eb2 B1— (K + 1)b2+ (K — 1)a2
Al = — Eb2
Cl -
(9)
—2Ea2b2
(K + 1)b2+ (K — 1)a2
The potentials (1) and (2) then become
Ai V. = (Er ± 7.) cos 6,
C J. Vi = (Bir + --i . ) cos 0
(10)
The lines of electric displacement are shown in Fig. 4.03a.
a
b
FIG. 4.03.
We may get the case of a conducting cylinder of radius a by letting
K = 1 in (9), so that 1 =E A1= C1= —Ea2 and B and the potential is
V . = E(r — 1 cos e r
(11)
The field is shown in Fig. 4.03b. We get the case of a dielectric cylinder of radius b by letting a = 0 in (9), so that
68
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Al =
b2E K+1 '
El = K + l'
§4.04
C1 = 0
and the potentials are vo
1.(7. _ K .
1b2) cos 0, K±1 r
2E V. — K ±ir cos 0
(12)
We notice that the field inside the dielectric is uniform. The displacement is shown, for K = 5, in Fig. 4.03c.
FIG.
4.03c.
4.04. Dielectric Cylinder. Method of Images.—Let us consider a dielectric cylinder of radius a under the influence of a line charge at r = b, B = 0. The potential due to the line charge alone may be obtained from 4.02 (2) by putting ro = b, 00 = 0. We must superimpose on this a potential, due to the polarization of the dielectric, which vanishes at infinity and is symmetrical about the x-axis. The final potential when a < r < b is therefore re,{ V° = :
nOn + Ann] [
cos nO — In b ± C1
(1)
n =1
Since the potential inside must be finite when r = 0 and symmetrical about the x-axis, it is of the form „rn cos nO + C2 V =./3 q ' 276,( n
(2)
=1
Setting V. = Vi when r = a, we have ()"
nB — a n
Setting Eva Vo/ar = Ea V,/ar or an--1 bn
and
C2 = —In b
C1
(3)
avo/ar = Kavi/ar when r = a, gives
n — an+1An = nKan-1.13.
(4)
69
IMAGE IN CONDUCTING CYLINDER
§4.05
Solving (3) and (4) for A. and B. gives
1 — K a2n
2 B. — (1 + K)nbn
A n = 1 + K nbn'
The potential outside then becomes
= 27qr€.1 n=1 1[(rb y ±
l lqa r
cos nO — In b
CI
(5)
and that inside becomes =
q
1r€v(1 -F
K)
l(ry n b cos nO —
b — Ci)
(6)
n =1
If we let
C=0=—
1 — K 1 +K
ln r +
K 1 ln 1+K
then (5) becomes exactly the expansion given by 4.02 (1) and (2) for the potential due to three line charges on the x-axis, one of strength q' at x = a2/b, one of strength —q' at x = 0, and one of strength q at x = b. Also, (6) gives the expansion due to a charge q" situated at x = b, where = -F — Ifq
and
2
1 ± Kg
(7)
Thus when an uncharged dielectric cylinder of radius a is brought into the neighborhood of a line charge q with its axis parallel to the charge and at a distance b from it, the additional potential in the region outside the cylinder due to its presence is the same as if it were replaced by a parallel "image" line charge q' located between q and the axis at a distance a2/b from the latter, plus a line charge —q' at the origin. The potential inside the cylinder is the same, except for the ac ,litive constant, as if it were absent and q were replaced by a charge q". Some authors write 2K instead of 2 in the numerator of q", in which case the potential inside the cylinder must be calculated as if all space were filled with the dielectric K. When a dielectric cylinder is introduced into any electric field produced by a two-dimensional charge distribution parallel to its axis, it follows from the last paragraph, since such a distribution can always be built up from line charges, that the form of the field inside the cylinder is unchanged but its intensity is reduced by the factor 2/(K + 1). 4.05. Image in Conducting Cylinder.—It follows from 1.17 (8) that as K co the lines of force become incident normally at the dielectric surface. This is the condition at the boundary of a conductor, so that we may get the law of images in an uncharged conducting cylinder by
70
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.07
letting K 00 in 4.04 (7), giving q' = —q. Thus when a line charge of strength q is placed parallel to the axis of an uncharged conducting cylinder of radius a and at a distance b from it, the additional field outside the cylinder due to its presence is the same as if it were replaced by a parallel line charge —q located between q and the axis at a distance a2/b from the latter plus a charge q on the axis. As will be noted from Fig. 4.05, the image position can be located by drawing a tangent to the dielectric or conducting cylinder from the point q. The line through this point, normal to —qq', intersects —qq' at q'.
4.06. Image in Plane Face of Dielectric or Conductor. Intersecting Conducting Planes.—If we let the radius of the cylinder become infinite, keeping the distance d = b — a of the line charge from the face constant, then the distance of the image q' from the face is a — a2b—' = ab-1(b — a) d Thus the image laws derived in the last two sections apply to a line charge lying parallel to the faces of a semi-infinite dielectric or conducting block. The image charge is the same distance back of the face as the actual charge is in front. For a conductor, q' = —q, and for a dielectric, q' and q" are given by 4.04 (7). It is evident from Fig. 4.06 that if two planes intersect at the origin at an angle 7r/m, where m is an integer, both planes will be equipotentials under the influence of line charges parallel to the intersection lying in the cylinder r = r0 and arranged with +q at 00, 27rm—' ± 00, 4rm1 00, . . . , 2(m — 1)7rm-1 0 0 and —q at 27rm-1— 00, 47rm—' — 00, . . . , 2,r — 00. 4.07. Dielectric Wedge.—Another solution of 4.01 (2) is obtained if n 0 by writing jn for n in 4.01 (4), (5), (6), and (7), so that 0. = A cosh no + B sinh nO R. = Ci er n = C1 2e±in r = C cos (n In r) D sin (n In r)
(1) (2)
This solution is periodic in In r instead of 0 so that in 4.01 (12) Rn not On is now the orthogonal function. These harmonics can be used to solve
§4.07
DIELECTRIC WEDGE
71
the problem of a dielectric wedge of capacitivity e2 bounded by the planes 0 = —a and 0 = a under the influence of a line charge q at 0 = 7, r = a 11♦ in a medium of capacitivity el as shown in Fig. 4.07. There are no cylindrical boundaries where the sine and cosine terms in (2) must vanish so n is not restricted to discrete values but may vary continuously which indicates the integral form of 4.01 (12). We write for (el — e2)/(€1 -I- €2) and choose potentials of the forms V1= (1 +0)fo [A(k)ek°
B (k)e-kej cos [k In (r / a)] dk-FC 0
V2= fo'[C(k)ek°+ D(k)eke] cos [k In (r/a)] dk + Co V3
=
JO
'[E(k)eke + F(k)e-h°1 cos [k In (r /a)] dk
Co
(3) (4) (5)
according to whether —a < 0 < a (3), a < 0 < y (4), or 7 < 0 < 27r — a (5). The constant Co can be chosen to make V zero at any specified point. For this point at r = a, 0 = 0, it has the value —(1 + 13)fo[A (k) + B(k)] dk The circle r = a passing through q is a line of force because aV /ar is zero there. Thus half the flux from q goes to r = 00 and half to r = 0 where there is a charge By Fourier's integral theorem, if two of the integrals in (3), (4), and (5) are equal to each other over the whole range of In (r / a), then their integrands are equal. Application of 1.17 (5) and (6) to the integrands of V1and V2 at 0 = a gives, after rearrangement,
D = 8e2kaA + B C = A + /3e-2kaB, Similarly joining V1 at 0 = —a to V3 at 0 = 2ir — a gives F = (gAe-2ka + B)e2k. E = (A + 13Be2ka)e--2kr,
(6) (7)
To meet the remaining boundary conditions at 0 = y, we write down an expression for the flux density originating anywhere in the 0 = 7 plane and evaluate it by Gauss's theorem, thus
4rav2 av3 = L ao ae el
)
ro
.' k[Cek'Y — De-h7 — Eek7 + Fe-k1 cos k In --r dk a
When both sides are multiplied by cos [t in (r /a)] dr and integrated from r = 0 to r = 00, the left side is simply q by Gauss's theorem because the integrand vanishes except near r = a and the right side is found by Four-
72
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.08
ier's integral theorem so that q = iireik[(C - E)eki - (D - F)e-k7] Also V2 is equal to V 3 at 0 = y so that (C - E)ek7 = -(D - F)e-k7 Elimination of D and F or of C and E from these equations gives C = E
ireik
'
D=F-
ire
(8)
The solution of (6), (7), and (8) for A and B gives A B - q[e'k(7-r) sinh kr - elk (7-7) Qsinh k(ir - 2a)] , (9) - 2a)] 2ifeik[sinh2 kr - 02 sinh2 where the upper signs go with A and the lower with B. In the special symmetrical case where 7 = 7r the potential in the dielectric wedge is ,
[k In (r /a)] - 1 dk (10) sinh k(ir - 2a)] The integrand, which is finite at k = 0 and falls off exponentially as k increases, can be plotted as a function of k and the integral then evaluated with a planimeter. If the only charge is to be at r = a, 0 = -y, then the charge at the origin may be canceled by addition to V1, V2, and V3 of the term cosh Ice cos rei jo k[sinh kir
1 + i3)
Vl = q(
iq In a [a(Ei — Es) —
7rEi]-1
(11)
If the cylinder r = b is at zero potential, then we must superimpose two solutions of the form just derived, one for q at r = a, 0 =- 7 and the other for -q at r = b 2/a, 0 = y. If the cylinders r = b and r = c are at zero potential, then we shall need discrete values of n in (2) and the potential function will be a series instead of an integral. 4.08. Complex Quantities.—Before taking up the general subjects of conjugate functions and conformal transformations, it will be well to review some of the more important properties of complex quantities. If z = x jy, it is clear that to every point in the xy-plane, usually called, in this connection, the z-plane, there corresponds one value of z. In polar coordinates, we have, using Pc 609 or Dw 408.04, z = x jy = r cos 0 jr sin 0 = (1) The magnitude of the vector r is known as the modulus of z and written izi. The angle 0 is known as the argument, amplitude, * phase or angle * This term, universally understood by mathematicians to represent the angle in this case, is also, unfortunately, in general use to specify the maximum fluctuation of any alternating quantity in connection with alternating currents. Thus, when the complex notation is used for such quantities, it becomes practically identical with the term "modulus" as used here.
§4.09
CONJUGATE FUNCTIONS
73
of z. When a complex number z is raised to a power n, we have zn = rnein°
(2)
so we may say that the modulus of zn is the nth power of the modulus of z and the argument of z. is n times the argument of z. When we take the product of two complex numbers, we get zzl = me; ("0
(3) so that the modulus of the product of two complex numbers equals the product of their moduli and the argument is the sum of their arguments. By putting zi1for ziin (3), we see that the modulus of the quotient of two complex numbers equals the quotient of their moduli and the argument equals the difference of their arguments. If we have zi = xi + jyi = f(z) = f(x jy), then the quantity zi = xl — jyi = f(x — jy) is called the complex y conjugate of z. That the latter relation holds if f(z) is an analytic function can be shown by expanding f(x ± jy) in a power series with real coefficients, for wherever ±j occurs to an even power the term is real and the choice of sign FIG. 4.08. disappears, whereas whenever +j occurs to an odd power the term is imaginary and the choice of sign remains as before. Thus we have Iz11 2 = xi + yi = zizf = f(x jy) • f(x — jy) This gives us the rule that to obtain the modulus of a function of a complex variable we multiply it by its complex conjugate and take the square root of the product. 4.09. Conjugate Functions.—Laplace's equation in two dimensions and rectangular coordinates is written
a2U a2U axe + ay2 =
(1)
Since this is a partial differential equation of the second order, the general solution must contain two arbitrary functions and it is easily verified by differentiation that such a solution is U = c1)(x jy) ''(x — jy) It should be noted that to be a solution of (1), and '' must possess definite derivatives where (1) holds, and hence are analytic functions in this region and capable of expansion in a power series (see Whittaker and Watson, "Modern Analysis," Chap. V). Since U is to be the electrostatic potential, it must be a real quantity and so the imaginary part of (13 must be equal and opposite to that of NP, so that, if c1)(x jy) = u jv, then '1,(x — jy) = w — jv, where u, v, and w are real quantities. Since
74
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.11
both 7 and are analytic functions, we may expand them in series; thus we have 43(x
jZ A„rn sin nO n=0
n=0
n=0
P (x — jy) = xli(re-2°)
A nr' cos nO
Ad-nein° =
jy) = 43(reig) =
ZB.rne—intl = ZB.r" cos nO — jZBnrn sin nO n=0
n=0
n=0
Since the imaginary parts of 43 and •If are equal and opposite over the entire range of 0, we see from 4.01 (14) that An = Bn so that the real parts of cf. and if are equal or u = w. Thus we have U = 2u. Let V be another real quantity such that V = 2v, and we have
U jV = 2(u -I- jv) = 240(x
jy) = f(x
jy)
(2)
The function V also satisfies Laplace's equation as may be shown by noting that the above expansion is in circular harmonics or by multiplying through by —j, giving jy)
V — jU = —jf(x jy) = F(x
Thus V is the real part of F(x jy), just as U is the real part of f(x We shall write W for U jV and z for x jy, giving
jy).
W = f(z)
(3) The functions U(x, y) and V(x, y) are called conjugate functions. 4.10. The Stream Function. Differentiating (3) with respect to x and y gives OW au .av , az —
ax = ax + 3ax = f (z)79X. = f/(z) ., aw _ au .av , az 5 = (z) ay ay + ay = f (z),
Multiply the second equation by j, add to the first, equate real and imaginary parts, and we have
av = au ax ay
au — = anddav ay ax
(1)
This is the condition that the two families of curves U(x, y) = constant and V(x, y) = constant intersect each other orthogonally. As we have seen, we may choose either set to represent equipotentials, in which case we call this the potential function. The other set which intersects this set everywhere orthogonally then represents the lines of force and is known as the stream function. 4.11. Electric Field Intensity. Electric Flux. Let us consider the derivative —
dW dz
dU-Fj dV dx-Fj dy
(a U / ax)dx+ (0 U/ ay)dy+jRav/ax)dx-F (a V / ay)dy] dx j dy
§4.12
75
FUNCTIONS FOR A LINE CHARGE
Substitute for a U/ax and aU/ay from 4.10 (1). dW = dz
=
(av/ay)(dx
j dy) j(aV /ax)(dx j dy) dx j dy
ay AV ay .ay = 73x —
(1)
ay
Thus if V is the potential function, the imaginary part of —dW/dz gives the x-component of the electric field intensity and the real part gives the y-component. Regardless of whether U or V is the potential function, the absolute value of dW/dz at any point gives the magnitude of the electric field intensity at that point. If dn is an element of length in the direction of maximum increase of potential and ds is the element of length obtained by rotating dn counterclockwise 4 7r radians, then we get from (1) -
dW dz
By =_- ay = an as
or
dW dz
= By = an
au as
(2)
according as U or V is the potential function. In the latter case, if we wish to know the flux through any section of an equipotential surface between the curves U1and U2, we integrate, giving by 1.10 (1), m
Flux =
f
av ds = el ul — ds ui as as = e(U2 —
ui an
U1)
(3)
Thus just as the difference of potential between any two points in the field is given by the difference in the values of the potential function at the two points, so the total flux passing through a line joining any two points in a field equals the product of the capacitivity by the difference in the value of the stream function at the two points.. If the surfaces V1and V2 are closed and all lines of force in the region they bound pass from one to the other, then, from 2.01, they form a capacitor. The charge Q on either is the total flux per unit length. From (3), this flux is the product of E by the increment [U] in U going once around a V-curve. Since the potential difference is V2 the capacitance and field energy per unit length are C —
[Q]
4U]
(4)
1 -172 — I Va — Vl Field energy = ICI V2 — V1 2= 4-€1 U2 - U1~ V2 (5) 4.12. Functions for a Line Charge.—Before taking up the methods available for finding the required f(x jy) to fit a given problem, let us consider a simple case where the form of this function is evident by inspection. In polar coordinates By Pc 609 or Dw 408.04, we have (1) z= x jy = r cos 0 jr sin 0 = U = --k(re)-1 In r from The potential of a line charge at the origin is 4.00 (2). Clearly from (1) this is the real part of ---k(re)-- i In z, so —
76
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.13
e ,q0 2 z = _ qnr W = U ± jV = qln q In (x + jy) _
=
2re
q In (x2 + y2)1 jqtan-1(y/x) 2re 276
(2)
is the desired function. To check this we observe that, as 0 goes from 0 to -21r, the function eV goes from 0 to q and so is the flux emerging from a charge q as required by 4.11. We may now write down f(z) when there is a line charge at ro, 00, or zo. The potential is
q FIG. 4.12.
U = -1q(7rE)--' In R = —.1q(7re)-1In [r2 + 4 - 2rro cos (0 — 00)] = ---1 1q(7re)-1In [(r cos 0 — ro cos 002 + (r sin 0 — r0 sin NM = —1q(71-€)-1In (A2 + B2)1 But from (2), substituting A for x and B for y, we see that In (A' is the real part of In (A + jB) so that the required function is
+ BY
W = y77 In (A ± jB) = 27eq In (rei° - roei 09 =-7 ---re In (z - zo) (3) The function for n line charges situated at z,, z2, . . . , z. is therefore n
W=
In (z - z.)
2--re 8
(4)
=1
4.13. Capacitance between Two Circular Cylinders.—We have already seen in 4.05 that the equipotentials about two equal and opposite line charges are circular cylinders. Let us therefore superimpose the fields due to charges 2re at y = a and -2re at y = -a. This choice of the charge q simplifies the coefficients. From 4.12 (3), the expression for W becomes, using Pc 645 or Dw 601.2 and 505.1, W = In z + .7 a = 2j tan-1 a = 2j cot-1
z - 3a
z
a
(1)
Solving for z and using Pc 601 or Dw 408.19, we have z = a cot
U + jV = -a sin(U/j) ± a sin V cos (U/j) - cos V 2j
Using Pc 606 and 607, we can now separate real and imaginary parts, giving
a sink U Y _ cosh U - cos V Eliminating V from these equations gives x-
a sin V
cosh U - cos V
x2 + y2 - 2ay coth U + a' = 0
(2)
(2.1)
§4.13
CAPACITANCE BETWEEN TWO CIRCULAR CYLINDERS
77
This may be written
x2 (y — a coth U) 2 = a2 csch2 U
(3) Thus, as anticipated, the equipotentials are a set of circles with centers on the y-axis, positive potentials being above the x-axis and negative
1
1 i
1
i
% %
)
I \ / • ` `. \
/
x
1
// a
/ ,
1
(b)
(a)
FIG. 4.13.
below. Eliminating U from (2) gives x2 — 2ax cot V + y2
a2 = 0
This may be written (x — a cot V)2 + y2 = a2 csc2 V
(4)
Thus the lines of force are also a set of circles, all of which pass through the points y = +a and y = —a on the y-axis. To determine the capacitance per unit length between the cylinders U = U1 and U = U2, it is necessary, from 4.11 (4), only to divide the charge 2re by the difference of potential U2 — U1. We are given the radii R1 and R2 of the two cylinders and the distance D between their axes, from which we must determine a, U1, and U2. From (3), R1= alcsch U11, R2 = *soh U21 and D = a(Icoth U11 ± tooth U21), taking the lower sign if U1 and U2 are both positive, giving one cylinder inside the other, and taking the upper sign if Uiis negative, giving one cylinder outside the other. We now write by Pc 661 or Dw 651.02, using the same sign
rule,
78
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.15
cosh (U2 — U1) = cosh U2 cosh U1± Isinh U2 sinh U11 = (bcoth U2 coth Uli ± 1)Isinh U2 sinh Ulf Now apply Pc 659 or Dw 650.08 to the ± 1 = ± ± term, giving cosh (U2 — U1) = [Icoth U1coth U2! ± i(coth2 U2 — csch2 U2) (coth2 csch2U1)]Isinh U1 sinh U21 (1coth U21 ± 1coth U11) 2 1csch2 Ui csch2 U2 2ICSCh U1csch U2I Putting in values of D, R1, and
R2
gives
cosh (U2 — U1) — ±
D2 — R? — RI 2R1R2
Thus the capacitance per unit length between the two cylinders is
D2 — R7 — C = 24cosh-1( -12R1R2
(5)
where the lower sign is taken when one cylinder is inside the other and the upper sign when they are external to each other. The two cases are shown in Figs. 4.13a and 4.13b.
4.14. Capacitance between Cylinder and Plane and between Two Similar Cylinders.—In Fig. 4.13a, let RI = D h co so that the outer cylinder circles through infinity and coincides, in the finite region, with the x-axis. Then we have, neglecting R2 compared with fel and D2, and h compared with 2R1, R1 D = 2R1 — h 2R1,
Ri + RI — D2 (Ri— D)(Ri + D) h 2R1R2
2R1R2
Thus the capacitance per unit length of a conducting cylinder of radius R with its axis parallel to and at a distance h from an infinite conducting plane is
C = 274cosh-1
R
(1)
Two similar cylinders with their centers at a distance D = 2h apart have half the capacitance per unit length given by (1) since they consist of two such capacitors in series. The resultant expression is in somewhat simpler form than that obtained by putting R1 = R2 and D = 2h in 4.13 (5)
C = VE(COSh.-1
D 2R
(2)
4.15. Conformal Transformations.—Evidently conjugate functions furnish a powerful means for solving two-dimensional potential problems, provided we can find the proper function. Before taking up methods of doing this, we shall investigate certain special properties of functions of a
§4.16 GIVEN EQUATIONS OF BOUNDARY IN PARAMETRIC FORM
79
complex variable. Suppose we plot values of z = x jy on one plane and values of z1= x1 + jyi on a second plane and let z be an analytic function of zi so that at least one point in the z-plane corresponds to each point in the zi-plane. Then as we move the latter along some curve, the corresponding point in the z-plane will also describe a curve, provided z = f(zi) is continuous, otherwise the second point may jump from place to place. If the z curve returns to the starting point when the z1curve does, then f(zi) is said to be single valued in this region of the z-plane. From the rule for quotients of complex numbers 4.08, we have
dz ds = = =h (1) dzl jdzil dsi where ds is the length of an element dz of a curve in the z-plane and ds1 is the length of the corresponding element dz1of the corresponding curve in the z1-plane.Thus the modulus of dz/dzimeasures the magnification of an element from a certain point in the z-plane when transformed to the corresponding point in the z1-plane. Let us draw the infinitesimal triangle made by the intersection of three curves in the zrplane, and let the lengths of the sides be dsi, Then the lengths of the sides of the transformed triangle will be ds = h ds1, Thus dsi:ds'I :ds'i' = ds:ds':ds", so that the ds' = h ds;., ds" = h two triangles are similar and the angles of the intersections of corresponding curves in the two planes are the same. Such a transformation is said to be conformal. Since the argument of the quotient of two complex numbers is the difference of their arguments, we see that the argument of (dz/dz1) is the angle through which the element has been rotated in transforming from one plane to the other. 4.16. Given Equations of Boundary in Parametric Form.—If f(x, y) = 0 is the equation of one of the desired equipotential boundaries and x and y can be expressed as real analytic functions of a real parameter t whose total range just covers the conductor, then there is a simple method of obtaining a solution of Laplace's equation fitting this boundary. Let x = fl(t) Then the desired solution is
y = f2(t) (1)
x jy = i(bW) jf 2(bT47)
(2)
This gives V = 0 as the conductor, for substituting this value gives exactly the parametric equation of the conductor with b U in place of t. Unfortunately, the number of cases where this method is useful is very limited. Among them may be mentioned the confocal conics and the various cycloidal curves. As an example, let us find the field on one side of a corrugated sheet of metal whose equation is that of a cycloid
80
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.17
and which forms one boundary of a uniform field. The equation of the surface is y = a(1 — cos 0) x = a(8 — sin 8) (3) j — so that z = a(bW — sin bW) aj(1 — cos bW) = a(bW giving x = a(bU — ebv sin bU), y = a(bV 1 — ebv cos bU). When V is large and negative x = ab U and y = d-abV, so that we have a uniform Substifield in the —y-direction of strength E = +9V /ay = tuting for b gives / TR V . -aE '— z = AcT + 3— 36 To find the field at any point, we differentiate, and
1=
,- cLE dW — —e_:kldw E dz dz
I
dW IE 1 — e-lawE = E 1 — 24 cos U e 2c 8. aE dz = To get the surface density on the sheet, we note that on the conductor V = 0 and y = a[1 — cos (U/aE)], and =
1 a\1 dW dz v-o— EE
This result gives the field on the side of the sheet with sharp ridges. The solution on the other side gives lines of discontinuity so that it is of no value there. 4.17. Determination of Required Conjugate Functions.—In most cases, the search for a function W which fits the given boundary conditions in the z-plane begins by looking for a transformation which reduces the boundaries to simpler shapes. If the new boundary conditions are unfamiliar, we endeavor to find a second transformation which will still further simplify the boundary conditions. Eventually, we should arrive at a situation where the solution can be written down by inspection. We then proceed backward along the steps we have come to the solution of the original problem. It is frequently possible to jump these steps and write f(W, z) = 0 by eliminating the intermediate complex variables. If this is impossible, those variables serve as parameters connecting TV and z. In manipulating these transformations, it is often helpful to visualize that portion of the z,-plane between the boundaries as an elastic membrane which possesses the property that no matter how we distort the boundaries the angle of intersection of any lines drawn on the membrane remains unchanged. The membrane may not separate from a boundary
§4.17
DETERMINATION OF REQUIRED CONJUGATE FUNCTIONS
81
but may slide along it and be expanded or contracted indefinitely. Vor example, suppose the boundaries in our problem are two nonconcentric nonintersecting circles or two intersecting circles or two of one type and one or two of the other intersecting orthogonally. The region with such boundaries may be transformed into a rectangle by 4.13 (1), the relation being
, z zi = in (1) z — ja 1 for W = U jV, since we are Here we have written zi = x1 jy attaching no electrical significance to x1and yiat present. From 4.13 (3) and (4), we see that, when — 00 < x < cc and — 00 < y < 00, then —0 < yi < 27r and — 00 < x1< 00. Thus (1) transforms a horizontal from y'-oo to y =a from from C X'. -IT x'=rt toroo tura)
Ait■
.4111. fox=-co
■ 41111/■ \A,
•11 from C' x'vt toy=-co T toy.-co from y'=-co to y=-a from I B
PIG. 4.17.
strip of width 2r in the z1-plane into the whole of the z-plane. Vertical lines in this strip, from 4.13 (3), go into circles given by
x2
(y — a coth x1) 2 = a2 csch2 x1
(2)
and horizontal lines go into circles passing through y = ±a, x = 0 given by 4.13 (4) to be (x — a cot yi)2 -F y2 = a2 csc2 yi (3) This transformation can be visualized by imagining an infinite horizontal strip of elastic membrane of width 2r being rotated counterclockwise to a vertical position in a z'-plane, so that points xi = 0, yi = 0, and xi = 0, yi = 27 go to AA' and BB' and then pinched together at y' = 00 and y' = — co . These points C and C' are then brought toward each other along the y-axis while the center of the strip is expanded horizontally. The lines CA, CB and C' A', C'B' are opened out, as a fan is
82
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.18
opened about the point C and C', respectively, until CA coincides with CB and C'A' with C'B'. The membrane thus expands to fill the whole z-plane, the infinitesimal arcs AA' and BB' being stretched into infinite arcs bisected by the x-axis. The operation, after the first rotation, is pictured in Fig. 4.17. If there are to be no discontinuities where CA folds against CB and C'A' against C'B', then the potential used in the horizontal strip in the 4-plane must be periodic in y,with a period 27. Problems involving line charges and rectangular boundaries or ones where different sections of such boundaries are at different potentials can now be solved by images. In other cases, it may he necessary to unfold the rectangle into a half plane by a Schwarz transformation, giving, in general, elliptic functions. 4.18. The Schwarz Transformation.—One of the most useful transformations is that which transforms the upper half of the zrplane, bounded by the real axis and an infinite arc, into the interior of a polygon in the z-plane, or vice versa. If the latter is finite, it will be bounded entirely by the deformed real axis yl = 0 of the zrplane. If not, portions of its boundary at infinity may be formed by expanding or contracting the original infinite arc of the zr-plane. To find the transformation that will bend the real axis of the zrplane into the specified polygon in the z-plane, consider the complex derivative dz
dzr
= Ci(zi — u1) 44(21 — u2)13' • •
(z, — un)P^
(1)
where /LI, u2, . . . , u. and 3 27 • • • , are real numbers, Cr is a complex constant and u. > u._, > • • > u2 > ur. As we have seen, the argument of the product of several factors raised to certain powers equals the sum of the products of the argument of each factor by its exponent, so that dz arg — = arg Ci arg (z — ur) ± • )3. arg (zi — u.) (2) d zi When dz, is an element of the real axis in the zrplane, we may write it dx1and we have dz dx j dy —, dy arg -d-4 = arg tan — (3) dx dxr This is the angle that the element dz, into which dzi is transformed, makes with the real axis y = 0 in the z-plane. When z1is real and lies between u, and ur+i, then (z1— (z, — u2), • • • , (zi — ur) are real positive numbers whose arguments are zero, and (z, — ur+i), (zi — 24+2), . . . , (zr — u„) are real negative numbers whose arguments are T. This gives, from (2) and (3)
83
THE SCHWARZ TRANSFORMATION
§4.18
Br = tan-'
= arg C1 + ((I,-÷2
i3r+2
± • • • +
gn)ir
(4)
Thus all elements of the xi-axis which lie between u, and u,..+.1 (Fig. 4.18) have the same direction after transformation and form a straight line, whose slope is given by (4), as shown. Similarly, the elements lying between ur4.1 and ur+2 form a straight line having the slope dy 0,.+1 = tan-' — = arg C1 + (13r+2 dx
Or+3
• • • ± i3n)r
(5)
The angle between these two lines is given by Or+1 — Or = —163r+1
We have now formed two sides of a polygon, and in a similar fashion we can evaluate the remaining (3's and u's to give us the desired vertex angles and lengths of sides of our polygon.
Ur.t
FIG 4.18.
Suppose that the field is to be on the upper side of the broken line in the z-plane shown in Fig. 4.18. Then any angle, such as cer+i, measured between two adjacent sides of the polygon on the side of the field is called an interior angle of the polygon, and since r - ar+7 = -Or+7 we have, to get this angle, to choose =
ar+ 7—1 7r
(6)
Substituting in (1) gives dz = Cl(ZI dzi
-ai -1
u1) 7 (Z1 — u2) 7
• • • (Z1 —
un ) r
(7)
Integrating this expression gives al Z = Cif [(Z
—
02
u1) 7 (Z1 — u2)7
• • •
dz1 + C2 = Clf(21) + C2 (8)
This is the desired transformation. From (4), we see that we may orient our polygon in the z-plane any way we please by giving C1the proper argument. The size of the polygon
84
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.19
is determined by the modulus of C1. The polygon may be displaced in any direction without rotation by choosing the proper value of C2. To show that we have measured a, on the correct side of the boundary, let us set z1= W so that the real axis yi = V = 0 represents an equipotential line. If ar < or, the field dW/dz = dzi/dz must be zero in the angle; and if a, > r, it must be infinite. Setting x1 = U = u, in (7), we see that this is true which verifies the result. 4.19. Polygons with One Positive Angle.—When the real axis is bent at only one point, no generality is lost by taking this point at the origin, so that ui = 0 in 4.18 (8) giving, if a > 0, z
+ C2
(1)
If we choose C2 = 0 and C1real, the polygon formed has its apex at the origin and is bounded by 0 = 0, 0 = a and by an arc at infinity. From 4.08, the modulus of z" is the nth power of the modulus of z, so that circles given by r1 = al in the z1-plane transform into circles r = a in the z-plane. Thus a problem involving the region bounded by two radii of a circle and the included arc can be reduced by (1) to one involving a diameter and a semicircular arc. When a = 2r, the real axis is folded back on itself and the upper half of the z1-plane opened into the full z-plane. Separating real and imaginary parts gives y = 2Cix1yi and x = Ci(xl — yl). Eliminating yi and xiin turn gives y2= —4C1x1(x — Cixi) and y2 = 4CIYI(x + CIA) which are the equations of two orthogonal families of confocal parabolas. Thus, if we plot the uniform field W = zion the z1-plane, we find it transformed in the z-plane into the field about a charged semi-infinite conducting plane. If we plot the field between a line charge and an earthed horizontal conducting plane passing through the origin in the z1-plane, then we find it transformed in the z-plane into the field between a semi-infinite conducting plane and a line charge parallel to its edge. When a = 3r/2, if we set W = z1, we have, in the z-plane, the field near the edge of a charged conducting 90° wedge whose sides coincide with the positive x-axis and the negative y-axis. The field between such a wedge and a line charge can be found in the same way as when a = 2r.
When a = r, there is, of course, no change in the coordinate system except that it is magnified by the factor C1. When a = fr, if we set W = z1, we find, in the z-plane that W represents the field in a 90° notch in a conductor, the positive x-axis and the positive y-axis forming the sides of the notch. In this case, cl U = x2 — y2 and C1V = 2xy, so that the equipotentials and lines of force are orthogonal families of equilateral hyperbolas. The field due to a charged filament parallel to the edge of the notch is obtained as when a = 27r.
-
§4.20
POLYGON WITH ANGLE ZERO
85
4.20. Polygon with Angle Zero. In this important case, the xl-axis is folded back parallel to itself and the upper half plane is compressed into the space between. In order that two parallel intersecting lines have a finite distance between them, it is necessary that the point of intersection be at infinity. Instead of 4.19 (1), we now have = Ci In zi ± C2 (1) If, as before, we take C2 = 0 and C1real, the origin zi = 0 is transformed to z = — 00 and the new origin is at 21= 1. Writing 21 = rieje., we see that when 01= 0, z = C1 ln r1, which is the real axis in the 2-plane, and when 01 = r, x = C1 ln r1and y = Cir, which is a line parallel to the real axis and at a distance Cur above it. Thus the upper half of the 21-plane is transformed into a horizontal strip in the 2-plane. Radial lines 01= constant are transformed into horizontal lines y = constant; the semicircles r1= constant go into vertical lines, of length Cur. Frequently, we have problems in which a configuration is periodic ; i.e.,the field may be split in identical strips. In such a case, the transformation (1) is useful. Suppose, for example, we have a charged filament between two parallel earthed conducting planes. Going back to the 21-plane, we see that this is reduced to a filament parallel to a single earthed plane. By images 4.06 and 4.12 (4), the required function is —
(ln zi ele° A 2re z i— e—i8° To make W = 0 when zi = +1, take C = — In (—eie.) so that we have = q in 21— eie. (2) 2re 1 — z1eiB0 Transforming into the 2-plane taking C2 = 0, Cir = a, and C100 = b, we have, substituting for 21from (1), W=—
-07
eiwa
eprbia
,y
eircz—Ava
In e_ ir(z+2b)/a eir(z+10/a 1 — er(z+3')/a Now using Pc 652, 660, and 645 or Dw 654.1, 655.1, and 702, we have sinh[1--n-(2 — jb)/a] W = — q In 2r€ sinh [1-.7r(z j5)/a] j tanh (4-71-2/a) cot (1-n-b/a) = — q ln 1 2re 1 — j tanh (1-71-2/a) cot (1--rb/a) = —iqtan-1[tanh (jrz/a) cot (irb/a)] (3) 'WE We can separate real and imaginary parts and get 2re V — sin (rb/a) sinh (rx/ a) tan (4) — cos (rb/a) cosh (rx/ a) + cos (ry/a) q 2reU sin (irb/a) sin (ry/a) tanh (5) cos (irb/a) cos (ry/a) + cosh (rx/ a) q W=—
2re
In
-
—
86
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.21
The z1- and z-planes are shown in Fig. 4.20. If the plates are to be at a difference of potential Uo, we need superimpose on this only a vertical field given by W' = —jUoz/a. The final solution will then be W" = W+ W' = — [jq/(7re)] tan-1[tanh (firz/a) cot (-7rb/a)[—jUoz/a (6) where U is the potential function.
I& WV,
z plane
z, plane
Fro. 4.20.—Line charge between earthed plates.
When a plane grating of parallel wires, spaced a distance a apart, forms one boundary of a uniform field, we can find the conjugate functions in the same way. It is necessary only to put +q at both z1 = j and zi = —j and proceed as before. We obtain, in the z-plane, a section of the field including one wire and bounded by lines of force proceeding from the wire to x = + co. It may be noted that if we plot a uniform field in the z-plane and transform back into the z1-plane, we get the field about a line charge already worked out in 4.12. z plane
z plane
z plane
y
y Z1=0
,=-00
z
Zi=0
z.z .#00
x
a -0 Fro. 4.21.
4.21. Polygons with One Negative Angle. Doublet. Inversion. It is now natural to investigate the meaning of Schwarz transformations with negative angles. The transition from positive to negative angles is clearly illustrated in Fig. 4.21. The ends of the real axis of the z1-plane are now brought together at an angle a at the origin in the z-plane, if C2 = 0 in 4.19 (1). The approximate form of the lines yi = constant in the z-plane is shown in the figure. The most important cases, by far, in this category are those in which a = --7r. Suppose, for example, we start with a uniform field W = z1. If U is the potential function, we may think of this field as produced by
§4.21
POLYGONS WITH ONE NEGATIVE ANGLE
87
an infinite positive charge at x1= 00 and an infinite negative charge at x1 = — . Inspection of Fig. 4.21 shows that the transformation in question brings these two charges infinitely close together on opposite sides of the y-axis in the z-plane. By definition, a two-dimensional dipole consists of two equal and opposite parallel infinite line charges, infinitely close together, the product of the charges per unit length by the distance between them being finite and equal to the dipole strength per unit length M. We can get the transformation directly from 4.13 (1) by writing a for ja and using Pc 769 or Dw 601.2; thus 1 — —a
z
W= — A- lIn tae
1 ± (1z
—>
aq= m 2rez
—
, 0 rez
(1)
—
whence
U—
m cos 0 2rer
V_ —m sin 0 2n-Er
also X2 +y2
MX
271-6U
—0 '
x2 -I- y2 +
My
271-€T7
= 0
(2)
Thus the equipotentials are circles tangent to the y-axis at the origin, and the lines of force are also circles but tangent to the x-axis at the origin. Another important case where a = —7 is obtained by letting C1 = a2 in 4.19 (1) giving a2 Z — (3) Zi Write this in polar coordinates, and separate real and imaginary parts, and we have rri = a2 and 0 = —01 Thus every point outside the circle r1= a in the z1-plane transforms into a point inside the circle r = a in the z-plane. If W = f(zi) is a solution of Laplace's equation, then W* = f(4), where zr is the complex conjugate of z1, is also a solution and this field is the mirror image of the other, giving r1= rf and 01 = —Of. Comparing the z-plane and the zr-plane, we have and 0 = 0i (4) rrr = a2 When such a relation holds, r, 0 and r r, Or are said to be inverse points and a is the radius of inversion. If the sum of the charges in the z-plane is not zero, then there must be an equal and opposite charge at infinity, on which the excess lines of force end, and which will appear at the origin on the z1- or zr-plane. Hence we have the rule for inversion in two dimensions.
88
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.211
If a surface S is an equipotential under the influence of line charges q', q", etc., at z', z", etc., then the inverse surface will be an equipotential under the influence of charges q', q", etc., at 4*, z','*, etc., plus a charge — Mq at the origin. This method is useful for obtaining, from the solution of a problem involving intersecting plane boundaries, a solution of a problem with intersecting circular cylindrical boundaries. In polar coordinates, the equation of the circle in the z-plane is u2 r2 — 2ur cos (0 — a) = R2 (5) where the center of the circle is at u, a and R is its radius. Multiplying through by rt2 and writing or for 0 and ar for a from (4), we have u2)7.512 (rrf )2 — 2rrfurr cos (Of — a t) = (R2 Writing a2 for rrf from (4) and rearranging give
a2u
rr2 + 2R2
Or
where
(Cr — ar) — R2 _ u2rcos 1*
212 — 2u1r1 cos (OP —
a4 u2
an = Ri — u?
azu — —1 2 and R1— — u2
(6)
a2R
Iry - u21
(7)
Thus the circle inverts into a circle. If the original circle passed through the origin, then = 1? so that lull Ri = co . Thus the inverted circle is of infinite radius with its center at infinity, which means a straight line. The nearest approach of this line to the origin is a2 a2 —2 1 2/11 — IR11 u R R and the equation of the radius vector normal to it is Or = «T. Conversely, a straight line inverts into a circle passing through the origin. 4.211. Images by Two-dimensional Inversion.—At this point, we shall pause in the discussion of Schwarz transformations long enough to give an example of two-dimensional inversion. We shall use the rules just derived to find an expression, in terms of images, for the field between an infinite cylindrical conductor, carrying a charge —q per unit length and bounded by the external surfaces of two circular cylinders intersecting orthogonally and a parallel infinite line charge of strength +q per unit length. In Fig. 4.211a, representing the z-plane, the charge is at P and the conducting surface is indicated by the solid line. From 4.21, we see that, if we invert about the point 0, the circles will become straight lines, and since this process is conformal, these lines will intersect orthogonally. For simplicity, we take the circle of inversion, shown dotted, tangent to the larger cylinder. Figure 4.211b shows the inverse system in the zr-plane. The latter system presents the familiar problem,
89
POLYGON WITH TWO ANGLES
§4.22
treated in 4.06, of a line charge parallel to the line of intersection of two orthogonal conducting planes. From 4.06, we know the field in the angle between the planes to be identical with that produced in this region when all conductors are removed, but we have, in addition to q at P', line charges -q, +q, and -q at P'1, P'2, and P3 respectively. The rules of inversion tell us that the surface inverse to these planes, which is the surface of our conductor, coincides with a natural equipotential in the field of charges +q, -q, +q, and -q at P, P1, P2, and P3, respectively, where P, P1, P2, and P3 are inverse points to P', P'„ and P3. We see that if CP = rc and BP = rb, then P1is on BP at a distance b2/rb from B, P3 is on CP at a distance c2/r, from C, and P2 ,
.p' )31• \ • 2 \ z plane
••
/
■
(b)
(a)
Fm. 4.211.—Two-dimensional inversion.
is at the intersection of BP, and CPI. Since P', Pi P2 and P', lie on a circle orthogonal to the lines Q'QI and Q'Q'2in the zf plane, then P, P1, P2, and P3 lie on a circle orthogonal to the circles in the z-plane. 4.22. Polygon with Two Angles.—From the large number of cases that come under this class, we shall select as an important example that in which the real axis is bent into a rectangular trough of width 2a. Taking a1 = a2 = it in 4.18 (7), ui = u2 = -a1, we obtain ,
,
dz A dzi (zf Integrating by Pc 126b or 127 or Dw 260.01 or 320.01 gives
(1)
z = A cosh-' (z CI = jA sin-1 ( z al) + C2 ai Taking C2 = 0 gives, if the second form is used, z = 0 when zi = 0. In order that z = fa when z1= + al, we must have a = ijkir or
jA = 2a/ir, giving z=
2az1 sin-1 -
—
7
C. i
or
z1 = a1sin —z 2a
(2)
90
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
0.22
The most useful application of this is obtained by applying (2) to a uniform field in the z-plane. Taking a = lir and writing W = z, we have zi = ai sin W ( 3) The vertical strip of the uniform field is unfolded as shown in Fig. 4.22a. Thus, if V is the potential function, the field in the z,-planeis the upper half of that due to a charged strip of width 2al, and if U is the potential, it is the upper half of that due to two semi-infinite coplanar sheets at a difference of potential r with a gap of width 2a, between them. Separating real and imaginary parts of (3), if Pc 615 or Dw 408.16 is used, gives xi = al sin U cosh V yi = al cos U sinh V Divide the first equation by alcosh V and the second by alsinh V, square, and add, and we have
x?
ai cosh' V
a? sinh' V
=1
(4)
This shows the curves on which V is constant to be confocal ellipses with major and minor axes 2a, cosh V and 2a1sinh V, respectively. In the same way, dividing by al sin U and al cos U, squaring, and subtracting, we have X?
ai sin' U
2
Y1 -
ai cos' U
1
(5)
The curves on which U is constant are confocal hyperbolas. We note that by giving V values from 0 to 00 and U values from —7r to —+.71- and r to lor in the lower half plane, the field due to the strip is represented everywhere in the plane. The hyperbolic lines of force are then discontinuous in passing through the conducting strip. On the other hand, by giving U values between --br and -A-71- and V values from 0 to — 00 in the lower half plane, we represent completely the field due to the two planes. The elliptic lines of force now are discontinuous in passing through the conducting planes. Suppose that in place of a single charged strip we have a great number of similar strips lying uniformly spaced and parallel to each other in the same plane. It is clear that we may obtain a typical cell of such a field by folding up the xi-axis in Fig. 4.22a at ul = -bbl and u2 = — b where bi > aland by applying (2), with bisubstituted for aland b for a, to (3), giving zi = al sin W = b1sin (brz/b) where 2b is the width of the cell in the grating. This is shown in Fig. 4.22b. Since x = ±a, y = 0 when V = 0, U = +fir we must have al = bi sin (jra/b), so that the transformation is W = sin-1
sin (rz/b)1 sin (fra/b) j
(6)
91
SLOTTED PLANE
X4.23
Although derived for the region above the x-axis, we notice that this formula represents the field equally well when 0 > y > - co, where choosing positive values of V requires that, when 0 < x < b, 37r < U < r and, when 0 > x > -b, < U < The strength of the field at y = co is --br /b, so that to get the transformation for a field of strength E' we must multiply by the factor 2bE'/ir. If we used this grating as the boundary of a uniform field of strength E extending to y = 00 , then we need only superimpose a vertical field of strength E' when E' = iE, U= - rr
U= 0
z plane
U= + f
;
I -I
1 pr I
I
u
I 1
I
I
LL
x
U=--
U= +
2
Fxo. 4.22a.—Transformation of semi-infinite vertical strip into upper half plane.
=+—
2cei
2a —p-1
2b U2
ur
4.22b.—Transformation of field of single charged flat strip into one section of field of charged grating composed of coplanar parallel flat strips.
which will cancel the original field at y = - 00, and add to that at y = + co giving , sin (4-rz/b)R 2b { z — sin= (7) sin (fra/b) jf The field intensity at any point is then given by I Iddz I
cos (irz/b) = .E -1 + 2 - [cos2 (frz/b) - cos2 (fra/b)]il
(8)
4.23. Slotted Plane.—To obtain the field in the neighborhood of a plane conducting sheet with a slit of width 2a in it, we may bend down that portion of the real axis lying between x = +a and x = -a and
92
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.23
expand it to enclose the whole lower half of the z-plane, at the same time drawing that area of the zrplane near the origin through the gap. By examining Fig. 4.23, we see that al = 2r, a2 = —r, as = 21r, ul = d-a1, u2 = 0, us = —as, giving, from 4.18 (7),
dz — dzi Integrating gives 4.18 (8) to be z
2— Zi
z,2
= CI(Zi ±
a,2
± C2
z l/
When z1 = al, we wish z = ±a. This gives C2 = 0 and 2a1C1 = a, so that the result is zi , al z= — (1) 2 al zi If we start with a uniform vertical field W = —Ezi in the zrplane and take 2a1= a, the transformation leaves the field at co in the z-plane
x
-4 Flo. 4.23.—Plane with slot forming one boundary of uniform field.
unchanged but it now terminates on a sheet with a slit in it and part of the field penetrates the slit. From (1), we have W = —Ez1 =
(2) ± (z2 — a2)1] where V is the potential function. If the imaginary part of the square root is always taken positive, then, to make the fields add above the x-axis and subtract below, we must use the positive sign in (2). The surface density on the sheet is given by eE ( x IdW + 1) — 2 \(x 2— a2)1— dz ,-0
= ei
(3)
where the sign is chosen so that the two terms add on the upper surface and subtract on the lower. The field between a charged filament and an earthed slotted sheet in the z-plane is easily obtained by (1) from the field between a charged filament and an earthed plane in the z1-plane.
RIEMANN SURFACES
§4.24
93
Writing the right side of (1) in polar coordinates and solving for x and y, we have a cos 0 () 4 ri + 2r1a1 a sin 01 (5) rl — al 2r1a1 Squaring and adding give x2 y2 a2 (6) (r/ + 2)2 + (7,2 1 a?)2 a 4rla1 Thus, if ri > al, the semicircle of radius r1and that of radius al/r1transform into the upper and lower halves, respectively, of the ellipse in the z-plane given by (6). The semicircle r1 = al flattens into the real axis between x = +a and x = —a. B
4.24.—Riemann surface.
4.24. Riemann Surfaces.—To visualize completely the possibilities of a given transformation, it is often valuable to use the concept of a Riemann surface. This is well illustrated by the example in the last article. Although we have already used up all points in the z-plane to represent positive values of yl equation 4.23 (1) gives us a value of z for every negative value of yi, i.e., when —r < 0 < 0 as well. From 4.23 (5), we see that if 0 < r1 < aland —7r < el < 0 we arrive in the upper half of the z-plane, and if al < r1 < 00 and —7 < 0 < 0 we arrive in the lower half. Thus there are two points in the z1-plane corresponding to each point in the z-plane. We can eliminate this double value by making the z-plane a double sheet. We must be careful, however, to connect these two sheets in such a way that, in tracing a continuous circuit in one plane, we trace a continuous circuit through corresponding points in the other. The central portion of such a surface, known as a Riemann surface, is shown in Fig. 4.24 at the right. Lines in the lower surface are dotted, and the sheets are spread apart to clarify the diagram When x2< ce, it is possible to pass only between A and B or C and D, and when x2 > a2it is possible to pass only between A and C or between ,
94
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.26
B and D. The circle r1 = ei where ei—+ 0 becomes a circle at infinity lying in and surrounding the AC sheet. It is possible to construct Riemann surfaces for a great many transformations. 4.25. Circular Cylinder into Elliptic Cylinder.—The last two articles show that the region outside the circle r1 = al in the z1-plane can be transformed into the region outside that part of the real axis lying between x = +a and x = —a, in the BD sheet of the z-plane, by means of the relation z= 2 a(zi ± zi a) (1) al Since, in the zi-plane we are restricted to a region outside the circle ri = al, we cannot cross the xi-axis where —al < xi < +a1. Consequently in the z-plane, we cannot cross the x-axis where —a < x < +a and the line joining +a and —a is said to be a cut in the z-plane. Electrically, this enables us to transform any field which, plotted in the zi-plane, makes r1= al an equipotential or a line of force into a field, plotted in the z-plane, in which the line joining x = +a and x = —a is an equipotential or a line of force. More generally, since from 4.23 (6) any circle r1 = b1where b1 > al transforms into an ellipse, we may obtain from the solution of any problem involving concentric circular cylindrical boundaries the solution of a problem involving confocal elliptical boundaries. If the circle in the z1-plane is taken eccentric to ri = a1in the proper way, it transforms into an airfoil. This is therefore the famous "airfoil transformation" used in aeronautics. 4.26. Dielectric Boundary Conditions.—With the aid of 1.17 (5) and (6) and the relations 4.10 (1), we shall now determine the boundary conditions that must be satisfied by the potential and stream functions when WI = Ui +jV1 = f1(z) and W2 = U2 + jV2 = f2(z) are functions representing the electrostatic fields on the two sides of a boundary where the capacitivities are ei and €9, respectively. Let Van and a/as represent differentiation normal to and parallel to the boundary, respectively. Then, from 1.17 (5) and 4.10 (1), if U is the potential function,
aui an
a U2 an
= €2-
or
3V1
el
as
= €2
a V2
as
(1)
If we take the zero stream line of W1 and W2 to join at the boundary, then we may integrate (1) along the boundary from this line to any other point V1, V2 and obtain in terms of the capacitivity (2) or K1V1 = K2V2 e1V1 = 62V2 in terms of the relative capacitivity. From 1.17 (6), we have = U2 (3) These are the boundary conditions when U is the potential function.
95
ELLIPTIC DIELECTRIC CYLINDER
§4.261
If V is the potential function, the boundary conditions are or K1U1 = K2U2 (4) €1U1 = e2U2 4.261. Elliptic Dielectric Cylinder.—Conformal transformations can be applied not only to cases involving equipotential or line of force boundaries, but also to many cases involving dielectric boundaries. The fact that such an operation preserves angles means that the law of refraction of the lines of force [1.17 (8)] will still be satisfied. Suppose, for example, our problem is to find the conjugate functions giving the field at all points when an elliptic cylinder is placed in a uniform field of force making an angle a with its major axis. Let the equation of the elliptic dielectric boundary be
V1 = V2
and
x2 971, 2
y2 ±— n2 = 1
(1)
The transformation of 4.25 immediately suggests itself as a means for deriving this elliptic boundary, represented in the z-plane, from a circular boundary r1= b, represented in the z1-plane. For simplicity, let us take al = 1 in 4.23 and 4.25, so that the z-plane represents the area outside the unit circle in the z rplane. Writing 4.23 (6) in the form of (1) and equating coefficients, we see that m n b2 a2 = - m m2 - n2 and (2) m-n It is often convenient, when elliptical boundaries are involved, to use confocal coordinates u and v instead of rectangular coordinates x and y. Such a system is shown in Figs. 4.22a and b. From 4.22 (3), we see that the relations are z = a sin w,
x = a sin u cosh v,
The transformation of 4.25 (1) then becomes z zi
=
1 je-iw = je-iu+v = -[z a
y= a cos u sinh v
(3)
zr1) or
c 2)i]-1 (4) (z2 a2)i ] = a[z - (z2 -
This equation shows us that at a great distance from the origin, where ri co, z = jaz1, so that a uniform field W = laEzitransforms into the uniform field W = Ez in this region. It is clear that the desired field may be formed by superimposing a vertical field of strength E sin a on a horizontal field of strength E cos a. These component fields are shown in Figs. 4.261a and b. Since the axis between the foci is an equipotential in a and a line of force in b, we see that in the zi-plane the situation is as shown in Figs. 4.261c and d, respectively. The field of Fig. 4.261c has already been solved in 4.03, from which we have, taking U' for the potential function and U.: = 0 on the unit circle,
96
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS z pla ne
Z
§4.261
pion
n
(a)
z, plane
(c)
(b)
z, plane
(d)
FIG. 4.261.—(a) and (b) show elliptic
cylinder, K = 9, m/n = 3, in uniform fields 2-4E. Equation (4), Art. 4.261, transforms (a) and (b) into (c) and (d), which involve only circular boundaries. Superposition of (a) and (b) gives (e) where cylinder is in field E at 45° to the axis.
(e)
§4.262
TORQUE ON DIELECTRIC CYLINDER
97
"r, = ijaE sin a(—zi = —ijaB'E sin a(zi zi -1) where, using (2), we have 2[(K + 1) (K — 1)b2] — (m n)(Km — n) A' — —b (K — 1) (K + 1)b2 (m — n)(Km + n) 2b2 m n B' (K — 1) (K 1)b2 Km — +n
(5) (6)
(7)
(8)
For Fig. 4.261d, we see similar harmonics must occur. Since T7:' = 0 when zi= eo, the solutions must be of the form = --aE cos a(zi A"zr1) = iaB"B cos a(zi z-fi) Since, from 4.26, we must have we have b2 + A" = B"(b2 + 1)
= U' and
and
(9) (10) = KV.' when r = b,
b2 — A" = KB"(b2— 1)
Solving for A" and B" gives A" _ b2[(K + 1) — (K — 1)b2] (m
n)(m — Kn) b2(K + 1) — (K — 1) (m n)(m Kn) 2b2 m n B" b2(K + 1) — (K — 1) — m Kn
(11) (12)
Superimposing and using the transformation (4), we have, outside the cylinder, We = + (A" cos a jA' sin a)zr9 = lajE[e-gc"-.) — (A" cos a + jA' sin «)eil = iEle-ja[z±(z2 —a2)4]+ (A" cos a+jA' sin a)[z— (e—a2)i]1 (13) Similarly, inside the cylinder, we have 1) = -aE(B" cos a — jB' sin a)(zi zr ( cos a . sin a \
= E(m n)
m Kn
3Km -1- nj
z
(14)
We notice that the field inside the cylinder is still uniform. The lines of electric displacement when the external field makes an angle of 45° with either axis of the ellipse are shown in Fig. 4.261e. 4.262. Torque on Dielectric Cylinder.—The torque T per unit length on an infinite dielectric cylinder perpendicular to a uniform electric field with which its major axis makes an angle a may be found from the last article and from 3.13 (5) and (6). Thus
T=
a [1.2e„(1 — K) f EE;cos 4, dv]
(1)
where 4, is the angle between the original field E and the final inside
98
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.28
field E, and cos cp are constant in v and may be taken outside the integral leaving f dv = irmn per meter. The relative directions and the magnitudes of E and Eiare given by 4.261 (14) with 1 or K replacing K.
(9W, = = = az K az 1 Simplification of the 4.261 (14) product and substitution in (1) give I E I IE;I cos (0 - = EEP (real part) = 1E(T)EN 1E(i)EN cos2 + i 2 ) T = ievE2(K - 1)7rmn(m n) ( aa m + Kn Km n _ re„E2(K - 1)2mn(n2 - m2) sin 2a n)(m + Kn) 2(Km E=
(2)
(3)
(4)
This torque acts to align the major axis with the field. 4.27. Polygon with Rounded Corner.—There are several methods of replacing the sharp corner in a Schwarz transformation with a rounded one. By one method, we replace the factor zr°-4 in
--1 a-1 dz ir (zi - u2)w •• dzi= z by [(z1 + X(zi - 1)/)](1r)-', where Ili.' > lun-1I > • • juzi j 1 and X < 1. By another method, we replace the factor zro-1by (z1
1)ir
+ X(zi -
In both cases, the argument of the new factor is zero when zi j 1 and is a - r when z1< -1, so that the faces of the polygon on either side of the region -1 < z1< +1 make the same angle with each other as if the factor zr°-1were used. Between z1= +1 and z1= -1, we now have a curve whose shape can be adjusted somewhat by means of the factor X. 4.28. Plane Grating of Large Cylindrical Wires.—We have already seen in 4.20 how the problem of a plane grating formed by cylindrical wires of small radius may be solved by taking for the cylindrical surfaces the natural equipotentials surrounding a grating of line charges. When the diameters of the wires become comparable with the spacing, this approximation breaks down completely. We may solve this case however by the method of the last article. Let us take the "cell" of the grating, outlined in Fig. 4.28a. From the figure, we see that the differential expression connecting z1and z is
dz (z1 + 1) 3 +X (z1 - 1)1 C dz1 - 1[(zi - 1)(zi 1)(zi al)] Cl AC1 ai)11 [(z1 1)(z1 Rzi - 1)(zi
*
(LOP
(1)
PLANE GRATING OF LARGE CYLINDRICAL WIRES
§4.28
99
We illustrate here a method we have not hitherto used for evaluating the constant Cl. Since z1 = rleith if we keep r1constant, we have
dz1 = jrieigi clOi= jz1 dei We notice that when r1 00 and Oi goes from 0 to r then z goes from y = 0 to y = b. Substituting jzidOi for dz1in (1) and letting zi co we have foal' dz = jC1(1 + X) f: dOi C1
,fib = jCir(1 + X)
or b
+A)
—
(2)
Frequently, this method can be used to evaluate a constant by an elementary integration of a special case, usually r1 —> 0 or r1 —> 00, before -- r
t
2b
(c)
(b) FIG. 4.28a, b, c.
(a)
performing the general integration. Integrating (1) using Pc 113 or letting z1 = (a1u2± 1)(1 — u2)--1and using Dw 140.02 gives
2b
1 1)i + tanh-1 z1 )+ + al zi + al When z = c, zi = +1, so that C2 is not real and Z —
c
—
+ A)
[t a nh-1 (
2bX tanh-1 ( 2 )1 or 7(1 + A) 1 + al
When z = jc, zi =
c
Zi
—
+2b 7(1 + X)
—
1, so that
tan-1 ( al 2— 1
C2
al + 1 2
— COth2
C2
(3)
[7c(1 (4) +A) 2bX ]
is not imaginary and or
—
2
1— cot2 r7c(1
L
X)]
2b
(5)
Solving (4) and (5) for (al — 1)/(al + 1) or eliminating al gives al — 1 rc(1 + A) rc(1 + A) irc(1 + A) — cot2 tanh 2 — cos2 al + 1 2b 2bX 2b cscrc(1 + rc(1 + A) A)— coth 2b 2bX
(6) (7)
100
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.29
To determine X in terms of b and c, we can plot the ratio of the left to the right side of this equation as a function of X and choose that value of X for which the ratio is unity. Having determined X, we now add (4) and (5) and obtain a1 = coth2[7c(1 2bX X) ]
cot2
[ 7C(1
] (8)
x)
2b
which determines al. The question of how close the 0 curve approximates to r = c has been investigated by Richmond,* \\ 1, and its distance from the origin has —— -4-4--/ been found to vary from c by less than 2 per cent if 2c < b. (d) To obtain the solution when the grating forms one boundary of a -a, uniform field (Fig. 4.28f), we can 0 -3•/ -------superimpose Fig. 4.28d and Fig. 2 4.28e where lines of force are dotted and equipotentials solid. For case — ------ 4.4 ).5 d, the desired function in the zr 4-1r plane is evidently, from 4.22 (3), re) W= Asin' zl where V is the potential function. -a, Since we wish iEb lines of force in _---- --> 2 the strip of Fig. 4.22a instead of / we must take A equal to *Eb/ir, giving ---Eb . W=— 2-ir in -1 z1 (f) (9) z planes (approximate) FIG. 4.28d, e, f.
s
Similarly for case e, we need the xraxis between —al and +1 at potential zero; so, shifting the origin,
we have Ebz 2 1+al — 1 . W = Z-- sin 27r al + 1 When these fields are superimposed, we get case f with a uniform field of strength E at x = + co and strength 0 at x = — co . The potential function, as in 4.22 (3), is V. Accurate drawings of such fields may be seen in the article mentioned. 4.29. Elliptic Function Transformations. Two Coplanar Strips.—The integral in 4.18 (8) of the Schwarz transformation often leads to Jacobi * Proc. London Math. Soc., Ser. 2, Vol. 22, p. 389, 1924.
101
ELLIPTIC FUNCTION TRANSFORMATIONS
§4.29
elliptic functions when the polygon angles are not integral multiples of r or when more than two fir bends are involved. Examples of these appear in Kober and other references. The functions for two similar coplanar parallel conducting strips with equal and opposite charges is a starting point for the solution of several important practical problems. The data given are the strip boundaries + a, + b and the potential difference 2 Uo. Figure 4.29 shows that four ir real axis bends in (a) form a rectangular box in (b) which encloses an area of the uniform field bounded by V = 0, V = V 1, U = Uo, and U = — Uo. Thus 4.18 (8) takes the form Dw 781.01.
f
w
Cl
[(z2
dz
a 2‘1 LI
b2) (z2
—
b\
1 _1
(1)
a)
C2 Sil
There is no additive constant if z = 0 when W = 0, so
z = b sn (Cow, b)
(2)
Special values of the elliptic sines are (HTF II, page 346) sn (K + jK', k) =
sn (K +
k) = k--+
(3)
where k = b/ a is the modulus, k' = (1 — k2)1is the complementary modulus, and K and K' are the corresponding complete elliptic integrals y
V
F U0
,1V1
/
17 —
b
I
0
17
—U0 —a
r
• VT
v=0 a (a)
in
1 \
r1
+U0
VI
FIG.
U
V=0
b 4.29a, b.
(b)
(Dw 773.1). Thus when z = a in Fig. 4.29a
C2(Uo + jV = K jK',
C2 = -u-
=
KU ' o
(4)
The final transformation may then be written
= b sn (KW, b) Uo a
(5)
When y = 0, on the strips Uo and — Uo or the strip V = 0 and the plane Vi,
aw az
Uo a z b) — sn-1 (b' - K az a
vo
TC-I{(x2
a2)(x2
V)]-11
(6)
102
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§4.30
For the two strips at potential ± Uo, b < I xl < a so that, as in 4.23 (3), b2)(a2 x2)1—i = ± U0(e1C)-1[(x2 (7) For a strip at potential 0 coaxial with a slit in a coplanar plane at Vi, x2)(a2 x2)1—} 0-8 = — Vi(eK')-1[(b2 (8) cr7, = V i(eK')-1[(x2 — b2) (x2— a2)]-1 It is evident that a and b are inverse points for the circle of radius r = (ab)i so that inversion of Fig. 4.29a about this circle leaves the equipotential strips Uoand — Uo, and hence their fields, unaltered. This is impossible if a field line crosses the circle diagonally, so the circle itself must be a field line of value iVi. iop
O
Z Vi
Vi
+ Uo
—ico
( c)
(d) FIG.
4.29c,
d.
The transformation (5) also solves the problems of two coplanar strips or a plane and strip (c) and a strip in a coaxial cylinder or a cylinder or strip coaxial with a slit in a plane (d) . For coplanar strips the capacitance per unit length is, from 4.11 (4),
C =
e[V] eK' = K 2 Uo
6K[(1— b2a-2)1] K (ba-1)
(9)
For strip and plane it is twice this, and for strip in slit it is C =
E[U]
V1
=
4e U 0 V1
4eK (ba--1) K[(1 — b2a-2)i]
(10)
For a strip (2b) in a cylinder [ro = (ab)i] or a cylinder [ro = (ab)i] in a slit (2a), the capacitance per unit length is twice (10). A logarithmic transformation, zi = In z, applied to (5) will give the fields of an infinite stack of strips charged alternately to potentials — Uo and + Uo, and includes a "strip line," which is a strip at potential U0 halfway between and parallel to two infinite earthed planes. Useful formulas for Jacobi elliptic functions of a complex variable appear in HMF, Dwight, Whittaker and Watson, Erdelyi et al., and elsewhere. Jahnke and Emde give some excellent graphs. 4.30. Unequal Coplanar Strips by Inversion.—The inversion method of Art. 4.21, applied to the results of the last article, can solve directly
§4.30
UNEQUAL COPLANAR STRIPS BY INVERSION
103
the unsymmetrical case of two unequal and oppositely charged strips, but the solution takes a much simpler form if we first solve the problem of a strip of width B at potential zero separated by a gap of A — B from a semi-infinite plane at potential V1 as shown in Fig. 4.30a. This is done by folding the real axis in Fig. 4.29a back on itself at 0 so that in 4.18 (7) al = 21r, ui = 0, and a2 = a 3 = • = r. Then if the
U=0
(a) Yi
U=0
h (b) FIG.
4.30a, b.
origin is kept at W = 0, 4.18 (8) becomes z = Czi. If z = B when zi = b, then C is B/b2. Substitution of zifor z in 4.29 (5) gives
z = B sn2 [KW (By]
(1)
'
The modulus is (B/ AP from 4.29 (3), for when W = Uo jVi, then KW/ Uo is K jK' and z = A. From 4.11 (4) and 4.29 (4) the capacitance per unit length between strip and plane is
C =
2e Uo V1
2eK[(B/A — (BMW} )]
(2)
Since the cylinder of radius R = (AB)i is at potential 4-V1, the capacitances between this cylinder and an enclosed strip of width B with one edge on its axis and between it and a semi-infinite plane whose edge is a distance A from its axis both equal 2C or twice that given by (2). Inversion of Fig. 4.30a about point z = w with an inversion radius w(2w < B) will solve the problem of two coplanar strips at potentials U = 0 and U = Uo when one strip is of width w, the inside gap is g, and
104
TWO-DIMENSIONAL POTENTIAL' DISTRIBUTIONS
§4.31
the distance between outside edges is h. That part of Fig. 4.30a lying outside the circle of inversion is shown in Fig. 4.30b, along with that part of the inverse system that lies inside it. From the inversion law w2
zi =
A—
z—w
W(W
g) g
B—
wh h—w
[ (w
k = (By A
(3)
g) (h
So that the transformation is, from (1) and Dw 755.1,
w(h — w) zi — w — h on' (KW/Uo,k)
(4)
From 4.29 (4) the capacitance between the strips is
C —
2E[V1]2€K[(1 — k2)1]
(5)
K(k)
Uo
Let the distance from the center of the iV1 cylinder of radius R to the edges of V = V 1be ci and c2 and to the near edges of V = 0 be d1 and d2 . y
U=0
i
V:=0 --1
Uo
V = Vo
Lx
FIG. 4.31
Then cidi = R2, c2d2 = R2, d1 = ci w, d2 = c2 h — g — w, and ci c 2 = g. With these relations h, w, and g may be eliminated in (3) so that R(ci R(d1 d2) c) k= (6)
R2+ cic2
R2+ cl1c12
The capacitance between the cylinder iV1 and the strip V1equals that between 4-V1 and the two semi-infinite planes V = 0 and is given by
C=
2E U0
iV1
LIEK(k) K[(1 — k2) 1]
(7)
4.31. Charged Circular Cylinder between Plates. The problem in which a conducting cylinder of radius c at potential Uo lies midway between two infinite parallel earthed conducting planes at a distance 2b apart can be solved by using 4.30 (1) and 4.28 (3). The solution is b. First Fig. 4.30a must be almost exact for c < ib but fails as adjusted so that Uo lies between +1 and —1 in Fig. 4.28b and W = 0 is —
§4.31
CHARGED CIRCULAR CYLINDER BETWEEN PLATES
105
at - a. Evidently the origin in Fig. 4.30a must be shifted a distance ai = (A + B) to the right where B = al - 1 and A is ai + 1. Thus 4.30 (1) becomes zi + al = (ai - 1) sn2 ( KW , k) Uo
(1)
The modulus, from 4.28 (6), is then k = (13\1 (al 1)1 - cos +1
firc(1 ± X) ]
2b
(2)
Substitute (1) in 4.28 (3), use the relations between en' u, dn2 u, and sn2u in Dw 755.1, 755.2, and 755.3, and note that tanh-1(jy) = j tan-1 y from Dw 722.3 and HMF 5.07.32 and that C2 = 0. Then 4.28 (3) becomes 2jb dn u en ) ± X tan7(1 + X) an k sn u 1 sn u = 2jb[cos-1(k sn u) + X sin-1(0n u)][7(1 + X)]-1
z -
(3) where u = (KW / U 0,k) . The parameter X must be chosen to satisfy 4.28 (6). From Fig. 4.31 and 4.11 (4) the capacitance per unit length is C -
e[V] 4eK[(1 - k2)l] K(k) Uo
(4)
This is a very close upper limit if c < 2 b, and the error is less than 1 per cent when c = 2 b. Numerical tables and formulas for the real and imaginary parts of en u, do u, and sn u are given in HMF, pages 574, 575. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans, by permission of the Cambridge University Press. 1. That half of the xz-plane for which x is positive is at zero potential except for the strip between x = 0 and x = a which is at potential Vo. The whole of the yz-plane is at zero potential. Show that the potential at any point for which x and y are positive is rV = Vo(tan-1
x — Y a
2 tan Y
x
Y
a
2. The positive halves of the x and y planes are conducting sheets. All the surface is earthed except the region near the intersection, bounded by the lines x = a and y = b, which is insulated and raised to a potential V. Find the surface density at any point on the sheet. 3. By inversion, establish the image law for a line charge in a parallel circular conducting cylinder. 4. A conductor is formed by the outer surface of two similar circular cylinders, of radius a, intersecting orthogonally. It carries a charge q per unit length. Show that q(2r 2— a') (47rar2)-1is the surface charge density, where r is the distance from the axis.
106
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
5. Show that when an elliptic conducting cylinder is charged, the ratio of the maximum to the minimum surface densities equals the ratio of the major to the minor axis. 6. A filament carrying a charge q per unit length passes vertically through a vertical hole of radius a in a block of relative capacitivity K. The filament is a distance c from the center of the hole. Show that the force per unit length pulling it toward the wall is [(K — 1) / (K + 1)]cq2/[27re„(a 2— 0)]. 7. Two fine fiber leads are to be run vertically through an earthed conducting circular cylindrical tube of radius a. Show that the spacing of the fibers must be 2(51— 2)la in order that, when charged to equal and opposite potentials, there will be no force on them. 8. An infinite circular conducting cylindrical shell of radius a is divided longitudinally into quarters. One quarter is charged to a potential V, and the one diagonally opposite to — VI. The other two are earthed. Show that the potential at any point inside is 2ax 2ay + tan ' (tan-1 a' — r' a' — 9. Consider the region of space between the cylinder x' + y2 = b' and the xzplane. All the curved boundary and that portion of the plane boundary for which a < Ix' < b is at zero potential. The part of the latter for which —a < x < +a is at the potential VO. Show that the equation of the lines of force inside is, when a < r b U = 2V ova' [ — ,.— 1 + 7. b' 11- .••1 n f"
cos nO
where only odd values of n are taken. 10C. Three long thin wires, equally electrified, are placed parallel to each other so that they are cut by a plane perpendicular to them in the angular points of an equilateral triangle of side (3)1c. Show that the polar equation of an equipotential curve drawn on the plane is r6
c6 — 2r3c8 cos 30 = constant
the pole being at the center of the triangle and the initial line passing through one of the wires. 11C. A long hollow cylindrical conductor is divided into two parts by a plane through the axis, and the parts are separated by a small interval. If the two parts are kept at potentials Vi and V2, the potential at any point within the cylinder is 1 2ar cos 0 V — V, tan-1 -(V1 +V2) + 2 a2 7.2 where r is the distance from the axis, and 0 is the angle between the plane joining the point to the axis and the plane through the axis normal to the plane of separation. 12C. An electrified line with charge e per unit length is parallel to a circular cylinder of radius a and inductive capacity K; the distance of the wire from the center of the cylinder being c. Show that the force on the wire per unit length is a2e2 K—1 K + 1 2re,c(c2— a2) 13C. A cylindrical conductor of infinite length whose cross section is the outer boundary of three equal orthogonal circles of radius a has a charge e per unit length.
107
PROBLEMS Prove that the electric density at distance r from the axis is e (3r' 67ra
+0)(3,2 — a2
filar) (374— a2 r2(9r4 — 307.2 + a4)
61ar)
14. A horizontal plane at potential zero has its edge parallel to and at a distance c from an infinite vertical plane at potential 17. Show that the charge density on the vertical plane is E(y 2 c 2)-1and on the horizontal plane — e(x 2— c2)-1, where x and y are measured from the line in the vertical plane nearest the edge of the horizontal plane. 15. Show that the capacitance per unit length between a flat strip of width 2c and an elliptic cylinder whose foci coincide with the edges of the strip and whose semimajor axis is a is 27re[cosh-1(a/c)]-1. 16. Show that the force of attraction per unit length between two similar parallel wires of radius a and carrying a charge +q and —q per unit length, respectively, is q2/[27ro(c2 — 40)1] where c is the distance between centers. 17. A plane grating is formed of parallel flat coplanar strips of width 2a with their center lines at a distance 2b apart. If the grating is charged, show that the equipotential surface which is at a mean distance b from the plane of the grating deviates from a true plane by approximately .028b cos' (.1-7ra/b). 18. Find, approximately, the field about a grating composed of parallel wires of radius a at a distance 27r apart, charged to a potential Uo and a parallel earthed plate at a distance b from the center of the wire, if a 1, - c0 < x < -1 and y = 0; when x1 < -1, - c0 < x < -1 and y = 2; when xi = 0, x = 0 and y = 1. Show that if y = 3.735, points x = -1, -1, 0, I Z, +1 all lie on a circle of radius 1 centered at x = -1, y = 1 in the z-plane. 78. A semi-infinite conducting sheet at potential Uo occupies the plane x = 0 from y = 0 to y = .0. The plane y = 0 is a conducting sheet at potential V = 0 except for a gap between x = -a and x = +a. Show that the force per unit length sucking the semi-infinite sheet into the gap is 2eUVra. 79. A circular cylinder of radius a centered at the origin is at potential U = U,. A set of four rectangular hyperbolic cylinders given by xy = +1,2 is at potential U = 0. Show from 4.31 that the transformation giving the field is --
KW z = 2b {j cos-1En k(—,k)]-1- jX Uo
[cn
KW 4)1} Uo
+ X)]-1
where the modululus k is cos [-kb-27rc2(1 + X)] and the parameter X is found from 4.28 (6) with a2 in place of 2c and b2 in place of 2b. Show that the capacitance per unit length is C = &K[(1 - k2)1][K(k)]-1
For good accuracy c/b 2-1. 80. A circular cylinder of radius a centered at the origin is at potential U = Uo. A parabolic cylinder given by y2 = 4p(x + p) is at potential U = 0. By folding the x-axis back on itself in 4.31 show that the transformation giving the field is z - - 8p[cos-1
sn
KW Uo
+ X sin-1(cn
KW)T
[r(1 X)]-'
PROBLEMS
119
where the modulus k is cos [71-2)-12a2 (1 + X)] and the parameter X is found from 4.28 (6) with 2a2 in place of c and 2b2 in place of b. Show that the capacitance per unit length is C = 2€K[(1 — k 2)1J[K(k)]-1. For good accuracy a/p 1. 81. The equation of a parabolic cylinder at potential U = 0 is y2 = 4p(x p). An infinite thin strip at potential Uo lies between x = 0 and x = a. Show by folding the x-axis in problem 61 back on itself that the complex potential function for the field between parabola and strip is 1 W = U o[K(k)]-2 sn-1{exp [iirpt(zi — al)],k), k = exp (-7rabp1) Show that the capacitance per unit length is eff[(1 — k2)1][K(k)]-2. 82. A rectangular bar at potential Uo is bounded by x' = ±b, y' = ±c and lies between earthed conducting planes at y = ±a. Form one-quarter of this system by applying a Schwarz transformation to Fig. 4.30a with a bends at x = 0, x = B, x = A and a fr bend at x = 1, where B b. The plane of the strip bisects the wedge angle making an angle a with each face. From Art. 4.30 show that the transformation for the fields is b ir/a 2a/r z = b{sn [KW, (
Uo a
Show that the capacitance per unit length between wedge and strip is C
=
20V1 Uo
(b/a) 7/a ]1} 2€K1[1 K[(b/a) 1""] —
84. The sharp edges of two identical infinite conducting wedges at potential zero, split by the x1-axis, face each other at a distance 2a apart. A strip at potential Uo occupies the x1-axis midway between them leaving a gap of width b on either side. Apply a Schwarz transformation to Fig. 5.30a with a bend a, between a and 2r, at the origin and 2a at z = A to form one-quarter of the field, so that if V = B / A
120
TWO-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and B (M,N) is an incomplete beta function, 1 zcatir)-1 }-1 zcalr)-1 b dz dz a o (1 — z)1 {fo (1 — z)1 cf.
1+
BRa/r), ()J
(ahr),n + 1] k2n+2}
rk2a/7(1 — k2)1{
an(4/r),( )]
Bk2 Ra/r)i (4)]
n=0/3q) + (a/lr),n + 1]
by HMF 6.6.1, 26.5.1, and 26.5.4. This gives k2 implicitly in terms of b/a. Show that the capacitance per unit length between strip and wedges is C=
4eVi Uo
4elf[(1 — k 2) 1]
K(k)
References The following books contain material pertinent to this chapter. H.: "Partial Differential Equations," Cambridge, 1932. Gives some interesting bipolar transformations. BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Has very complete and detailed treatment of all methods. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Gives extensive, well-illustrated treatment in Chap. X. ERDELYI, A., W. MAGNUS, F. OBERHETTINGER, and F. G. TRICOMI: "Higher Transcendental Functions," 3 vols., McGraw-Hill, 1953. Vol. II has extremely useful elliptic function formulas. (HTF) FLUGGE, S., "Handbuch der Physik," Vol. XVI, Springer, 1958. Pages 40 to 82 give two-dimensional fields including many conformal transformations. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Berlin, 1927. JAHNKE, E., F. EMDE, and F. Loscif.: "Tables of Higher Functions," McGraw-Hill, 1960. Numerical tables and graphs of many functions. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Follows Maxwell and gives examples. KOBER, H.: "Dictionary of Conformal Transformations," Dover, 1952. The most extensive collection. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Considers images inversion and conjugate functions and gives good field drawings. MILNE-THOMPSON, L. M.: "Jacobian Elliptic Function Tables," Gives all needed manipulations formulas and numerical tables of sn u, en u, do u, and Z(u). MORSE, P. M., and H. FESHBACH: "Methods of Theoretical Physics," 2 vols., McGrawHill, 1953. Many conformal transformations scattered through both volumes. NATIONAL BUREAU OF STANDARDS: "Handbook of Mathematical Functions," U.S. Bureau of Commerce, 1964. Good for all functions. Has both formulas and numerical tables. (HMF) ROTHE, R., F. OLLENDORF, and K. POHLHAUSEN: "Theory of Functions," Technology Press, 1933. Gives theory with important practical applications. RYSHIK, I. M., and I. S. GRADSTEIN: "Tafelin," Deutscher Verlag der Wissenschaften, 1957. Formulas for most functions. Very good on elliptic functions. WHITTAKER, E. T., and G. N. WATSON, "Modern Analysis," Cambridge, 1920. Gives very lucid treatment of elliptic functions. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and references. BATEMAN,
CHAPTER V THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS 5.00. When Can a Set of Surfaces Be Equipotentials?—At first glance, one might think that the class of three-dimensional potential distributions in which there is symmetry about an axis could be obtained by rotation of a section of a two-dimensional distribution provided that, in so doing, the boundaries in the latter case generated the boundaries in the former. This is not true in general. We shall now find the condition that a set of nonintersecting surfaces in space must satisfy in order to be a possible set of equipotential surfaces. Let the equation of the surfaces be
F(x, y, z) = C (1) Since one member of the family corresponds to each value of C, if it is to be an equipotential, we must have one value of V for each value of C, so that V = f(C) must satisfy Laplace's equation. Differentiating results in
a2v = f"(C)(-ac)2 -I- (C)a2C
ac av = f ,(C)a--i, etc.,
axe
ax
OX
O X2
etc
Substituting in Laplace's equation gives
a2v v2v = a2v — aye axe
a2 v az'
= f"(c)(vc)2 + r(c)v2c = 0
giving V 2C (VC)2
f " (C)
(2) f'(C) The condition then that the surface F(x, y, z) = C can be an equipotential is that V2C/(VC)2 can be a function of C only. By integration of (2), we can now obtain the actual potential. Since f " (C)/ f'(C) = d[ln f'(C)]/ dC, we have
4.(C) dC = — in [f'(C)] + A' or
f'(C) = Ae-fcc) dc Integrating again gives
V = f(C) = A f e-f cc) dC dC B 121
(3)
• 122
• c
-
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.01
The constants A and B can be determined by specifying the values of the potential on any two of the surfaces given by (1). 5.01. Potentials for Confocal Conicoids.—As an application of the formula just derived, we shall now show that any one of the three sets of nonintersecting confocal conicoids, given by the equation 112 22 x2 ± —1 (1) c2 + 0 a2 b2 where c > b > a and — c2 < 8 < 00, is a possible set of equipotential surfaces. To get a picture of these surfaces, let us vary 8 over the given range. For the range —a2< 8 < 00, every term in (1) is positive so that it represents an ellipsoid. When 0 = 00, we have a sphere of infinite radius, and when 0 = — a2, the ellipsoid is flattened to an elliptical disk lying in the yz-plane. When 0 passes from —a2 3 to —a2 — 3, we pass from the region of the yz-plane inside the disk to that outside. The latter is one limiting case of the hyperboloid of one sheet which (1) represents when —b2 < 0 < —a2. When 0 = —b2, the hyperboloid of one sheet is flattened into that region of the xz-plane which includes the x-axis and lies between the hyperbolas cutting the z-axis at z = + (c2 — b2)1. (3 to —b2 — 3, we pass to the region of the When 0 passes from —b2 xz-plane on the other side of these hyperbolas which is the limiting case in which the hyperboloid of two sheets, represented by (1) when —c2 < 0 < — b2, is flattened into the xz-plane. When 8 = —c2, we have the other limiting case in which the two sheets of this hyperboloid coalesce in the xy-plane. Thus one curve of each set passes through each point in space; and, since it can be shown that the three sets are orthogonal, we can apply to them the theory developed in 3.03 for orthogonal curvilinear coordinates which leads to ellipsoidal harmonics. The latter are too complicated to be treated here, although later we shall treat the special cases of spheroidal harmonics. To return to our problem: Let x2
Mn — (a2 0)n
Y
2
(b2 + 0)n
Z2
( 2 + 0)n
and
1 N — a2 +0
1
1
b2 +0 c2 ±0 With this notation, (1) becomes M1 = 1, and differentiating this we have 2x ao a0 2x M2 = 0 or a2 0x M2(a2 0)' etc. aX so that 2 (1 2 4M2 — 4 (v0)2 =
a e)2 (1+ +
=
M2—M z
§5.02
CHARGED CONDUCTING ELLIPSOID
123
Differentiating again gives
a20
2
a x2
M2(a2 + 0)
2x
2 M2(a2 +
0)
00
a2
M2(a2 + 0)2 ax 4x2
2x
12
2x
0 m3 (a2 + 0)2
4x2
ax
8x2M3
m2(a2 + 0)3 m2(a2 + 0)3 + (a 2 0)2mg, etc.
Adding similar expressions for y and z gives
8M3 8M2M3 - 2N
v20 = 2N
M2 Mi
MI M2
Substituting in 5.00 (2), we have
N V202N .11/2 (V0)2 _ M2 4 - -2-
1( 1 so 43(0) - 2 a2
1
1
) (2)
b2 ± 0 c2 + 0
This proves that such a set of equipotentials is possible. We now find the potential by 5.00 (3) to be
V = A f e[(a2
0)(b2
0)(c2
d0 B
(3)
This is an elliptic integral given by Peirce 542 to 549 with x = - 0. The constants A and B may be taken real or imaginary, whichever makes V real. 5.02. Charged Conducting Ellipsoid.—If we choose V = 0 when 0 = CO, 5.01 (3) takes the form V = -A f 'e[(a2
0)(b2
0)1-1 de
0)(c2
(1)
If we choose V = Vo when 0 = 0 then, substituting in (1) gives
-A = V 0 ) fo 'eRa2
0)(b2 +0)(02 + 0)]-1 de
(2)
The field at infinity due to this ellipsoid, if its total charge is Q, will be Q/(47rEr2). We see from 5.01 (1) that as 0—* co, x2 + y2 +z2 = r2 -> 0, and so a0/ar -> 2r giving
ay ar
2A ay ae A ae ar = r3 2r = r2 =
4rEr2
(3)
Hence The capacitance of the ellipsoid is, from (2),
Q Vo
8/rE A
, = 8re{f [(a2 vo
0)(b2
0)(c2
4,e(a2 - b2)IFRa2 -b2)1(a2 - c2), sin I (1
0)]-i dB c2a 2)11
The surface density is given by OVI T7 to
o = -e(VV)0-0 = -e(— 490
v vi
)o-o
(4)
124
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.04
From (1), (OV/00)0-0 = A(abc)-1and, from 5.01 (1.1), IVOI = 2M2-1 so that Q (x2 y2 ,_ (5) = 4irabc
+
5.03. Elliptic and Circular Disks.—The capacitance of an elliptic disk, obtained by putting a = 0 in 5.02 (4), is still an elliptic integral. To get the surface density, we write 5.02 (5) in the form a 2z2y a 2y2 Q (x2 c4= 471-bC a2 b4 Now let a -> 0, and the terms involving y and z can be neglected. Since both x and a are zero, the first term must be evaluated from 5.01 (1) where 0 is put equal to zero, giving Q = 4.71-bc\
y2
z2 --3
b2
c2
(1)
The capacitance of a circular disk is obtained by putting a = 0 and b = c in 5.02 (4), giving by Pc 114 or Dw 186.11 1-1 = 8 tan-lei (2) -1c/0 7e0 = C = 8ire[f 0-1(b2 0) 0 b Letting p2 = y2 -I- z2 and b = c in (1), the surface density on each side is Q
(3) p2)i The potential due to such a disk given by 5.02 (1) with a = 0 and b = c is 4irb(b2 -
)= 2V0 1b 2V 1 oi tan_ V =( 2 Oi Putting in the value of 0 obtained by letting r2 = x2 4. y2 + z2, and b = c in 5.01 (1) gives 2 Vo 52)2 4b2x2P1-1) V = — tan-1(2ibIr2 - 52 + [(r2 -
a=
(4)
This problem can also be solved by oblate spheroidal harmonics (5.271). 5.04. Method of Images. Conducting Planes.—An application of the test of 5.00 shows that in no case involving more than one point charge can we obtain the potential from the analogous two-dimensional case. Nevertheless, two of the methods used in such cases can also be applied to three-dimensional problems. One of these is the method of images. Any case in which the equation of a closed conducting surface under the influence of a point charge can be expressed in the form n
a .■ 1
§5.05
PLANE BOUNDARY BETWEEN DIELECTRICS
125
where r is the distance from q to any point P on the surface and r, is the distance from some point s on the other side of the surface to P can be solved by the method of images. We shall consider only simple spherical and plane surfaces. It is evident from symmetry that a single infinite conducting plane or two such planes intersecting in the z-axis at an angle ir/m under the influence of a single point charge q in the xyplane can be solved by locating point images in the xy-plane at the same places as in the two-dimensional case shown in Fig. 4.06. Adding up the potentials due to the point charge q and its point images gives exactly the same potential V in the region between the intersecting planes as would be obtained by adding up the potentials due to the charge q and the equal and opposite induced charge distributed over the planes. Hence we can determine this induced surface density o on the conductor by finding —69 V/On on the plane to be replaced by the conductor. In the case of a point charge q at a distance a from an earthed conducting plane from 1.14, and 1.16 (1), the induced density at P is
0- —
—aq 27rr3
(1)
where r is the distance from q to P. From 2.17, we know that whatever the actual distribution of charges in the space separated from q by the earthed plane this cannot affect c on the side facing q. 5.05. Plane Boundary between Dielectrics.—Since the case of a uniform line charge parallel to the plane face bounding two dielectrics may be considered as built up of equal point charges uniformly spaced along the line and since the images can be built up in the same way, it is plausible to suppose that the same image law holds in both cases. Let the relative capacitivities of regions of positive and negative z be K1 and K2, respectively. Consider any configuration of charges in dielectric K1whose potential, with no dielectric present, would be given by
Vva, = f(x, y, z) so that, if only K1were present, it would be, from 1.06 (3), 1 y, z) V1 = — K1f(x, The potential of an image in the z = 0 plane would be = Ci f(x, y, —z) The law of images of 4.04 suggests that when K2 is present the fields in K1and K2 should have the forms V1 = rif(x, y, z)
Ci f(x, y, —z)
(1)
126
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.06
and (2)
V2 = C2f(x, y, z)
When z = 0, V1 = V2 and KiaVdaz = K2OV2/az from 1.17 (5) so that 1 + K1C1 = K1C2
and
1 - K1C1 = K2C2
solving and substituting in (1) and (2), we have 2 f(x y, -z)] VI = —[f(x y Ki "Ki ± K2 "
(3)
2 V2 - Ki K2f(x, y, z)
(4)
and
Referring again to 1.06 (3), we see that if we have an actual charge q at P1in K1, then the field in K1is the same as if the whole region were occupied by K1and we had an additional charge q' at the image point P2, and the field in K2 is that which would exist if the whole region were occupied by dielectric K. and there were only a charge q" at P1, where q,
K _ K1 K1+ K2q 2
r K 2K„ 1 K2q
and
(5)
For calculating the field the choice of K. is immaterial but all authors take K1, 1, or K2. We use K1. 5.06. Image in Spherical Conductor. We saw in 4.05 that the cylinder p = a is an equipotential under the influence of a line charge q at p = b and a second line charge -q at p = -I-a2/b. Likewise we shall now see that it is possible to have the sphere r = a at zero potential with a point charge q at r = b and a point FIG. 5.06. charge q' at r = a2/b, and we shall determine q', which does not, as in the two-dimensional case, equal -q. The potential V' at r = a due to q' is, from Fig. 5.06, a2 471-E V' = q,(a4b-2 2a3b--1cos 0)-1 = ba-lq'(a2 +52 - 2ab cos 0)-i = ha-T-1g' —
The potential at r = a due to q is q/(471-€R). In order that the sphere be at zero potential, these must be equal and opposite, giving q,
= _aq
The potential at any point is then b2 - 2br cos 60-i - a(b2r 2 V = (41-€)-icr2
(1) a4
2a2br cos 0)--i] (2)
The surface density on the sphere will be
av\ CT
=
ar
2 b2 qra-b cos 0 b(b - a cos 0)1 aR3 a 47raR3 q (3) R3 -
— 44
;5.07
EXAMPLE OF IMAGES OF POINT CHARGE
127
We can evidently raise the potential of the sphere to any desired value V by adding the potential due to a point charge q at the center where q = 47reaV
(4)
If we wish the field between a point charge q and a conducting sphere having a total charge Q, we need only add to (2) the potential due to a charge Q aq/b at the center. We find that there is not, as in the analogous two-dimensional case, a simple image solution for the dielectric sphere and the point charge. We shall have to use spherical harmonics to solve this problem.
5.07. Example of Images of Point Charge.—As an example illustrating ev, the last three articles, let us locate the -q. aq -ce images in the case where the earthed b conductor consists of a plane sheet, / // /:/ /:,,AV:// o /f /:/•-erbq lying in the yz-plane with a spherical / / / boss of radius a centered at the origin, q r•//, // // and the region below the xz-plane is filled _ // // with a material of relative capacitivity ',/ ' //,/ K. The point charge q lies at xo, yo, zo / / , = V. To locate the and 4 + FIG 5.07. required images, we complete the boundaries with dotted lines as shown in Fig. 5.07. For the field above the xz-plane, the potential can be found by superimposing that due to the original charge plus seven images located as follows: —
q at xo, yo, zo q at —xo, yo, zo
aq —
a)2 (
a)2 at ( xo,
—
(a)2 yo, 7) zo
b
aq
75-; at —
(a)2 (ay (ay xo, yo, zo
q' at xo, —yo, zo q' at — xo, — yo, zo
— aqi at (Ix 412 ( a)2 b b °' b Y°' -6 z° aq' a)2 (a)2 --b— at 4-6 xo, — yo, zo
where, from 5.05 (5),
, 1 —K q — 1 + Kg Below the xz-plane, in the dielectric, the potential can be obtained using only the images above the xz-plane and substituting q" for q where, from 5.05 (5), _ 2q q1 + K
A section of the conductor by the xy-plane together with a projection of the images upon this plane is shown in Fig. 5.07.
128
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.08
5.08. Infinite Set of Images. Two Spheres.—Frequently, it is impossible to get equipotentials of the desired shape by any arrangement of a finite number of point charges. In some cases, however, one surface can be made an equipotential by some interior configuration of point charges. Then, by a suitable arrangement of image charges, the other surfaces may be made equipotentials, at the same time, however, distorting the first surface. By a third set of images, the latter is restored to its original shape at the expense of the other surfaces, and so forth. If, because they become more remote or grow smaller or tend to cancel each other, the effect of each successive group of images diminishes, we may get as close an approximation to the exact solution as we wish by taking enough of them.
V=0 (a)
(6)
FIG. 5.08.
This method can be used to find the self- and mutual capacitances for two spheres. Let the radii of spheres 1 and 2 be a and b, and let the distances between their centers be c. From 2.16, the coefficient cii is the charge on 1 and c12 that on 2, when 2 is grounded and 1 is raised to unit potential. By interchanging a and b in c11, we can obtain c22. The two cases, when c < (b — a) and when c > (b + a), are shown in Figs. 5.08a and 5.08b, respectively. In these figures, m = a/c and n = b/c. First, we put sphere 1 at unit potential by placing a charge q = 4rea at its center 0'. We then put 2 at zero potential by placing an image q' = —217rEab/c = —4irEna [see 5.06 (1)] at a distance b 2/c = nb to the left of 0. Next, we restore 1 to unit potential with the image -T aq' +471-Emna = (c — nb) — (1 — n2 ) at a distance a2/(c — nb) = ma/(1 — n2) to the right of 0' and then q
restore 2 to zero potential by placing an image
r, =
— bq" c — ma/(1 — n2
— )
-T 4remn2a 1 — m2 — n2)
(
▪•▪ §5.081
DIFFERENCE EQUATIONS. TWO SPHERES
129
at the proper distance from 0, and so forth. We note that the size of each successive image decreases so that we are approaching the exact solution. Adding up the charges on 1 gives m 2n2 n cii = 4rea 1+ — m2± •I • • (1) —1m — n2 (1 — n2)2 where the upper sign refers to case b. in case b, c12 = 4re[—na —
Adding up the charge on 2 gives, mn2a
(2)
1 — n2 — m2
In case a, from 2.17, c12 = —cii. From symmetry, we can write down, in case b, mn m 2n 2 c22 = 47rEb[l (3) 1 — m2 (1 — m2)2 — n2 In case a, c22 is of no importance. From 2.20 (6), the force of attraction between the two spheres will be, since V2 = 0, acli F V2 == 2 ac
47rEa V2 mn {(1_n2)2
m2n2[2(1 —n2) — m2] [(1 — n2)2 — m92
••
(4)
If we wish the potential at any point P, we must add up the potentials at P due to each of the image charges. If the spheres are at potentials V1 and V2 where V2 0, we shall have, in case b, a second set of images, for we must now start with a charge qi = 4reaV1at the center of 1 and q2 = 4n-Eb V2 at the center of 2. The sizes and positions of the second set of images can be obtained by interchanging 0 and 0' and a and b in the values for the first set. 5.081. Difference Equations. Two Spheres.—The formulas of the last article are not very convenient for precise calculation unless m and n are small. Better formulas, involving hyperbolic functions, can be obtained by getting the general relation between successive images and then solving the resulting difference equation. Let us designate the nth image in 1 by qn so that the initial charge at its center is q1= 471-Ea and the image of qn in 2 by pn. In Fig. 5.08, the distance from 0' to qn is sn and that from 0 to pn is rm. Then, using lower signs for case a and upper for case b, we have b2 — bqm — _ (1) pn = —
c + Sn + sn + abqn apn _ c(c sn) — b2 c — rm +a2(c T s,) a qm+i, _ sn) se.) b qn sn) — b2 c(c
c
q.+1 Sn+1 =
c
+ a2 rm —
(2)
-
(3)
130
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.081
Eliminating c snfrom (2) and (3), we have qn s7L qn = -d2 q.+1 qn+i
so that
qn-i qn
= C qn-1 a22 qn
_ b
a n
(4)
-
Eliminating s. from (2) and (4) rearranging give b2 1 1 = ,2 _ a2 1 (5) ab qn qn+ qn-i This is known as a difference equation of the second order with constant coefficients. The general method of solution is to substitute 1/qn = divide through by u"--1, and solve the resultant quadratic equation for u algebraically. If the two solutions are u1 and u2, the solution of the difference equation is
1
q.
(6)
= At? ± Bu4
where A and B are to be determined by the terminal conditions. The present equation is specially simple because the coefficients of 1/q.+1 and 1/qn_i are the same. Comparing with Dw 651.03 and 651.04 or Pc 669 and 671, which give sinh (n cosh (n
1)0 sinh (n - 1)0 = 2 cosh 0 sinh nO 1)0 + cosh (n - 1)0 = 2 cosh 0 cosh nO
we see that a solution of (5) is 1 - = A cosh na qn
if we choose cosh « = ±
c2
B sinh na a2
(7)
b2
(8) 2ab To evaluate A and B, let us write down for the first two images 1 1 = _= A cosh a± B sinh a 421-Ea qi c2 _ b2 1 ± + - (2 cosh a + A cosh 2a + B sinh 2a 3 421-Ea _ = 1 = ab 4rea q2 Multiply the first of these equations by 2 cosh a so that it can be written in terms of cosh 2« and sinh 2«, and solve for A and B, and we have _ 1 b + a cosh a B - . (9) 4reab sinh a A = -1-zlreb Substituting in (7) and using Dw 651.01 or Pc 660, we have 1 b sinh na ± a sinh (n - 1)a = (10) 42reab sinh a qn
§5.082
SPHERE AND PLANE AND TWO EQUAL SPHERES
131
Adding the charges on 1 then gives 02
c11 = 4reab Binh a E [(b sinh na ± a sinh (n — 1)a]-1
(11)
n=1
where the lower sign refers to case a (Fig. 5.08a) and the upper to case b (Fig. 5.08b). In case a, from 2.17, c12 = —ca. In case b, to get c12, we must determine p„. Taking the upper signs and eliminating c — from (1) and (2), we have 1 a 1 b 1 — c c Substituting for 1/q.+1and 1/q„ from (10), using (8) and Dw 651.06, we have 1 = c sinh na (12) 47reab sinh a p. so that adding up the charges on 2 gives 4reab sinh
csch na
C12 =
(13)
n=1
We can write down c22 from (11) for case b, by symmetry, thus: c22 = 4n-eab sinh a E [a sinh na b sinh (n — 1)a]-1
(14)
n=1
5.082. Sphere and Plane and Two Equal Spheres.—A special case of some interest is that of a plane and a sphere. We can get this case by letting d c = b —> co in case a or c — d = b —> co in case b, where d is the distance from the plane to the center of sphere 1. In either case, n —> 1, m —> 0, and m/11 — n1 = a/d so that we have OD
c11 = 41-ea( ±
a2 2d
4C12
—
a2
•
csch na (1)
= 4irea sinh n=1
where d = a cosh a and a is negligible compared with b. In either case, In the case of the sphere and the plane, the force is C12 = C11. 8ad ? acii = —27rea2 1 a 2)2 v.{2d2 + (4d2 F = 2 Od
= 2n-eVfE[csch na(coth a — n coth na)]
}
(2)
n-1
In the case of two equal spheres of radius a at a distance c apart, the formulas of 5.081 are somewhat simplified by writing in terms of
132
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.10
= ia where, setting a = b, we have, by Dw 652.6 or Pc 667, cosh g =
c
2
+ cosh a)] = 2—a
(3)
In terms of 0, 5.081 (11) becomes, setting a = b,
c11 = c22 = 41-ea sinh
csch (2n — 1)I3
(4)
n= 1
and 5.081 (13) becomes c12 = — 4rea sinh OE csch 2n0
(5)
n =1
5.09. Inversion in Three Dimensions. Geometrical Properties.—We have already seen in 4.21 that if, in a plane, we draw a circle of radius K about the origin and then draw a radial line outward from its center, any two points on this line, distant r and r' from the origin, are said to be inverse points if rr' = K 2 (1) In 4.21, the circle represented a section of a cylinder, but it can just as well represent a principal section of a sphere. Thus for every point of any surface in space there corresponds an inverse point, and the surface formed by these inverse points is said to be the inverse surface. If the equation, in spherical polar coordinates, of the original surface was f(r, 0, 0), then the equation of the inverse surface is f(K 2/r, 0, 0). We proved in 4.21 that straight lines invert into circles lying in the same plane and passing through the center of inversion, and vice versa, whereas circles not passing through the origin invert into circles. Hence, in the present case, planes invert into spheres passing through the center of inversion and spheres not passing through the center of inversion invert into spheres. Since, in 4.21, we arrived at the laws of inversion by a conformal transformation, we know that angles of intersection are not altered by inversion. It is evident from this that when a small cone of solid angle dSt with its vertex at the origin cuts out area elements dS and dS' from the surface S and its inverse surface S', the angle 0 between the axis of the cone and the area elements is the same, so that
dS _ r2dO cos 0 _ r2 (2) r" dSt cos 0 — r' 2 dS' 5.10. Inverse of Potential and Image Systems.—We shall now show that it is possible to formulate laws for the inversion of electrical quantities such that we can obtain from the solution of a problem with a certain boundary surface the solution of a second problem with the inverse boundary surface.
§5.101
EXAMPLE OF INVERSION OF IMAGES
133
Consider Fig. 5.10 in which P', R', and Q' are the inverse points of P, R, and Q, respectively, and 0 is the center of inversion. Then a charge q at P gives a potential V= q/(4rePQ) at Q, and a charge q' at P' will produce a potential V' = q' /(47EP'Q') at Q'. Triangles OQ'P' and OPQ are similar since K 2= OP • OP' = OQ • OQ', and 0 the included angle of both is a. The necessary relation between the potential V at Q before inversion to the potential V' at Q' after inversion is then
V' q' PQ q' OP V q P'Q' q OQ' In order to make this relation useful, we must find a suitable law for the inversion of charges. It was shown in 5.06 that a sphere of radius K is at zero potential under the influence of a charge q at r = b and a charge jq'I = --FKIql/b at r' = K 2/b which is the inverse point. If the sphere is to remain at zero potential after inversion about itself, which will interchange the charges, the law of inversion of charges must be
q' _ K K _ _ OP' q b OP K
(1)
We give the ratio the positive sign because we wish the inversion of charge to leave its sign unchanged. Substituting this value in the equation for the inversion of potential, we have the relation
V' K OP KOQ V OP .OQ' OQ' = K
(2)
This shows that V' = 0 if V = 0 provided OQ is finite, which means that if any surface is at zero potential under the influence of point charges a, a9 • • • , qn at Pi, P2 • • • , P,, which are at finite distances, not zero, from the center of inversion, then the inverse surface will be at zero potential under the influence of the inverse charges qc, q;, . . . , at P'„ P2, . . . , P. The qualification is necessary since (1) does not provide for inverting charges when OP is zero or infinite. It also follows from this rule that if any problem can be solved by images, the inverse problem can also be solved by images. 5.101. Example of Inversion of Images.—Let us now compute the force acting on a point charge q placed in the plane of symmetry at P at a distance b from the point of contact of two earthed spheres of radius a. Clearly, the two spheres can be obtained by inverting the system of planes shown in Fig. 5.101a. Taking the circle of inversion, shown dotted, tangent to the planes simplifies the computation. Before inver-
134
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.102
sion, the potential at P' due to the images alone is co 1 V'= = — 27re
(
q' 4na
— 1. )n—=—
ql ln 2
8rea
n = 1
' of The field of the images has a neutral point at P' so that the value V, V' inside a circle of radius S' around P' is constant if terms in S'n are neglected when n > 1. Thus inversion by 5.10 (2) gives the potentials at
+4a > .< 4a - 4 4a-►
#q .1
-
Q.
f#.7'
0 ..cil
6'
/ -1-'• •
9
Inverse System (a) Fia. 5.101.
P and at a distance 8 above P of the images that lie in the spheres on a circle through P and the point of contact to be, respectively,
2a
Vb = -6- V p,
V6+5
+6 2a
-
b
, 2ao , 172, ',z--,Vb — -p--V p
From 5.10 (1), q' inverts into 2aq/b so that the force on q is
F = qE =
lim 5—. 0
a) _ _ aq2
q(Vb
+
a
b
aireb 3
ln 2
(1)
5.102. Inversion of Charged Conducting Surface.—Let us now consider a conducting surface S charged to potential 1/K, which produces a surface density a, and let Q be any point on this surface. From 5.10 (2), the potential at the corresponding point Q' on the inverse surface AS' is V'Q= 1/0Q' . Since the potential at Q' due to a negative charge 4re at 0 is —1/0Q', it is clear that we can make the potential over the inverse surface everywhere zero by superimposing the potential due to such a charge. Reversing the procedure, we have the useful rule that, if we can solve the problem of a conductor at potential zero under the influence of a positive charge 4re, then we obtain by inversion, with this charge as a center, the solution of the problem of the inverse conducting surface raised to a potential —K—'.
-
THREE-DIMENSIONAL HARMONICS
§5.11
135
For the inversion of the surface density at P, we have, from 5.09 (1) and (2) and 5.10 (1), the relation
• = cr
dS K . (OP) 2 _ (OP) 3 q OP (OP')2 K3
q'
K3
—
dS'
(1)
(OP') 3
5.103. Capacitance by Inversion.—As an application of the rule of 5.102, we shall compute the capacitance of a conductor formed by two spheres of radii a and b which intersect orthogonally. Clearly, the inverse system is that shown in Fig. 5.103a, in which two earthed planes,
/1"qf I I a - /
/
%\
\ \
1/ I
-q'.
11 1
/
/
/ \
\
/
-q .
\
/
//
/
I
I
I
I
\ \
/
/
1
// /
/
.471
\
/
\
\
\
\
ab .
/
/
/ /
Inverse System la)
(61 FIG. 5.103.
intersecting at right angles, are under the influence of the point charge q' = 47re. This field can be replaced by that due to the image charges shown. The desired conductor, obtained by inversion, will be at potential V = —1/K = —1/2a. By 1.10 (1), the charge on this surface equals the sum of the image charges, qi, q2, and q3, giving, by 5.10 (1), = qi q2 q3= 49re i —a + ab L 2a 2a(a2 b2 )1 and the capacitance is Q
C
r
= Q =, (a ± b)(a2 b 2)i — ab] ± V "L e
b
(1)
The potential around the conductor can be computed directly from the image charges ql q2, q3or by inversion of the potential due to image charges alone in the region in the angle of the planes. 5.11. Three-dimensional Harmonics.—We saw in 5.00 that it is not, in general, possible, by rotating the orthogonal set of curves representing a normal section of a two-dimensional field, to obtain a set of three-dimensional equipotential surfaces. It is, however, possible to generate ,
136
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.11
in this way a set of surfaces which, together with the planes intersecting on the axis and defined by the longitude angle, form a complete set of orthogonal curvilinear coordinates which can be treated by the method of 3.03. When the axial section of our three-dimensional boundary problem gives a two-dimensional boundary solvable by a conformal transformation, this provides a method of finding a coordinate system in which the original boundary conditions are very simple. The problem then becomes that of obtaining a general solution of Laplace's equation in this coordinate system. Let us suppose that u1 = fi(x, y) and u2 = f2(x, y) are conjugate functions in the z-plane, given by z = x ±k
= f(ul + ju2) = f(u)
(1)
Then, by 4.11 (1), we have
, dz du= f (u)
or
dx ± j dy = I (u)(dui± j due)
Multiply this by the conjugate complex, and we get
dz 12 (dui + dul) du
ds1 = dx 2 + dy2 = 1
If this system is rotated about the y-axis, we have, for the element of length,
ds2= ds? + x2(4)2 =
dz 2 du (dU!
+ dui) + x2(dcb) 2
(2)
Comparing with 3.03 (1), where dcl, = dui, we see that h1 = h2 =
dzi
dui
_ 3 = .._
and
h
It
so that, from 3.03 (4), Laplace's equation is dz1-21- a i an + ±a( all + a2 V 0 (3) x Idu Laui\ xauif ' au2 xau2 j ' aq52 = We can immediately split off the last term by the method of 4.01 by assuming a solution of the form (4) V = U(uI, u2)c13(95) Dividing through by V and setting the last term of (3) equal to -m2, we have for cf the form, as in 4.01 (6), cI,= Aiefro + Bie-fro = A cos mcp + B sin mcp
(5) The partial differential equation to be solved for U(ui, u2) then becomes
1— a (au) 1 a (al — -I- — x— x aui x814 x au2 au2
- m2 dz 2
x du U = 0
(6)
137
THREE-DIMENSIONAL HARMONICS
§5.11
The difficulty of solving this equation depends on the form of x(u i, u2) and of Jdz/duJ. In many important coordinate systems, including all those treated in this chapter, x has the form x
= g1(u1)g2(u2)
(7)
and, when 4.11 (1) is used,
dz 2 du
ax _ .ax 2
aui
"au2
( ax
)2
au'
aX )2
au2
= 912g2
gIg;2
(8)
Assuming a solution of the harmonic form U(ui, u2) = Ui(ui) U2(u2)
(9) and substituting (7), (8), and (9) in (6), we find that, when divided through by U1(u1) U2(u2), each term contains only one variable so that, putting terms containing u1equal to a constant +(s m2c) and those containing u2 equal to — (s m2c), we obtain the total differential equations 1 d drli) M2[ (12 s=0 (10) gl dui g1 + c ,1 2 1 d (dU2) C] s=0 g2 U2 dug g2 dU2 m2[ g2) ]
(
The remainder of this chapter will be concerned with the solving of these equations and with applying these solutions and (5) to potential problems. We obtain spherical polar coordinates from 4.12 (2) by interchanging x and y so that
gi = eui = r and g2 = sin u2 = sin B (12) If we put c = —1 and s = n(n 1) in (10) and (11) and write in terms of r and 0, these equations become 5.12 (2.1) and 5.14 (2). We get oblate spheroidal harmonics from 4.22 (3) giving and g2 = cosh u2 = (1 + '2)1 (13) gi = a1sin u1= ci(1 — 2)1 If we put c = +1 and s = —n(n + 1) in (10) and (11) and write in terms of t and 1 these equations become exactly 5.271 (2). We obtain prolate spheroidal harmonics from 4.22 (3) with x and y interchanged, giving ,
and g2 = sinh u2 = (n2 — 1)i (14) 91 = a1cos ui = c2(1 — t2)i If we put c = +1 and s = —n(n + 1) in (10) and (11) and write in terms of t and n, we get the standard form of Laplace's equation in prolate spheroidal coordinates referred to in 5.28. We obtain cylindrical coordinates by rotating the transformation z = u so that we have gi = ui = p = k-1v and g2 = 1. If, in (10), we put c = 0, s = —k2, m = n and divide through by k 2, we obtain
138
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.111
Bessel's equation [5.291 (3)]. The same substitution in (11) gives 5.291 (2). 5.111. Surface of Revolution and Orthogonal Wedge.—In the particular case where m 0 and Eq. 5.11 (7) is satisfied, we can obtain the solutions of 5.11 (10) and 5.11 (11) in a simple form. Let the conjugate functions to be rotated be w1 = fi(x, y) and w2 = f2(x, y). Then let us choose new orthogonal coordinates u1and u2, such that (1) = F1(wi) gi(ui) = (A le", (2) g2(u2) = (A2 cos U2 + B2 sin u2) = F2(w2) We have from 5.11 (7), by hypothesis, using p for the radius of rotation of any point, p = gi(ui)g2(u2) = Fl(w1)F2(w2) (3) If we set me = 1 and s = 0 in 5.11 (10) and 5.11 (11), we can easily verify by substitution that solutions of these equations are U1= C win
(4)
and U2 = C2g7
(5) The differential equations are of the second order so that there must be a second solution of each. Since the method of finding this from the known solution will be useful in other cases besides the present one, we shall work out the case with general coefficients here. Suppose that y = v is a particular solution of 2y d — Maxy = 0 (6) dx dx 2 where M and N are function of x. Substitute y = vz in this equation, and write z' for dz/dx, and we have, after eliminating v by (6), the result
dz' v— dx
dx,
ivi
, v)z = 0
Multiply by dx, divide by vz', and integrate ln z' ln v2 f M dx = C or
dz = By-2e–f mdx dx Integrating gives
z= A+ Bfv-26–f M dx dx so that
y = v(A B fv-2e– f " " dx) (7) In the present case, putting 5.11 (10) and 5.11 (11) in the form of (6) by carrying out the first differentiation and multiplying through by
SPHERICAL HARMONICS
§5.12
139
U1,2, we see that M1,2, has the form 2
M1,2 — ,
_ d In
gi,2
g1,2
dui,2
Using for v the solutions of (4) and (5), we obtain the second solutions from (7), setting A = 0, to be dui U1 = D19i 2.+1
(8)
U2 = D2V3 f 92du_!, L 2m,
(9)
and
Thus a solution of Laplace's equation is given by
/ du1 du2 V = gi l + Di f gi u1 g'2" C2 ± D2 grn+i cos (mch -I- 1) (10) This form of solution is particularly useful when the surface of a charged conductor is formed by a figure of rotation and a wedge with its edge on the axis of rotational symmetry. A simple example will illustrate an application of this solution. A charged infinite conducting wedge of exterior angle a has a spherical conducting boss of radius a intersecting both faces orthogonally. Here, taking the center of the sphere as the origin and the faces of the wedge to be 95 la, = we + have p = r sin 0, gi = = r, g2 = sin u2 = sin O. All the boundary conditions are satisfied by taking m=
, DI = (1 —
1+ la a ,
C2 = 1, D2 = 0, and = 0
Thus we obtain
V = Cirfl —
ay +1 74, (sin 0)a cos a
(11)
This gives V = 0 on the surface of the wedge and sphere and agrees with 4.19 when we are far from the origin. 5.12. Spherical Harmonics.—When the boundary conditions of a potential problem are simply expressed in spherical polar coordinates, it is useful to have a general solution of Laplace's equation in this system. To get this solution, we proceed in exactly the same way as in 4.01. In terms of the distance from the origin r, the colatitude angle 0 measured from the z-axis, and the longitude angle measured about the z-axis from the zx-plane, Laplace's equation, from 3.05 (1), takes the form
a av ar ar
1 av 12 a av sin 0 y.) =0 sin 0 ae (. u sin2e a o2
We wish to find solutions of the form V = ROI) = RS
(1) (2)
140
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.13
where R is a function of r only, 0 a function of 0 only, and 4,a function of 4) only. The function S = 04,is called a surface harmonic, and the function 0, when 4 is a constant, is called a zonal surface harmonic. Substituting RS for V in (1) and dividing through by RS, we have 1 d ( 2dR\ _L 1 a ( 1 agS sin 0°S + ae S sin e ae s sing 0 a¢2 - 0 dr\ r dr The first term is a function of r only, and the other ones involve only the angles. For all values of the coordinates, therefore, the equation can be satisfied only if 1 as 2 — -K 1 sin °as) ± a ( 00 S sing 0 act.2 S sin o 00
1 d 2dR) K (2.1) R dr r dr ) The solution of the second equation is easily seen to be R = Arn Br-n-1 + (3) where K = n(n + 1). Substituting this value of K in the first equation and multiplying through by S give 1 0 sin1
a (sin as)
1 as 2 sing 0 002
n(n + 1)S = 0
(4)
Equation (2) therefore takes the form
Br-"--1)S„
V=
(5) It is evident that this is a solution of Laplace's equation only if the same value of n is used in both terms, hence the subscript on Sn. Any sum or integral with respect to n of terms like (5) is also a solution. In the special case where n = 0, (4) becomes
(. aso ae
sm n — sin 0— si
e ae
)
a,s0
2 =0
(6)
In 6.20 it is shown that either U or V satisfies this equation, provided j sin 43) tan 10] U jV = F[(cos (7) Thus every conjugate function of the last chapter yields two solutions of Laplace's equation in three dimensions by substitution of cos yt, tan 1-0 for x, and sin ct, tan 10 for y, and multiplication by A + Br-1. Especially useful solutions obtained by choosing W = zI'n and W = In z are, respectively, V = (A + Br-1)(C cot'n 1-0 D tan'n 1-0) cos (mci, + S.) (8) V = (A + Br-1)(C In tan i0 + DO) (9) 5.13. General Property of Surface Harmonics.—Before obtaining solutions of (4), we may derive, by Green's theorem, a useful property of
15.131
POTENTIAL OF HARMONIC CHARGE DISTRIBUTION
141
S,,. Let us again write down 3.06 (4), giving 9*) dS (PV2c13 — (1)V2T) dv = fs(1a(13 );-z— 11-
(1)
Let us take NI, = r'n8. and = rnS„ so that V21. = V2NY = 0 and the volume integral vanishes. Taking the surface to be a unit sphere and writing dig for the element of solid angle, we have
a* an
a (rmS,n) = mrm—iS„, = mS„, ar
and similarly a1/an = n8n. Substituting in (1) gives
fo (n8„8. — mS„S„,) d 1 = (n — m) feS„S„, dig = 0 so that if n m we have the result foS„8„, d it = 0
(2)
5.131. Potential of Harmonic Charge Distribution.—Let us suppose that, on the surface of a sphere, we have an electric charge density, cr,,, which is finite, continuous, and of such a nature that we can choose a small area, AS, anywhere on it so that, over AS, an — an is negligible compared with the mean value, an, of an over AS. This charge gives rise to a potential, V0, outside the sphere and the potential, inside it. Applying Gauss's electric flux theorem [1.10 (1)] to a small box fitting closely this element of the shell, we have =
avi av)
(
ar
aT
aft
(1)
Let us consider that an is such that 1 a0+2 2n + 1 rn+' "
E Vo =
(2)
Then, since V° = V when r = a, we must have EVi —
1 2n + 1 an--1811
(3)
Substituting (2) and (3) in (1) gives an
(4)
= Sn
In studying 5„ we shall see that when 0 enters as FT (cos 0), it meets the conditions imposed on anat the beginning of this article. By using these equations and 1.06 (6) we can evaluate two useful integrals.
' dS
SAS
_ 4/r rnS„ 2n + 1 an-1'
sR: dS =
4ir an+2S. 2n + 1 rn+1
(5)
142
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.151
The angles in S, are the coordinates of the terminal point of R, or R0. Thus by superposition, the potential due to any surface distribution which can be expanded in the form (6)
cr = So ± Si + S2 + • • •
is given by
0113-1 ± a
= 120
/20
-a 5
±•••I
if r < a
Vo = Lig-So ±
if r > a (7) r 5 +•'• r r 3 5.14. Differential Equations for Surface Harmonics.—The variables 0 and 40, in the differential equation, 5.12 (4), for the surface harmonics S = 043, may be separated by the process already used. Substitute 043 for S in 5.12 (4), and divide through by 0,10/sin2 0 giving sin e 0d (. de) sin 0 —
dO
dO
1 d2c1) — 4)(1 2
n(n + 1) sin2 =0
For all values of 0 and 95, this equation can be satisfied only if sin 0 d ( 0 sin sin 0a -cr o
n(n + 1) sin2 0 = K1
-)
and
1 d2c13 dq52
= —K1
If we put K1 = m2, the solution of the second equation is easily seen to be
(13 = C cos m¢,
D sin ing5
(1)
except when m = 0, when it becomes c13 =
N
Putting K1 = m2 in the first equation and multiplying through by 0/sin2 0 give 1 d ( dO m2 [n(n + 1.) sin 0 ) 0=0 (2) Sin.2 0 sin 0 dO dO This is the differential equation for 0. 5.15. Surface Zonal Harmonics. Legendre's Equation.—Before considering a more general solution of 5.14 (2), let us solve the most important special case in which V is independent of 4, so that t is constant and, from 5.14 (1), m is zero. Equation 5.14 (2) then becomes, when A is written for cos 0, d
[
d0„ (1 — I.42) -1+ n(n 1)0„ = 0
(1)
This is Legendre's equation, and its solutions are surface zonal harmonics. 5.151. Series Solution of Legendre's Equation.—To obtain a solution of 5.15 (1) in series, let us assume the solution 0„ = /ariir (1)
SERIES SOLUTION OF LEGENDRE'S EQUATION
§5.151
143
Substituting this in 5.15 (1) gives 1{r(r — 1)argr-2+ [n(n + 1) — r(r + Martel = 0 To be satisfied for all values of A, the coefficient of each power of A must equal zero separately so that (r + 1)(r + 2)ar4.2 + [n(n + 1) — r(r + 1)1a,• = 0 (r + 1)(r + 2) (r + 1)(r + 2) ar = n(n + 1) — r(r + 1)a1.42 (n — r)(n + r + 1)(4+2 (2) To expand in increasing powers of we notice that if a, = 0 then ar-2 = ar_4 = • • • = 0 and from (2) a_1and a_2 are zero if ao and al are finite, so that all negative powers of A disappear. Hence, if we choose ao = 1 and use the even powers, we have the solution pn = 1
n(n + 1) 2n(n —
1)(n + 3)
µ4• 2)(n4 If we choose al= 1 and use the odd powers, we have gre
—
12
(n — 1)(n + 2) g 3!
(3)
(n — 1)(n — 3)(n + 2)(n + 4) 5 A — 5! (4)
A complete solution of 5.15 (1), if —1 < µ < + 1, is
On = A nP n Bnqn regardless of whether n is an integer or a fraction, real or complex, provided the series converges. Recurrence formulas for pn and qn may be obtained by subtracting pn±i from pa_i, giving pn-l
[(n + 1)(n + 2) — n(n — 1)µ2 2! (n + 1)(n + 4 — n(n — 3) (n — 1)(n + 2)µ4 3! (n — 1)(n + 2) 3 = (2n + 1)/2(./ 1.1 + • • • ) 3! = (2n + 1)Aqn
— p7H-i =
)
(5)
By a similar procedure, we obtain (n + 1)2qn+1 —
(6) = (2n + 1)APn Differentiating (4) and adding n dqn_i/c112 and (n + 1) dq.÷i/c111 give
(n — 2 + n + 3)n(n + 1) 2 p 2! (n — 4 + n + 5)n(n — 2)(n + 1)(n + 3)ih4 4! • • • 1 n(n + + n(n— 2)(n + 1)(n + 3 ---- (2n + 1)(1 2! 1)/42 4! (2n + 1)p.
nq',1+ (n + 1)q;,±1 =[2n + 1
) µ4
144
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.152
By a similar procedure, we obtain np4, = —n(n + 1)(2n + 1)qm
(n
(8)
5.152. Legendre Polynomials. Rodrigues's Formula.—If n is a positive even integer, the series in 5.151 (3) evidently terminates and has -En + 2) terms and may be written in
[( 71)!P pm = (— 1)22n n!
1)r—in
2n[-En
2r)! µ 2* (n — 2r)]! [En + 2r)] (2r)
r=0
In this case, we define the polynomials P,,(µ) to be (-1)1"n! 2'1(i) q2Pn
(1)
If n is a positive odd integer, the series in 5.151 (4) evidently terminates and has -(n + 1) terms and may be written —1)
qn =
1)1(n-1)2n-1{[En—
1)1!) 2
n! r =0
2r ± 1)!1,2,4-1 (n 2"[En — 2r — 1)] ! [En ± 2r ± 1)] !(2r In this case, we define the polynomial P,,,(µ) to be (_1)1(n—i)n t Pn(P) 2„-1{{En — 1)1! }2qn
1) !
(2)
Legendre's polynomial, given by P n(A) where n is a positive integer, in ascending powers ofµ by (1) and (2) may be written in reverse order by substituting s = in — r in (1) and s = En — 1) — r in (2), both giving the result m
(2n — 2s)! un-28 1)8 2n (s!) (n — s)!(n — 2s) l
P.(µ) =
(3) s=o where m = in or i-(n — 1), whichever is an integer. An expression for Pn(A), known as Rodrigues's formula, may be obtained by writing (3) in the form m
Pn(P)
n! (2n — 2s)! „2a 1 2n1(1)8s!(n — s)! (n — 2s)! '21 8=0
n
1 dn 2nn ! d pn
( 1)8
nI „ 2n-2e (n — s)!'”
§5.154
RECURRENCE FORMULAS FOR LEGENDRE POLYNOMIALS
145
The last summation is the binomial expansion of (p2— 1)n so that
1 cln (4) 2r —Tzrdr tn(A2 — 1)n Equations (3) and (4) are valid solutions of Legendre's equation [5.15 (1)] whatever the range of the variable A. In prolate spheroidal harmonics, we have 0 < µ < . For very large values of A, the highest power outweighs all others; so we have (2n)! Pn(11) A—. —÷ (5) co 2n(n!)21A Pn(P) =
5.153. Legendre Coefficients. Inverse Distance. The polynomials of 5.152 are also known as Legendre's coefficients, the reason being evident from the following considerations. The reciprocal of the distance between two points at distances a and b from the origin, where b > a, when the angle between a and b is 0 and A = cos 0, can be written: —
1 _ (a2 + b2 —
=
a' —b22abAy
R
1 l 1 a' — 2abiA 1 • 3(a2 — 2abily b[ 2 b2 -1- 2 4 b2 ••• a 3122 1(4 _ 1 i_ 5/13 — 3p(ay = 4 -42 \b/ 1;[ • • b 2 \b —
-
u
We see that the coefficient of (a /b)Th is exactly the expression for P, (µ) in 5.152 (3) so that we may write 2 (1) =— bl[Po(A) (bPI(P) (0 P20.0 + • • • — R 1
We shall refer to this expansion many times in solving problems. 5.154. Recurrence Formulas for Legendre Polynomials. If n is an odd integer, we may substitute for pn_i, pn_Fi, and qn in 5.151 (5) from 5.152 (1) and (2) and obtain, after dividing out the factor —
{[-En — 1)]!12 n!(-1)i(n-") the result nPn_i (n 1)P.+1 = (2n 1)AP.
(1)
An identical expression is obtained, when n is even, from 5.151 (6). We omit the argument of these polynomials when there is no ambiguity in doing so. If n is an even integer, we may substitute in 5.151 (7) for pn, en,_„ and q'n4.1from 5.152 (1) and (2), divide out the factor 2n(2n 1)[(n)!]2 (-1)inn!
146
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.155
and obtain = (2n + 1)Pn P41 An identical expression is obtained, when n is odd, from 5.151 (8). The integral of P.(.4) is given by integrating (2), the result being Pn(P) diA =
Pn+1 Pn-1 2n + 1
(2)
(3)
The derivative of Pn(u) is given, by adding successive equations of the type of (2), to be P'n (g) = (2n - 1)P,1 +(2n - 5)P,3
•••
(4)
Another useful expression for the derivative may be obtained by differ1by (2), giving entiating (1) and eliminating = /AP:,
(n
1)Pi,
=
or
nPn_i (4.1)
Eliminating Pn_1 and P,',±1between these equations and (2) and combining with (1), (2), or (3) give the following equivalent forms:
P'.
1)(I.IP„ P.±1) _ -n(tiPn - P.-1)
_ (n
1 - 112
1-µ2
i,n„ ,
n(n + 1) P„_1 n-1 — Pn+1 -n(n 1) 1 - /22 2n 1 1 —µ 2
UI)
(5)
5.155. Integral of Product of Legendre Polynomials.—In using Legendre polynomials, the integral of their product over the range 0 = 0 to r, orµ = -1 to +1, is important. We saw in 5.13 (2) that
j_ i r-n(-L)P.(A) dµ = 0 If m = n, we substitute for one
11 [Pn(a)]2
(1)
if m 0 n
P. from 5.152 (4)
1 f+t =— Pn(o —I cit-( 12 2fin!
dun
1)n dpi
Integrate the right side n times by parts, letting u be the first term and dv dr(;2 2 — 1)n the second each time. Since always contains the factor dAr ().,2 1)n—r, v will always be zero when the limits are inserted and so the product uv drops out and we have finally +1
f-1 [p..}2 dµ _
(-1- rid,.Go — 1)n
dA
)
2nn ! J
eV
But from 5.152 (3), dnP,,(p) _ (2n) !n !— (2n,) !
di‘n
2nn !n !
2nn !
(2)
TABLE OF LEGENDRE POLYNOMIALS
§5.157
so that [p.(/1)]2
_
22(n
(nn)1)2i
147
- p2)
(2n — 1)!!.1 (2n) !!
0 sin2n+i 8 dB
(2.1)
Integration by Pc 483 or Dw 854.1 gives
r+,
2
[Pn(A)]2 (3) 2n + 1 J --1 5.156. Expansion of Function in Legendre Polynomials.—Any function that can be expanded in Fourier's series in the interval —1 < < +1 can also be expanded in a series of Legendre polynomials in the same interval and by a similar method. Assume the expansion
= aoPo(A) + aiPi(u) + • • • + a,,P.(A)1)+ • • • Multiply by Pm(u), and integrate from µ = —1 to +1. By 5.13 (2), all terms vanish except the mth term; so we have, from 5.155 (3). am = 1(2m + 1)f+11f(p)Pm(A) du
(2)
Note that if AA) = 0 when —1 < µ < +1 then am = 0. This means that if we have an expansion in Legendre's polynomials equal to zero, the coefficient of each term must be separately equal to zero. As in the case of a Fourier's series, at a discontinuity this expansion gives half the sum of the values of Au) on each side. A formula for am which is frequently more convenient than (2) can be obtained by substituting Rodrigues's formula [5.152 (4)] in (2). Thus am
2m + l +1.. — 2) m = ( -1)' 2„,÷ Im ! f_1 (A)dinadin
Integrating this by parts repeatedly, always letting the first member be u and the second dv, we find that luv1+1 is zero and f +1u dv alternates in —1 sign until finally we are left with am
2m + I-SI-lc/1(1)(1 = 2"1+1M _
difn
A2)17' cipt
(3)
If the derivatives of f(g) are simple, this gives usually an easy integration. 5.157. Table of Legendre Polynomials. A table of values of P.(12) can be computed from 5.152 (3) or (4). The values of n < 9 are —
Po(A) = 1,
Pi(A) = it,
1 P3(m) = — 2(5/13— 3/2),
P2(ti) = EU' — 1) (35m4 _ 301.0 3) P4(A) 8
148
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Pb(A) =
(63/46— 70/.4 3
15/4
8
,
P6(Y) =
§5.158
(231/.46 — 315124 + 105/22 — 16
5)
(429/47 — 693125+ 3150 — 35/4) 16 6930/.44 — 1260442 + 35) (6435/.48 — 12,012/46 P801) — 128 P7(A)
Other useful values of P„(i.L) are (n odd)
Pn(0) = 0
(n even)
Pn(0) = (-1)inl
(any n) (any n) (any n) (any n)
Pn(1) = 1
3 5 • • (n — 1) 2.4.6 • • n
Pn( — = (-1)P.(14)
P'n (0) = —(n P'„(1) = in(n
1)Pn+1(0) 1)
[from 5.154 (5)] [from 5.15 (1)]
The values of Pn(A) for 0 n 7 are shown in Fig. 5.157. 5.158. Legendre Polynomial with Imaginary Variable.—We shall have occasion, in oblate spheroidal harmonics, to deal with Pn(j0 where j = (-1)1 and 0 < 00. Substituting for A in 5.152 (3), we have 13, (
=
m )in: A‘J s= 0
(2n — 2s)! 2" (s) !(n — 8)!(n — 2s)!'
2s
(1)
where m = ?tn or -En — 1), whichever is an integer. A similar substitution in 5.152 (5) gives P„(j0
(2n)!
2.(n!)2
(-1)in?-.
(2)
§5.17
CHARGED RING IN CONDUCTING SPHERE
149
5.16. Potential of Charged Ring.—If V is symmetrical about the x-axis and its value is known at all points A on this axis and if this value can be expressed by a finite or convergent infinite series involving only integral powers of x, then the potential at any point can be obtained by multiplying the nth term by P„(cos 0) and writing r for x. The result holds for the same range of values of r as the range of x in the original expansion. Let us apply this to a ring carrying a total charge Q (Fig. 5.16) giving 471-EV4 = Q(c2 +
x2
0
2cx cos a)-I
Fm. 5.16.
Expanding this by 5.153 (1) gives X C
VA
a)
= Lircc n =0
x< c
VA =
(C os
4irEc
a)
n =0
The potential at any point P at r, 0 is given by r > c, or e
a, r = c
VP
r
= 471-€c
n+1 P„(cos a)P„(cos 0)
n =0
r < c, or 0
a, r = c
Vp
P„(cos a)P„(cos 0)
4/rEc
(1)
n=0
Other examples of this method will appear at the end of this and subsequent chapters. 5.17. Charged Ring in Conducting Sphere.—If the value of the potential due to a given fixed charge distribution is known in a certain region, then the values of the potential when an earthed conducting spherical shell is placed there can be found. Expand the original potential in spherical harmonics, and superimpose a second potential, similarly expanded, due to the induced charge such that the sum is zero over the sphere. The latter should vanish at infinity if the original distribution is outside the sphere and be finite at the center if it is inside. As an example, let us find the potential at any point inside a spherical ionization chamber of radius b, if the collecting electrode is a thin concentric circular ring of radius a. Let us set a = fr and take r > a in the last problem, inserting the value for P.(0) from 5.157 and writing 2n
150
§5.18
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
for n, since only even powers remain, we have for the potential due to the ring alone 4/1-ea
'In
1 3 • 5 • • (2n. — 1)(aYn+i P2n(cos 0) \ r) 2 • 4 • 6 • • • 2n
n=o The potential of the induced charge must be finite at the origin; so it has the form = 4rea l
A r2nP (cos 0)
2n 2n
n= o
But Vi V,.= 0 when r = b, and from 5.156 (2) we may equate the coefficient of each Pn(cos 0) separately to zero, so that 1 3 5 • • • (2n — 1) 1 (aYn+1 2.4.6 • • 2n b 2n\bf
A
= — ( —1)n
and if a < r < b or r = a, 0 'yr, the potential becomes
v= Q 47rea
n= o
( _ 1),,(2n — 1)!! / a)2n+1 (2n) !!
a 2.+1 r 2n
-6
]P2.(cos 0) (1)
If r < a or r = a, 0 0 Fr, we have
= ___1( _ 1)„(2n
— (2n) !!
4rea
ryn
(ayn+102n —
\bj
]P 2n (cos
0)
(2)
n=o
Care should be taken in the application of this method to fields produced by a distribution of charge on conductors. In general, in an actual case, the field of the induced charges will cause the inducing charges to redistribute themselves and invalidate the result. 5.18. Dielectric Shell in Uniform Field.—We now compute the field inside a dielectric shell of internal and external radii a and b, placed in a uniform electrostatic field of strength E. As in the last problem, we may consider the potential outside as that due to the original field Er cos 0 plus that due to the polarization of the dielectric. The latter must vanish at infinity so that it can contain only reciprocal powers of r. Furthermore, the only surface harmonic appearing in our boundary condition at infinity is Pi(A) = cos 0 so that the potential outside must be of the form A = (Er + —2) cos 0 In the dielectric of relative capacitivity K, since r is neither zero nor infinite, we include both terms, giving
OFF-CENTER SPHERICAL CAPACITOR
§5.19
V2 =(Br
151
+ —1 cos 0 r2
In the cavity, the potential must be finite so our only choice is V3 = Dr cos 0 The four boundary conditions necessary to determine A, B, C, and D are 2' = b,
=K aviav2 r = a,
2A
Or
Or
Or
V2=
A C Eb -I- 172 = Bb ± p
or
VI = V2
V3
,av2 av3
.n.— =
UT
Or or Solving these equations, we obtain
E — — = KB — b3 C
or
2KC
(2)
0
Ba + — = Da a2
KB —
2KC a3
(1)
(3)
=D
(4)
9KE
D=
(5) 9K — 2(1 — K) 2[(a/b) 3— 1] Looking at the expression for V3, we see that this is the strength of the electric field inside the shell. 5.19. Off-center Spherical Capacitor.—As an example of boundary conditions involving surface harmonics, let us compute, approximately, the charge distribution on the inner shell of a slightly off-center spherical capacitor. The formulas derived by images in 5.08 require many terms to give accuracy in this case if the inner radius a is nearly equal to the outer radius b. Let us choose the origin at the center of the inner surface; then the approximate equation of the outer surface is
r=b
cP1(.1) (1) where c is the distance between centers and A = cos 0. This is obtained from 5.153 (1), the reciprocal of which may be written b = r[1 cr-1P1(1)]-1= r — cPi(A) when terms in c" are neglected if n > 1. Since here the boundary conditions involve both Po (A) and Pi(A) and since both r = 0 and r = 00 are excluded from the field, the potential must be of the form
B
V =A+—
(Cr
—NA) r2
where C and D are small correction terms of the order of c. ary conditions are r=a r=b
cP1(.1)
(2) The bound-
V=A+
(C a ± I a 3 B c P (Cb V2 = A +— T4)]
Li)P101)
152
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.21
where the products Cc and Dc have been neglected. From 5.156 (2), we may equate the coefficients of P0(u) and Pi(A) separately to zero, giving
A + — — Vi = 0 a A + — V2 = 0
Ca + - = 0 a2 Bc D - + Cb + — = 0 b2
Solving, we have B —
ab(V1— V2)
b—a
C=
'
abc(V1. — V2) (b — a)(b3— a3)'
a4bc(V i— V2) (b — a) (b3— a')
We may now obtain the surface density
aV
0. = —E(—) = ar
eab(Vi — V2) (1
b—a
3c cos 0) a2 b 3— a3 \
where terms in cn are neglected. We notice that if we integrate this over the surface of the sphere, the correction term drops out so that the capacitance is the same as if the spheres were concentric, if only terms in c are included. 5.20. Simple Conical Boundary.—We have already seen in 5.131 how the potential due to any charge distribution on the surface of a sphere can be expressed in spherical harmonics. It might also be pointed out that if the value of the potential V on the surface of any cone of revolution, whose equation is 0 = a, can be expressed in the form /3nr--n-1) V = /(Anrn where n is an integer, then, at any point inside the cone (Anrn + Bar—n—i)P„,(cos 0) V P,(cos a) —
(1)
for it can be seen by inspection that this solution satisfies Laplace's equation and the boundary conditions. 5.21. Zonal Harmonics of the Second Kind.—The second solution of Legendre's equation given by the infinite series in 5.151 (3) or (4) is called a zonal harmonic of the second kind and is denoted by Qn(A). We Alan define it by expressions similar to 5.152 (1) and (2) to be 22.-1. Qn(A) = (-1)1(..+1)1[ En — 1)M (n odd) (1) P. n! run)!122n Qn(A) = (-1)int " ' (2) (n even) n! qn These hold for —1 < µ < +1. Although at present we are dealing with solutions of Legendre's equation in which p = cos 0, we shall find later, in using spheroidal harmonics, that we need solutions in which p2 > 1. This requires the
LEGENDRE FUNCTIONS OF THE SECOND KIND
§5.212
153
expansion 5.151 (1) with negative values of r. We may write 5.151 (2) in the form (n — r)(n + r + 1) (r + 1)(r +2) ar ar+2 (3) We observe that if ar.+.2 = 0, ar+4 = a7+6 = • • = 0 and if a_,.+1 = 0, = ct_r+5 = • • --- 0. But (6+2 is zero if an is finite, and a_n+i is zero if a,1is finite. If we let an =2n!/[2n(n!) 2], we get 5.152 (3), a polynomial defining Pn (A), but if we take a_n_1 = 2n(n!) 2/(2n + 1)! we obtain, by writing r = —n — 3, r = — 5, etc., in (3), the coefficients in the series 2n(n!) 2[ 1 (n + 1)(n + 2) 1 (2 .(i) = (2n + 1) ! An-FI 2(2n + 3) An+3 03
2n(n!)2 (n + r)!(n + 2r)!(2n + 1)! (2n + 1)!LJ r!(n!)2(2n + 2r + 1)!
—n-2r--1
r=0
Writing s for r to agree with the notation of 5.152 (3), we have (in(A)
*1(n + s)!(n + 2s)! — 2n .ZJ s!(2n + 2s + 1)!A
(4)
a =0
This series evidently converges when A2 >1 and defines Qn(u) in this range. When fl is very large, the smallest negative power in (4) outweighs the remainder so that we have
(2.01)
2n(n!)2
(5)
(2n ± 1)
5.211. Recurrence Formulas for Legendre Functions of the Second Kind. The following formulas connecting different orders of Q, (µ) can be obtained from 5.151 (5) and (6) by substituting for pn and q>, the values of 5.21 (1) and (2) in exactly the same way as 5.154 (1), (2), (3), (4), and (5) were derived. —
nQn_i + (it + 1)Qn+1 = (2n + 1)4. Q'n+1— = (2n + 1)Qn
= Qn +i Qn(A) 2n + 1 J + • • • + p(1 — Q'„ = (2n — 1)Q,„.1+ (2n — 5)(20_8 (n + 1)(AQ. — Qn4-1) 1— u —
(1) (2) (3) (4) (5)
5.212. Legendre Functions of the Second Kind in Terms of Legendre Polynomials. A useful expression for Qn (A) may be obtained from Legendre's equation if we know that Pn(A) is a solution. In 5.111 (6) —
154
§5.212
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and (7), we saw that if v is one solution of the differential equation 2y dx 2
dy M— dx
(1)
Ny = 0
where M and N are functions of x, then a second solution is y = v(A
Bfv-2e— fm dx dx)
(2)
In Legendre's equation [5.15 (1)] v = P„(A) and M = —2A(1 — 1.12)-1 so that 5M diA = In (1 — 2) and e—I m dµ = (1 — /22)-1giving (2) the form d11 (1 — ) [P.(A)] 21 To determine the constants A and B, let n = 0 and n = 1, and integrate by Dw 140 or Pc 48 and by Dw 152.1 or Pc 57, and expand the logarithm by Dw 601.2 or Pc 769. This gives Q.(ii) =
P.(A){A
Bf
1 + — A + B( + 3 1 Qo(A) = A ± —B ln 1 -44 5 + 2 3 +-115 A — Qi(A) —
=
2
ha l + 1—
B = B(-1
142 ±
3
+ 1A2
• • )
5
From 5.21 (2) and (1), we see that Wiz) = qo and Qi(A) = — pi, and comparing the values of qo and pl. in 5.151 (4) and 5.151 (3) we see that B=1
A = 0 and so that the general expression for Q„(A) is
and in particular
dA Qn(A) =Pa(A).) (1— 42)[Pn(14)]2 Qo(A) = 1 In
(3)
1+A
(4)
and 1 1±A WA) = ? 1n 1 _ A
1
Applying 5.211 (1), we obtain 1 1 ± A 3A 1 —(3/42 — 1) In 1 A -2 (22(11) = 2 —AQi(A) — 2 —Qo(A) = 4 1 ± I.L 1 = -P2(.1) In A 2 1— By repeated application of 5.211 (1), we get 1 +A 2n — 1 in l_ /I 1 . n P.--101) Qn(A) = 1 2Pn(A)
(5)
3 2Pi.(
)
(6)
2n — 5 3(n _ i)Pn-3(ii) — • • • (7)
This holds for 42 < 1. The general solution of Legendre's equation is 0 = AP.(A) + BQn(A)
15.214
LEGENDRE FUNCTION OF THE SECOND KIND
155
If we write A' = A - iB In (-1) and substitute in (7), we have 0 = A'Pn(p) + BrP n(A) in 2
+ 2n -1 P„_1(/.1) - • • 1 1•
So that when j.c2 > 1, we may define Qn(i.t) by the equation 1 2n - 1p 2n - 5 Qn(A) = -12Pn(A) In A + — 1 1 • n n-1 \Aj — 3(n - 1)A
( — • •
'
(8)
Putting n = 0 and n = 1, expanding the logarithm by Dw 601.3 or Pc 770, and comparing the values of Qo(A) and (21(12) with 5.21 (4), we see that the formulas agree. 5.213. Special Values of Legendre Functions of the Second Kind. The actual values of Qn(A) may now be found easily from 5.212 (7) or (8), but it is frequently convenient to know certain important special values for real arguments as follows: (n even)
Qa(0) = 0
(n odd)
5' Qn(0) = (-1)4(n+1)2 41 63 :
(any n)
(1) n 1)
Qn( --A) = ( -1)n+I(24) Q„(1) = co Q„( co ) = 0
(2) (3) (4)
5.214. Legendre Function of the Second Kind with Imaginary Variable.—In connection with oblate spheroidal harmonics, we shall have to deal with Qn(ji-) where j = (-1)1and 0 < < 03. No difficulty is experienced when ?- > 1, for substituting jl forµ in 5.21 (4) we have Qn(i0 = (—P1+12'.2 s =o
(-1).(n s!(2n
2sn s)! 2s '+ 1)! '
(1)
If we make the same substitution in 5.212 (8), the result is ambiguous owing to the fact that the logarithm term is multiple valued. Using Pc 770 and 780 or Dw 601.2 and 506.2, we may write In
±1 1
1 1 .( 1 - 23 ± - — 50 + •
= -2j cot-1I-
If we expand the coefficient in 5.212 (8) by 5.158 (1) and substitute the above series for the logarithm, we find the result is identical with (1) so that a consistent value has been chosen and we may write Qn(j1') = —j cot-1-Pn ii-) (
. 2n 5 . 2n-1 n Pn-1(3° -3(n -1)Pn-3(3)
• • • (2)
where the arc cotangent ranges from 0 to a. We may use this expression to define Qn(k) over the whole range - < < co since it has no peculi-
156
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.22
arities at /' = 1. We note that Qn(j0 is finite when ;- is finite and that (n even) Qn(,/ 0) = — 1'1-P.( •
P in
—
2
1 • 3•5• • •(n—1) (3)
24• 6• • n )(n+1)2 4 • 6 • (n — 1) (n odd) (2„(j • 0) = (-1 1•3 5 • n )
(4)
When is very large, the smallest negative power in (1) outweighs the remainder so that we have 2n(n!)
(2n
1 ( j)"+1 1)!
(4.1)
An expression that is sometimes useful is obtained by differentiating 5.212 (3) and eliminating the integral from the resultant equation by 5.212 (3). Substituting for A gives
P.(ft)V,(.10 — PW0Q.Ci0 = (1+ ?'2)-1
(5)
5.215. Use of Legendre Function of the Second Kind in Potential Problems.—The most important applications of zonal harmonics of the second kind occur in connection with spheroidal harmonics. Owing to the fact that Qn(A) is infinite when 1.1 = 1, they are used as spherical harmonics chiefly in problems where conical boundaries exclude the axis from the region of the electrostatic field. Let us consider the case of two coaxial cones. Let the potential be zero over the cone B = )3, and let V„ = M(Anrn Bnr—n-1) over the cone 0 = a. Then, at any point in between,
v=
(A,,rn Bnr—' 1)[(2.(cos i3)Pn(A) — Pn(cos S)Qn(A)] Pn(cos a)Qn(cos 0) — Pn(cos g)Qn(cos a)
(1)
where /A = cos 0. It is evident by inspection that this solution satisfies the boundary conditions. A particular case of some interest is that in which one cone is at potential zero and the other at potential V1. We then have, since Po(A) = 1,
V1
A Qo(cos 13) — Qo(cos a)
ri()
Qa(cos t3) — Qo (cos a)
—
A0
Also Qo(cos 0) = 4 In [(1 + cos 0)/(1 — cos 0)1 = — In tan 40 from Dw 406.2 or Pc 578 so that the potential has the form tan 13)
V = V1(ln tan 13)/(ln tan tan 40
(2)
5.22. Nonintegral Zonal Harmonics.—For many problems involving conical boundaries, we find harmonics with integral values of n inadequate. In such cases, we must expand in terms of harmonics with values
§5.23
157
ASSOCIATED LEGENDRE FUNCTIONS
of n so chosen that the roots of P n(kt) or Qm (A) are zero on the cones in question. Many of the equations already used, such as the recurrence formulas for P„(/.4) and (2,,(A) are equally valid for nonintegral values of n but the definitions and derivations need modification to fit this case. A convenient expression for Pn(A) good for all values of n is the series — n) • • (r — 1— n)(1-
(n + 1)(n + 2) • • • (n + ( — (r!) 2
2
1.4Y (1)
r=0
This series converges for A = cos 0 except at 0 = a. An example of this type of harmonic is given in 5.26, 5.261, and 5.262, if we take the charge on the axis of the box so that m = 0. When v is not an integer, Pp(A) and 13,(— A) are independent solutions of Legendre's equation and are connected with Q,(µ) by the relation
(2,(A) =
ir[cos
virP,(A) 2 sin
— P,(-12)]
(2)
V7
When n is integral, the (2.(u) of 5.21 is the limit of Q,(1) as v n. 5.23. Associated Legendre Functions.—We have seen in 5.12 and 5.14 that a solution of Laplace's equation, in spherical coordinates, of the form Rect. is possible when 5.12 (3) 5.14 (1)
R = Arn + Br —n-1
=
C cos m(j) +
(1) (2)
D sin mq5
and 43 is a solution of 5.14 (2) which becomes, when /.1. is written for cos 0, ni2
di.4[
(1 — µ2) dµ
dA ]
+ [n(n + 1)
1—
12 2
0 = 0
(3)
To obtain a solution of this equation, we start with Legendre's equation, obtained by putting m = 0. Differentiating the product composing the first term, we may write Legendre's equation 2y dy (1 —µ2)d µ-2µ + n(n + 1)y = 0
d
We know solutions of this to be y = Pn(A) and y = Qn(A) this m times, and write v for dmy/di.en, and we obtain
Differentiate
dv (1 — A2)— d 2v — 2A(m + 1)- + (n — m)(n m 1)v = 0 41
(4)
Write w = (1 — /22)Pnv or v = (1 — 112)—onw, and this is (1 —
d2w
dw
ZAT., En(n + 1)
2
I"1.,2 1v ( =0
1
4.1)
This equation is identical with (3), when the first differentiation in (3) is
158
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.23
carried out, so that a solution of (3) is 'fly
e = w = (1 — A2)1mv = (1 —
Since y is a solution of Legendre's equation, a complete solution of (3) is
GO
0 = ATT(A) +
(5)
where if —1 < /./ < +1, PT (A) and QT (µ) are defined by
PZ(1.1) = (1— P2)1m
d'nP„(A)
Q;n
m
i(A) = (1—µ2)1m
(6)
.(12)
(7)
c/Aen
For —1 < µ < +1 Hobson introduces the factor (-1)m on the right. Whenµ is imaginary or real, but greater than unity, we define these functions by the equations PT (µ) = (.12 —
(27:(1 ) = (µ,2 —
(8)
d imµ
1)imdm(0z)
(9)
These functions are known as associated Legendre functions of the first and second kind. We can compute them from the expressions for P„ (µ) and Q.(A) already obtained by using (6) and (7). Forµ real but less than unity, (6) and (7) give /1(A) = 15(1 — /42)1 Pi(A) = (1— /42)i 1:10.0 = -(1— 1.42)1 (71.13— 31i) = 3(1— Pi(A) = -V(1— ih2)(7A2 — 1) ITO = 3 (1— 1) /1(A) = 105(1 — A TAL Pa(m) = 4(1— ii2)1(5A2 MA) = 15(1 — ;12) P,10.0 = 105(1 — p2) 2 Q1( 1) = (1 —
2)1(1In 1 +
2
1
in 1+ Q1 1) = (1 — 112)i01, 2 1— (
(A (A) = (1 —
A
+ 1
/.1 2 )
— + 3;12 1—
(10)
512 3A '3) 2)(3 2 in 1 + + (1 112)2
For higher values of m and n, use the recurrence formulas of 5.233. For Az real but greater than unity, the values given by (8) and (9) can be obtained by substituting in the above formulas for the first factor (I.12— 1)im instead of (1 — /.12)1m and by writing, in addition, in the logarithm term of Q',,n(A) the factor /./ — 1 instead of 1 — A. For higher values of m and n, use the recurrence formulas of 5.233. For the imaginary values of the argument, we have, from 5.157, 5.214, and (8) and (9),
§5.231
P1(j0
INTEGRALS OF PRODUCTS OF ASSOCIATED FUNCTIONS
= ji
cot'
Qi(k) (21(j0
Pi(k) = — 3(1 +
Q2(k) =
Pi(k)
= —3(1+
—1
= (1 ± .* 2)1(cot-1 Q2(if) = i.7[(31.2 + 1) cot'
= i(1 .2)4 P2(k) = — 01' 2 + P1(..g)
3.2)
Qi(j•
1-2) (11) — 3fl n2 + 2)
+ 0)1(33" cot-1g.
= j(1
159
i-2) 3 cot-1
1 + 1-2
(1± i-2)2
In the formulas for Q:(k), the arc cotangent ranges from 0 to a as ranges from ± ,c) to — . co, we have, using 5.152 (5) and 5.21 (5) in (8) and (9), When a
r. (
(2n)! „, 2'in !(n — m)!/1 (_1).n!(n m) !2. Q:(A)--> (2n ± 1) t,.+1
(12) (13)
When m = n, insertion of 5.152 (3) in (6) leaves only the s = 0 term and yields the solution used in 5.111 (11). P'„n,(cos 0) = (2m — 1)!! sin" 0 An integral, useful for all values of n. when x
1-Pz(x) = (n + 1)(n
(14)
1, is
2) • • (n m)E[x + (x2— 1)i cos ckin cos m4 dqh (15)
Substitution in (3) and two integrations by parts prove that it satisfies Legendre's equation. The constant factor is verified by letting x —> 00 so that the integral can be evaluated and the result compared with (12). 5.231. Integrals of Products of Associated Functions.—Equation 5.13 (2) shows us that, if n n', f +1t.f 0*2VT(A)P:;(A)(A cos m4
B sin mck)(A' cos m'4) B' sin m' cliz c14) = 0
(1)
Because of the trigonometric products occurring, this integral is also zero if m is an integer, and m m', by Pc 359, 360, and 361 or Dw 435, 445, and 465, regardless of the value of n and n'. To get the result when n = n' and m = m' we must determine the integral of the square of P„'n(A) over a unit sphere. Substitute from 5.23 (6) and 5.152 (4) for P',.n,(A) and rearrange the factors and we have, integrating by parts,
160
§5.231
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS 1
1
+1
[P7,',(2)]2 (n1) 2
22n
=
J-1
u dv dnmdn+m 1)In di.o+.(112 — 1)nid[ d
f +1 [(112
(-1)ni
2
r+1 d
„
Cln+m
i'mdihn+.(112
2".(n!)2 _1 Idtd_
(A2 —
1)n]
Cln+m-2
1)"]}4d 1.01- m-2 (u2— 1)n]
We now integrate repeatedly by parts, the general form of u and v being U=
d8-1 [ ( 2
(
Mdte+M
48-1 °.4
—
1)n ],
V=
d m+n— a („2 dttm+n—s'
1)n
The product uv always vanishes at both limits since u contains the factor 112— 1 when m j s and v when m < s. Thus after m + n integrations by parts we obtain +1 da+m 5+1( 1 _ ti2)n {dn+m 2 (2) (12 1)d} [PT(A)P = 22n(n!) 2 I d µn+,, J-1 The second factor in the integrand is evidently a constant, A being eliminated by the differentiations. Keeping only the highest power of A, by substituting /22for 142 — 1, we see by inspection that its value is [2n(2n — 1)(2n — 2) • • • (n — m + 1)](n m)! = •
(2n) !(n + m)! (n — m)!
The equation now becomes, using 5.155 (2.1) and 5.155 (3),
[13n(µ)]2 1
J-1
m)! (2n) ! f +1 1— A2)n dµ (n — m)! 22"(70)2 1 (
= (n
= (n (n MI)) 1:11[13n(g)P
— 2n 2 + 1(n (n — ±m) ! (3)
From (1) when n n', +1
P'n'.( A)P;',%(11) dµ = 0
(4)
When m m', another integral relation that is sometimes useful is
f +1 1 P7(A)PV(A) = 0 1 — ;12
(5)
To obtain this formula, write 5.23 (3) with 8 = y, m = m and with 8 = y', m = m', multiply the first by y', the second by y, subtract and integrate from —1 to +1. To integrate when m = m', transfer the middle term of 5.233 (4) to the right side, square, write n — 1 for II throughout, multiply by n m, and eliminate 437+,113"„t_72. from the result by using the square of 5.233 (7) multiplied by n — m. When integrated from —1 to +1, all terms not involving 1 — /12 are integrated by (3) and cancel
§5.232 ASSOCIATED FUNCTIONS WITH IMAGINARY ARGUMENT
161
out, leaving +1(07,
i
+1(07_0 2 d/I
n
)2
_1 1 - p2
n - m _1 1 - /12
(n
m)! f +1(0,:02dau
(2m)!(n - m)! _1 1 - /.42
Substitution of 5.23 (14) for e: and integration by Dw 854.1 give 1. + V: (A)12 , (n + m) ! (2m - 1) !! 1 (fl ± m)! 'r sin 0 dO = (6) j_ 1 1 -i.L 2 u'm = (n - m)! (2m,)!! o m (n - m)! In dealing with vector potentials, we shall need to use the orthogonal properties of the surface vector function of cos 0 defined by PZ(A) = (1 - µ2)1P7'(µ) ± (1)m(1 - iz2)-IP'."(A) 0 40 Integration of the scalar product of two such functions gives r-Fi ri .1 1PZ(A) • 137:(A) dil = j 1 [(1 1.42)13 7' (14)PP
(7)
-
+ M 2(1 - 112)-1PT(A)P;(1-)1 dil
Substitution for the second term from 5.23 (3) and rearrangement give ±i{ H. R i - 2>p,,,,p,,, + 4[0 - ,u2)P71P4 dp + n(n + 1) 5 1 PT137; dp
r
J-1
„
11PT.P7,1 dii = n(n ± 1)f 11 PT,Pi; diA 1)f By (1) this is zero if n / p and by (3) if n = p it is f +1 n. 2n(n ± 1) (n + m)! (8) P n(-4) • PIZ (A) dA - 2n + 1 (n nt)!
--- [(1 -
p2)1';,7=7]+: + n(n +
-
6.232. Associated Functions with Imaginary Argument—In connection with oblate spheroidal harmonics, the functions P':(j?-) and (27,'(k) occur, where ranges from 0 to 00. Series for these functions can be found by applying 5.23 (8) and (9) to 5.158 (1) and 5.214 (1). It will be noted that the resultant series for P',n,(j0 has only zero or positive powers of ;- so that P7(j • 0) is finite and PZU • 00) is infinite. The series for QZ(k) contains only negative powers of so that (27,(j • 00) is zero. We shall obtain an expression analogous to 5.214 (5) which is sometimes useful. From 5.212 (2) we have (27(14)
=1:1(11){A +Bf (1 - il2d)11[PZ(11)]2}
Eliminating the integral from this equation and the equation obtained by differentiating it and substituting ..)* for ,u, we obtain d
jB
PZ(ii")— di.Q.7:(:70 - (27(i0d—.P7,(in - 1 + r
(1)
162
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.233
Letting ;.
co and using 5.23 (12) and (13) we find that (n + m) B = ( 1)m (n m)!
(2)
5.233. Recurrence Formulas for Legendre Associated Functions. We have seen in 5.154 and 5.211 that Pn(,u) and Qn(A) have identical recurrence formulas. The signs in the corresponding formulas for 0: = AP: (A) + depend on whether —1 < µ < +1 or whether > 1 or is imaginary. We shall now obtain these formulas, the upper sign referring to the case At = cos 0. Differentiate 5.154 (4.1) m times, multiply through by sinn'+' 0 or (i.t2 — 1)i(m+1), and then by 5.23 (6) and (7) or 5.23 (8) and (9) we have = (m + n + 1)[ ± (1 — /22)]10',,n + (1) 1.4 2)r ( m+ 1), Now differentiate 5.154 (2) m times, multiply by [+ (1 — and we have, by 5.23 (6) and (7) or 5.23 (8) and (9), — Oa±i = (2n + 1)[ ± (1 — ,u2)]10:
(2)
Subtract this from (1), and we obtain = (In — n)[ ± (1— 1/2)]18,11 + 1107+1 (3) Write n — 1 for n in (1), and eliminate 0:,±1.and 0,7_1by using (3) and (3) with m — 1 substituted for m, respectively. Dividing the result through by — (1 — ,u2), we get 07.±2
0:+1-T 2m4.4± (1 — /22)]-10: ± (m + n)(n — m +
=0
(4)
This is the recurrence formula in m. Now multiply (1) by m — n and (3) by m + n + 1, and subtract. Writing m for m + 1 in the result, we have (m — n — 1)0:+1+ (2n + 1)A0: — (m + n)0:_1 = 0
(5) This is the recurrence formula in n. Differentiating 5.23 (6) and (7) or 5.23 (8) and (9) and using the above formulas, we have [± (1— 112)POZ = d mg± (1— 112)]-107: = _TEm n)(n — m + 1)01:-1+10:+1 (6) =± ± (1 — /22) ]-4 0: 4-. (m n)(n — m + 1)0;i-1 We sometimes wish to express [-I- (1 — ,u2)]-40: in terms of Legendre functions. To do this, write m — 1 for m in (1), divide through by [ ± (1 — /12)]1, substitute for iz[± (1 — A2)]-107,1from (4), and write n — 1 for n in the result, giving 2m[+ (1 — A2)]-10: =
+ (m + n — 1)(m + n)07,1:.-i (7) Writing m — 1 for m in (2) and substituting in the last form of (6) give (1 — 1.42)9:' = ±
T- (m n)(n — m + 1)(2n + 1)-1(0,7+1 OZ-1)
NEUTRAL POINTS AND LINES
§5.235
163
Substitution of (5) in the last, first, or middle term gives, respectively, (1 — A2)0':' = ± (n 1)1.4017: T- (n — m 1)0;4, = ± (2n + 1)-1[(m — n — 1)neT4-1 (n = 1 nµ8n ± (m n)(31,7_.1
1)(m
+ n)07-1] (8)
5.234. Special Values of Associated Legendre Functions.—With the aid of the recurrence formulas of the last article, especially 5.233 (4) and 5.233 (6), combined with 5.157 and 5.213, we obtain the following relations m — 1) F":(0) = (-1)4(n—r )1 • 3 5• • • (n (n m even) 2 • 4 6 • • (n — m) (n m odd) P':(0) = 0 • Q7(0) =(-1)t(n—m+1~ 2 4 • 6 " • (n m 1) (n m odd) 1. 3.5 • • • (n — m) Q:(0) = 0 (n m even) —
[— dr PZ(A)]= PZ-Fr(0) de.
[d— Cit:' A7(1)]
1:1
= (27±r(0)
P7( — µ) = ( -1)n±mPZ(A) (— = (-1)n+ni+1(27(A)
5.235. Neutral Points and Lines.—The nature of neutral points is clarified by taking one as the origin for a spherical harmonic potential expansion. Inside a sphere so small that it includes no charge, the potential must be everywhere finite and so of the form A„,„rnP":(cos 0) cos m(4) — (5,n)
V =
(1)
n--0 m=0
At a neutral point, VV is zero so that as r 0, aV/ar —> 0, and Amt is zero. Neutral points may be classified according to the highest p value for which all Amn's (n < p) vanish at r = 0. Thus for a neutral point of the pth order, p is the largest integer for which, for all values of 0 and 0,
a2V ar r-+O art NO
av
aPv
0
ary
(2)
Sufficiently close to a neutral line or point of the pth order , p+1
V
rP÷' Z A.,p-1-1P7F1(cos 0) cos (m0 + S.)
(3) m=o The potential at the neutral point is AN. The equipotential surfaces intersecting there are given when r is small, by equating the summation to zero. For an axially symmetrical field, m is zero so the surfaces are = A00 +
164
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.24
circular cones of half angles a given by P2 (cos a) = 0. From 5.157, for a first-order point, a is 54°44' and 125°16'; for a second-order point, a is 39°14', 90°, and 140°46', etc. The simple neutral line occurs when all but the m = p + 1 term are zero. In this neutral line p + 1 planes intersect, the angles between adjacent planes being 7r/(p + 1). 5.24. Biaxial Harmonics.—It is sometimes convenient to be able to express a surface zonal harmonic P.(cos 0') in terms of general surface harmonics S,,(0, 0) referred to another axis. Let the two axes intersect at the origin and let the coordinates of the 8'-axis, referred to the 0, j system, be 0 = 0 and ¢ = 0. We must then determine the coefficients in the expansion , 4. 1
m=n
A„,P',;(cos 0) cos m4
P„(cos 0') =
(1)
m= o Multiply both sides by P:,(cos 0) cos s4,, and integrate over the surface of a unit sphere. On the right side, by 5.231, all terms go out except the S. terms. By 5.231 (3), this becomes
A,,,f [PT (cos 0)12 cost me dS =
2r (n + m)! 2n + 1 (n — m)! A'n
(2)
on the left side, we must evaluate the integral fsP.(cos 0')S„(0, 4,) dS This identical integral appears as the coefficient of bn when we find the potential V,, at the point 0' = 0, r = b due to a distribution of charge density S.(0, q5) on the surface of a unit sphere. We have solved this problem in 5.131 so that we can evaluate the integral by equating it to the coefficient of bn in that solution which must hold for all values of b < 1. Thus we have
4.71-6Vp = f NdS = f S„(0, cp)(1 b2— 2b cos O') —'3 dS s PQ When expanded by 5.153 (1), the integrals of the products of the 'orm Pr(cos 0')S„(0, 4)) go out by 5.13 (2) leaving only the term 47r€Vp = bnfsPn(COS 0' )Sn(0, 4)) dS But we have already evaluated VP in 5.131 (5) in the form 47-€Vp —
4r b1S„(0, O)]P 2n + 1
In the present case, the coordinates of P, referred to the 0-axis, are 0 = 0, cfr = 0 and so S,,(0, 4)) = P7 (cos 0) and equating coefficients
§5.26
NONINTEGRAL ASSOCIATED LEGENDRE FUNCTIONS
165
of b" in the two expressions for Vp gives
is
P„(cos 0')/:';',1(cos 0) cos mct, dS =
F';,'(cos 0) 2n + 1
(3)
Equating (2) and (3) and solving for A„„ we get Am
— 2(n—ni) •Pz(cos 0) (n + m)!
(4)
when m = 0, (2) and (3) become by 5.155 (3) and 5.131 (5) 47rAo — f Pn(cos 0')P,,(cos 0) dS 2n+ 1 s Substituting in (1) gives A of [Pn(cos 0)]2 dS =
47/3„(cos 0) 2n + 1
m
Po(cos 0') =
(2 — S° )
(n — m) ! (cos e)Pr, (cos 0) cos mcb (5) (n + m)!Pmn
m=0 where Sni ° = 1 if m = 0 and gn =0 if m 0. This symbol is called the Kronecker delta and is more generally written ST where 3,7 = 1 if m = n and 3: = 0 if m n. 5.25. Conical Boundaries.—The potential inside a cone produced by an arbitrary potential distribution over its surface is found by choosing separation constants in 5.12 (1) so that R and (bare orthogonal functions. Thus, if n is jp — -I, then K is —p2— and R becomes, from 5.12 (3), R, =
riP—t
= r-424. cos (p In r + 82,)
(1)
The product RpRp, dr then becomes cos (74 + 3„) cos (p'1,G + 6„,) d,& where ¢ is In r and leads to Fourier's series or integrals in In r. There now appear in the 0-factor Legendre functions of order jp — known as cone functions and treated by Hobson, Heine, and others. Hobson gives the series 4292 + 12 (1T± = 1 ± 22 2 12)(42 ± 32) (1 _T_ (4p2 • • • (2) 22 . 42 2 The upper sign solution, infinite atµ = 1, is useful outside a cone, and the lower sign one is useful inside it. The function is not periodic. No assumptions regarding n were made in deriving 5.23 (6) so that
Prp-A(±1.0 = (1 —
a"P
ate,
(3)
5.26. Nonintegral Associated Legendre Functions.—As mentioned in the last article, when working with conical boundaries we may use Pn (12) or Qm„ (i.,) where n is not an integer. To use these functions, we
166
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5261
need expansion in series of Legendre functions that include all values of n for which OVA) = 0. For this purpose we must have formulas analogous to 5.13 (2) and 5.231 (3). Let OVA) = y and TAIL) = y' be solutions of 5.23 (3) such that (1) 07, (Ao) = 0,7(.10) = 0 Thus we have in ,,)d2 — 2 dY + [n(71 + 1) 1 2 .1Y = 0 (1— 42 Adg — ,dy' 2 dy' _L + 1) 1 m 2 1y, 0 (1
122) d/12
d/1
—
Multiply the first by y' and the second by y, and subtract, and we obtain
r (1
_dy' ] ydµ)] + (n — n')(n n' 1)yy' = 0
Integrate from go to 1, and we get
r
i
J
=
0":01)07,(A)
(2)
if) eZdcle 2)iddelf (en (1— (n — n')(n n' + 1)
From (1), we have, if n X n',
r
i
J 0:GI) 0:
(1.1)
A.
dµ= 0
If n = n', we must proceed as follows. Let n — y = y' (ay' /on')An' in (2), and we get dy'
,dy _ ,ay'
Since at A =, Ao,
+
Oy' Oy' A
,
an''
— y,aY' — —
(3) = An'. Substitute aY
apn,An
t
= 0, we have, as n n',
fo [43
7(4)]2 diA
(1 —p.g)(ae:ao;,4\ 2n + 1 \ ap. an „_„.
(4)
To calculate values of ae'n ,lan, one may use series such as 5.151 (1) or definite integral expressions which may be found in works cited at the end of this chapter. 5.261. Green's Function for a Cone.—We shall now solve the problem of an earthed cone B = a under the influence of a point charge q at r = a, 0 = 13, ¢ = 00. By a point charge, we mean one having dimensions too small to measure physically but mathematically not zero so that the field-intensity and potential functions are mathematically bounded. Our boundary condition V = 0 on the cone will be satisfied automatically if we use a series of Legendre functions in which we choose such values of n that Pm „ (go) is zero where go = cos a. Clearly, from
§5.262
167
GREEN'S FUNCTION FOR A CONICAL BOX
5.23 (1), (2), and (6), with such a choice of n, solutions of Laplace's equation which are finite at r = 0, r = co , agree at r = a, and have the proper symmetry about cko are ra
aln+1 P7,;(1.) cos m(cb — 950) V =2=1.,,,,,(r — i
(2)
n =0
To determine the coefficients Amn, we use a new variable (45' = sb — 95o and proceed, as in 4.07, to write #VWar — 3V0/ar, multiply both sides = 2r and from by Mkt) cos pq5' dµ d4,', and integrate from e = 0 to = go to A = 1. Then, by Pc 361 or Dw 445, all terms on the right are out unless m = p and, by 5.26 (3), they are also out unless n = 8. Thus we have, multiplying through by a2, 2w+ f 1 ay, avo r_adµ d¢' Pni) cos p¢i( a2.1
or — Or
o
+1
2r
= aA„„(2s + 1) f [1:10.0i 2 f cos2p¢' d¢' 0
(3)
The integrals on the right are evaluated by 5.26 (4) and, except when p = 0, by Pc 363 or Dw 440.20. To evaluate the left side, we note that a2 = —a2 sin B d0 de = —dS. The field is continuous across the sphere r = a, except over the infinitesimal area AS at 0 = 0, 4, = 400 or 01 = 0 where the charge is located. Thus OVi/Or = aVo/ar except over AS and the integral vanishes elsewhere. We take AS so small that we may write Pis' (Ai) for Pf(i./), where AI = cos )3, and 1 for cos pe. We note that on the inner face of AS, #Vi/#r = --av/an and that on the outer face of AS, aVo/or = ay/an so that the left side of (3) becomes, by Gauss's electric flux theorem [1.10 (1)] —
ay dS = eq-Pf(i.q)
Pf(A1)
(4)
or p and n for s to fit the notation of Solving (3) for A„, and writing Asa mfn (1) and (2), we have
Am,, —
— (2 — 31)q 27rea(1 —
) n
.913;7 (L)aPT [aµ
1
(µ)
an
=1,0
(5)
where 3„, ° = 1 if m = 0 and S;)„ = 0 if m 0. 5.262. Green's Function for a Conical Box.—Suppose that the charge q lies between the earthed spheres r = d and r = c and inside the earthed cone 0 = a, so that c < a < d and 0 < # < a. We must then superimpose on the Green's function for the cone a potential that will be zero
168
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.27
on the cone and will give a resultant zero potential when r = d and r = c. Since both r = 0 and r = co are excluded, such a potential will have the form of 5.261 (1) or 5.261 (2) where the terms (r/a)n or (a/r)n±' are replaced by Cnrn + Dnr-n-1. When this is added to 5.261 (1) and we put r = c, the result must be zero so that + Cne
„
=
(1)
When we add it to 5.261 (2) and put r = d, the result must also be zero 60 that n+1
Ccdn
dl 7 1 = 0
(2)
4 ?
Solving (1) and (2) for C'n and Dfl, and adding the new potential to 5.261 (1) give, if r < a,
a 2n-1-1
TT =
Amnan
c 2n+1
d2n-F1 d2n+1)
rn
c2n+1) n+l P7:(A)
cos
— 00) (3)
n m=0
Similarly, if r > a, we have, from 5.261 (2), A
V° =
a2"-F1— c2n+1
-fnan(c2n1
d2n+1) rn
d2n-I-1)
rn+1 P7(-1) cos ni(4)
(4)
n m=0
If, in addition, the planes 4. = +7 and q = 0, where 7 > cbo > 0, are at zero potential, and if 7 = 7/8 where s is an integer, we can obtain the potential by the method of images, superimposing solutions of the type (3) and (4). If 7r/ s, we would need to use nonintegral harmonics of the form sin (mT-0/7). This would introduce in A.,„„ the factor 2r/-y, and in (3) and (4) the factor sin (mr4)0/-y) sin (mr0/7) would replace cos m(4, — cbo). 5.27. Oblate Spheroidal Coordinates. Common geometrical forms occurring in electrical apparatus are the thin circular disk and the thin sheet with a circular hole. None of the coordinate systems so far studied gives these forms as natural boundaries except the confocal system of 5.01 and 5.02. In this system, we can specify a single value of one coordinate that gives exactly the desired surface, and no more, for the whole range of the remaining coordinates. Our problem is greatly simplified by the axial symmetry which gives oblate spheroids instead of the general ellipsoids. We shall now investigate the solution of Laplace's equation in such a system that introduces the functions called oblate spheroidal harmonics. In 5.01 (1), let the longer semiaxes b and c be equal and let y = p cos 41 and z = p sin gb. We then have the equation —
§5.27
169
OBLATE SPHEROIDAL COORDINATES X2
P
a2
2
b2
0
—1
(1)
(b2 a 2)0? = c?81. Then we have In this equation, let a2 = confocal oblate spheroids when —a2 < 0 < 00 or 0 0, we may neglect 1 compared with 2in 5.27 (5) and have P ai
c13-(1 — t2)i =
—
§5.272
CONDUCTING SHEET WITH CIRCULAR HOLE
171
Solving for t gives ci-oo
p2)—i = cos 0 =
x(x 2
(8)
Thus (5) becomes 5.23 (5). From 5.27 (4), we now see that x
ci cos 0
ci
(9)
Referring to 5.23 (12) and 5.23 (13), we have A IP":(.i")
c,-> o
A 2Tn
and
-81.Q:(.i)
B2 ci--) 0
(10)
Thus (6) becomes 5.23 (1). By means of (8) and (10), we can frequently choose, at once, the proper oblate spheroidal harmonics to use in a given problem from a knowledge of the form of the solution for the corresponding problem in spherical harmonics. 5.272. Conducting Sheet with Circular Hole.—Let us now consider the case of an infinite thin plane conducting sheet, with a circular hole, which either is freely charged or forms one boundary of a uniform field.
--------
In order to have our coordinate continuous in the region of the field, we take 0 < t < 1 and — 00 < < + co , 3having the same sign as x. The equation of the sheet is t = 0, as shown in Fig. 5.272. Let us work out the case where such a sheet at potential zero forms one boundary of a uniform field. At x = co, the field will still be uniform and undisturbed by the hole, and at x = — cc the field will be zero. Equations 5.27 (4) and (5) simplify when 2—÷ co, since 1 can then be neglected compared
172
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
with j giving r 2 = x2 + p2
§5.273
,
cif
c?re = c2ir
2t2 + 43-2
Ixl/r = Icos 01. Now the equation so that ?' —> ± r/ci and t = x/(ci?-) of the uniform field at x = co is V = Er cos 0 so that t can enter as cos 0 only which permits only Pi(a), and hence only in = 0, n = 1. Then the potential from 5.271 (7), (5), and (6) is (1)
V = Pi(E)[A/Pi(i0
B'Qi(jOi Writing in the values of P, and Q, from 5.23 (11), we have
V=
g-— 1)]
B'(?'
To evaluate A' and B', consider = ± 00. When /' = ficient of B' is negligible and jA'r Er cos 0 = — cos 0 ci
(2) 00, the coef-
Eci = jA'
giving
(3)
when = — 00 the constant term is negligible and jA'r +.21-Hr 0 = cos 0( ci ci
giving
7r-B' = —jA'
(4)
Or
Eci
B' =
(5)
If the edge of the hole is given by p = a, we have, from 5.27 (5), since both ?' and t are zero there, ci = a, giving for the potential V = aEEk —
1 .7r
cot—'
— 1)]
(6)
To get the charge density on the plate, we have ay av = — tax x-0 +€ ag
(7)
From 5.27 (5), when t= 0, a?- = ±(p2— a2)i and also cot-1 [ ±
a2)1
(p2
a
]
1
=2
,
-T cos--
so that from (6) and (7) cr
—2
l[cos---1 p
+ (p2
a ay ]}
(8)
where we take the plus sign on the upper side of the plate and the minus sign on the lower side. 5.273. Torque on Disk in Uniform Field.—The case of an uncharged conducting disk of radius a whose normal makes an angle a with an electric field E which was uniform before the introduction of the disk
§5.273
TORQUE ON DISK IN UNIFORM FIELD
173
may be considered as the superposition of two cases, the first being a field E cos a normal to the plane of the disk and the second a field E' = E sin a parallel to the plane of the disk. To solve the latter case, we may use oblate spheroidal harmonics. In this case, in order that the coordinates will be continuous in the region of the field, we shall take 0 < oo and —1 < t < +1. The coordinates are then as shown in Fig. 5.273. As in the last article at r = co, --* r/a and t —> ± cos 0,
FIG. 5.273.
having the same sign as x. At infinity, where the field is still undisturbed, we have V., = E' p cos 4, = E'a[(1 — E2)(1 ± 3-2 )]1cos 4, (1) Since 4) enters only as cos 4., we must have m = 1 in 5.271 (4), (5), (6), and (7). Since n j m, the smallest value of n that can occur is n = 1. Furthermore, when r = oo , E and g- can enter only as in (1) which is given by the factor (1 — A2)1'n in 5.23 (6) and (7). We can therefore not have n > 1 since the differentiation in 5.23 (6) and (7) would then bring in t and i-as factors. From 5.23 (11), Q1(.i) i = (1± i-2)1(cot -1?" 1 ±r 2) and from 5.23 (10), PI() = (1— e)I Since the potential is finite when E = +1, the factor (21(t) is excluded and the potential must be
V' = Pi(t)[APi(90 + BQi(ini cos 4, ... (1 — 2)i[j A(1 + 0)i + B(1 + i-13 -1
1 ± r 2)] cos 4. (2)
174
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.273
When /- = 0 V' = 0, therefore jA = —}7rB. When ?" = 00, the last two terms in are negligible compared with the first term, and equating to (1) 2E'a 1 B=— E'a = Or 2 so that the final expression for the potential is ± 1 ±i_2) cos q5
V' = 2r—IE'a[(1 — 2)(1
(3)
Substituting p from 5.27 (5), this may be written ± 0)cos q5
V' = 27-1Eup(tan-1
(4)
Since the component E cos a of the field perpendicular to the disk is not disturbed by the presence of the uncharged disk, its potential is given by V" = Ex cos a, so that, superimposing on the solution (4) gives V = V' + V" the form
2p 7
) sin a cos
V = E[-(tan—'
r
1± is given in terms of p and x by 5.27 (4), (5).
x cos a]
(5)
whe e r We could get the torque from (5) by finding the surface density a whence the force per unit area would be lcr2le, etc. An easier method, however, is to observe that the surface density induced by E cos a produces no torque when acted on by E sin a because the lever arm is zero. Hence, the entire torque must be that due to cr', induced by E sin a, and acted upon by E cos a. From (4),
=
[acT,xT'1=0 _
Jr
2eE' p
1
1 -
r tLi + 1 + -2) 2 ax}r=0 cos cA so that c31'/ax = (at)-1. When t = 0, x/(a) But from 5.27 (4), ?" = a2e = a2 — p2 so that from 5.27 (5), 4€E sin a pcos (6) cr' = (a2 (
For the element of area p dp d4), the lever arm is p cos 4,, and the field is E cos a so that the total torque is, including charge on both sides,
T=
f
8€E2 sin a cos afa
p3cost 4 (a2 p2 ) dcA dp
Since 2 sin a cos a = sin 2a, we have upon integrating by Dw 858.3 or Pc 489 a 3d
T = —4€E2 sin 2af
(a2P
P
§5.274
POTENTIAL OF CHARGE DISTRIBUTION ON SPHEROID
175
Integrating again by Dw 323.01 or Pc 159 gives p2)1 - 861,3E2 sin 2« (7) 3 o 5.274. Potential of Charge Distribution on Spheroid.—Let us suppose that, on the surface of an oblate spheroid, = l.o, we have an electric charge density an of the nature described in the first paragraph of 5.131. This charge gives rise to a potential V. outside the spheroid and Vi inside it. Applying Gauss's electric flux theorem, 1.10 (1), to a small box fitting closely an element of the spheroid gives 1 p2)1 - a2(a2 T = -4€E2 sin 2«[3(a2
-
cr_L, E
1avi - ( an
avA
r 1(avi
an ) 8
L112\
n
avo
n
i-91-0
(1)
Let us consider that crn is such that eVo =j( 1)1n((:: 1 " m)) ;(1+ a)P:(il- o)Q:(j;-)87,
(2)
where
ST = C.,„Pinn() cos m(q5 - 0.) (3) This choice makes V. finite at /• = 00. In order that VVibe finite when t- = 0, = 0, and Vi equal V. when i- = -,a, Vi must be of the form (n - m) ; EVi = ,i(-1)1n (n ± m ) (1 + il)Q:(ko)P7(..081:
(4)
Substituting (2) and (4) in (1) and using 5.232 (1) we obtain an=
Setting
A" =
(5)
{172 1}ro877
j(-1)m (n E (n
- m)!(1 4_ a) in)!
(6)
we can, with the aid of (2), (4), (5) and 1.06 (6), evaluate two useful integrals, which are
S: dS = A „PmW-01Q-(j08'n n n ny froh2R.
I
AST
jr0h2R-i dS
= A.,„Q:(ji-0)13:(jOS: (7)
Now consider a surface density which is a superposition of densities of the form 00r so that the potentials are superpositions of potentials of the form of (7). Thus we have = IMItylS; =
ZhylC„PgE) cos s(cb - Os)
To determine the coefficients C.., we multiply by
P:(E) cos m(ct, - 0.)hih3d dcb
176
§5.275
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
and integrate over the spheroid. Exactly as with (5), all terms go out except those for which m = s and n = r. For these terms, we use 5.24 (2) and so obtain, when m 0, c
2
r+1
= 2n + 1 (n — m)! 1 27rci (n + m)! 1
o
„
opm( ) cosm(0_,,m)hili3 d n
c/q5 (8)
The potentials due to the distribution a are then given by
3o
v.
=
N 7,,,,QT(j0P:(t) cos m(cf, — (15.)
(10)
n=0 m=- 0
where, by combining (4) and (8),
Mmn
= j( _ 1).(2 — 31)(2n +
(n — m)!12 . L (n m)! J n g 0)
41rEc1
2T-
-1 0 f-F1f
aPZ () cos m(0 — q5m)hih3A dO
where SI = 1 if m = 0 and SI = 0 if m 0. Nmn
(11)
From (4) we have
Pm (ko) = m- ran •
(12)
Cg 0)
The 2 — al factor is required because if m = 0 the integration with respect to cb used to obtain (8) introduces a factor 27 instead of 7. 5.275. Potential of Point Charge in Oblate Spheroidal Harmonics. We can use the results of the last article to obtain the potential due to a point charge at Eo, (bo. By a point charge, we mean one whose dimensions, although too small to measure physically, are mathematically not zero so that the field-intensity function and the potential function are everywhere bounded. We take the charge density a everywhere zero except in an area S, at 4 = 00, E = Eo, which is so small that Pr() has the constant value Pm ,, (to) and cos m(4 — 0.) = cos m(4) — cpo) = 1. The integral in 5.274 (11) then becomes rhih3 d dck = Pn (to) f dS = P:(o)ffc
(to)
The coefficient of 5.274 (11) now becomes
M. = ./j(2
61)q(
2n + 1)m 4reci
By 5.274 (12) we have 2n +
Nmn
(n
m)
(2"1(./NP7(to)
ir(n — m)!12.nm„,.0,
= j(2 32q ( 1)m 4reci L ( z + m)!J i-n""1-
(to)
(1)
(2)
177
PROLATE SPHEROIDAL HARMONICS
P.28
The potentials due to q are then given by
.=zz
v
n=0 m=0
With the aid of these formulas, we can now get Green's function for regions bounded by surfaces of the oblate spheroidal coordinate system, using the same methods as with the spherical harmonics. 5.28. Prolate Spheroidal Harmonics.—A common form of spark gap consists of two coaxial rods with rounded ends which can be well approxi-
FIG. 5.28.
mated by the two sheets of an hyperboloid of revolution. We also find cases involving elongated conductors which resemble prolate spheroids. These surfaces are natural boundaries in the prolate spheroidal system of coordinates since a specified value of one coordinate will give one of these surfaces, and no more, for the entire range of the remaining coordinates. The solution of Laplace's equation in this system leads to functions called the prolate spheroidal harmonics. We proceed exactly as in 5.27 except that we now let the two shorter semiaxes a and b in 5.01 (1) be equal and let x = p sin 4) and y = p cos 0. We then have the equation p2
z2
b2 + + c2 + 0
=.-.1
178
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.281
= (c2 b2)02 = eV. We have now confocal prolate spheroids Let c2 when —b2 < 0 < 00 or 1 < .12 < co, where 4 = n2, and we have confocal hyperboloids of two sheets when —c2 < 0 < —b2 or 0 < E2 co let us, as a first approximation, drop the terms containing v--' and V-2. This gives the approximate differential equation , CIV2
R= 0
(1)
184
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.296
The solution of this equation is (2)
R=
We now insert this trial solution in Bessel's equation 5.291 (3) and consider R' to vary so slowly with v that d'R' /dv 2, v1dir /dv, and can be neglected compared with dir /dv and v--'R', and we obtain
2 dR' c 27 +R'= 0
R' = Cv-i
Or
so that, from (2) our asymptotic solution becomes R = Cv-le±29
(3)
We now see that the largest term neglected was of the order v--1. J„(v) and Y „(v) must be real linear combinations of the two solutions given by taking the plus or minus sign so that they must be of the form Jfi(v) --> Av-i cos (v + a)
(4)
t)-> 00
By-4cos (v
Y fi(v)
fi)
(5)
To find how A and a depend on n, put (4) into 5.294 (1) and 5.294 (2) co, J,' = J, _1 and which give, as v = —J,+1, respectively. This which is satisfied by afi -= gives the relation an±1= an T+ and shows that A does not depend on n. Because n need not be an integer, we may write n = in (4) and compare it with 5.31 (2) which shows that -y = —fir and gives 2
J fi(v) —> () cos (v 1 — 1 )
(6)
where terms in v-im have been neglected if m j• 3 and n is real. To get Y „(v) substitute (6), with v and — v for n, in 5.293 (4). The result gives 0/0 when v is integral, but replacement of numerator and denominator by their derivatives with respect to v gives, when v = n, Y fi(v) —> co
2 y sin (7, — 1-n7r —1-7r) 2
7r1)
4
(7)
Thus both Bessel functions vanish at infinity. From (6), (7), and 5.293 (10) we find that for the Hankel functions 2)1e,(fi-Inr-ir), 4e-3(u—in.—I.-) — 2 ) HO (V) (8) irV v--) 5.296. Integrals of Bessel Functions. In 5.261, we made an expansion in spherical harmonics satisfy the condition V = 0 on the cone 0 = a by choosing only such orders n of the harmonics en'(cos 0) as made 0„m(cos a) = 0. To determine the coefficients in this expansion, it was first necessary, in 5.26, to evaluate the integral of the product of two such HT (v)
—
185
INTEGRALS OF BESSEL FUNCTIONS
§5.296
harmonics Over the range of 0 from 0 to a. In the same way, if we are to get an expansion in Bessel functions that meets the condition V = 0 or E = 0 on the cylinder p = a, we must evaluate the integral of the product of Ri,(k,p) and Ra(k,p) over this range, where 1c and k g are chosen to meet the boundary conditions. Let u = .1?7,(k,p) and v = Rn(k,p) be two solutions of Bessel's equation. Then from 5.291 n2 pkt) c
1cli ap, PA
P
+ (k; — -2 )u = 0
P
+ k 2— - - v=0 0 — dp\pirp) + ( g p2
Multiply the first by pv and the second by pu subtract, and integrate, and we have
u
a d (pT du) d ( dv\-1 d a[vir k ,2i)i puv dp = — — dp\PTpij P o aP Integrate each term on the right side by parts, and the integrals cancel, leaving
(4
-
5
-
k:)f p uv dp
—
dvy Puddo = —a[k„Rn(k ga)R'„(k,a) — k gRn(kpa)1=e;,(k ga)]
( du --\Pvdp
This is zero if
R„(kpa) = Rn(k,a) = 0
(1)
k(kpa) = Rak,a) = 0
(2)
kpak(kpa) BR„(k,a) = k gagi (k,a) BR.„(k,a) = 0
(3)
or if or if Thus if 1c
k g, we have the result
foapfen(k,p)Rn(k,p) dp = 0
(4)
If R„(kpa) = Rn(k,a) = R„(kpb) = Rn(k,b) = 0 we have, since
the result
fabf(x) dx = fobf(x) dx — foaf(x) dx fobpRa.(k,p)R.(k gp) dp = 0
(5)
To evaluate the integral when k, = k, multiply Bessel's equation [5.291 (3)] through by v2(dRn/dv) dv giving
odl:endORA dv \ dv
ORA2 d + v\ dv )
v
2„ dR„ dv v
_ n
—
186
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.296
Integrate from 0 to a using integration by parts on the first and third terms. We thus find the following expression to be zero; 2 2 av(dR„ v2 ,, n2R! a v2(dRI a ) dv+ f av(--) dR dv+ R2 2 " dv o Jodv 2 0 dv
r
Canceling the second and third terms and solving for the fifth give a a 1 v2[d civ(V)]2 (1) 2 — n 2 )[Rn(v)]2 v[R„(v)J 2 dv - 2
Jo
(6)
0
Substituting for the derivative from 5.294 (2) gives
fav[Rn(v),2 dv ==1-a2ia2{[Rn(a)]2 { [Rn(a)]2
[R,H-1(a)121 - naRn(a)R,,+i(a)
[R"-i(a)P1 - naRn(a)Rn-1(a) (7) In dealing with the vector potential we shall have occasion to make use of the orthogonal properties of the vector function defined by R„(k,p) = R' (k p) ± k R (k p) (8) P e pP P The integral of the scalar product of two such functions from 0 to a is
a J:
R„(k„p) • Ra(k gp)p dp a n2 = f [R'„(k„p)k(k,p) + kpicQp2Rn(k,,p)Rn(k gpd p dp
(9)
With the aid of (1), (2), (3), (4), (5), (6) of Art. 5.294 we may write this as the sum of two integrals of the form of (4). Thus tfoaRn+I(kpp)R.+1(kaP)P dP
foa Ra-i(kpp)Ra-i(k gP)p dp
(10)
Evaluate each integral by the formula for f puv dp already given and add the results, then eliminate the derivatives by 5.294 (1) and 5.294 (2), cancel terms not involving the nth order, and combine the resultant n + 1 and n - 1 orders by the same formulas. These operations give, if v = k„a and v' = k ga, foa Rn(k,p) • R„(k,p)p dp = (14-k:)-1[vRn(v')R:,(v)-v'R„(v)K,(V)]
(11)
Thus, if k, k g, the integral vanishes under conditions (1), (2), or (3). But, if k, = k g, evaluation of each integral in (10) by (7) gives for their sum (a/k„)R'„(k„a)R„(k„a) (12) -i[a2 -(n/k„)9[R,,(k,a)]2 + i[aR'n (kpa)]2 Thus a surface vector function of p and 0, one component of which vanishes at p = a, may be written as a sum of terms of the form Rn(ko) sin (nit, +
5n)
GREEN'S FUNCTION FOR CYLINDER
§5.298
187
5.297. Expansion in Series of Bessel Functions.—Consider a function f(v) which satisfies the conditions for expansion into a Fourier series in the range from v = 0 to v = a and which fulfills one of the following boundary conditions: (a) f(a) = 0. This case arises, if f(a) is a potential function, when the boundary is at zero potential. (b) f'(a) = 0. This case occurs when the boundary is a line of force. (c) af' (a) + Bf(a) = 0. This case reduces to (a) if B = co and to (b) if B = 0. An example of its use is given in 11.08. The function f(v) may be expanded in the form f(v) = ZA,,1„(1.4v)
(1)
r =1
where the values of 14 are chosen so that in case (a) J.(ura) = 0, in case (b) J (Ara) = 0, and in case (c) Ara4(Ara) = 0. To determine A,., multiply both sides of (1) by v./.(Asv), and integrate from v = 0 to v = a. By 5.296 (4), all terms on the right vanish except Asf v[J „(A„v)12dv so that a
vf(v)J,,,(Aav) dv (2)
fo v[J„(p4v)P dv We can evaluate the lower integral by 5.296 (7) giving Aso ov[J„(12,v)12 dv = pc-2 f x[J„(x)]2 dx JO
1 = cl21{4,(1A.coi2
{efn±j(Asa)12 —
na
J (ti.a)./.±i(Aaa) (3)
In case (a), substituting (3) in (2) gives
2 vf (v)./.(11,v) dv [aJ n±i(gsa)12/0 In case (b), substituting (3) in (2) gives 2 — (a2 _ n2i.c2)binoisa)12f0 vf(v)Ja(p.v) dv As =
(4)
(5)
In case (c), substituting (3) in (2) gives
a vf(v)Jn(A,v) dv (6) [a2 + (B2 — n2)/4 ./.(Asa)Pio 5.298. Green's Function for Cylinder. Inverse Distance.—We shall calculate by the principles of the last few articles the potential when a point charge q is placed at the point z = 0, p = b, = ¢o, inside an earthed conducting cylinder. By a point charge, we mean one whose As —
2
[
188
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.298
dimensions, although too small to measure physically, are different from zero, so that the field-intensity function and the potential function are everywhere bounded. From 5.291 (6), a solution that vanishes when z = 00, is symmetrical about the cP (Po plane, gives V = 0 when p = a, and is valid for positive values of z, is CO
Arae —A,'.1.(14P) cos s(g5—
V=
o)
(1)
r=1s=0
where Ar is chosen so that Js(pra) = 0. Y8(pr p) is excluded because it is infinite on the axis. From symmetry, the whole plane z = 0 is formed by lines of force except at the point charge itself. To complete the boundary condition in this plane, we shall consider (aV / az) 0to be zero except in a small area AS at cp = cpo, p = b. Differentiating (1) and setting z = 0 give
av\Jo
ii„A„..1.(Arp) cos s(4) — 4)o)
\az
r = 1 s =o
We determine A„ in this expansion as in 5.296 and 5.297 by multiplying through by p4(1. 0p) cos p(iP — q5o) and integrating from p = 0 to p = a and from 4) = 0 to 4) = 2r. By Dw 858.2 or Pc 488 and 5.296 (4), all terms on the right disappear unless p = s and q = r. In the latter case, we see by Dw 858.4 or Pc 489 that the 4) integration introduces a factor 7 on the right if s > 0 and 2r if s = 0, so that we have, by 5.297 (4),
,,
(2 f f az op. ,, Arp) cos s(4) — 4)o) dp d4) (2) [ctsia+ (Ara)] 2 where 62 = 1 if s = 0 and 62 = 0 if s O. In the z = 0 plane, the area AS in which (a V/az)0 0 is taken so small that in it Js(prp) has the constant value Ja(prb) and cos s(4) — cko) = 1. The integral then becomes
A„ =
J.Carb)
—
ff
—
av aZ 0
dqS
an
— dS — 1.1.(11,b) dp = .18 (Arb)Asf an 2E
from Gauss's electric flux theorem [1.10 (1)] it being remembered that only half the flux passes upward. Substituting for the integral in (2) gives g)J,,(prb) q(2 A, (3) 27E,Ura 2[Js+I(Ara)]2 —
—
Thus we obtain, for the potential,
V=
(2
271-Ea2 r=1 a=0
Vs)e—Adzi
Cub),1(AL P)
s(4) il,[4+1(mfa)]2 cos
rko)
(4)
GREEN'S FUNCTION FOR CYLINDRICAL BOX
§5.299
189
This is Green's function (see 3.08) for a circular cylinder. If the coordinates of q are p = b, z = z o, and 4, = 00, we should substitute Iz - zo l for iz! in the above formula. If the charge is on the axis, all but the first term of the s-summation drops out and Jo(Arb) = 1. When a -> 00 in (4), it gives the potential of a charge q in unbounded space. From 5.295 (6) J„(Ara) oscillates sinusoidally as Ara -> oo so that its zeros are uniformly spaced at intervals 7r = Ar+ia - Ara = a Ap. and so that, when J.(.1,,.a) is zero, Jn+I(Ara) is (lirpira)-1. Thus (4) becomes V-
27rect2
(2 - g,) cos s(4, - 950)e-rAmi'-'.1 J„(r Ailb).1,(r 8=0
If k = r OA then as Ai..(
47reR
47re
co this takes the integral form
0 and a
,
=
a2 Aps
r —1
(2 - (52) cos s(4, - 4)o)f e-kk-z01.1.(kb)J.(kp) dk
(5)
8=o
where R2 = (z - zo)2 + P2 ± b2— 2pb cos (4) - cbo). gives
If b = z o = 0, this
zz)-i =r y eki'lJo(kp) dk
( p2
(6
With z, zo, and 00 zero, write R for p in (6) and compare with (5). Thus OD
Jo[(p2 + b2 - 2pb cos OA = Z(2 - Vs)Js(P)Js(b) cos .34,
(7 )
s=o 5.299. Green's Function for Cylindrical Box.—Superimposing potentials of the form of 5.298 (4) according to the principle of images 5.07 and 5.101, we may make the planes z = 0 and z = L at zero potential as well as the cylinder p = a. If the coordinates of the positive charge q are z = c, p = b, and 4) = 4>o, there are positive images at z = 2nL c and negative images at z = 2nL - c where n has all integral values from - .0 to + 00. If z < c, the factor involving z in the resultant potential is eA,(2nL-c+z) n =0
Z e—p,(2nL—e—z)
n =1
n =1
= 2(0—Ar c
e-271)1,1
e—ur(2nL-1-e-1-2) n =0
eAreZ
sinh mrz
n =1
n=0
2[ (e-Pre - e+mre)Ze-2"i"
emrc] sinh
n=0
Summing the series, by Dw 9.04 or Pc 755, multiplying numerator and
190
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.301
denominator of the sum by eArl, and putting over a common denominator give 2 sinh ar(L — c) sinh kirz sinh Substituting this value in 5.298 (4) gives, when z < c, the potential
(2 _
V= q rea2 r=1 a=o
sinh Ar(L — c) sinhtiTzJ .(m,b)J .(mTP) cos s (4, — (bo) (1) sinh prL fi(Ara) P
When z > c, substitute L — z for z and L — c for c. If the charge is on the axis of the cylindrical box, drop the summation with respect to s in (1), retaining only the s = 0 term. If, in addition, the planes 4, = 0 and 4, = 4,1, where 0 < (Po < 4)1, were at zero potential, and if chi= 2r/n, where n is an integer, we could get the Green's function by superimposing, according to the principle of images (see 5.07), 2n solutions of the type of (1). 5.30. Bessel Functions of Zero Order.—In the important case where we are dealing with fields symmetrical about the z-axis, the potential is independent of 4) so that Bessel's equatio4 becomes, from 5.291 (3), d 2R 1 dR dv 2
vdv
R=0
(1)
and the solution [5.293 (3)] becomes v2 V 4 + 22 24(202
Jo(v) = 1
V6
(2)
26(3!)2
This series is evidently convergent for all values of v. As with Jn(v), Jo(co) = 0 but Jo(0) = 1. Equation 5.293 (5) becomes, when n = 0, Yo(v) = —2[Jo(v)
In av
v2
v4(1 + 1)
v6(1 ± + i)
22
24(2D2
26(3!)2
•••
(3)
where In a is —0.11593. 5.301. Roots and Numerical Values of Bessel Functions of Zero Order.—If we plot the values of Jo(v) and Yo(v) given by 5.30 (2) and (3), we get the curves shown in Figs. 5.301a and 5.301b. We see that they oscillate up and down across the v-axis. It can be shown that both Jo(v) and Yo(v) have an infinite number of real positive roots. The same is true of J.(v) and Y„(v). As we have seen in finding the Green's function for a cylinder, the existence of these roots is very useful as it makes it possible to choose an infinite number of values of lc which will make J.(kp) or 177.(kp) zero for any specified value of p. Many excellent tables exist which give numerical values, graphs, roots, etc., of Bessel functions. Care should be taken in observing the notation used as this
§5.302 DERIVATIVES AND INTEGRALS OF BESSEL FUNCTIONS
191
varies widely with different authors. Asymptotic expansions provide an easy method of evaluating Bessel functions of large argument.
.0.,
Jo (x) D.5
A PA■ Jow AI PAlb, . wir ./0,4x.
r
0
.10,(x) Og
5 FIG. 5.301a.
10
1..) Yoi(x)
lk() AlSW-Alk
0.5
0
0.5
_
5.301b.
5.302. Derivatives and Integrals of Bessel Functions of Zero Order. Putting n = 0 in 5.294 (2) and (5), we have Yo(v) = —Y1(v) (1) and JO(v) = —J j(v) From 5.294 (7) and (8), we have
fovvJo(v) dv = vJ i(v)
and
fovvYo(v) dv vY i(v) +27r-1 (2)
There are several definite integrals involving Jo(v) which will be useful. From 5.30 (2), using Dw 854.1 or Pc 483, we have e)
(2n)! (2n)! 22 n(n)2
(-1)nv2n
Jo(v) =
22n(n )2
n=o
= n=o
= 14( — 1.)n2V2n (2n)! 2
3 • • • (2n — 1) 2 4 • • • 2n
n
(2i vZ f: coent dt n∎O
192
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.303
Interchanging integration and summation symbols and using Dw 415.02 or Pc 773, we obtain
Jo(v) =
( 1)n(v cos t)2' dt = 1 - f cos (v cos t) dt (2n) ! 7 0
7r o n=0
so that 1 cos (kp sin t) dt cos (kp cos t) dt = (3) o o We can show easily, by using the trigonometric formulas for sums and differences of angles and integrating by parts, that an expression that satisfies the recurrence formulas 5.294 (1), (2), and (3), is 1 Jo(kp) = -
1o Jn(v) = - f cos (nt - v sin t) dt
(4)
This evidently reduces to (3) when n = 0. Direct substitution in 5.291 (3) and integration by parts show that
In(v) -
(1v)n
r(i)r(n +)
o
cos (v cos 0) sin2n 0 do
(5)
satisfies Bessel's equation if n > Comparing the value of (4v)-nJ„(v) given by (5) as v ---> 0 with that given by 5.293 (3) fixes the constant. It can be shown that 5.298 (6) still holds if we write jz for z, thus
c-ikzJo(kp) dk =
(1,2
z2)---1
Jo whence, equating real and imaginary parts, we have, when p2 > z2,
o. cos kzJo(kp) dk = (p2 f
-
(6)
sin kzJo(kp) dk = 0
(7)
cos kzJo(kp) dk = 0
(8)
sin kzJo(kp) dk = (z2 -
(9)
When p2 co from those for J„(v) in 5.295 by substituting jv for v. The results are good to the same order, i.e., neglecting v-1 compared with v—i. Putting jv for v in 5.295 (6),
§5.324 GREEN'S FUNCTION FOR A HOLLOW CYLINDRICAL RING
199
writing e— },A for j-1, using the exponential form Dw 408.02 or Pc 613, and neglecting the e-v term, we have
I n(v) = j—nJ n(iV)
11evineijnr (-1 27111 j
Evaluating the last term, n being an integer, by Dw 409.04 and 409.05 or Pc 609 we have 1 1 /n(v) ev (1) From 5.32 (8) and 5.295 (8), we have
K,(v)
(7r ) e-v 2v
(2)
Although we have derived these equations assuming n to be an integer, we see, by substitution in 5.32 (1), that they hold for any value of n. 5.323. Integral of a Product of Complex Modified Bessel Functions. In Chap. X, when we calculate the power dissipated by eddy currents, we frequently find it necessary to evaluate the integral of the product of a modified Bessel function R„ °[(jp)ix] by its conjugate complex Rn°[(-jp)ix], where (j)i = 2-i(1 j) and ( -j)i = (1 - j). In cylindrical problems, n is usually an integer, and in spherical problems it is usually half an odd integer. We may evaluate this integral, for any value of n, by using the equations of 5.296. Let us take
k, = -j(jp)i = (-jp)i and k g = j(-jp)i = (jp)I u = R4-Ai0x1 = v = Rnt A - :70x1 = inn( Kf -A:70xl = (- D' int(i0x1 - i0xl = i0x] Substituting these values in the equation preceding 5.296 (1), we obtain, writing x for p, foax.14,[(jp)441??,[(- jp)ix] dx = f xl??,(k,x)M(k,x) dx = raj'73-i[jil??,(k,a)R?: (k ga) - (- j)iRg(k,a)R?: (k,a)] = (2p)-1[kpal??,(k,a)14,_1(k,a) k gaMi(k,a)RL,(kpa)] (1) = (2p)-1[kpar??,_2(k,a)M,_1(k ,a) k ,al??,_2(k ,a)RLi(kpa) - 4(n. - 1).1??,_1(k,a)Mn_1 (k ,a)] 5.324. Green's Function for a Hollow Cylindrical Ring.—As an example of the use of modified Bessel functions, we shall find the potential due to a small charge q at z = c, p = b, and cp = q50 inside the hollow conducting ring whose walls are given by z = 0, z = L, p = d, and p = a where a > d. The special case, when d = 0, gives us the cylindrical box which we have already solved in terms of Bessel functions in 5.299.
THREE—DIMENSIONAL POTENTIAL DISTRIBUTIONS §5.324
200
Since neither /.(v) nor Km(v) has any real roots, it is evident that we shall need a combination of the two to get a function that is zero for a given value of p. Clearly, such a function is
(k, s, t) = lf,„(ks)I.(kt) — .I.(ks)K.(kt)
(1)
Since, in general, this function is zero for only one value t = s of t, we must use different functions for the region near the inner face and that near the outer, but they must have the same value at p = b, where the regions meet. We can now write down, by inspection, two functions that are zero over the conducting surfaces and identical when p = b. These are
d p co and is defined by G8 (v) = fo'e—v cc.sh cos vt dt = K;8(v)
(6)
Below a certain value of v, which increases with v, both functions are periodic and oscillate as sin [v ln (iv)] and cos [v ln (iv)] as v approaches
209
MIXED BOUNDARIES
§5.39
zero. The integral with respect to v of v—' times a product vanishes unless the factors have the same v so that an arbitrary function of v suitable for expansion in Fourier series can be expanded in terms of R;(kp). The three-dimensional dielectric wedge problem is solved by methods similar to those used for the two-dimensional wedge in Art. 4.07. 5.39. Mixed Boundaries. Charged Right Circular Cylinder. Many three-dimensional boundaries met in practice do not fit any separable coordinate system; thus the methods so far used are not applicable. For example, if a charged circular disk lies inside a coaxial circular cylinder, a system of coordinates fitting the boundaries can be generated by rotating the last part of Fig. 4.22b about the y-axis. Laplace's equation, 4.11 (6), in this system cannot be separated, and the surfaces, from 5.00 (2), are not equipotentials. Clearly the boundaries fit into the cylindrical system, but at y = 0, V is a constant only over the disk while between the disk and the cylinder a V/ay is zero. Thus this becomes a mixed boundary value problem. It can be solved (see Art. 5.42). There are many cases where V or a V/an is known on part of a coordinate surface and nothing over the remainder. One of these is the charged solid right circular cylinder. A possible approach to this is to express the charge density as a series having arbitrary constants adjustable by 3.15 (3) to give a uniform potential inside the cylinder. When magnified the edge looks like a right-angle wedge for which 4.18 (1) gives dz/dW = WI so that W = zi and dW/dz = z-1. Thus the series will diverge at the edge and converge slowly elsewhere unless the infinity appears explicitly in a. A series of Gegenbauer or Jacobi polynomials with the proper weighting factor has some advantages but complicates integrations and numerical work. If the cylinder is bounded by z = ±b and p = a, then simple forms for the charge densities on the end 0-, and on the side a3 are —
N
M
A n(1 —
Qe =
b-2.Z2) 71-1
(re =
n=0
B„,(1 — a-2p2)m--1
(1
)
M=0
Note that at the edge where z = b and p = a only the Ao and Bo terms survive. Thus the matching of cre and a, there requires that (2)
IAA o = alBo
There are two ways in which 3.15 (3) may be used to get An and B„,. One is to calculate 32PV(z,0)/8z2P on the axis, set z = 0, and require that [a2PV(z,0) 4:92pz
= z=0
Op V 0
(3)
210
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.39
where bop = 0 if p 0 and 150„ = 1 if p = 0. This and (2) yield p ± 2 equations which may be solved for N M + 2 of the low-order terms in (1). Alternatively the potential in the plane z = 0 due to (1) may be expanded in even powers 2p of p and coefficients of all but the p = 0 term equated to zero. For a given b/a ratio the number of significant digits carried fixes the best values of M and N, which are those that give potentials nearest Vo at pole and equator where maximum deviations occur. Fewer a coefficients can be found for the more remote point than for the nearer one. If too big values are chosen for M and N, large terms alternating in sign appear, which reduce the number of significant digits in the sum. All the values of An and B„, found from (1) must be used in subsequent calculations as omission of one may change its neighbors so they are no longer correct. If the polynomials already mentioned replace (1) (see Taylor, J. Research, Natl. Bur. Std., Vol. 64B, pages 142 and 143), An and B,,„ decrease rapidly as n and m increase, so that dropping the last terms has little effect on the rest. The chief labor is finding the derivatives in (3). The axial values of the expressions for the potential of a charged ring in the text and problems of this chapter must be integrated over the charge distribution of (1). It is usually necessary to expand some of the functions in series in order to get fractional integrals of the form tabulated in Vol. II, Chap. XIII, of "Integral Transforms" (see bibliography). A hypergeometric series usually results. By summing these and using recursion formulas, the simultaneous equations of (3) are set up and then solved for A„ and B,„. Numerical values of b/a have been calculated (J. Appl. Phys., Vol. 33, page 2966) for 0.125, 0.25, 0.50, 1, 2, 4, 8, which show that for the extreme values 0.125 and 8 only one coefficient can be found for the more remote surface. The coefficients for b = a appear in the table below. Clearly for practical purposes the solution is exact when b = a, as the potentials V„ and V, and the surface deviations Ab and Da at pole and equator show. Charge density coefficients for unit potential when b = a.
A0 = Al = A2 =
A5 = A4 =
A5 = A5 = Be, = B, = By
=
0.55941519 0.24032463 —0.46123818 0.71795706 —0.67534061 0.34357563 —0.07271528 0.55941519 —0.35462716 1.4624910
= —3.9209295 = 6.2012014 B5 = —5.6962826 B6 = 2.8184609 B7 = —0.5816914 B3 B4
V9 = 1.0000002 V. = 1.0000001 Surface deviations Lb = —0.0000004b Act = —0.0000002a
The potential outside the sphere passing through the cylinder edges is found by the second integral in 1.06 (6) to be
211
CYLINDRICAL BOUNDARIES
§5.40
V=
aVo 26
s=0 p=0
( —1)P(Ce CS) (2p± 2s) !a2Pb23 P2s+2p(COS 0) 4P(p!) 2(2s)!r 28+2p+1
(4)
Ce = aB(p + 1, m ± -3-)B,„,
4)A n (5) Cs = bB(s +, n n =0 where B(x,y) is a beta function. Note that at a great distance (4) takes the form V = Q/ (4rer). The capacitance is then Q/V o. We get an empirical interpolation formula for C by adding a correction term containing two adjustable constants to the exact known capacitance of a circular disk given in 5.03 (2). This curve, which passes through the seven calculated points with an error of 0.2 per cent or less, is 7.76] farad (6) C = [0.708 0.615( X 10-10a m=o
a
It is clear that this method applies to other geometrical forms such as the spindle formed by two finite circular cones fitted base to base. It also works when such objects are placed in uniform fields as seen in the foregoing Taylor reference. 5.40. Mixed Spherical, Spheroidal, and Cylindrical Boundaries. Often different boundaries of an electric field belong to different coordinate systems. No method given so far handles this case. The formulas needed to relate the four systems, oblate, prolate, spherical, and cylindrical, naturally split into two groups: those relating the exterior harmonics, which satisfy Laplace's equation and vanish at infinity, and those relating the interior harmonics, solutions that are finite at the origin. The former depend on a set of Fourier transforms due to J. C. Cooke (see Monatshefte fur Math. Vol. 60, page 322, 1956; and Zeit. fur Math. and Mech. Vol. 42, page 305, 1962). These transforms have been modified to fit the notation of this chapter. Note that Hobson, HTF, HMF, and others insert (-1)m in the right side of 5.23 (6) and (7), so in their notation the sign of m in the exponent of j in the following formulas should be reversed. To shorten the formulas Rmn is written for (n — m) !/(n m)!. —
10 rilf„,(tp)In+I(tc) cos (tz) dt = j'-m(27rc-')IR,„„Q7(j)P7( ) 10 't--11C„,(tp)/7,±i(tc) sin (tz) dt = j 2-'n(i7.c-91RmnQ7,','(,g)P7(E)
(2)
1:
tnK„.(tp) cos (tz) dt = j"-ri(47)(n — m)!r-71-1137(cos 0)
(3)
10
talc„,(tp) sin (tz) dt = jn-'n-i(ir)(n — m)!r-n-iP7(cos 0)
(4)
fo-t —IK„.(tp)J.+1(tc) cos (tz) dt = jn+m(firc-1) 11?..Q7(n)PT() Jo y
(1)
t-ilc„,(tp)J,,+I(tc) sin (tz) dt =
(5) (6)
212
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.40
m is odd, if m is even and in (2), (4), and (6) n In (1), (3), and (5) n —1 < < +1 and 0 < < 00 in (1) and (2), but the reverse is true when 0 < < 1 and — 00 < < 00. The interfocal distance in the spheroids is 2c as in Arts. 5.27 to 5.281. Additional formulas needed for combining spherical and spheroidal forms are found by substitution of the power series 5.32 (1) and 5.293 (2) for /„±i(t) and J.±1(t) in (1) and (5) or (2) and (6), giving the integrals the form of (3) or (4). Thus CO
\ 28+1
(2'nn(i0P7(E)
)
=
P'2ns(cos 0)
(7)
s= in 28+1
(8)
P;s(cos
QZ(n)P„(0 = s= in B
( —1)"(2s — m)!(n m)! —
(it + 2s+ 1)!!(n — m)!(2s — n)!!
(9)
In (3) or (4), in may be written as a Neumann series involving in+I(t) and J.+#(t) by Watson (1), page 138, and HMF 9.1.87, page 364. The resultant integrals are those in (1) and (5) or (2) and (6). Thus n+1
P',,'"(cos 0) =(-1)"'"1-8 jC.Q;n8(jilP;n8 a)
r
(10)
s=in CO
= a .r n
Cmns—
—1)nCnoisQ2s(n)P78(t)
(2s + 1)(n + 2s — 1)!!(2s — m)! m)! (n — m) !(2s — n)!!(2s
(11)
(12)
From these a set of formulas for combining oblate and prolate boundaries may be derived. Let the interfocal distances and the E coordinate in the oblate case be 2c1and Eland in the prolate case 2c2 and E2. In (7) write ci for c and Eifor E, and in (11), (c2/ci)n±1 (c1/r)n+1for (c/r)n-", p for s, and then 2s for n. Elimination of spherical harmonics gives \ 28+1
27(3.0Ptnn(1) =
—1)m+P+ni-1emnspei)
(
8
Cmnsp
—
(2772(0) P 273(6)
(13)
= in P=8
(n
(2p + 1)(n 2p — 1) !!(2p — m)! m)!(2s 2s + 1)!!(n — m)!(2s — n)!!(2p — 2s)!!(2p
m)! (14)
§5.40
213
CYLINDRICAL BOUNDARIES
A similar operation on (8) and (10) gives 28+1
(21.:(77)P7(E2) =
Q'2;(3.0P'2n,(Ei)
(15)
8 = in P
In the same way for nonconfocal coaxial spheroids of the same type 23+1 1) n+'+Pemnsp0 V2np(A-2) P''p(E2)
QZUNFnnal) =
(16)
s= in P=s 23+1
(711)PV El) =
Cinnsp(7 C1)
V2np(n2)P7p(E2)
(17)
s=in P=s
The interior harmonics, like (1) to (6), depend on Bessel function formulas due to Cooke. In Hobson's notation the right sides of (18) to (23) should be multiplied by (-1)m. The relations are CO
\4- 1.1-m(2n+ 1)(n — ViC) (n m)! 7r
cos (tz)I,,(tp)
jn+I(cOPT(Z)P'nn(i0 (18)
nm
=
.ji'±nv(tr)n[(n
m)!]-1P7(cos 0)
(19)
n=m
l) (n—
( —1)m(2nn -F 2rtc = (—) n=m
( -F m) !
m)!
Jn±i(ct)PZ(OPT(n) (20)
sin (tz)/„,(tp)
(—j)m-E1(2n 1)(n — m)!
611
(n
m)!
n=m+1 in+1(c0P7(
(tr)"[(n
m)!]-1PZ(cos 0)
OPT, (3.0
(21) (22)
n=m+1
r Vtc
LJ
— 1)'n(2n + 1)(n — ) (n m) ! m !Jn+I(ct)P7(E)P7(n)
n=m+1
(23) In (18), (19), and (20) m n is even, and in (21), (22), and (23) m n is odd. The interior equations analogous to (7) to (12) are now polynomials instead of infinite series. They may be derived by writing out
214
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.40
the left sides in products of powers of its coordinates, then changing over to the coordinates of the right side by 5.27 (4) and (5) or 5.28 (4) and (5) and regrouping to form harmonics. Alternatively write in±i(ct) and Jn+1(ct) in (18) and (20) or in (21) and (23) as power series in ct by 5.32 (2) and 5.293 (3), replace (On by (r/c)"(tc)' in (19), and equate coefficients of (tc)n. Thus we get Tin
2a c) PZ(cos 0) (71.
P7.(E)Pinn(i0 =
(24)
s S
in 28
— 1) in—aDnins() P;(COS 0)
PZMPZ(n) =
(25)
s=S
Dmns —
m)! (n 2s — 1)!!(n (2s + m)!(n — m)!(n — 2s)!!
(26)
in
- PT(cos 0) (cY r
:7--nG.n8P2. ()P;(jr)
(27)
(
(28)
8=8
in
=
) P78(17) 1)1n—sGmn.P72n3(t
3=8 limns
(2s
(4s + 1)(2s — m)!(n m)! m)!(n + 2s + 1)!!(n — 2s)!!
P9)
Here S equals im or 4 (m + 1) and is an integer if n is even and an integer plus if n is odd, and s increases by unit steps. Equations (30) to (33) derive from (24) to (29) in the same way that (13) to (15) derive from (7) to (12). Thus in
\28
P7(3.0P7`(E1) =
1 ) 8—PFninsP ci) P2p(n)P2p(E2)
(30)
s=S p=P in
3
28
P7(n)P7(Ei) =
(-1)InF,anspC-1 (
8=8 p=p in
Cl
)
P2np(ii-)P7p(E2)
\28 .in( -1)8Fmnsp() ci P2np(.7?.2)P79(E2)
P'„n(in)PZ(Ei) =
(31)
(32)
s=S p=P
in
3
P',7(n)P7(E1) =
/ 1 ) 1n—PF mn8 P
2s Cl C1
P2np(712) P7p(E2)
(33)
s=8 p=P
F„,s, —
(4p + 1)(n + 2s — 1)!!(2p — m)!(n m)! (2s + 2p + 1) !!(2p m)!(n — m)!(n — 2s) !!(2s — 2p) !! (34)
§5.41 SPHERE IN CONCENTRIC, COAXIAL, EARTHED SPHEROID
215
Here S and P equal km or 4(m + 1) and are integer if n is even and integer plus 2if n is odd and s and p increase by unit steps. There are two more formulas that will prove useful in working problems. The first is IT II (6), page 29, and the second is derivable from HTF (25), page 159, by expansion of the integrand binomial and integration of the real part. In both P'.(cos 0) has been multiplied by ( —1).8 to change it from Hobson's notation to that of 5.23 (6) and (10). Thus
:e-gq,a(tp)tn dt = (n — m)!r-.-43,7,i(cos 0) I ( c)
(-1)i(8--no (n — s)!(s m)!!(s — m)!
P',7(cos 0) = (n m)!(.c)n 8
(35) P)
(36)
= In
where S is n or n — 1 and s m is even so s jumps by two in successive terms of the summation. 5.41. Sphere in Concentric, Coaxial, Earthed Spheroid. Suppose that the potential inside and on the surface of a sphere of radius a is —
)2n
P2Th(cos 0)
V =
(1)
n=0
and that outside it there is a concentric coaxial spheroid 1 o (or no) with interfocal distance 2c, at potential zero. It is desired to find the potential between them and the charge densities. Let the interior and exterior potentials of the charge distribution on the sphere in the presence of the spheroid be, using 5.40 (27) and 5.40 (10) for the oblate case, N
An(-)2n P 2.(cos f
Vin =
a
)
n=0
N
n
( — ).(4s± 1) ( 2n) !(a/ c 2s 1) ! ! (2n — 2 s)2)1;!P2s(k)P28q)
An ( 2n 1 ±
(2)
n=0 8 =0
=
N
ayn+1 P 2n(COS
V ex n=0
N =
+ 1)(2n + 2s — 1)!!(c)2 n+1 ( — 1)8(2n) !(2s — 2n) !!
a
3) Q28 00P28(t) (
n=0 s= n
In the prolate case 5.40 (28) and 5.40 (11) are used. The potential Vin of the charge density on the spheroid 1 = N (or n = no) must exactly cancel Vex there and thus is given by (3) when —Q28(.go)P2800/P28(..go)
216
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
§5.41
replaces Q28(j0. Thus N
Vin
—n)11), !!(02''+ip )n (2)n1(2Fs 2s2 .(3 0 ) p2s(A.)p2s(E) Q22:( A„i((2s±os1(2 -
= n -= 0 s = n
(4)
V:„ must equal The sum of the coefficients of P2s(j0P2(E) due to V,„„ its original coefficient as given by (1), which is (2) with B. written for An. Thus we have N 1 simultaneous equations, one for each value of s from 0 to N inclusive, to be solved for A 0 , • • • , AN, which are 0
A0000 + A iCoi + _2 A C_ 02 + • • + ANC6N = 7 . BnCOn
1
A 0C +
om n=0
A2C12 + • • • + ANCON =
B.0
(5)
n =1
tAT
A CN ± A1CN1
A2CN2 ± • • • ± ANCIVN —
BNCNN
It is evident from (2) and (4) that when s < n only the sphere's charge, and when s n only the spheroidal charge, contributes to the coefficients of P2e(i0P2.(E) so that (ays 1)(2n)! ( —1)n(4s (6) Csn = (2n + 2s +1)!!(2n — 2s)!! c (2n)! a )2n + (2n + 1)(4n — 1) !!(c)2n+1(22.(A-0) — 1) "[ (2n — i)! 2c j(2n)!! a P2n(3To) (7) c\2n+1 1)(2n ± 2s — !!(22sW-0) j(2s ± s n (8) ( —1)8±1(2n)!(2s — 2n)!!P2s go) ct) s c to get Va and in (4) as modified by step 3 to get V,. Then V = Vs ± V,
(11)
Another form for the potential may be more convenient. Once the A„'s are known, the sphere can be replaced by a set of multipoles of known magnitude by (2). In the Green's function 5.298 (4), z — zo is written for z, b for a, a for b, and 0 for ctn. Then, by differentiation with respect to zo and b, the potential in an earthed cylinder of each multipole may be found, and the sum of these, weighted according to their magnitude, gives V for a sphere centered at z = 0, p = b, ¢ = 0. If the sphere is replaced by a spheroid, the problem can be set up in the same way. By 5.40 (7) or (8) the spheroidal harmonics are expressed as a sum of spherical harmonics, which are then treated as in this article up to (7), at which point the spherical are converted back into spheroidal by 5.40 (10) or (11), and so forth. Before the advent of digital computers the numerical work involved in such methods would have been too laborious to be practical, but with these devices, solutions of any desired accuracy can be found. Problems Problems marked C are taken from the Cambridge examination questions as reprinted by Jeans by permission of the Cambridge University Press. 1. Two similar charges are placed at a distance 2b apart. Show that a grounded conducting sphere placed midway between them must have a radius of approximately b/8 to neutralize their mutual repulsion. 2. The radius vector from the center of a conducting sphere of radius a, carrying a charge Q, to a point charge q is c. When the system is placed in a uniform field E along the direction of c show that, in order that no force act on q, Q must have the value qa3 2c' — ' — c2[471-€E(1 2(c)a) — c (co — a c 3. An infinite conducting plane forms one boundary of a uniform field of unknown strength X. A known charge q, when placed at a distance r from the plane, experiences an unknown force F. It is found that if a hemispherical piece of conducting
224
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
material, of radius a, is placed flat against the plane opposite the charge the force is unchanged. Show that q2 [ 2r2 1 qr6 X — F— 4r2 tre (r 4 — a4) 2 2/re(r,— 0 4. A hollow conducting sphere of radius a is half filled with a dielectric. On the axis of symmetry at a distance a/3 from the plane dielectric surface, a charge experiences no image force. Show that the capacitivity is 1.541e„. 5C. An infinite conducting plane at zero potential is under the influence of a charge of electricity at a point 0. Show that the charge on any area of the plane is proportional to the angle it subtends at 0. 6C. If two infinite plane uninsulated conductors meet at an angle of 60° and there is a charge e at a point equidistant from each and distant r from the line of intersection, find the electrification at any point of the planes. Show that at a point in a principal plane through the charged point at a distance r31 from the line of intersection, the surface density is
( 4/1-r2 4
) 71
7C. What is the least positive charge that must be given to a spherical conductor, insulated and influenced by an external point charge e at distance r from its center, in order that the surface density may be everywhere positive? 8C. An uninsulated conducting sphere is under the influence of an external electric charge. Find the ratio in which the induced charge is divided between the part of its surface in direct view of the external charge and the remaining part. 9C. A point charge e is brought near to a spherical conductor of radius a having a charge E. Show that the particle will be repelled by the sphere, unless its distance from the nearest point of its surface is less than -la (e /E)1, approximately (take e E). 10C. A hollow conductor has the form of a quarter of a sphere bounded by two perpendicular diametral planes. Find the images of a charge placed at any point inside. 11C. A conducting surface consists of two infinite planes which meet at right angles and a quarter of a sphere of radius a fitted into the right angle. If the conductor is at zero potential and a point charge e is symmetrically placed with regard to the planes and the spherical surface at a great distance f from the center, show that the charge induced on the spherical portion is approximately —5ea3/(7rf 3). 12C. A thin plane conducting lamina of any shape and size is under the influence of a fixed electrical distribution on one side of it. If of is the density of the induced charge at a point P on the side of the lamina facing the fixed distribution and o2 that at the corresponding point on the other side, prove that al — 02 = co where oo is the density at P of the distribution induced on an infinite plane conductor coinciding with the lamina. 13C. A conducting plane has a hemispherical boss of radius a, and at a distance f from the center of the boss and along its axis there is a point charge e. If the plane and the boss are kept at zero potential, prove that the charge induced on the boss is —e(1
a2
a 2)1
14C. Electricity is induced on an uninsulated spherical conductor of radius a by a uniform surface distribution, density a, over an external concentric nonconducting spherical segment of radius c. Prove that the surface density at the point A of the
PROBLEMS
225
conductor at the nearer end of the axis of the segment is —crc 2a2 (c a)(1
AB)
where B is the point of the segment on its axis and D is any point on its edge. 15C. Two conducting disks of radii a, a' are fixed at right angles to the line that joins their centers, the length of this line being r large compared with a. If the first has potential V and the second is uninsulated, prove that the charge on the first is 872Ear 2 V (r2r2 — 4aa') 16C. A spherical conductor of diameter a is kept at zero potential in the presence of a fine uniform wire in the form of a circle of radius c in a tangent plane to the sphere with its center at the point of contact, which has a charge Q of electricity. Prove that the electrical density induced on the sphere at a point whose direction from the center of the ring makes an angle v, with the normal to the plane is
—c2Q sec3 477-2a
2 .1.
r
(a2
c2 sec2 G — 2ac tan cos 0)-1 do
17. A stationary ion with charge q and mass m is formed in a highly evacuated conducting bulb of radius a at a distance c from its center. Show that the time for this ion to reach the wall is
t = q-'1c(47rmEa 3) 1(K — E)
where V = (a2 — c2)a-2
K and E are complete elliptic integrals of modulus k. 18. An earthed conducting sphere of radius a has its center on the axis of a charged circular ring, any radius vector c from its center to the ring making an angle a with the axis. Show that the force sucking the sphere into the ring is
Q2E(c2 — a2)k 3 cos a 1672Ec2a2 sin3 a(1 — k2)
where k2 —
4a2c2 sine a
a4
c 4 — 2a2c2 cos 2a
E is the complete elliptic integral of modulus k. 19C. If a particle charged with a quantity e of electricity is placed at the middle point of the line joining the centers of two equal spherical conductors kept at zero potential, show that the charge induced on each sphere is —2em(1 — m + m2 — 3m3 4m4) neglecting higher powers of m, which is the ratio of the radius to the distance between the centers of the spheres. 20C. Two insulated conducting spheres of radii a, b, the distance c of whose centers is large compared with a and b, have charges Qi, Q2, respectively. Show that the potential energy is approximately (87rE)-'[(a-' — b 3c-4)(2!
2c-,(21(22
(b-,— a3c-4)(a
21C. Show that the force between two insulated spherical conductors of radius a placed in an electric field of uniform intensity E perpendicular to their line of centers is 127rEE2a6c-4(1 — 2a3c-3— 8a 5c-5 + • • • ), c being the distance between their centers. 22C. Two uncharged insulated spheres, radii a, b, are placed in a uniform field of force so that their line of centers is parallel to the lines of force, the distance c between
226
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
their centers being great compared with a and b. Prove that the surface density at the point at which the line of centers cuts the first sphere a is approximately eE(3
6b3c-3
15abac-4
28a2b3c-5
57a3b3c-6 + •
•)
23C. Two equal spheres each of radius a are in contact. Show that the capacity of the conductor so formed is 87rect In 2. 24. Two spheres of radii a and b are in contact. Show that the capacitance of the conductor so formed is 47reab[
C-
a +b
2C' -I-
a (a
b
b)
4'(a
b)]
where C' is Euler's constant 0.5772 and gz) = r'(z)/r(z). 25C. A conducting sphere of radius a is in contact with an infinite conducting plane. Show that if a unit point charge is placed beyond the sphere and on the diameter through the point of contact at distance c from that point, the charges induced on the plane and sphere are Ira (a) - — cot —
and
7ra — cot (— c c
-1
26C. Prove that if the centers of two equal uninsulated spherical conductors of radius a are at a distance 2c apart, the charge induced on each by a unit charge at a point midway between them is
Z
(-1)' sech na
where c = a cosh a
27. Two equal spheres, of radius a, with their centers a distance c apart are connected by a thin wire. Show that the capacitance of the system is 8rea sinh
Z(- 1)n+l csch n13 n=1
where cosh )3 = c/a. 28. Two equal spheres, of radius a, with their centers a distance c apart, are charged to the same potential V. Show that the repulsion between them is 277-€V2 Z(-1)n+I(coth p - n coth n13) csch n0 n=1
where cosh 13 = 2c/a. 29C. An insulated conducting sphere of radius a is placed midway between two parallel infinite uninsulated planes at a great distance 2c apart. If (a/c) 2 is neglected, show that the capacity of the sphere is approximately 47rEa[1 + (a/c) log 2]. 30C. Two spheres of radii r1, r2 touch each other, and their capacities in this position are c,, c2. Show that c1= 47T-er2(f2Zn-2 f3 Zn-3 f4Zn-4 + • • •) where f = r, (r1
227
PROBLEMS
31C. A point charge e is placed between two parallel uninsulated infinite conducting planes, at distances a and b from them, respectively. Show that the potential at a point between the planes which is at a distance z from the charge and is on the line through the charge perpendicular to the planes is z Sre(a
b)[
2a
2b
,v (2a — z 2a 2b
2b +z 2a 2b
*(2a 2b — z)] 2a 2b
where 4/(z) = [r'(z)]/[r(z)]. 32. An electron, charge e, mass m, traveling horizontally in a high vacuum with a velocity v, must pass over an uncharged horizontal dielectric plate of length d. Show that, if it comes within a distance a of the front edge of the plate, it will be drawn into it by image forces before clearing the back edge, where, if edge effects are neglected, a' —
(K — 1)e 2d 2
2m2rae„(K
1)v'
33C. A point charge is placed in front of an infinite slab of dielectric, bounded by a plane face. The angle between a line of force in the dielectric and the normal to the face of the slab is a; the angle between the same two lines in the immediate neighbourhood of the charge is 0. Prove that a, 13 are connected by the relation
sin i:13 =
( 2K y sin 1 K
Za
34C. Two dielectrics of inductive capacities K, and K2 are separated by an infinite plane face. Charges e,, e2 are placed at points on a line at right angles to the plane, each at a distance a from the plane. Find the forces on the two charges, and explain why they are unequal. 35C. Two conductors of capacities CI, c2 in air are on the same normal to the plane boundary between two dielectrics with coefficients k,, k,, at great distances a, b from the boundary. They are connected by a thin wire and charged. Prove that the charge is distributed between them approximately in the ratio ki
[4,7re,,
c2
k, — k 2
2b(ki
kz)
2k 2 (k,
k2)(a
421-E„ k 2[ ci b)1
k, — kz
2a(ki +kz)
2k,
(ki -Fk2)(a b)]
36C. A conducting sphere of radius a is placed in air, with its center at a distance c from the plane face of an infinite dielectric. Show that its capacity is zlireoa sinh
aZ
(
K — 1 yi cosech na K +1
where cosh a = c/a. 37. Show that the charge induced on an earthed conductor consisting of two similar spheres of radius a, intersecting orthogonally, by a charge q in their plane of intersection, at a distance 3a2-1 from the axis of symmetry is (1 /3 — 2/51)q. 38. A charge q is at a distance b from the center of a circular hole of radius a in an infinite earthed flat conducting sheet and lies in the same plane. Show that the charge density induced on the sheet at a distance r from the charge and c from the center of the hole is —q(a 2— b2)-1/[27-2r 2(c2— 39. Show that, if in the preceding problem the charge is distributed around a ring of radius b concentric with the hole and coplanar with the sheet, the induced charge density at P is —q(a2— b2)1/[221-2(c2 — b2)(c2— a2)1].
228
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
40. Invert the preceding problem, apply Green's reciprocation theorem, and show that the potential at a point P on a sphere, part of whose surface is occupied by a thin conducting bowl at potential Vo, is (2V0/7) sin-,(cos a/cos 0) where 0 is the angle, measured at the bowl end of the rotational symmetry axis, between this axis and P and a that between this axis and the edge of the bowl. 41. Starting with the preceding result find the first term in the spherical harmonic expansion for the potential of a freely charged spherical howl all elements of which are at a distance a from the origin where its radius subtends an angle /3. From this show that its capacitance is 4ea(0 + sin (3). 42. Invert the result of problem 39 and find the charge density induced on a spherical earthed conducting bowl by a uniform charge distribution —0-1 over the remainder of the sphere. By superimposing a uniform spherical surface charge of density al show that the charge density at P on the inside or outside, respectively, of a freely charged conducting spherical bowl at potential Vo is
=
EV 0
sin a
.
sin a sin 0
Zra [ (sin2 0 — sin 2 a)i
=
eV o
where the angles are defined as in problem 40. 43C. A conductor is formed by the outer surfaces of two equal spheres, the angle between their radii at a point of intersection being 27/3. Show that the capacity of the conductor so formed is 2rae(5 — 3-14), where a is the radius of either sphere. 44C. An uncharged insulated conductor formed of two equal spheres of radius a, cutting one another at right angles, is placed in a uniform field of force of intensity E, with the line joining the centers parallel to the lines of force. Prove that the charges induced on the two spheres are +1171-ea2E/2. 45C. A conductor is bounded by the larger portions of two equal spheres of radius a cutting at an angle 7/3 and of a third sphere of radius c cutting the two former orthogonally. Show that the capacity of the conductor is 47re 1c + a (I — -131) — ac[2(a2
c2)-1 — 2(a2
3c2)-+
(a2
40)-111
46C. A spherical shell of radius a with a little hole in it is freely electrified to potential V. Prove that the charge on its inner surface is less than ieVS/a, where S is the area of the hole. 47C. A thin spherical conducting shell from which any portions have been removed is freely electrified. Prove that the difference of densities inside and outside at any point is constant. 48C. Prove that the capacity of an elliptic plate of small eccentricity e and area A is approximately Ay( 1 e4 e 6) Se(-— 64
64
49C. An ellipsoidal conductor differs but little from a sphere. Its volume is equal to that of a sphere of radius r, its axes are 2r(1 a), + 2r(1 0), 2r(1 y). Show that neglecting cubes of a, 0, -y, its capacity is 4orer[1 + ( A) (a2 +32 + ,2)]. 50. An oblate conducting spheroid with semiaxes a and b has a charge Q. Show that the repulsion between the two halves into which it is divided by its diametral plane is (12[1671-e(b 2 — a2)]-1In (b/a), where b > a. 51C. A prolate conducting spheroid, semiaxes a, b, has a charge Q of electricity. Show that repulsion between the two halves into which it is divided by its diametral plane is Q2[167e(a 2 — b2)]-' In (a/b) where a > b. Determine the value of the force in the case of a sphere.
229
PROBLEMS
52C. A thin circular disk of radius a is electrified with charge Q and surrounded by a spheroidal conductor with charge Qi, placed so that the edge of the disk is the locus of the focus S of the generating ellipse. Show that the energy of the system is [Q 2 L BSC + (Q Qi) 2L SBC]/ (87-ea), B being an extremity of the polar axis of the spheroid and C the center. 53C. If the two surfaces of a condenser are concentric and coaxial oblate spheroids of small ellipticities S and S' and polar axes 2c and 2c', prove that the capacity is 471-Ecc'(c' — c)-2[c' — c ;(Sc' — o'c)], neglecting squares of the ellipticities. Find the distribution of electricity on each surface to the same order of approximation. 54C. An accumulator is formed of two confocal prolate spheroids, and the specific inductive capacity of the dielectric is K1/6.), where (.7) is the distance of any point from the axis. Prove that the capacity of the accumulator is 2T-2E„K1 In
b, a+b
(al +
where a, b and al, b, are the semiaxes of the generating ellipses. 55C. A thin spherical bowl is made by the part of the sphere x 2 +y2 + z2 = cz bounded by and lying within the cone (x/a)' + (y/b)2 = (z/c)' and is put in connection with the earth by a fine wire. 0 is the origin, and C, diametrically opposite to 0, is the vertex of the bowl; Q is any point on the rim, and P is any point on the great circle arc CQ. Show that the surface density induced at P by a charge E placed at 0 is
Ec CQ 47rabI OP 2(0P2 — 0Q2) 1'
where I =
do f Jo .r (a2 sin2 0
b 2cos' 0)1
56. A charge q is placed at a distance c from the center of a spherical hollow of radius a in an infinite dielectric of relative capacitivity K. Show that the force acting on the charge is (K — 1)q' 4,rf„c 2
n=0
n
n(n ± 1) (,)"+1 K(n 1) a
57. An earthed conducting sphere of radius a has its center on the axis of a charged circular ring, any radius vector c from this center to the ring making an angle a with the axis. Show that the force sucking the sphere into the ring is Q2
4n ec 2
a)2n+1 Z(n + 1)P, Ei (cos a)P„(cos a)( 0
58. The spherical coordinates of a circular ring are a, a. A sphere of relative capacitivity K and radius b has its center at the origin. If the ring carries a charge of linear density 7, show that the potential between the sphere and the ring is
—ZP„(cos a) sin a l
2f„
re = 0
n(K — 1)b 2n±' a
an(n(K + 1) + 1)r"÷,
Pn (00s 0)
59. That portion of a sphere of radius a, lying between 0 = a and 0 = it — a is uniformly electrified with a surface density a. Show that the potential at an external
230
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
point is 2n-FI
cur {a
cosa + Z
—
e
7'
,, a
4n + 1
[P2n+i(COS a) — P2n_i(COS ah
P2n(P)
60. A circular ring of radius a and charge Q is used as a collector at the center of an earthed spherical photoelectric cell of radius b. Show that the field strength at the spherical surface is 00
Q
•
4re
1)n
n=0
3 • • • (2n — 1) 4n + 1()2n P 2o(CoS b2 b 2 • 4 • • • 2n
61. A point charge q is placed at a distance b from the center of two concentric earthed conducting spheres of radii a and c, where a < b < c. Show that when a < r < b, the potential is =
V
b2n+1
q
41T-EZ bo+1(a 2n-1-1
a270-1)
c2n -1-1 c2n+1)
r"
1.1-1
P ,t (cOs 0)
n=0
62. An infinite flat conducting plate is influenced by a charge q at a distance b from it, and there is a solid dielectric hemisphere of radius a flat against the plate opposite the charge. Show that the surface density induced on that part of the plate lying outside the dielectric is
b (b2
K) (2n + 1)(2n + 2) 1• 3 • • • (2n +1) 1 + (2n + 1)(K + 1) 2 • 4 • • • (2n + 2)
(-1)n+1(1
7.2)1
n=0
4n+3 a b2'+2r 2'±3
63. Show that the potential at any point due to a circular disk of radius c raised to a potential V1 is (2V1 /71-)Z( —1)”(2n + (c/r) 2"+1P2n(cos 8) where r > c. Find the value for r < c. 64. Two similar rings of radius a lie opposite each other in parallel planes, so spaced that the radius of one ring subtends an angle a at the center of the other. Show that when they carry charges Q and Q', the repulsion between them is
QQ' z 4rea2
n=0
un l • 3 5 • • • (2n + 1) (sin (si a) 2"±2P2.÷1 (cos a) 2 • 4 • 6 • • • 2n
65. The inside of a conducting sphere of inner radius a is coated with a uniform layer of dielectric of inner radius b. Show that the force on a point charge q, in the cavity at a distance c from the center, is CO
(72
4re,,
n=0
nc2"-11(K — 1)(n + 1)a2"+' [(K + 1)n + 1]b2n+1 1 b 2n+1 [ RK + 1)n + K]a2"+' (K — 1)702n+11
66. A point charge q is placed at a distance c from the center of an earthed con-
ducting sphere of radius a on which is a dielectric layer of outer radius b and relative capacitivity K. Show that the potential in this layer is Nk
47ro,,c
n=0
c"
(2n + 1)b2n+1(r" — Pn(cos 6) + 1)n + l]b2"+' (n + 1)(K — 1)a2n+11
PROBLEMS
231
67C. A conducting spherical shell of radius a is placed, insulated and without charge, in a uniform field of electric force of intensity F. Show that, if the sphere is cut into two hemispheres by a plane perpendicular to the field, these hemispheres tend to separate and require forces equal to tirca'F' to keep them together. 68C. A spherical conductor of internal radius b, which is uncharged and insulated, surrounds a spherical conductor of radius a, the distance between their centers being c, which is small. The charge on the inner conductor is Q. Find the potential function for points between the conductors, and show that the surface density at a point P on the inner conductor is 3c cos 0) (1 a' b3 — a3
Q
where 0 is the angle that the radius through P makes with the line of centers and terms in c2 are neglected. 69C. The equation of the surface of a conductor is r = a(1 SP„), where is very small, and the conductor is placed in a uniform field of force F parallel to the axis of harmonics. Show that the surface density of the induced charge at any point is greater than it would be if the surface were perfectly spherical by the amount
(5
[ 3neF.5 [(n 1)Pn+1 2n +1
(n — 2)P„-”
SP.) is 70C. A conductor at potential V whose surface is of the form r = a(1 surrounded by a dielectric (K) whose boundary is the surface r = b (1 + nP„), and outside this the dielectric is air. Show that the potential in the air at a distance r from the origin is
KabV
(2n
{1 b r
(K — 1)a
+1)3a.b2.+1 (1
(n 1)a2n+,] P.} (K — 1)nb'[nb 2n+1 (K — 1)(n + 1)a2n+3 nK)b 2n+3
n
where squares and higher powers of S and n are neglected. 71C. The surface of a conductor is nearly spherical, its equation being r = a(1
nSa)
where n is small. Show that if the conductor is uninsulated, the charge induced on it by a unit charge at a distance f from the origin and of angular coordinates 0, 4 is approximately
fL
+ \f/ (--Ynsn(o, )1 95
72C. A uniform circular wire of radius a charged with electricity of line density e surrounds an uninsulated concentric spherical conductor of radius c. Prove that the electrical density at any point of the surface of the conductor is L[i
102P2 + 1 . 304P4 52 a
2c
92 • 4 a
. 3 . 506P6 132 2• 4 • 6 a
73C. A &electric sphere is surrounded by a thin circular wire of larger radius b carrying a charge Q. Prove that the potential within the sphere is [ 4r€,,b
1 ± Z(— 1)n 1
1 + 4n 1 + 2n(1
K)
1 • 3 • 5 • • (2n— 1)(r)2np 2n b 2 4 6 • • 2n
232
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
74C. A conducting sphere of radius a is embedded in a dielectric (K) whose outer boundary is a concentric sphere of radius 2a. Show that, if the system is placed in a uniform field of force F, equal quantities of positive and negative electricity are separated of amount 367a2c„FK/(5K + 7). 75C. A spherical conductor of radius a is surrounded by a uniform dielectric K, which is bounded by a sphere of radius b having its center at a small distance y from the center of the conductor. Prove that, if the potential of the conductor is V and there are no other conductors in the field, the surface density at a point where the radius makes an angle 0 with the line of centers is, approximately, E,KVb
a[(K — 1)a
b][
1±
6(K — 1) ya2 cos 2)b' 2(K — 1)a3 (K -
76C. A shell of glass of inductive capacity K, which is bounded by concentric spherical surfaces of radii a, b, (a < b), surrounds an electrified particle with charge q which is at a point Q at a small distance c from 0, the center of the spheres. Show that the potential at a point P outside the shell at a distance r from Q is approximately
q
1
2c(b 3 — a3)(K — 1) 2
471- E
r
2a3(K — 1) 2 — b 3(K + 2)(2K + 1)
cos 0 r2
where B is the angle which QP makes with OQ produced. 77. Show that the attraction between a sphere of radius a and relative capacitivity K, and a point charge q at a distance c from its center is (1 — K)q 2 n(n + 471-E„c2 .4 1 (K n=1
1)
(a)2'.+1 1)n c
78. A conducting sphere of radius a is supported by an orthogonal conducting cone whose exterior half angle is a. Show that, if this system is charged, the potential is A (r" — on+ir-.-1)P.(cos 0) where 0 < n < 1 if I-7r < a and P,,(cos a) = 0. 79. Two conducting coaxial cones whose half angles, measured from the positive axis, are a and 13 intersect a conducting sphere of radius a orthogonally. If the system is charged, show that the potential outside the sphere between the cones is
A (r" — a 2n+ir-n-1)[Q„(cos (3)Pn (cos 0) — P.(cos ()Q„(cos 8)] where n is the smallest number for which Q,,(cos 0)Pn (cos a) — P„(cos 13)(2n(cos a) = 0 80. An oblate dielectric spheroid whose surface is 3•0 is placed in a uniform electric field E parallel to the axis C = 1. Show that the resultant field inside is uniform and that its strength is —E/ l(K — 1ko[(1 31) cot-' 10 — 1-0] — 81. The spheroid of the last problem is placed in a uniform field E normal to the axis C = 1. Show that the electrical intensity inside is uniform and of magnitude 2E 2 + (K
—
1
)3'0[(1
cot-230
—
"()]
82. A prolate dielectric spheroid whose surface is no is placed in a uniform electric field E parallel to the axis C = 1. Show that the resultant field inside is uniform and that its strength is —E/1(K — 1)no[(1 — no) coth -1no + no] —
PROBLEMS
233
83. The spheroid of the last problem is placed in a uniform field E normal to the axis t = 1. Show that the electrical intensity inside is uniform and of magnitude 2E 2 + (K — 1)no[( 1—
coth-i n o + nol
84. A spheroid of relative capacitivity K is placed in an electric field E, its axis of revolution making an angle a with the field. Show that the torque acting on it is iirEo(K — 1)m2nE(E — E2) sin 2a where n is the semiaxis in the direction of the axis of revolution and m is the semiaxis normal to it. For an oblate spheroid, El and B2 are the results of problems 80 and 81, respectively, where 3 0 = n(m2 — n2)-1. For the prolate spheroid, El and B2 are the results of problems 82 and 83, respectively, where no = n(n2 — 85. An earthed conducting disk of radius a whose equation is 3.= 0 is influenced by a point charge q at 1, 3-0, 0. Show that the potential due to the induced charge on the disk is CO
=
q
27r2ea
(4n + 1)Q2o.W-o)Q2.(i0P2n(t) n =0
86. Show, using 5.275 and Green's reciprocation theorem, that the potential due to a charged ring whose equations are 3- = 3 0, t = to is CO
= i Q 4 2n + i)Pn(7o)Pnctoui?-)Pn(z) , (
47rea
n=0
when 3' > 3 o and the charge on the ring is Q. Write the expression valid when t- < 87. An earthed conducting disk of radius a is influenced by a ring carrying a charge Q whose equation is 3- = 3 0, t = to. Show that the potential due to the induced charge is
V; =
27r 2ea
(4n + 1)Q2nU3"o)P2n(E0)(22n(j0P2n(t) n=0
88. The upper and lower halves of an oblate spheroidal shell = 3 o are insulated from each other and charged to potentials + Vo and —V0, respectively. Show that the potential at any external point is V = VO
1 • 3 • • (2n + 1) (4n + 3)Q2,,+I(/3-) 2 • 4 • • • [2(n + 1)1 (2n + 1)(2211-1(3?-0)
(-1)n2
n
1,\ n41 \
Write down the potential for the region inside. 89. Show, by the method used in 5.274, that if the density of electric charge on a prolate spheroid n = no is Gr(t, 95) then the potential inside is
V =ZZM „,„F,;(n)P7,(t) cos m(4,
—
Om)
n=0 7n=0
where M.t. is (
-
1)m(2 — 47reC2
(2n + 1)[ (n — m)12
+1 /21-
PN t) cos m(4
Q"'(no) f
(n
m)!
—1 0
—
O„,)hih3 dt
234
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
where hiha are given by 5.28 (6) and (7). Write the analogous formula for the potential outside. 90. Using the results of the last problem, show that the potential due to a point charge q at Ea, no, 00 is given by
4n-qcC2
I TP',;( t)(2,7(n)PZ(no)P,7((to) cos m(95 — tfro)
(2 — 5?„)( 1)'n(2n
(n —m)I (n +m)
n=0m=0
when n > no. Write out the result when n < no. 91. Starting with the potential inside an earthed cylindrical box of radius a due to a point charge q on the axis and using Green's reciprocation theorem, find the potential on the axis of a ring of radius b coaxial with and inside the box. Hence, show that the potential anywhere inside the box due to this ring is
q2 Irea
sinh µ kz sinh ihk(L — c)Jo(w.b)Jo(iikP) sinh µ k L p.k[Ji(p.ka)] 2
k=1
where z < c and the coordinates of the bottom and top of the box and the plane of the ring are, respectively, z = 0, z = L, and z = c. Take Mk so that Jo(Aka) = 0. Show by 3.08 (2) that the force toward the bottom of the box is q2
21rea2
k =1
sinh ihk(L — 2c)[Jo(Akb)] 2 J 1(Aka) sinh I.Lk L
92. The walls of a conducting box are given by z = ±c, p = a. The halves above and below the plane z = 0 are insulated from each other and charged to potentials -I- Voand — Vo, respectively. Show that the potential anywhere inside is given by V = +V 0{1
2 a
sinh [Atk(c— 1z1)]Jo(AkP)} µk sinh AtkcJi(Aka)
k =1
where Jo(µka) = 0 and the sign is that of z, and also by v = v o[z + 2 xk /o(nirp/c) sin (nirz)] I../ n/o(nra/c) c c n=1
93. The walls of a conducting box, which are given by z = ±c, p = a, are earthed with the exception of two disk-shaped areas at the top and bottom bounded by p = b, which are charged to potentials + Voand — Vo, respectively. Show that the potential inside is given by 2bV0x.k sinh 1(Ihkb)J 0(1.4kP) a2 ;Lk sinh tikc[J1(Aika)] 2 k=1
where Jo(Aka) = 0, and also by 2bV0 ,--■ c
(
n=1
1
Ko(mra/c)10(nrp/c) — .10(thra/c)Ko(nrp/c) (nr . Torz —b sin —
)'
lo(nra /c)
235
PROBLEMS when p > b and when p < b by Vo — [z 2b
(-1)'
K o(n7ra/c)Ii(nlrb/c)
/o (nra/c)Ki(mrb/c) (or
Io — p
/0(Thra/C
n=1
Torz]
sin — c
94. A semi-infinite conducting cylindrical shell of radius a is closed at one end by a plane conducting plate normal to its axis and at the same potential. Show that the image force on a charge q placed on the axis at a distance b from the plate is c's
q2 22rea 2
k= 1
e-Akb
2
where .1-o(Pka) = 0
[ Ji(Aka)
95. The infinite conducting cylinder p = a is filled below the z = 0 plane with a dielectric of relative capacitivity K. There is a charge q on the axis at z = b. Show that the potential above the dielectric is K
e-Akk-b1 271-c,,a 2
K
k=1
—
1
Jo(Ahp)
le-mk(z+"AdJi(1.4ka)J2
where Jo(mka) = 0. Find the potential in the dielectric. 96. The surface of a conductor is generated by rotating a circle of diameter a about one of its tangents. Show that the capacitance of the conductor is
87rea Z [J I(Aka)]-1.10. e -Ak..l.h dct, = 8rea k=1
80, 0 (pka)[JIGika)]-1
k=1
where .1-o(Aka) = 0 and So.o(uka) is a Lommel function. 97. The conductor in the last problem is placed, uncharged, in a uniform field E parallel to its axis of rotational symmetry. Show that, in polar coordinates, the potential at any point is e 2Ea2 7 1 =k
8 j0( Ak
a 2r
1 sin 0)
coo
(Aka)1 2
where Jo(uka) = 0. 98. The conductor in the last problem is rotated to make E normal to its axis. Show that, if J o(pka) = 0, the potential at any point is 2Ea2 r
e Aka
k=1
..Ji(Aika 2r-1sin 0) [J2(ihka)] 2
COO v
COS cb
99. Show that the potential due to a ring of radius a carrying a charge Q is Q — f K o(ka)I0(kp) cos kz dk
if p < a
Q Io(ka)Ko(kp) cos kz dk 272e o
if p > a
272€ 0 or
236
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
100. A ring of radius a carrying a charge Q is coaxial with an infinite dielectric cylinder of radius b. Show that the potential in the dielectric is Q
2 -„ 7 0NY(k)/0(kp)K o(ka)
cos kz dk
where 4,(k) = [1 kb(K - 1)0kb)Ko(kb)]-1• 101. The earthed toroidal ring generated by rotating a circle of radius a centered at p = b about the coplanar z-axis lies halfway between the plane z = c at potential Ec and the plane z = -c at -Ec. Observe that ring multipoles at p = b, z = 0, which give fields normal to the planes and zero potential halfway between them, have potentials a"Vi /aer where V is the potential of a ring dipole of strength M given by
2,c2
nit
nrb
nrb) (mrp) mrz
r
-
nr0(7)K0(1rP) sin 717z
sin — or
n=1
n=1
according as p < b or p > b. Hence show that a potential function that gives V = 0 b) 2 = a 2 is on the ring at 2m lines p = p,, z = +z, where 0 < s < m and (p, -
V = Ez -
(A in + A 2n 3 +
mrb) (nir . mrz / 0 -- sin c c c
+ A.„,n2'n+2 )K0(
• • •
72=1 This holds if p < b.
A, = E
If p > b, interchange p and b.
VII
•
"
V-r-1,1Z1Vr+1,1
"
V12
•
•
V,-1,2Z2Vr+1,2
•
•
•
The coefficient A, is given by
•
Vml
V11 V21
•
•
Vm2
V12 V22
'
" •
VInl
V m2
V1„,1 2,o • • • V„.„,
Ps
0 and r > a the potential is given by —
C(rm+1 - a'"'-1-3r-m-2) cos 0 sinm 0 cos m4) where m is the smallest number for which cos ma = 0.
238
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
112. A charged conducting body has a deep rectangular hole whose boundaries are given by x = 0, x = a, y = 0, y = b, and z = 0. Show that, far from the opening, the potential is given by b2 P
V = C sinsin sinh (a2 b a
ab
113. The walls of an earthed rectangular conducting tube of infinite length are given by x = 0, x = a and y = 0, y = b. A point charge is placed at x = xo, y = yo, and z = zo inside it. Show that the potential is given by (9nzaz TrE
zbe)1.-lz - z01 ab
(m2a2 n2b2)le
V=
sin
mix°
n=1 m =1
a
sin
nirx a
sin
mryo b
miry
sin — b
114. The walls of an earthed conducting box are given by x = 0, x = a, y = 0, y = b, and z = 0, z = c. A point charge q is placed at xo, yo, zo. Show that the potential inside the box is given by miry sinh A„,„(c - 20) sinh A mnz nrxo nrx miry° sin sin sin sin a A„,„ sinh A „„tc a
4q V =— eab
n = 1 m =1
(m2a2
n2b2)17,.
where A,„„ -
and z < z0. If z > z0, we interchange z and z0. Show
ab
that the z-component of the force on the charge is
F, =
eab
nrxo miry° csch A„,„c sinh Amn(c - 2z0) sine a sine n=1 m=1
For F„, substitute c for a, a for c, x0 for z0 and z0 for x0 in F, and Am,,. For F y, substitute c for b, b for c, Yo for z0 and z0 for yo in F, and 115. Show that the potential of a dipole M parallel to the 0-axis at b, a 0 is ,
M
n
4reb 2 4./ n=0 m =0
when r < b.
(2 (51)(m
-
n
If r > b, write (m
1) (n - OnFn÷, (cos a)P„m(cos 0) cos m4' (n m) b n)(b/r)n+i l,"_1(cos a) for
(m - n - 1)(1)P"'+1(cos a)
b
116. Show that the torque exerted by a sphere of radius a and relative capacitivity
K at the origin on a dipole M at r is, if the angle between M and r is a, T = 111(r sin a)-1C,,,,,[(n
, (cos a) + 1)P;
-
n - 1) cos a/37,, (cos a)]
where C„,n is a summation operator given by
C„,„
(K -1)M 4re,,r2
n+1
z (2
n =0 m
0° )a2n-1-2 (m - n - 1)(n - m) !n /7+,(cos a) m)! n 1)94"+2(n
-
(nK
239
PROBLEMS Show that the radial force is
F = — Mr-2(n + 1)C„,,,[cos a PZ(cos a) + (n — m
1)137_0 (cos a)]
117. Rotate the transformation of Art. 4.13 about the x-axis letting ui = U and u2 = V and set up 5.11 (6) for this case. Substitute (cosh /hi — cos u2)1Ui(ui) U2(22) for U in 5.11 (6) and show that the differential equations for U2 and U1 are d 2U2
du; — —n
d2 U2
2U2,
du?
coth uidUl
(n2 1 4
duo
m2 ) Ul = 0
sinh 2 ui
The latter is identical with 5.23 (3) if n — is written for n and cosh u1 for z. 118. The torus generated by the circle of radius b given by ui = uo in the last problem is charged to a potential Vo. Show that the potential outside it is 2Vo(cosh u1— cos u2)2 n=0
1)Pn_i(coth uo )Pn _1(cosh ui)cos nu2 ( —2), (2n (2n + 1)!!(sinh u0)1P.-1(cosh u0)
where cosh uo is c/b, c being the distance from the center of uo to the line ul = 0. 119. Find the total charge on the torus of the last problem by comparing the potential at u1 = 0, u2 = 0 with q/(42ror) and thus show that its capacitance is 871-€21b(sinh uo)1 n =o
1)Pn_1(coth uo) (2n + 1)!!P,,_1(cosh uo)
( —2)'(2n
120. The torus of the last problem lies in a uniform field E parallel to its axis of rotational symmetry. Show that the potential of the field is
V = Ex + (cosh ul— cos u2)1
A „P„_1(cosh ui) sin nu2
n=1
where, if a is defined by 4.13 (3), U = uo,
A,, —
4(2n + 1) ( —2)nnaEoPn_1(coth uo) (2n + 1)!!(sinhuo)1P,.-1(cosh uo)
121. The torus of problem 118 is placed in a uniform electric field E normal to its axis of rotational symmetry. Show that the potential outside it is 00
E cos 514/3+ (cosh Ul — cos U2)1 Z , A.Pnl_i(cosh ui) cos nu2] n=0
where
An =
(2n — 1)(2n + 1)( —2)n+lb(sinh uo)1Pn1(coth uo) (2n + 1)!!P;;_1(cosh uo)
122. Rotate the transformation of 4.13 about the y-axis letting u1 = U and u2 = V, and set up 5.11 (6) for this case. Substitute (cosh ui— cos u2)4/1(ui)U2(u2) for U in 5.11 (6) and show that the differential equations for U2 and Ui are d'Ui = (u + 1 2 ui, duo 2
d 2U2 dU2 du; ± cot u2 ±[n(n + 1) du2
The latter is identical with 5.14 (2) with u2 in place of O.
m2
sin2 u2
]U2 = 0
240
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
123. The two spheres u1 = uo and ui = —uo of radius b generated in the last problem are aligned with a uniform field of potential Ez so that their potentials are Vo and — Vo, respectively. Show that the potential outside them is CO
Ez + (cosh u1— cos u2)1
A n sinh (n DuiPn(cos u2 ) =0
A n = 21[V0 — (2n + 1)Eb sinh uo](e(2n"uo — 1)-1 where cosh uo is c/b, c being the distance between the centers of the spheres. 124. If the spheres in the last problem are uncharged, show that 2n + 1 e(2.1-1),40 — 1 [
V 0 = Eb sinh uo n=0
1 n=0
e(2'+1)u,
—
1
125. Show that the force between the spheres in the last problem is
F = 7c
— (n + 1)A”+1 — 21aE]
A n[(n +
n=0
where An is found by substitution of Vo from problem 124 into A n in problem 123. 126. The potential on the surface of a sphere of radius a due to external sources is f(0), where 0 is the polar angle. Show by expanding as in 5.156 and summing the harmonic series under the integral by comparison with 5.16 that, when r < a, 1 ,r
- -
Cr
ar/ ad¢ rkf(a) sin a da d V(r, 0) = -7 ar _„ sin a sin 0 cos 014 [a 2 +r 2— 2ar(cos a cos 0
Jo
127. Solve the problem of a dipole normal to and halfway between two parallel earthed conducting planes. Now by inversion show that if two conducting spheres of radius a are in contact in a uniform electric field E parallel to their line of centers, the charge density is
cr
=
271- 2EE
sec' a Z(-1)"n2K0 (nr tan a) n=1
where a is measured about the point of contact from a diameter passing through this point. 128. The surface of an oblate spheroid is maintained at a potential
V=
BnP2n(t) Q2a0) n =0
Inside it a concentric sphere of radius a is at zero potential. Show that in the region between the two the potential is N V
n
= n=0s=-0
[02. 028
2a+1
(-1)nA nC,„()
a
r
a
a
—
+1 ]P2.(COS 0)
PROBLEMS
241
where A nis given by 5.41 (5) with the following values of C„, s
n
c8'
(4n — 1)!!(ar iQ2n(iro) (2n)! \c P2.(i3"(1)
(-1)"(2s)! 2s + 1)!!(2s — 2n)!!
j(2n
8
BnC3n.
and with the right side given by n=0
129. The potential V of a concentric sphere of radius r is maintained at the value BnP2n (COS 0) n=0
Inside it is an oblate spheroid 3 = 3-0at potential zero with an interfocal distance 2c. Show that the potential between spheroid and concentric sphere is N
n
V =
A.C3.[ P23(j! °)Q23(i0
P23(j3)
Q2.0r0)
n=03=0
where An is found from 5.41 (5) with s 5
— I ) n (48 + 1)(2n)!
P2g(ko)
(2n + 2s + 1)!!(2n — 2s)!! Q2,4 ji-0)
a
= (-1)8
C`,, =
(
(T+Ij(2n
+1)(4n — 1)!!
c
(2n)!
(c)2 n (2n)! P2.(ko) ] a (4n — 1)!! Q2n(73'o)
(ay-EV( —1)'(2s + 1)(2n + 2s — 1)!! c
(2n)!(2s — 2n)!! 8
and the right side of 5.41 (5) is
C.nBn. n=0
130. In the last problem if the sphere is at potential V0, then Bo = Vo and Bn = 0 if n 0. If the concentric spheroid inside it is a disk of radius c, then •o = 0, so P2.(i3-0)/Q23(i3-0) is 2j/7r by 5.214 (3). Show with the aid of 5.27 (6), 5.214 (5), and 5.214 (3) that the charge density on the disk is N
2€ —
prct
n
r r (-1)'(2s)!!A.C..P2,(E)
n=03=0
(2s — 1)!!
131. The surface of an oblate spheroid is maintained at a potential
V
=
ZBnP2n(t) n=0
242
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
Outside it a concentric sphere of radius a is at zero potential. Show that in the region between the two the potential is N
a
V= n=0 s=n
'
1
1P2,(cos 0)
A„C,,„[0 2' — a
where A nis found by solving 5.41 (5) with the following values of C..: s < n,
Can =
C„,„ s > n,
Can
( — 1)n(2n + 2s (2s)!(2n
— I)!! 2s)!!
(2n)!(c/a)2n+' P2n( jr0) [ (4n — 1)!! (2n.) ! 3(4n + 1)!! ( —1)'(2s)!j(c/a)2°+' P2.(sii-o) (2n + 2s + 1)!!(2s — 2n)!! (22.(31•o)
( —
=
—
1)n
and with the right side given by n=s
132. The upper and lower halves of a conducting sphere of radius a are insulated
and charged to potentials + Vo and —V0 . It is surrounded by a concentric coaxial oblate spheroid of interfocal distance 2c at potential zero. Without the spheroid the external potential of the sphere is a)2n-I-2
=
Bn(
P2n+1(COS
(4n + 3)(2n (2n + 1)(2n
B„ — ( 1)'
f
n=0
1)!! 2)!!
Show that with the spheroid in place the potential between them is P2s (i A.G.{
V2 =
n=08=n+1
0
)P2.(ir)
Qu(N-o)
—
(22.(ii-432(t)
where Anis found by solving 5.41 (5) and Can is given by s n + 4,
c.n =
(a)2n+1 ( —1)nj(4s + 1)(2n + 1)! (2n + 2s + 2)!!(2n — 2s + 1)!! c
2n + 2 c 2n+2 Q2s (jr0) (2n + 1)! a Pu(ii-o) ( —1)°(2s +1)(2n + 2s)!! (c)2'+2 (22.W-o) C.. = . P2.(71- 0) 3(2n ± 1)!(2s — 2n — 1)!! a
On (4n ± 1)!![ j02ft+1 c
Note that s is an integer plus zso right side of 5.41 (5) involves Co1„, C1.5n, etc. 133. An uncharged conducting sphere of radius a lies at a distance b from an infinite line charge of uniform density q where b > a. Take the origin at the center of the sphere, and write the potential of the line charge in circular harmonics by 4.02 (2) using gs for 0, p for r, On = 0, and V = 0 when p = 0. Express this in spherical coordinates by writing r sin 8 for p, so it has the form Zr.S.(0,0) where n 0. Add to this the induced charge potential —M(a 2n+,/rn+98.(0,0), which cancels it at r = a. Sum
243
PROBLEMS
by 4.02 and show that the total resultant potential is 192(19 2r 2 +a' sin' 9 — 2a2br sin 8 cos 0)“fr V = q In (br) 2afr(b 2 r 2sine B — 2br sin B cos 0)
Show that the attractive force on the sphere is F=
q2a2 sin sm-' re,,b(b 2 — a 2)
134. If the conducting sphere in the preceding problem is replaced by a dielectric one e, show that the total potential outside it is (€ — e0a2n-f-1
Vo =
2re,,
n
n=1
[no
(n
sin" 0
1)€,]rn+2
bn
cos no
and write down the potential inside. Show that the force on the sphere is F=
— q2
(2n
—
2)!!n(e — e„)
LI (2n — 1)!![ne
n=1
(n
(ay-E1
1)e,] b
135. A sphere of radius a at potential Vo lies at the origin between two infinite parallel earthed planes spaced a distance b apart, and its center is a distance c from the nearest plane. This is one section of the field produced by an infinite set of pairs of V o and — V o spheres at a distance 2c apart similarly aligned along the z-axis with pair centers spaced at intervals 2b. The potential between sphere and planes is
V=
Z
(ap;Ov+IPp(cos On)
CP
p=0 n=0
where rn is the distance from the field point to the center of the nth sphere and 8„ is the angle at the center of the nth sphere between r„ and the z-axis. With the aid of 5.44(2) and 5.298 (6) write down a'V/az' on the axis for all pairs except the pair that includes the origin. Sum this for all but the center pair. Add to the sum the a3v /az. term inside the sphere at z = 0 due to it and its nearest image. Thus as in Art. 5.41 show that the P equations, one for each value of s from 0 to P, to be solved for C2, are, after dividing out s!ci-s, +.+1 3 08 V 0
= P=0
f(v) =
CP[ a"
1 ao-,4-1 (1 2 avP+'+' v2
;19 r cot nv)
a P+*+1 f C (2b) (4b -) [1 — ( -1)9+13-(P 4- s + 1)
where ;In) is the Riemann zeta function tabulated in HMF, page 811. Note that as b 00 only the first two terms in the bracket survive, so the result applies to a single sphere at a distance c from a plane. The capacitance is 47reaCo/Vo.
136. A sphere of radius a at potential Vo lies halfway between two earthed parallel planes at a distance 2b apart. Observe that the potential in the region 0 < r < b
244
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
which is Vo on the sphere is
a)2n+1 V = Vo -1-Ze n =0
r
02n —
iP2„(cos 8)
a
where the (r/a) 2" term arises from the induced charge on the planes. Expand V at z = b when p is small in the form ZD,p2., so that there are N simultaneous equations D, = —Vohs, one for each value of s from 0 to N — 1, to be solved for C.. With the aid of MO, page 50, 5.298 (6), and HMF 22.10.12 show that
D„ —
(-1)'
[(2s)!112
N [ x-A (2n + 2s) !a2"+' n= 0
(2n) !b2"÷2'+'
Cn n
(2n) !b2"-". (2n — 2s) !a2"
Show that the capacitance is 4ireCoa/Vo. 137. Obtain an expression for the potential in the last problem when r > b by writing down the Green's function V i(p,z) for a charge between two planes by 5.324 (10), shifting the origin to halfway between the planes, taking a 1/"/az2" as r —> 0, and comparing this with 82"(1/r)/az2" in MO, page 50, as r 0. Thus N 21-2n (a)2n-l-l
V— V o n=0
(2n)! b
)2n1(.0
en s=0
[(2S ± 1)7 p]
2b
cos
(2s
± 1)7z 2b
Note that this expression takes the indeterminate form 0 In p as p —> 0 except at z = 0 where it is infinite as it should be. 138. The potential at a spherical-surface of radius c due to a concentric cylinder inside it having the charge density of 5.39 (1) is given by 5.39 (4) when r = c. If the spherical surface conducts, its charge density must cancel this potential and give a potential V1 when r < c like 5.39 (4) when r-( 2"+2P+0is replaced by —r2n-1-2p/c4n+4p-1-1. Show that the coefficients an and bn can be found by solving equations 5.39 (2) and 5.39 (3) provided V V 1is used in place of V. Thus show that the capacitance is still found from 5.39 (4) with s = p = 0 but with the new C, and C, values.
References The following references will be found useful in connection with this chapter. A. ELECTRICAL BUCHHOLZ, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Elaborate discussions of spherical caps, toroids, and many other cases. DURAND, E.: "Electrostatique et Magnetostatique," Vol. I, 1953; Vol. II, 1966, Masson et Cie. Well-illustrated treatment of solutions in separable coordinate systems and of graphical and numerical methods. FLUGGE, S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. G. Wendt gives a clear discussion of three-dimensional potential problems, including numerical and graphical methods. GEIGER-SCHEEL.: "Handbuch der Physik," Vol. XII, Berlin, 1927. GRAY, A.: "Absolute Measurements in Electricity and Magnetism," Vol. I, Macmillan, 1888. Extended treatment of images and inversion. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extended treatment but omits Bessel functions.
PROBLEMS
245
J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Treats images, inversion, confocal surfaces, and spherical harmonics. MORSE, P. M., and H. FESHBACH: "Methods of Theoretical Physics," 2 vols., McGrawHill, 1953. Discusses the functions and methods of this chapter and tabulates solutions of Laplace's equation in separable coordinate systems. PLANCK, M. K. E. L.: "Theory of Electricity and Magnetism," Macmillan, 1932. Gives excellent treatment of confocal coordinates. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives considerable material on potential theory. THOMSON, W.: "Papers on Electrostatics and Magnetism," Macmillan, 1884. Solves many problems and gives numerical tables on spheres and spherical segments. THOMSON, W., and P. G. TAIT: "Treatise on Natural Philosophy," Cambridge, 1912. Treats spherical harmonics. VAN BLADEL, J.: "Electromagnetic Fields," McGraw-Hill, 1964. Treats approximately dielectric cubes and similar problems not in other books. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and literature references. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. X, Leipzig, 1930. MAXWELL,
B.
MATHEMATICAL
H.: "Electrical and Optical Wave Motion," Cambridge, 1915. Applies the functions of this chapter. BATEMAN, H.: "Partial Differential Equations," Dover, 1944. Important applications of toroidal and other special coordinate systems. BRITISH ASSOCIATION: "Bessel Functions, Part I," Cambridge, 1937. Extensive numerical tables of Jo, Ji, Yo, Y1, Jo, I,, K o, and K,. BRITISH ASSOCIATION: "Bessel Functions, Part II," Cambridge, 1952. Numerical tables of J,,, Y,,, I,,, K„ for 2 < n < 20. CARSLAW, H. S.: "Mathematical Theory of the Conduction of Heat in Solids," Macmillan, 1921. Bessel functions and spherical harmonies are applied to problems in heat conduction having boundary conditions identical with electrical problems. COPSON, E. F.: "Theory of Functions," Oxford, 1935. ERDELYI, A., W. MAGNUS, F. OBERHETTINGER, and F. G. TRICOMI: "Higher Transcendental Functions," 3 vols., McGraw-Hill, 1953. Very complete and concise descriptions of functions with formulas. (HTF) ERDELYI, A., et al.: "Integral Transforms," 2 vols. McGraw-Hill, 1953. A large collection of all types of integral transforms. (IT) FLETCHER, A., J. C. P. MILLER, L. ROSENHEAD, and L. J. COMRIE: "An Index of Mathematical Tables," 2 vols., Blackwell, Oxford, 1962. GEIGER-SCHEEL: "Handbuch der Physik," Vol. III, Berlin, 1928. GRAY, A., G. B. MATHEWS, and F. M. MACROBERT: "Bessel Functions," Macmillan, 1922. Includes physical applications, numerical tables, and problems, as well as mathematical developments. HARVARD UNIVERSITY COMPUTATION LABORATORY: "Tables of Bessel Functions of the First Kind of Orders 0 to 78," 11 vols., Harvard, 1946 to 1950. HossoN, E. W.: "Spherical and Ellipsoidal Harmonics," Chelsea, 1955. Includes almost everything known about the subject. JAHNKE, E., F. EMDE, and F. Loser': "Tables of Higher Functions," McGraw-Hill, 1960. Formulas and numerical tables with outstanding illustrations.
BATEMAN,
246
THREE-DIMENSIONAL POTENTIAL DISTRIBUTIONS
F. M.: "Spherical Harmonics," Dutton, 1927. Includes physical applications and a chapter on Bessel functions. MAGNUS, W., and F. OBERHETTINGER: "Formulas and Theorems for the Special Functions of Mathematical Physics," Chelsea, 1949. A fine collection of formulas for the functions of this chapter. (MO) MCLACHLAN, N. W.: "Bessel Functions for Engineers," Oxford, 1934. Many applications. MORGAN, S. P.: "Table of Bessel Functions of Imaginary Order and Imaginary Argument," California Institute of Technology, Pasadena, 1941. NATIONAL BUREAU OF STANDARDS: "Handbook of Mathematical Functions," U.S. Bureau of Commerce, 1964. The best collection of formulas and numerical tables for the functions of this chapter. (HMF) RIEMANN-WEBER: "Differentialgleichungen der Physik," Vieweg, Braunsweig, 1925. RYSHIK, I. M., and I. S. GRADSTEIN: "Tafeln," Deutscher Verlag der Wissenschaften, 1957. A large collection of useful formulas. WATSON, G. N.: "Theory of Bessel Functions," Cambridge, 1922. The standard treatise on the subject. It includes some numerical tables. WEBSTER, A. G.: "Partial Differential Equations of Mathematical Physics," Hafner, 1950. WHITTAKER, E. T., and G. N. WATSON: "Modern Analysis," Cambridge, 1935. MACROBERT,
CHAPTER VI ELECTRIC CURRENT 6.00. Electric Current Density. Equation of Continuity.—If we touch two conductors A and B, charged to potentials VA and VB, respectively, to two points in a third conducting body, we have seen in 1.00 that electric charge flows from one to the other until the potential of A equals that of B. We observe two phenomena associated with this transfer; one is the heating of the conductor and the other is the existence of a magnetic field in the neighborhood, while the flow is taking place. The second phenomenon will be considered in the next chapter. The rate of transfer of electric charge between A and B at any instant is called the electric current, which is therefore defined, in any system of units, by
I = dQ di
(1)
When Q is measured in coulombs and t in seconds, I is measured in amperes. If, by some mechanical electrostatic means, such as a moving insulated belt, we transfer, continuously, a charge from the point of contact of B to that of A at such a rate that the difference of potential VA — VB is kept constant and at the same time cool the conductor so that its temperature has a fixed value, we find that the current and the magnetic field remain constant. The latter, therefore, need not enter into our considerations of steady currents. Electric current, at any point, is evidently a directed quantity. If, at the point P in a conducting medium, we take an element of area dS, which is normal to the direction of the electric current at P, and if the current flowing through this element of area is dI, then we define the current density at P to be
.
dl = dS
(2)
When a steady current flows, the amount of electricity entering any element of volume must equal that leaving it. Thus, the surface integral of the normal component of the current density over this elementary volume is zero. This gives, by Gauss's theorem, 3.00 (2),
fsi • n dS fvV • i dv = 0 Since this holds for all volume elements V • i = div = 0 247
(3)
248
ELECTRIC CURRENT
§6.01
This is known as the equation of continuity, and a vector which satisfies this equation everywhere is said to be solenoidal. 6.01. Electromotance.—In the experiment considered, the belt is an agent, known as an electromotance, which exerts a mechanical force on the electric charges carried by it just sufficient to overcome the electrostatic forces between A and B. The work in joules which the belt does in moving 1 coulomb of positive charge from B to A, after eliminating friction losses in the driving mechanism, etc., measures the magnitude, 6', of the electromotance in volts. If the force driving the belt is increased, it will accelerate until the additional charge conveyed increases VA — VB so that the electrostatic forces on the charged belt just balance the driving force. If the contact between A and B is broken, the belt stops, since it cannot drive against increased electrostatic forces, and work against belt friction ceases, so that the electromotance e exactly equals VA - VB. In order to maintain an electric current, at ordinary temperatures, it is necessary to have some source of electromotance in the circuit. In the present case, the location of the electromotance is perfectly clear. It is distributed along the belt between A and B. In chemical sources of electromotance, such as storage batteries and primary cells, it is located at the surfaces of the electrodes. In thermocouples, it lies in the interface between the members. When a whole metallic circuit is placed in a varying magnetic field, the induced electromotance, discussed in Chap. VIII, may be distributed over all the elements of the circuit. When a dynamo supplies power to a circuit, the electromotance is spread over its coils. In all these cases, if the circuit is opened and the difference of potential across the opening is measured by some device drawing no current, such as an electrostatic instrument or a potentiometer which is screened from induced effects, the result equals the sum of the electromotances around the circuit. We usually define the electromotance around any given path in a different way. We defined the electrostatic field intensity at a point as the force acting on a stationary unit positive charge placed there. The nature of the force was not specified, except that it was taken proportional to the charge. As we shall see in subsequent sections, this measurement is not affected by the electrical resistance of the medium, as the latter, like viscous friction, acts on moving bodies only. Thus, if we examine the force acting on a charge on the belt, we find it to be zero, since the mechanical and electrostatic forces balance. In the part of the circuit where the current is flowing, only the electrostatic forces are present. Thus the line integral of E around the circuit just equals the potential difference between A and B, which, as we have seen, equals 6'. Thus we find 6 = j'E • ds (1)
OHM'S LAW RESISTIVITY
§6.02
249
which defines the electromotance around the path of integration. It is sometimes instructive to consider the electric field intensity E, as made up of two parts, viz., the electrostatic part E', which can be derived from a scalar potential, and for which the line integral around any closed path vanishes; and a part E" due to the electromotance, which is solenoidal in nature and does not necessarily vanish when integrated around a closed path. The intensity E" helps to visualize the case of a distributed electromotance but is not so satisfactory when the electromotance is located in a surface layer, since, in the latter case, it is infinite in these layers and zero everywhere else. The value of the integral of E" will be different around different paths. This is illustrated, in the example given, by the fact that if the path is chosen entirely on the belt or entirely off it the integral is zero. In many cases, it is possible to place barriers in the path of integration so as to exclude the sources of electromotance, such as the belt in the example given. The line integral of E over all permitted closed paths is then zero so that we can use a scalar potential throughout this region. Even when the electromotance is distributed, a barrier may exist such that, for permitted paths, the integral of (1) is zero. In this case, we are unable to distinguish between E' and E" and can use the terms electromotance and potential interchangeably. This is the sense in which E appears in the next article except in Eq. (4). We may write, in this case, E = —VV = Ve (2) 6.02. Ohm's Law. Resistivity.—If, in the experiment of 6.00, all physical conditions, such as temperature and working efficiency, are kept constant and the electromotance is increased we find that when equilibrium is reached the current has increased in the same ratio. This is Ohm's law. For a perfectly efficient machine the ratio of the electromotance between A and B to the current flowing is called the electrical resistance RAB between these two points, and so we have RAB =
VA V )3
e AB
[ AB
[ AB
(1)
Let us consider an elementary cylinder at the point P in a conducting medium whose ends, of area dS, are normal to the direction of the current at P and whose walls, of length ds, are parallel to it. The electromotance between the ends is then OE as) ds = VE • ds, and the current flowing through them is i dS. Assuming that Ohm's law applies to this cylinder, we may write (1) in the form VE • ds Ivel cos a ds RP =
where
a
.
s dS
i dS
is the angle between VC and ds.
If we take ds numerically
250
§6.04
ELECTRIC CURRENT
equal to dS, we write T for Rp and call it the resistivity or specific resistance of the medium at the point P. Thus
Ivel
(2)
=. COS CC 7,
If T is to be independent of the direction of the current, then a = 0 so that the potential gradient and the current must be in the same direction. In this case, we say the conductor is isotropic. The conductivity y is defined as the reciprocal of the resistivity T, so that for an isotropic medium (2) becomes ye i= = leYre (3) T
For a closed path in a conducting medium, we have from (3) and 6.01 (1) .95'i • ds = -ye
(4)
6.03. Heating Effect of Electric Current.—We mentioned in 6.00 that the passage of an electric current heats a conductor. To keep the temperature of the conductor constant, we must remove this heat by some cooling device. If no changes are taking place in the conductor, then the energy which we thus obtain must enter the system in some way. Evidently, we do mechanical work against electrostatic forces in carrying the charges from B to A on our insulating belt. From the definition of potential, this work is W = Q( VA — Vs), so that the power supplied is P dW =
=
, dQ A — v B)(— )= dt
V Wks
(1)
Substituting from 6.02 (1), we have P = PABRAB
(2)
If I, V, and R are all in esu or if they are all in absolute emu, then P is in ergs per second. If I is in amperes, V in volts, and R in ohms, which is the practical mks system of units, then P is in watts. A complete table of these units appears in the Appendix. The heat that is generated in this manner is called Joule heat. 6.04. Steady Currents in Extended Mediums.—In 6.00 (3), we showed that if, in the steady state, no charge is to accumulate at any point in a conducting medium, the divergence of the current density i must be zero at all points. In 6.02 (3), we found that Ohm's law, in an isotropic medium, requires that i be proportional to the potential gradient and inversely proportional to the resistivity. Combining these two equa-
§6.04
STEADY CURRENTS IN EXTENDED MEDIUMS
251
tions, we have V • i = — V • T-1V V = 0
(1)
If the medium is homogeneous, T is constant throughout it, so that this becomes (2) vzV = 0 Comparing (1) and (2) with 3.02 (2) and (3), we see that these equations are identical with Laplace's equation and that the reciprocal of T, called the conductivity, plays exactly the same role as the capacitivity does in electrostatics. It follows that all mathematical technique used in electrostatics also applies here. The tubes of flow here bear the same relation to the equipotential surfaces as the tubes of force do in electrostatics. At the boundary between two conducting mediums, both the potential and the normal component of the current density must be continuous so that, from 1.17 (5) and (6), the conditions are
1 ay' = 1 av" T" an
(3)
T' an
(4)
= V"
These two equations determine completely the relations between the potential gradients on the two sides of the boundary, so that if the relative capacitivities of the two mediums also differ and are not proportional to the conductivities, another variable is required to satisfy this additional condition. By taking a charge density o- on the boundary, this can be done, giving, by (1),
av1— K2aV 2 = _(Kin an an
o- = K1 Eu
K2T2)in
(5)
Anyone observing the magnetic forces between conductors carrying currents, which is the subject of the next chapter, might be led to believe that the current distributions computed by solving (1) or (2) would be erroneous owing to the mutual displacement of current elements by magnetic interaction. Such is not the case. It is true that, in isotropic nonferromagnetic conductors, there is always an increase of resistance in a magnetic field, sometimes known as the "longitudinal Hall effect." The resistance of those portions of an extended conductor where the magnetic field is most intense is therefore increased relative to that of other portions, producing a decrease in the relative current density. Thus when a steady current flows in a cylindrical conductor, the current density is slightly larger near the axis. If this conductor is placed in an external uniform transverse magnetic field, which adds to the field of the current on one side of the cylinder and subtracts from it on the
252
ELECTRIC CURRENT
§6.05
other, we obtain a lateral displacement of the current in the same direction as the force on the conductor. This does not change the orientation of the equipotential surfaces and does not imply any transverse force on the current carriers. Although the change of resistance can be measured, the resultant current displacement is too small to observe at ordinary temperatures. The usual transverse Hall effect changes the orientation of the equipotential surfaces in either direction, depending on the material, without, in this example, affecting the current distribution. It is possible, however, to arrange a circuit so that the latter is changed. This effect can be observed only by using very strong magnetic fields and sensitive instruments. Thus, we may conclude that the distortion of the current distribution in extended conductors caused by magnetic interactions is negligible, and the results obtained by fitting solutions of (1) and (2) to the given boundary conditions may be considered rigorous, as far as such effects are concerned. Since, as we have seen, heat is generated by the passage of electric current and since, in general, the resistivity depends on the temperature, there will actually be deviations from the solutions obtained where heavy currents heat those portions of the conductor where the current density is highest. The solutions are therefore strictly true only when the temperature coefficient of resistance of the medium is small, when the current density is small, or when the current has been circulating only a short time. 6.05. General Theorems.—The general theorems derived from Green's theorem in Chap. III may now be restated in a form suitable for the present subject. 1. If the value of the potential V is given over all boundaries of a conductor as well as the size and location of all sources or sinks of current inside, then the value of V is uniquely determined at all points in the conductor. 2. If the normal component of the current density is given over all boundaries of a conductor as well as the size and location of all sources or sinks of current inside, then the value of the potential difference between any two points in the conductor is known. 3. If the resistivity of any element in a conductor is increased, then the resistance of the whole conductor will be increased or remain unaltered. 4. If the resistivity of any element in a conductor is decreased, then the resistance of the whole conductor will be decreased or remain unaltered. To these we may add another theorem which we shall then prove. This is 5. The current density in a conductor distributes itself in such a way that the generation of heat is a minimum.
§6.06
253
CURRENT FLOW IN TWO DIMENSIONS
To prove this, let us suppose that there is a deviation from the distribution given by Ohm's law [6.02 (3)], the additional current density being j. In order that there may be no accumulation of electricity, j has to satisfy the equation of continuity V • j = 0. Applying 6.03 to an element of a tube of flow in the conductor, we have, for the heat generated in it, from 6.02 (3) and 6.03 (2),
dP
=R— --11717.
dS
s =T dS N_ d
vv
j) 2 dv
where dS is the cross section of the elementary tube and ds is its length. Integrating gives P
1 (V V) 2 — 2j • VV TP]dv = fV[r
(1)
Applying Green's theorem [3.06 (3)] to the second term, where j = Vc1) and NI/ = V, we have fv j • V V dv = — f VV • j dv
f Vn • j dS
The first term on the right is zero since V • j = 0. The second term is zero since the total electrode current is fixed. Thus, the first and third terms in (1) remain. The first gives the rate of heat generation when Ohm's law is followed, and the third, being positive, shows that any deviation from this law increases the rate of generation of heat. 6.06. Current Flow in Two Dimensions.—As pointed out in 4.00, there are, strictly speaking, no two-dimensional problems in electrostatics, since all cylindrical conductors are finite in length and there is no way to terminate the electric field sharply at the end, there being no known medium whose capacitivity is zero. In the case of current flow in conductors, we have, however, many rigorously two-dimensional problems, since the current may be confined to a finite region by an insulating boundary. All cases of flow in thin plane conducting sheets are of this type. We thus have available all the methods of Chap. IV for treating these problems. The most powerful of these is that of conjugate functions. We saw in 4.09 that a solution of 6.04 (2), when the potential is a function of x and y only, is U(x, y)
or
(1)
V(x, y)
where
W = U
V = f(z) =
iY)
(2)
254
§6.07
ELECTRIC CURRENT
The current density i at any point is, from 4.11 (2) and 6.02 (2), if V is the potential function, .
=
tau — T as
(3)
l ay 1 101W =— au—— + T as T an T dz
(4)
1 dW
dz
l ay = T an
=——
If U is the potential function, this becomes
.
=
the same sign convention being used as in 4.11 (2). Thus, if the conductor is bounded by the equipotentials U1 and U2 and by the lines of force V1 and V2, the current flowing through it will be Va / = fTT'
i ds=
_11 f v'aU ds= 1 f v2aV ds _ V2 — VI T vi an T vi as
By Ohm's law, the resistance of the conductor is
R = U2 — U1lIU2 — Uil = T IV 2 — V1I
(5)
If the equipotentials U1 and U2 in (5) are closed curves, then the electrostatic capacitance between the electrodes, in vacuo, is, from 4.11,
C —
1U2 —
(6)
UiI
where [V] is the integral of V around Ui or U2. So that if C is known, we may find the resistance between U1 and U2, if the resisting medium fills the same space as the electrostatic field would, by combining (5) and (6), giving
R = T61
(7)
Thus, from 4.13 (5), we see that the resistance per unit length between two parallel cylindrical electrodes of radii R1and R2 with a distance D between their centers is
R = —cosh-1( -F D 2ir
—
(8)
2R IR 2
where T is the resistivity of the medium between them, and we use the positive sign if they are outside each other and the negative sign if one is inside the other. 6.07. Long Strip with Abrupt Change in Width. Let us apply the theory just developed by computing the current distribution in a long straight strip of conducting material of uniform thickness whose half—
§6.07
LONG STRIP WITH ABRUPT CHANGE IN WIDTH
255
width changes abruptly from h to k. The region near the junction is shown in Fig. 6.07b. Such a boundary can evidently be obtained by bending the real axis in the zrplane, the angles being 7r/2, 37r/2, 0, 37r/2, and 7r/2 at —1, —a, 0, +a, and +1, respectively. To have a finite distance between the points —a and +a after the bending, it will be
necessary to put the origin in the z1-plane at z = — j co in the z-plane. Putting these values for the a's and the u's in 4.18 (7), we have dz = dzi
(z? — a2)i
cz1
— 1)i
[(4 —
— a2)]+
ca2 z 1[(4 — 1)(z1 — a2)11
(1)
We shall evaluate the constants c and a by the method of 4.28 before performing the main integration. As in 4.28, when r1is constant, we have dzi = jrieh do t = jz1 dot. When r1is a very small constant and 61goes from 0 to 71" in the ziplane, then y is a large negative constant and x goes from +h to — h in the z-plane. Thus, substituting in (1), we have
-h
r2e2jej
dzz — 3
o
J+h
giving h = 4jc7ra. stant so that
— 1
del
= iCi a dOi
Similarly, when r1—* 00, y is a large positive con-
k
f+k
a2y
1r
dz — jc[ - .
i le2 j9i — a2dB = + jc f del ' lv e et _ 1 1ri—,.., o 0
256
§6.07
ELECTRIC CURRENT
giving k = -Ti jcir.
Solving for c and a gives and
a=
2jk
(2)
c = — 7r
a2)/(1 u2) for Putting these values into (1) and substituting (u2 zi and 0/4, in the first and second terms of (1), respectively, we may integrate by Pc 49 or Dw 120, giving
— z = -{k tan-' rzi (1 7r
a2)]1
h tan-1 [a
—
(1 — a2)i
f
No integration constants are added since this gives, when y, = 0 and x i = + a, z = +h, and when yi = 0 and xi = +1, z = +k. The electrical conditions require that we have a different line of flow on each boundary. Clearly, if we take W = In zi or zi = ew, where U is the potential function, then the line of flow V = 0 passes from xi = 00 to x, = 0 in Fig. 6.07a and the line of flow V = ir passes from xi = — 00 to xi = 0. In Fig. 6.07b, therefore, these lines follow the right and left boundaries of the strip, respectively, from top to bottom, giving a total current I = 7r/s in the strip. Substituting for zi, we have 2
z = -[k tan- 1
(w o _ a2y 1—
e2W
1 — e2 wil \ a2 )
h tan-1 a( e2w —
(3)
where the real part of z is positive when the real part of z, is positive. If s is the resistance between opposite sides of a one meter square cut from the same piece as the strip, then the resistance of the length yh of the wide part of the strip is Rh = slyh1/2k and that of the narrow part alone is Rh = Siyhi/2h. When they are joined as in Fig. 6.07b, the resistance between the ends is not Rh + Rh but there is an additional resistance OR due to the distortion of the lines of flow near the junction. When yk >> k and yh >> h, equipotentials are parallel to the axis in Fig. 6.07b. Let us compute OR in this case. On the y-axis, V = ir so that e 2W = — e2U and (3) becomes, writing j tanh-1u for tan-1ju, x = 0 and
y =
e 2U + tanh-1 ( 1 + e2U
Let U = and 3.3
h tanh-1al
+ e2a27]
(4)
co, then ew is very large and from Pc 753 or Dw 4 2
yk = 7r (k
2e 2u, tank-1 2e 2u,
+
-
2e2U1 ± 1) a2 1 — h tank 1 a2e2U1 a2
CURRENT FLOW IN THREE DIMENSIONS
§6.08
257
Using Pc 681 or Dw 702 and neglecting 1, a, and a2 compared with ew give yk = —k In ir
2k Ui
7r
4e2ui 1 — a'
1
h In
1
+ a\ —
+1(k ln 4k2 ir — h2
h ln
k+ k—h
2kU 1
+A
Let U = U2 -- — co, then e2U is very small and, as before, Yh =
=
2 r
+ e 2U2
2a2 k tanh-1
l[k In 1+a r 1 a
h In
—
=
+ k In k ir k
+ e 2U2) 1
' 2a2 + e2u2
4a 2 1 (1 — a2)e2u2]
k+h
2h U2 1(
r
a2(2
h tanh-1
a(2 + e2u2)
h
h In 24 k ) L'i2h2
2h U2 + r
B
Subtracting and solving for U1 — U2 give
ui _
-
= ryk 7yh 2k 2h
B\ ir (A Vk 2h )
By 6.06 (5), R=s
U2
2\k
B\ + sy k + l Y h i h ) 2k 2 h _ OR + Rk Rh
Substituting for A and B, solving for AR, and simplifying give [h2
OR —
k2 _ h2] k2 k h + 2 ln ln 4hk hk k—h
(5)
Other examples of this method will be found in the problems at the end of this chapter. 6.08. Current Flow in Three Dimensions. If the entire volume between two electrodes is filled with a uniform isotropic conducting medium, then we may obtain the current distribution and the resistance between the electrodes from the solution of the electrostatic problem for the capacitance between the same electrodes when the medium is insulating. In both cases, the equation to be solved is —
(1)
V2-17 = 0
In the electrostatic case, the boundary conditions on the electrode a in vacuo are, from 1.15 (1),
V
= v.
Q„ — f ev
av dSa
s an
(2)
258
§6.09
ELECTRIC CURRENT
In the current case, they are, from 6.01 (2), and 6.02 (3),
lay
V = vaIa = — f s —T on uoa
(3)
The equipotential surfaces correspond exactly in the two cases, since the boundary conditions are identical. From Ohm's law, the resistance is Vb Vat Te,, R = 117b — Val (4) = TEv C IQ.I where C is the capacitance in vacuo in the electrostatic case. If an electrostatic case can be found such that the walls of a tube of force are identical in shape with the insulating boundary of a conductor of resistivity T and the equipotential ends of the tube have the same shape as the perfectly conducting terminals of this conductor, then the resistance between these terminals can be computed by (4) from the "capacitance" of the tube of force. By the capacitance, we mean the ratio of the charge on an end to the difference of potential between the ends. 6.09. Systems of Electrodes. Two Spheres. Distant Electrodes.— When n perfectly conducting electrodes are immersed in a homogeneous isotropic conducting medium of resistivity T, we may find all the relations between the currents entering and leaving the electrodes and their potentials by the methods of Arts. 2.13 to 2.18. It is necessary only to write the current I$ from the sth electrode in place of Qs, in the electrostatic case and to multiply the capacitances by (rEs)-1and the elastances by rev.
Thus, to find the resistance between two spheres of radii a and b, internal or external to each other, having a distance c between their centers, we have, from 2.15 (1), since I1= —/2, V1 — V2 = rev(811 — 2512 + 822)/1
so that the resistance is R—
V1 — V2
Il
Tev(Sii — 2812 + 820
In 5.08, we computed by images the values of case. In terms of these, by 2.16 (2), we have R=
rev
C11 + 2 C12 + C22
C11C22 — .12
C12,
(1) and c 22 in this
(2)
If one sphere is inside the other, this formula is of little use and the method of 5.19 can be used.
§6.10
259
SOURCE AND SINK IN SOLID SPHERE
If two electrodes are at a great distance from each other in an infinite conducting medium, we have, from (1) and 2.18 (1), the result 6'
R T
Co
T(
— 27rr 1 ±Cb
(3)
where r is the distance between them and Ca and Cb are the electrostatic capacitances of a and b alone. 6.10. Source and Sink in Solid Sphere. Spherical Bubble.—The problem of the calculation of the potential when a current I flows from a source to a sink in an insulated sphere of resistivity T indicates how spherical harmonics apply to insulating boundaries. Superposition of the potential of a source I at P1and a sink —I at the center on that of a sink —I at P2 and a source I at the center gives the potential with a source at P1 and a sink at P2 and simplifies calculation. Let the source of the first pair be at 0 = 0, r = b, which is a distance r1from the field point. Add to the potential of this pair a solution Vsof Laplace's equation which makes a V/ar zero at r = a and has no poles if r < a. From 5.153, if b < r < a,
_ 1
TI (1 _1) + vs _
TINF
47r\ri
47r—J1_7-
r/
b: n 1+ (n
n =1
1)rnbn Pn (cos 0) nen-El
(1)
The Vs term may be written, with the aid of 5.153 (1), Vs
=
•
rnbn
f r rnbn dr)
,,IVi 271-1-1
0 a2n+1
n =1
r)
TI F a br/i
1
a
f ( barl
Jo
1)dri
a) r (2)
where the distance from P to the source image at 0 r c. =
= 0, r =
a2b—' is
(r2 +a4b_z— 2a2rb--' cos 0)1
(3)
Insertion of (3) in (2), integration by Dw 380.111, and substitution of the resultant expression for Vs in (1) give ,r/[
1
V 1 =7r — — 4 r
1
r—
a
1
(br'i a— In
a2 — br cos 0) 2a2
( (4)
Equation (3) gives cos 0 in terms of a, b, r, and ri. In the general case where the radius vectors to the source, sink, and field point are a1, az, and r, two potentials, one with a source at a1, sink at center, and one with sink at az and source at center, are superimposed. Equation (4) then gives T/11 r1
1+ a r2
a azr
1 (a2 + airD 2 — 0.21 In 2r2)2 — 4r2 ] a (a2 a
(5)
260
ELECTRIC CURRENT
§6.11
Here r1, rz, ri, and r2 are the distances from the field point to the source, sink, image source, and image sink, respectively. When source and sink are at opposite ends of a diameter, r1 = rz = r2, and al = az = a. Thus Eq. (5) becomes V=
T/[ 1 r1
1 rz
1 in ri(1 + cos al)] 2a r2(1 + cos az)
(6)
where al and az are the angles at which r1and rz intersect the diameter. Any tube of flow in this case is a surface of revolution whose equation can be found by equating to the current I' in the tube the integral of the normal current density over any cap bounded by it. It is simpler to use a spherical cap centered at the source for the r1terms in (6) and one centered at the sink for the rz terms. Thus
Ii = — Ti f
ri
dS = 1(1 + —)f sin 01 del 2a 0
Since the fluxes from source and sink add, integration and the addition of a similar expression for I2 give
I' = I[(1
2ria-1)(1 — cos a l) + (1 + ir2a-1)(1 — cos a2)]
(7)
If is the obtuse angle between r1and rz, this simplifies to
I' = I[1
i(ri
rz)a-1cos 0]
(8)
A solution similar to those given applies to the case of flow from a point source to a point sink in an infinite homogeneous conductor which has a spherical hollow of radius a in it. The equation analogous to (1) for a source at b and a sink at infinity has direct powers of r in the first term since a < b. The perturbation potential of the hole vanishes at infinity and so contains reciprocal powers of r. In the equation analogous to (2) the second and fourth terms cancel, so that =
riar 4/rbLri
f dr jr rr;. -
(9)
where 7-'1is still given by (3). Integration and addition of a similar term for the sink yield a result like (5) except that the last term is replaced by 1 az(1 — cos 09)(air'i — ln a ai(1 — cos 01)(azr
a' — air cos 60 a2— azr cos 02)
(10)
6.11. Solid Conducting Cylinder.—To illustrate the application of Bessel functions to insulating cylindrical boundaries, we shall compute the potential anywhere inside a solid circular conducting cylinder of length 2c, radius a, and resistivity T when the current I enters and leaves
SOLID CONDUCTING CYLINDER
§6.11
261
by thin band electrodes at a distance b on either side of the equator. We take the width of the band too small to measure physically but mathematically not zero so that the current density and potential functions are everywhere bounded. Taking the equatorial plane at potential zero, we see that a solution of the equation of continuity which is finite on the axis is, from 5.292 (3) and 5.32, V = ZAnI o(k,,P) sin k„z
(1)
Since z = c is to be an insulating boundary, we must make a V/az = 0 there. Since cos (2n + 1)471- = 0, we satisfy this boundary condition by taking (2n + 1)r kn = (2) 2c To determine A„, we differentiate (1) with respect to p by 5.33 (4), put r = a, multiply by sin k9z, and integrate from 0 to c. By Dw 435 or Pc 359, the only term left on the right is that for which k, = kn so we have, writing n for p,
cav sin knz dz = A „I i(k„a) o ap
knc
sin2 knz d(k„z)
—
0
We use Dw 430.20 or Pc 362 for the right-hand integral. On the boundary, p = a, a V/ap is zero except in the area AS, at z = b, covered by the electrode, which is so small that we may give sin knz the constant value sin knb therein. From 6.02 (3), setting 2ra dz = dS, the left-hand integral becomes sin knb
aira
T As
i dS =
2ra
sin knb
Solving for An gives A =
2TI sinknb (2n + 1)7r2a I i (knd)
(3)
The desired solution is given by (1), (2), and (3). The boundaries of the tubes of flow, in this case, are surfaces of revolution. The equation of the tube through which a current I' flows can be obtained by integrating the current density — (1/T)(0V/az) over a disk bounded by the tube and equating it to — I', giving 1 = -I- T.
2rknA„ cos knz f I o(knp)P dP n=0
262
§6.12
ELECTRIC CURRENT
Integrating by 5.33 (5), we have, substituting for A. from (3), /' =
k.b cos k„ili(ko)
(4)
1)/i(kna)
(2n
era n=0
where is given by (2). 6.12. Earth Resistance.—Geophysicists sometimes investigate the structure below the earth's surface by observing the distribution of potential on the surface when current is passed through the soil between two or more surface electrodes. Let us investigate the simplest case, in which to a depth a the resistivity is T1and below this depth it is T2. We shall use the method of 5.303 to find the distribution about a singlepoint electrode. The case of two or more electrodes may then be found by superposition. The potential due to the electrode alone may be written down from 5.303 (1), by substituting for q, from 6.08, the value 2TEL It should be remembered that 2/, in this case, corresponds to 1 in 6.08, since all the current flows in half the field. Then, we have V
2rr
=
2r
fo
~
J o(kp)ekizi dk
(1)
owing to the electrode alone. As in 5.303 in the region of T1, we shall superimpose a second potential due to the discontinuity at z = a that may contain both the e—k' and ekz terms since z is finite in this region, thus we have V1 =
1.(k)J o(kp)e-kz dk
f %11(k),10(kp)e±kz dk
f
J o(kp)e-k' did
(2)
Since the earth's surface z = 0 must be a line of flow and since the last term already meets this condition, the remaining terms must satisfy it independently so that, if 8171/az = 0 when z = 0, —4,(1c)
I'(k) = 0
(3)
The potential in the region T2 must vanish at infinity and so can have only the form
V2 = Tlif
27r o
(4)
0(k),10(kp)e-k' dk
At z = a, the boundary conditions are, from 6.04 (3) and (4), V1 = V2
and
1 avi T1
az
1 al72
az
§6.13
CURRENTS IN THIN CURVED SHEETS
263
Substituting from (2), (3), and (4) and canceling out Jo(kp) give 43(k) (eka
e—ka — T2e — k
a
,24 (k)(eka
e—ka)
e
ka\
)
e(k)e—ka = 0 rie (k)e-ka = 0
(5) (6)
Eliminating e(k) and substituting in (2), we have, on the earth's surface where z = 0 and setting (71 — 72)/(T1 + T2) = 0, the result
=
ri
oe-2ka Jo(kp) dk 27r- Jo 1 + Oe-2ka —
(7)
Expanding the denominator by Pc 755 or Dw 9.04 and interchanging the integral and summation, we have V8 = 2[ I
f
Jo(kp) dk
2nka o (k p) did
(-1)n0nf n=1
Substituting for the integral, from 5.298 (6), gives
V'
=
ri/[1
2a Lp
2N n=1
(-1)n0n
(4n2a2
p2)1
(8)
Now if we take the earth's surface as the xy-plane and the current I passes from an electrode at x = +b to one at x = — b, the potential at the surface is, since p+= [(x — b) 2 y 2]+ and p_ = [(x b) 2 y9+1, Ve =
ri/ { 1 L7 p+ — p_
+
( —0)n[(4n2a2 +
-i]} — (4n2a2 pl)
(9)
n=1
The case for any number of layers has been worked out by Stefanesco and Schlumberger, Journal de Physique, Vol. I, p. 132, 1930. 6.13. Currents in Thin Curved Sheets. If a thin curved sheet of uniform thickness can be developed into a plane, then, by so doing, the current distribution in such a sheet can be reduced to two dimensions and 6.06 applied. If, however, the surface cannot be so developed, other treatment is necessary. From 3.03 (4) and 6.04 (2), the equation o f continuity in orthogonal curvilinear coordinates is —
a (h2h2 av) aui h1 au,
a (h3hi av) au2 h2au2
a eih2 av) = 0 011,3 au3
(1)
Let us suppose that u1and u2 are a set of orthogonal curvilinear coordinates drawn on the surface and let u3be a distance measured normal to it. The surface is so thin that the current distribution on each side is the same, making a V/au3= 0. The thickness of the surface is uniform so
264
ELECTRIC CURRENT
§6.14
that h3 is independent of u1and u2. Under these conditions, ( 1 ) takes the form + a (hi av\ = 0 a th2 (2) aui \hi aui) au2\h2 au2) This is the equation that we must solve to get the current distribution. 6.14. Current Distribution on Spherical Shell.—From 3.05 ( 1 ) , letting V be independent of r, we get 6.13 (2) for a spherical shell to be sin 01(sin
ao
Ba aog
+ago gao = 0
(1)
where 0 is the colatitude angle and 95 is the longitude angle. Although a spherical shell cannot be developed in a plane, it is possible to represent every point of a spherical surface on an infinite plane in such a way that angles are preserved. This process is known as stereographic projection and closely resembles inversion. A plane is taken tangent to the sphere at one end of a diameter. A line passing through the other end 0 of this diameter and the point P (Fig. 6.14) intersects the plane at P' which is called the projection of P. Let be a solution of the equation of continuity in the plane, where, from 4.01 (2), it takes the form
a (aT) ,
r r ar ar
a24,
u
802
(2)
In projecting the system of curves, which represents, onto the sphere, the longitude angle will be unchanged. From Fig. 6.14, we see that r and 0 are connected by the relation r = 2a tan 01 = 2a tan 10 The equation that the projected curves satisfy is obtained by substituting 0 and 4) for r and 95 in (2).
. a dO a r— = r— — = sin ar
dr
a8
so that, from (2), sin 0 (in
a
ao
+
a ae
11—
adz
=0
This is identical with (1) so that the stereographic projection of a solution of the equation of continuity in a plane gives the solution of this equation for a thin uniform spherical shell.
§6.15
SURFACE OF REVOLUTION
265
We found, in 4.09, that both U and V are solutions of V' V = 0 where U + jV = f(x + jy) = f(r cos cp ± jr sin q5) if f(z) is an analytic function. It follows from the preceding that if
U + jV = f[2a tan 4e(cos 0 + j sin 0)]
(3)
then both U and V are solutions of the equation of continuity on the surface of a sphere of radius a and that if U is the potential function then V is the stream function, and vice versa. From the laws of inversion, it follows that lines in the plane project into circles through 0 on the sphere and circles project into circles. As an example, let us find the potential at any point on a spherical shell of radius a and surface resistivity s when a current I enters at 0 = a, 0 = ,}ir and leaves at 0 = «, 4 = —47. From 4.13 (2.1), writing from 6.08, e„s I for the charge which, in 4.13, is 2/7- 6„, we have for the potential U at any point on a plane sheet 2,7r u x 2 ± y2 ± b2 r2 ± b2 = coth 2by 2br sin 4, s/ But from Fig. 6.14, we have r = 2a tan 40 so that coth
1 — cos a cos 0 27U tan2 -0 + tan2 4-a = — sin a sin 0 sin ct, 2 tan 40 tan 4« sin 0 s/
(4)
In a similar fashion, the equation of the lines of flow is, from 4.13 (4), cot
27- V
s/
=
r2 — b2 cos a — cos 0 — 2br cos 4,sin a sin 0 cos 4'
(5)
6.15. Surface of Revolution.—As another example, let us solve 6.13 (2) for the surface of revolution formed by rotating the curve y = f(z), where Az) is a single-valued function, about the z-axis. Clearly, on such a surface, any point P can be located by the orthogonal coordinates z and 4), where y6 is the longitude angle about z. In 6.13, let ul, which lies along the surface, equal z numerically and let u2 = 4,. From symmetry, h1and h, are independent of .45; so 6.13 (2) becomes
h2a(h2 av) i_ 8 2v hiaz h, az F 802 = °
(1)
-
To solve this equation, let us take a new variable u that is zero when z is zo and satisfies the relation
a _ h2 a au — hi az
Or
z h1 u= f — ,.. dz zo n2
(2)
266
ELECTRIC CURRENT
§6.15
Then (1) becomes 32V
32V
au 2 + ao 2 = 0
(3)
From 4.09, a solution of this equation is known to be either U(u, 0) or V(u, 95) where U jV = F(u jcP)
(4)
To evaluate u in terms of given quantities, we must calculate h1 and h2. The equation of the surface is p = f(z)
so that
dp = (z) dz
(5)
and we have ds2 = dp2
dz2
p2dcp2 =
(z)]2 +11 dz2
{f(z)l 2d(P2
giving hi =
(z)}2
1) 1,
h2 = f(z)
(6)
FIG. 6.15.
Putting these values in (2), we obtain u=
z
fo
{[i (OP ± f (z)
1)1dz
(7)
Since an increase of 2nir in cp returns us to the same line on the surface, it is necessary, in order to have single-valued solutions for the potential, that F(u j(I)) be periodic in with a period 27. The solution is identical in form with the solution for a cylindrical surface which can
§6.16
LIMITS OF RESISTANCE
267
be unrolled into a flat strip. If one end of the surface is closed, as in Fig. 6.15, then f(z) is zero at this point and, from (7), the equivalent cylinder will extend to u = — .0. If one end is open, as in Fig. 6.15, then the equivalent cylinder will terminate at some positive value of u and the boundary conditions on the edge of the cylinder will be the same function of as on the edge of the surface illustrated in the figure. 6.16. Limits of Resistance.— The general theorems of 6.05 frequently enable us to compute limits between which the resistance of a conductor must lie, even when we cannot compute the rigorous value. To get a lower limit, we endeavor to insert into the conductor thin sheets of perfectly conducting material in such a way that they coincide as nearly as possible with the actual equipotentials but, at the same time, enable us to compute the resistance. In this case, we know, from 6.05, that the
result is equal to or less than the actual resistance. To compute the upper limit, we insert thin insulating sheets as nearly as possible along the actual lines of flow, in such a way as to enable us to carry out the resistance computation. We know, from 6.05, that this resistance equals or exceeds the actual resistance. For example, the resistance between two electrodes of a given shape lies between that of two circumscribed electrodes and that of two inscribed electrodes. As a specific example, let us compute the resistance between perfectly conducting electrodes applied at the ends A and B of the horseshoe-shaped conductor of triangular section, shown in Fig. 6.16. To get an upper limit for the resistance, we insert infinitely thin insulating layers very close together which require the current to flow only in straight lines and a semicircle. The length of such a layer, from the figure, is 2c + ir(b x). The area is x dx so that the resistance is dR. —
T[2c
ir(b x dx
x)]
268
ELECTRIC CURRENT
§6.17
All these layers are in parallel so that the upper limit of the resistance is, if Pc 29 or Du,91.1 is used, given by (1), x dx R'
y
oa 2c + id) + rx 1f (f dR 1 „) 1 (T 2c + r(a + b)]-1 — iorka — (2c +
rb)
In
2c + irb
(1)
To get a lower limit of resistance, we insert perfectly conducting sheets across each leg of the horseshoe at M and N. The current in each leg then is uniformly distributed, and so the resistance of the legs is 4cr/a2. In the semicircular part, the conductor has the shape of a triangular tube of force in the two-dimensional electrostatic field given by W = In z. This gives, for the stream function and the potential function the values, respectively, (2) and V= 0 U = In r The resistance of the strip at y in Fig. 6.16 is then, from 6.06 (5),
d n = rir(ln ba±2by dy) i The strips are in parallel so that, when Pc 426 or Dw 620 is used,
/ r \la a +b )-1 — 71-7.1?;' = U dn) = rr (2 fo Inb ± 2y 4 - bln (a + b) — blnb — a The legs and the curves are in series, so that Ri =
T
[4c a2
ir
b ln (a + b) — b ln b — a]
(3)
6.17. Currents in Nonisotropic Mediums. Earth Strata.—Taking the divergence of 6.02 (3), we have v . (-yvV) = —17 • i = 0
(1)
If the medium is not isotropic, the conductivity in different directions varies so that -y cannot be factored out. If, however, the medium is homogeneous, the conductivity in any direction remains unchanged throughout it. In this case, we may choose a set of rectangular axes as in 1.19 (7) and 3.02 (5) so that (1) becomes 7x
a2v1-+ aw 1-yzaw= 0 Ty ax2
ay2
+
8z2
(2)
CURRENTS IN NONISOTROPIC MEDIUMS
§6.17
269
Owing to the method of formation of the soil it frequently happens that the conductivity in the horizontal direction is greater than that in the vertical direction. If we take the z-axis vertical, (2) becomes
,92-v ) -rh( ax2 + ay2 +
az2 = 0
(3)
If we choose a new variable defined by (4)
U = -1 ) Z = aZ yv
this becomes
a2v ax2
a 2v ay2
a2 v
at e
=
o
(5)
We have all the methods of Chap. V available for solving this equation but we must translate the boundary conditions into the x, y, u system and fit the solution to them, or vice versa. Suppose we have a spherical electrode of radius R half-buried in the earth. The boundary condition is then that V = Vo when x2 + y2 + z2 = x2 + y2 + (u/a)2 = R2. Thus, in the x, y, u system, our boundary condition is that for a spheroid whose equation is u2 x2 ± y2 =1 R2 a2R2 From 5.02 (1) and (2), taking a2 = b2 = R2 and c2 = a2R2, the solution is
v = v 0[f
(a2
d0 0)(c2
°' 0)1]/[. 0 (a2 ± 0) (c2
Using Pc 114 or letting x = a2 + 0 and using Dw 192.11, we have, since a > 1, 0)]1 -1[(c2— a2)/(c2 V = Vo tanh tanh-1[(c2 — a2)/c2] But from 5.28 (1) and (2)
c2 + 0 = (c2
a2),12
so, on substituting for a and c, this becomes V = Vo[tanh-i (7h —
7,\1]-1
j
tanh-1
1
(11]-1 smh-i (n 2 — 1)-1 = Vo[ cosh-1 ,.yv
(6)
270
ELECTRIC CURRENT
§6.18
and from 5.28 (4) and (5) u = az = R (a2— x2
1) (1
y2 = R2(a2
2) (712
1)
Eliminating we have a 2z2
X 2 + y2
= 1 (7) 1)n2 R2 2 R2(a2— 1) (n2— 1) The smallest value of n is on the electrode where n =a/(a2 — 1)1. Hence, the denominator of the first term is always larger than that of the second so that the equipotential surfaces are nonconfocal oblate spheroids. On the earth's surface, where z = 0, the equipotentials become (a
'Yu ]4 V = V o[cosh-1 MT sinh-1 R[ 7v (x2 ± y2)
(8)
If R is very small, we have a point electrode. Writing the angle for the arc sinh and putting a constant times the current I, for remainder, (8) becomes V = 61(x2 + y2)--i (9) These curves thus have the same form as for an isotropic medium.
(a)
( b)
(c)
FIG. 6.18a, b, c.
6.18. Vector Potential. Flow around Sphere in Tube.—So far only scalar potentials have been used for current flow. With axial symmetry the vector potential whose curl is io is easier to use for it simplifies the boundary conditions and gives 22rpAo = I for the equation of a tube of flow carrying a current I where p is the radius of a circular tube section. Formulas needed appear in Chap. VII if B is replaced by the current density io. An instructive example is the calculation of the current flow in a solid circular conducting cylinder of radius a when a concentric spherical nonconducting obstacle or cavity partially obstructs the flow. All the formulas needed for solution are in Art. 5.40 except the expression for A given in 7.06 (3). The approach is illustrated by Fig. 6.18a, b, and c. The flow (a) plus the uniform flow (b) yields the flow (c) around the sphere in the tube. If the current density far from the sphere is io, then the
§6.18
VECTOR POTENTIAL. FLOW AROUND SPHERE IN TUBE
271
vector potential of this flow is -io p from 7.02 (8). The sources on the sphere surface required to maintain the circulation of Fig. 6.18a produce the internal and external vector potentials N (A0)in
r)2n-f-1
=
N
PL,+i(cos 0),
C)2n-I-2 (110) =Dn(7,lin+1(cos 0)
n=0
n=0
(1) The physical nature of the sources needed to maintain the circulation in (a) is more evident if iois replaced by B, in which case they are obviously coaxial current loops that lie in the cylinder and spherical shell. The vector potentials of (1) are single-valued even functions of z. Values of Dn must be found such that, in the presence of the cylinder, the internal current density in (a) cancels that of the uniform flow in (b). In cylindrical coordinates (A ,) ex may be written, by 5.40 (3), N (- 1)".2Dna2'1-2r
(14 - q5) =
rn! (2 )
t2n-"Ki(tp) cos (tz) dt
(2)
n=0
Evidently from Fig. 6.18a, since no net current traverses the cylinder, A is zero at both p = 0 and p = b, so the internal vector potential of the sources in the wall that cancel (2) must be, using 5.40 (19), N
(Aib )*, =
f;(ta) 'Alt(at
t n )D 2n ) n2
—
1 7 ( ( -
2n
(tp)
cos (tz) d(ta)
n= 0
N
= n=0
+s2Dnr2s + 1
(_ s=0
r(2n)!(2s + 2)!as+114„ fcos
/1 (2n + 2s + 2) 2n + 2s + 3
(3)
The integral /'(2n 2s + 2) is that of 5.42 (4) but with first- instead of zero-order Bessel functions and is tabulated (Phys. Fluids, Vol. 7, page 635, 1964) and will be written I„,. Inside the sphere only the harmonic --ijorPI (cos 0) or --iio p, which must cancel the uniform flow, can appear. So if s 0, the coefficients of r28-"Ph+1(cos 0) must vanish. Thus each value of s gives a linear relation between the coefficients Dn. So as in 5.41 a set of N 1 simultaneous equations must be solved for Do, D1 , . . . , DN. The value of the right side is zero for all but the first equation, for which it is — The values of Cs„ in 5.41 (5) are, from (3) and (1), S
n
=
1)n+'21 n. (2s + 1) !a28-1-1'
1
( —
7r (2n) !
C nn
c2n-1-1
2/nn 2)!a2n+1 r (2n) !(2n
(4)
272
ELECTRIC CURRENT
§6.19
The above procedure works well for oblate and prolate spheroids including the disk. Extensive numerical values are given in the reference cited. Some are given below to show the interrelation of various factors. The D„ of this section equals iocCu/(4n + 3) in the reference. c/a
Do
D1
D2
D3
0.1
0.0500399 0.153298 0.277673
—0.0000001 —0.000037 —0.000870
0.000001 0.000045
—0.000003
0.3 0.5
6.19. Space-charge Current. Child's Equation.—So far, we have considered currents in conductors where the net charge is zero. Let us now consider currents in which charges of only one sign are present so that the net charge is not zero. We must assume the motion of the charges sufficiently slow so that the formulas of electrostatics are valid. In this case, Poisson's equation [3.02 (3)] must be satisfied so that V'2 V = —
E„
(1)
where V is the potential and p the charge density. The charges are supposed to be similar and associated with a mass m and to acquire their energy entirely from a superimposed field. Thus, if their velocity is v and charge q, their energy at a point where the potential is V will be my' = 2q(Vo — V)
(2)
where Vo is the potential of their point of origin. The current density at any point is given by = pV
(3)
The simplest case is the one in which the charges are freed in unlimited quantity at the plane x = 0 and accelerated with a total potential Vo to a plane x = b. At the surface x = 0, charges will be freed until there is no longer an electric field to move them away so that the boundary condition there is
(an ax
0
(4)
Here, all the velocities are in the x-direction so that, eliminating p and v from (1) by means of (2) and (3), we have 3217 axe
[
m
c, 2q(V o — V)
§6.19
273
SPACE-CHARGE CURRENT. CHILD'S EQUATION
Multiplying through by dV/dx and integrating from V = Vo and dV/dx = 0 to V and dV/dx, we have (dVy 4ir m(Vo — V)11 E,L 2q
(5
)
Taking the square root of both sides, integrating from V = Vo, x = 0 to V = 0 and x = b, and solving for i, we have = 4€„(2q\iVol 9 Vrr, b 2
(6)
This is known as Child's equation. It shows that, with an unlimited supply of charges at one plate, the current between the plates varies as the three halves power of the potential. Such a current is said to be "space-charge limited." We see from (6) that the space-charge limitation is much more serious for charged atoms than for electrons, because of their greater mass. In practice, we frequently have the emitter in the form of a small circular cylinder and accelerate the charges to a larger concentric cylinder. In this case, we use cylindrical coordinates, and if I is the total current per unit length of the cylinders, (3) becomes I = 2irrpv
(7)
Writing (1) in cylindrical coordinates by 3.05 (2), eliminating (2) and (7), and writing V for Vo — V, we have m y d2V dV = dr 271-E„\2qV
r dr'
p
and v by
(8)
The direct solution of this equation in finite terms is difficult, if not impossible. We may, however, obtain a solution in series as follows: Let us see if the assumption that q, m, and V enter into this solution in the same way as into (6) leads to a solution of (8). Try the solution _1
87r€,(2q\I 9 m) r/32
(9)
and see if 132 can be determined to satisfy (8). Substituting (9) in (8) gives the equation
d20 313d-y2 where
(12 + odo + 02 _ d-y
1=0
(10)
274
§6.19
ELECTRIC CURRENT
This equation can be solved in the regular way in series which gives =
g 72 ± Twy3 44 074
..
(12)
Here a is the radius of the inner cylinder. Tables of i3 as a function of r/a have been published by Langmuir. Problems The problems marked C in the following list are taken from the Cambridge examinations as reprinted by Jeans, with the permission of the Cambridge University Press. 1. A cylindrical column of length land radius a of material of resistivity r connects normally the parallel plane faces of two semi-infinite masses of the same material. Show that, if R is the resistance between the masses, 7/
11-7
r
_ — < R< ,rat 2a
— / In (1
♦ era/l)]
Observe that, when 1 = 0, both limits give the exact value. 2C. A cylindrical cable consists of a conducting core of copper surrounded by a thin insulating sheath of material of given specific resistance. Show that, if the sectional areas of the core and sheath are given, the resistance to lateral leakage is greatest when the surfaces of the two materials are coaxial right circular cylinders. 3C (Modified). Current enters and leaves a uniform circular disk through two circular perfectly conducting wires, of radius a, whose centers are a distance d apart and which intersect the edge of the disk orthogonally. Show that the resistance between the wires is 2(shr) cosh-1(d/2a), where s is the resistivity. 4. Two electrodes each of small radius ö are situated at a distance 2a apart on a line lying midway between the edges of an infinite strip of width r and resistivity s with insulated boundaries. Show that the resistance between them is, approximately, s
ln
sinh 2a 5
5. Instead of being situated as in the last problem, the two electrodes lie symmetrically on a line normal to the edges of the strip. Show that the resistance between them is, approximately, s ln (2 tan a) 8
6. The electrodes are situated as in problem 4 but the edges of the strip are perfectly conducting. Show that the resistance is, approximately, s - ln
tanh
7C. A circular sheet of copper, of specific resistance si per unit area, is inserted in a very large sheet of tinfoil (so), and currents flow in the composite sheet, entering and leaving at electrodes. Prove that the current function in the tinfoil, corresponding to an electrode at which a current e enters the tinfoil, is the coefficient of i in the
275
PROBLEMS imaginary part of
cz so - si In cz - a2 so ± si
roe -ln (z - c) 2.71-
where a is the radius of the copper sheet, z is a complex variable with its origin at the center of the sheet, and c is the distance of the electrode from the origin, the real axis passing through the electrode. 8C. A uniform conducting sheet has the form of the catenary of revolution y2 z 2 = c2 cosh2 (x/c). Prove that the potential at any point due to an electrode at xo, yo, zo, introducing a current C, is
c,
constant - — In cosh 47r
x - xo c
zzo yyo R y2 + z2)( + zDik
9. The edges of two thin sheet electrodes are held in a circular cylinder of resisting material of length L. Both electrodes and the axis of the cylinder lie in the same plane and their edges are parallel to, and at a distance c from, this axis. Take c small compared with a, the radius of the cylinder. If r is the resistivity and I? is the resistance between electrodes, show that Tir
711-
. 2L cosh-' (a/c) < R < 2L smh-' (a/c) 10. Show from Art. 4.29 that the resistance in the last problem is given rigorously by
R-
2TK(c2a-2) LK[(1 — c4a-4)1]
11. Show from Art. 4.30 that if, in the foregoing problems, the edges penetrate to different depths, so that the distance of the cylinder axis to one edge is c1 and to the other c2, then the resistance is
R
rK(k) LK[(1 - k 2)11
where k -
a(ci a2
c2)
cic2
Here K(k) is a complete elliptic integral of modulus k. 12. A semi-infinite plate of resistivity r is bounded by x = 0, x = 1, and y = 0. Two perfectly conducting electrodes are applied to the edge, so that their contact areas are bounded by x = 0, x = 1, z = b, z = a and x 0, = 1, z = -b, z = -a where b < a. Show from Art. 4.29 that the resistance between these strips is
R=
2rK(b/a) 1K ([1 - (b/a)9+ }
13. The contact areas in the preceding problem are of unequal width. The distance between their near edges is q, far edges h, and the width of one is w. Show from 4.30 that the resistance between them is
R=
rK(k) 1K[(1 - k 2)1]
where k = [
hg (w g)(h - w)
14. A long slab of resistivity r, thickness 2b, and width l has two perfectly conducting strip electrodes of width 2a opposite each other on opposite faces, at right angles to
276
ELECTRIC CURRENT
the sides. Show from problem 61, Chap. IV, that the resistance between them is R—
2rK(e-ra/b) 1K[(1 — c 2n.191]
15. A long strip of width 2b has a hole of radius c bored on its center line far from the ends. Using the results of 4.28, show that the hole increases the resistance by an amount equivalent to adding AL to the length of the strip where AL = (1 + X)-12b7r-iln cos [-brcb-1(1 + X)] and the parameter X must be found from 4.28 (6). 16. The curved surface of a right circular cylinder of radius r, length d, and resistivity r is given a perfectly conducting coat. Two thin coplanar saw cuts are made, so that the two halves are connected only by a strip of width 2b whose center line is the cylinder axis. Show that the resistance between the two halves of the conducting coat is rlf[(1 — b 4r-4)1] R= 2dK(bzr-2) where K(k) is a complete elliptic integral of modulus k. 17. There is a dipole source of current of moment M in a sphere of radius a and resistivity r with an insulated surface. Take the sphere center as the origin and the = 0 axis parallel to the dipole which is located at r = b, B = 00, and 0 = 0. Show from problem 115, Chap. V, that the potential when b < r < a is V=
n TM V v (2 — 5!;,.)(n — m)![(b)n+1 474) 2
n=0 rn=0
m — 1)!
(n
(n +
r
157_1(tio)PZ(p)
cos mc5
where ,uo = cos 00, A = cos 0, and r > b. If r < b, replace (b/r)n+IP7_1(20) by — (r/b)n[(n — m 1)/(n m)1/37,z4.1(Ao)• 18. A solid conducting cylinder of resistivity r is bounded by the planes z = ±c and the cylinder p = a. A current I enters at p = a, 0 = 0, z = 0 and leaves at p = a, 0 = lr, z = 0. Find the potential anywhere inside the cylinder and show that, if the electrodes are small spheres•of radius r, half embedded in the surface, the resistance between them is 2rz (2 ir 2a
n=0 nt=0
en)/2m+1,[nrr(a — r)/c] n/2.,„+1 (nira/c)
19. A perfectly conducting rod of radius b is partially insulated by a semiconducting bushing of radius a and length 2c where it passes through a perfectly conducting thin sheet at z = 0. If the total leakage current is I, show that the current density on the rod when —c < z < +c is CO
I 47 2ab
.k
(2 — SD cos (nrzlc) n[I i(nrci/ c)Ko(nirb /c) + I o(nrb /c)Ki(nra/c)!
71=0
20. Current enters an infinite plane conducting sheet at some point P and leaves at infinity. A circular hole, which does not include P, is cut anywhere in the sheet. Show that the potential difference between any two points on the edge of the hole is twice what it was between the same two points before the hole was cut.
277
PROBLEMS
21. An insulator has the form of a truncated cone of height h, base radius al, and top radius a2. The base rests on a metal plate, and the top supports a concentric metal post of radius a3. If the specific surface resistance is s, show that the surface resistance between the plate and the post is s
f[h2
27r
(al — a2)211in (at) + in (a2)} ce3 a2
al — a2
22. On a spherical conducting shell, current is introduced at the point 0 = a. = 0 and taken out at the point 0 = a 4) = 7, where 0 is the colatitude angle and the longitude angle. Show that the potential on the surface is of the form ,
A In
1 — cos a cos — sin a sin 0 cos 1 — cos a cos 0 + sin a sin 0 cos
C
23. The ends of two metal rods of radius b are embedded in a sphere of insulating material of radius a whose body resistance is very high but whose surface resistance is s. If the axes of the rods were extended, they would intersect at the center of the sphere at an angle a. Show that the surface resistance between the rods is - cosh ,
a a) sin b 2
24. A current flows in a thin shell whose equation is (z/a) 2 (p/b)2 = 1, where p2 = x3 +y2 and a > b. Show that the potential V on the shell is either Ui or U2 where U7 + jU2 = f(a j0), tan rp = y/x and
a = tanh-, [a4
(a2
bz —
b2)1 sin-1
(a 2 — b2)z2] 2
[z(a2 — b 2 )1 a2
The function f(a j(p) is periodic in 4 with a period of 27, and we must also have a V/Oa = 0 at z = ±a or a = ± .0 if there are no sources or sinks of current there, otherwise av /act, = 0 there. 25. If we have an oblate spheroidal shell carrying current instead of the prolate one treated above, so that a < b, show that the preceding considerations apply except that a = tanh-1 ([a4
+
bz ) 03 — az)z91
(b 2 — a2)1 [z(b2 — a2)1] sinh-, b a2
26. Consider the surface of the earth as plane and its resistivity to be To except for a region in the shape of a semi-infinite vertical cone with its apex at the surface which has the resistivity Ti. A current I is introduced at the surface at ¢ = 0, at a distance r = a from the apex. Show that the potential at any point r < a outside the cone is given by
Vo = gar)-4-2 Am(p)
cos mOfo [Am(P)P7v_1(A) Bm(P)P,1-1( -- i4)1 cos (p In ra-0 dp m= o (ri — T2).8.(P) = TI(Ti — T2) (2 —
S =
T2S' — TIS P7;_1(0)[TI — T2 a[Prp (cos c)] / a (cos a)
a[Pz,_+( —cos a)]/a(cos a)
s =
— 1)m(TIS — T28')]
P7p_4(cos a) P72,_1(—cos a)
278
ELECTRIC CURRENT
27. A thin circular loop current source I is coaxial with, and immersed in, a solid cylinder of resisting material of radius a and length L with perfectly conducting earthed ends. Show that the potential distribution in the cylinder is, if z < c, T/ sinh Akz sinh mk(L — c)Jo(mkb)Jo(µkP) sinh pkL ra2 L-/ Pk[Jo(pka)] 2 k= 1 where Ak is so chosen that Ji(Ihka) = 0 and the loop is at p = b, z = c. 28. Two rings made of wire of small radius d are coaxial with an infinite circular cylinder of resisting material of radius a and embedded in its surface to a depth d. Show that, if mk is chosen so that Ji(Aka) = 0, the resistance between them when their centers are at a distance c apart is, approximately, — e-Ak(-2d)ie-Akd
R= as
=1 k
29. A conducting circular cylinder with inner and outer radii a and b is cut on one side longitudinally, the edges of the cut being maintained at potentials of 1V0 and —1V0, so that current flows around the cylinder. Show that the electrostatic fields, when r < a and r > b, are given, respectively, by the transformations Vo W = — ln (a ± z)
and
W
Vo
7
ln
z
V Z +
b
30. A current I is introduced at one pole of a thin conducting spherical shell of radius a and uniform area resistivity s and removed at the other. Show that the electrostatic potential at any point inside the shell is
V =
s/ 2r
n=0
r 4n + 3 (2n + 1)(2n + 2)(a
2n+1
P2.-0(14)
31. A wire of radius a and resistivity T1is coaxial with the z-axis. The medium outside the wire is insulating except for that portion lying between z = —c and z = c, which has a resistivity T2. If T2 >> rishow that the ratio of the resistance across the gap of a perfectly insulated wire to that of the actual wire is, approximately, = + 32cTI 1r 3cts 2
n=0
( — 1)° K1[1(2n+ 1)7ra/cl (2n + 1)3 Ko [i(2n 1)ra/c]
32. The corners of the wider strip in Fig. 6.07b are cut off at 45° so that the half width of the strip increases linearly from h to k. Show that the presence of the tapered section near the center of a long strip increases its resistance over the sum of the resistances of the straight sections by an amount
rhk
(h 2
k 2) tanh-1 -
(k2 — h2) tan-1 -
hk In
— 8h2k 2
33. An infinite slab of resistive material bounded by z = a and z = —a with perfectly conducting surfaces has a spherical cavity of radius 1 centered between its faces on the z-axis. From 6.18 the vector potential for unit current density far from the
279
PROBLEMS cavity is — -1.p.
Assume as in 6.18 that inside the cavity it has the form
A4,, = ZCtr20'+1/1.+1(cos Bo) + n=0
(1 — 4)rsn-2P2n+1(cos 8 = -
where r, is the distance from the sth image sphere to the field point and B, is the acute angle between r, and the z-axis. Express AO,. in cylindrical coordinates by 5.40 (35) and (36), expand in powers of p, and require that the coefficient of p2n+1be zero if n 0 and a if n = 0. Thus show that the N 1 equations to be solved for C„ are N i-(2/ + 2n + 3)(2/ ± 2n 2)! 4;, = (2n + 2)!(2/)!(2a) 2H-2n+2 l-0
where i-(x) is the Riemann zeta function. Show that the vector potential in the current region is given by N
Ao. = ZC'nr."-2PL+1(000 re = 0
O.)
a=—
For a complete discussion of this problem see Paul Michael, Phys. Fluids, Vol. 8, page 1263, 1965. References Some material on the subject matter of this chapter will be found in most of the electrical references of Chap. V.
CHAPTER VII MAGNETIC INTERACTION OF CURRENTS 7.00. Definition of the Ampere in Terms of the Magnetic Moment.— As mentioned in the last chapter, electric currents flowing in neighboring conductors exert forces on each other known as magnetic forces. This magnetic interaction was carefully investigated by Ampere. One of his experiments showed that, when two wires carrying equal currents in opposite directions lie sufficiently close together, their power of interacting magnetically with other circuits is destroyed. Let us now construct several small plane loops of wire, twisting together the thin wires by which the current is to enter and leave the loops so that any magnetic forces observed will be due to the loop alone. Keeping them at distances great compared with their dimensions, we find that, when carrying steady currents, the forces and torques that they produce on each other are in every way like those between electric dipoles, provided that we orient the normal to the plane of the loop as we would the axis of an electric dipole. Thus, using three small loops carrying fixed currents, we can, by taking them two at a time and measuring the forces and distances between them, determine by 1.071 (6) the products m1m2 = A, M 1M3 = B and M2M3 = C and so find Pr? = AB/C, 31= AC/B, and 2/3 = BC/A. If the experiment is done in vacuo and 1/E is replaced by = 47r X 10-7, then M for each loop equals the product of its area in square meters by its current in amperes and is the magnitude of its "magnetic moment." Thus an ampere may be defined as that current which, flowing in a small plane loop of wire, gives it a magnetic moment equal to its area. In a medium other than a vacuum the forces and torques are different and the factorµ which replaces is called the permeability of the medium. 7.01. Magnetic Induction and Permeability.—If a conductor, carrying a current, when placed in a certain region, experiences a magnetic force we say a magnetic field exists in that region. We shall map the magnetic field by means of a small exploring loop just as we could map the electric field by means of a small electric dipole. When free to move, the loop assumes a certain orientation. We shall define the positive direction of the magnetic field to be normal to the loop and directed as a right-handed screw moves when rotated in the sense in which the current traverses the loop. As in the electrostatic case, we can measure the strength of the magnetic field in terms of the torque acting on the loop. 280
§7.01
MAGNETIC INDUCTION AND PERMEABILITY
281
Thus, we define the magnetic induction or flux density B to be a vector in the magnetic field direction whose magnitude, in webers per square meter, is the torque in newton meters on a loop of moment one whose axis is set normal to this direction. This torque, and hence B, depends on the medium in which the experiment is performed. We define the relative magnetic permeability Km of a substance to be the ratio of the magnetic induction when all space is occupied by the substance to its value at the same point in vacuo, the configuration of all circuits and the magnitude of all currents being the same. The permeability A is the product The experiments on which the definition of A and B depend involve no theoretical difficulty in the case of liquid and gaseous mediums. It is obviously impossible, however, to manipulate a loop of wire in a solid medium. If all space outside the small region in which our loop is situated is filled with one liquid and this region is filled with another, then we find that the measured value of B is independent of the shape and size of this region only when the liquids have the same permeability. Thus, to determine B and A at a point in a homogeneous isotropic solid, we excavate a cavity at P and fill it with a liquid such that the measurement at P is independent of the shape and size of the cavity. The value of B andµ at P, so determined, is the same as before the cavity was created. In 7.24, by using the boundary conditions derived in 7.21, we shall describe an experimental method of defining B andµ which includes the case of magnetically anisotropic crystals. Suppose that we take a great number of small loops, each carrying a current I and, without changing the area, fit them together into a mesh always keeping the direction of the induction due to adjacent loops parallel. From Ampere's experiment, the magnetic effect of all loop boundaries, except those forming the outer edge of the mesh, disappears, so that the resultant magnetic effect is exactly the same as if the current I circulated around the boundary alone and, hence, is independent of the shape of the surface of the mesh. But we know that the magnetic induction due to any elementary loop of area dS is identical with the electric field due to an electric dipole of moment AEI dS. The magnetic induction due to the whole circuit is therefore the same as the electric field due to a uniform electric double layer of strength AE/, whose boundary coincides with the circuit and which we discussed in 1.12. This magnetic equivalent of a double layer is called a magnetic shell. If we look for this shell by following a magnetic line of force produced by an electric circuit, we find that we pierce the same surfaces repeatedly, without arriving at any magnetic discontinuity corresponding to the electric double layer. Thus, the magnetic shell has no physical existence and is merely a convenient device of use in computing magnetic fields. Furthermore, this
282
MAGNETIC INTERACTION OF CURRENTS
§7.02
shows that there are neither sources nor sinks of a magnetic nature corresponding to the electrostatic charge so that divB=V•B= 0
(1)
The work required to take a unit electric charge from one face of a uniform electric double layer of strength Ae/ to the other, by a path passing around its edges is i.4/, by 1.12 (3), since the difference in the solid angle subtended by the layer at the two ends of the path is 4r. Thus, if the electric intensity is E, we have 5E • ds when ds is an element of the path. As we have seen, the magnetic induction due to a current I is everywhere the same as the electric intensity due to a uniform electric double layer of strength AEI, having the circuit as a boundary, so that, in vacuo, f B • ds = Ad", and in a medium of uniform permeabilityµ we would have (2)
JCB • ds = µI
But from 6.00 (2), we have I = fi•ndS where i is the current density and S is the area inside the path s, thus we have f B • ds = Ali • n dS. We can transform the left side by Stokes's theorem [3.01 (1)], giving x B • n dS = fsi • n dS
(2.1)
It is easy to see that this holds for any path, even if it includes only part of the current; for in that case either that part of the shell due to currents not encircled will be crossed twice, once in the positive and once in the negative direction, or it will not be crossed at all. In either case, it contributes nothing to the integral. Thus, this relation holds for the integrands alone, giving in a region of uniform permeability, V x B = µt
(3)
7.02. Magnetic Vector Potential. Uniform Field.—It is well known that the divergence of a vector which is itself the curl of another vector is zero and so satisfies 7.01 (1). Thus, we may define a new vector A, which we shall call the magnetostatic vector potential as a vector whose divergence is zero and whose curl is B, thus (1)
VxA=B V•A=0 The standard formula for curl curl A now becomes V x [V x = V(V • A)
—
(V • V)A =
—
V 2A
Substituting in 7.01 (3), we have, in a region of uniform permeability, pm = (2)
§7.02
MAGNETIC VECTOR POTENTIAL. UNIFORM FIELD
283
On writing out the components, this is iv2A z N 2A, + kVA, =
iiy
kiz) (3) Thus A, Ay, and Az all satisfy Poisson's equation [3.02 (3)]. We have found a solution of this equation in 3.09 (1) to be
As =
f iz dv jy r
A y= A r dv, A 47t-
r
A= A
dv
dv jv r
(4)
Adding components gives 471- jv r
(4.1)
In the region outside the wire, i is zero and since, in a thin wire of crosssectional area dS, we have iz dv = i dS dx = I dx, we may write, summing up the vector components,
A = p, 47r
ds r
(5)
Physically, this is a more satisfactory concept than the magnetic shell since it depends only on the configuration of the circuit, the current strength, and the location of the point of measurement and involves no artificial discontinuities. It will give the correct value for the line integral of B for any path. Evidently, we can consider the vector contribution to A by any element of the closed circuit as parallel to that element. The vector potential in rectangular coordinates associated with a uniform magnetic induction B in the x-direction has only two components Ay and A, and their values are Ay = — aZ.B Az = (1 — a)yB (6) where a is an arbitrary number. Obviously, the x-component of the curl is B and the other components are zero. In spherical coordinates, the vector potential of a uniform field parallel to the 0 = 0 axis is A4, Br sin 0
(7) In cylindrical coordinates, the vector potential of a uniform field parallel to the z-axis is Ao = ABP (8) In oblate spheroidal coordinates, the vector potential of a uniform field parallel to the = 1 axis is Am ciB = 23 (9) In prolate spheroidal harmonics, it is c2B Am=2Pi()Pi(,)
(10)
284
MAGNETIC INTERACTION OF CURRENTS
§7.04
The forms of A given in the last five formulas are not, of course, the most general forms that give the required B. We could add to each the gradient of a scalar without affecting B. 7.03. Uniqueness Theorems for Magnetostatics.—In 3.10 we found just what information concerning the interior and boundary surface of a region is necessary to determine uniquely the electrostatic potential inside it. This proof applies without change to the magnetomotance discussed in 7.28, provided there are no electric currents within the region which would make it multiple valued. The vector potential requires a different proof. We shall show that, if the location and magnitude of all fixed currents inside a closed surface are specified as well as the value of the tangential component of either the vector potential or the magnetic induction over this surface, then the value of the magnetic induction everywhere inside it is uniquely determined. The contribution of internal sources given by 7.02 (1) and (5) is unique. Suppose there were two internal values B and B' having identical external sources and boundary values and derived from A and A'. If V • A = V • A' = 0, then V2A' — '7 2A = — Ai and V x V x (A — A') vanishes in v so that putting
IP =
= A — A'
in 3.06 (5) gives
fv(B - B02 dv
si(A — A ') • [(B — if) x = f =1
dS,
(1)
If, on 8,nxA=nxA'orBxn=B'xn, then the surface integral is zero so that in either case B = B' throughout v because (B — B') 2is positive. If B — B' is zero throughout the volume, then A and A' can differ therein only by the gradient of a scalar. By 3.10, if the gradient of a scalar is zero over S, it vanishes in v so that if A — A' vanishes over S it vanishes in v and A is uniquely determined therein by its value over S. 7.04. Orthogonal Expansions for Vector Potential.—In electrostatics, after finding the potential due to a fixed charge distribution, we superimposed suitable perturbing potentials to satisfy the conditions at dielectric or conducting boundaries. To use the same method with vector potentials, it is necessary that, having found by 7.02 (4.1) that part of the vector potential due to the given current distribution, we know suitable forms of perturbing potentials to superimpose to meet magnetic boundary conditions. The direct method of solving Laplace's equation for the vector potential is not easy for, where Laplace's operator is applied to a vector, it operates not only upon the magnitudes of the components of the vector, but also upon the unit vectors themselves, as illustrated in 10.05 (2) and (3). Except in rectangular coordinates, this gives rise to a set of three
§7.04
ORTHOGONAL EXPANSIONS FOR VECTOR POTENTIAL
285
simultaneous partial differential equations whose solution may be very complicated. We are led, therefore, to look for a simpler method. We have seen [3.02 (1)] that in free space where there are no electric charges the divergence of the electric field is zero and, since from 1.06 (4) it is the gradient of a scalar, its curl is also zero. Similarly, from 7.01 (1) and (3), in free space, where there are no electric currents, the divergence and curl of the magnetic induction are zero. We would expect, in such regions, to write expansions for the two fields in terms of orthogonal functions, in the same form. In Chap. V, we obtained these expansions by solving Laplace's second-order partial differential equation. This equation was broken up into three total differential equations, each involving a single coordinate, connected by two indices. Thus, each term in the expansion involves two indices and six integration constants. As just pointed out, owing to the mathematical similarity of electric and magnetic fields, we should expect that part of the vector potential which contributes to the magnetic induction to have the same number of indices and constants. This potential should be derivable from three scalar potential functions because, in rectangular coordinates, each component of the vector potential satisfies Laplace's equation. These components are not, however, independent, but are connected by the relation V • A = 0, so that only two independent scalar functions, at most, can be used. Furthermore, as just pointed out, when properly chosen, only one of these will contribute to the magnetostatic field. The general expression for the vector potential, giving zero divergence, is A =VxW
(1)
where W is a vector which should be derivable from two scalar potential functions. To fit boundary conditions, when dealing with eddy currents and electromagnetic radiation, we shall find it convenient to split W into two components, normal to each other, each of which is derivable from a different scalar potential function. Thus we shall write W = ufri u x VW2
(2)
where V2 W1 = V2 W2 = 0 and u is an arbitrary vector to be chosen so that V2A = v2(v x W) =V x (O2 w) = 0 (3) This holds when V2 W is replaced by V2 W CW unless C depends on x, y, z. Now it is easily verified, by writing out in rectangular coordinates, that v2uw1 = uv2w1or uV2 W1 217 Wi and V 2 (u x VW,) = u x V (V2 W2) (4) where u = r, i k, or r, so that these are suitable choices of u. From the similarity of B and E, we suspect that B, like E, can be obtained from a single scalar potential function. This surmise is confirmed by examining ,
286
MAGNETIC INTERACTION OF CURRENTS
§7.05
the contribution of W2 to A. Thus we have u = .i, or k, V x (u x VW2) = u(V2W2) - V(U • VW2) (5) u = r, V x (u x VW2) = u(V2W2) — V(u • VW2) — VW2 (6) Since V2W2 = 0, the part of A derived from W2 is the gradient of a scalar and contributes nothing to B in the magnetostatic case. If u1, u2 and u3are orthogonal curvilinear coordinates, if the u specified above lies in the direction of u1, and if WI is of the form U(u1)F(u2, u3), then B . A = 0. This is proved in Lass, pp. 48 and 57. In the following articles, we shall use the theory just developed to obtain solutions of the equation pm = 0 (7) in the form A = ulU11U12U13 u2 121U22U23 u3 U31U32U33 where ul, u;, and u; are unit vectors in the directions of the coordinates u,, u2, and u3, and Ursis a function of u8only. The solution should be in such form that the vector potential anywhere inside a volume containing no sources and bounded by a set of surfaces, on each of which one of the coordinates is constant, can be calculated when the value of its tangential components on these surfaces is given. at; 7.05. Vector Potential in Cylindrical Coordinates.—In Chap. V we found that the general solution of Laplace's equation in cylindrical coordinates could be built up of a sum of terms involving, except in particular cases, Bessel functions. We now wish to find analogous solutions for the vector potential which possess orthogonal properties on the surfaces of a right circular cylinder so that we can express the tangential components of the vector potential thereon as a sum of such solutions and thus determine its value at any interior point. Choosing the Bessel function solution of VW' = 0 given in 5.291 and setting u = k, the W of 7.04 (1) becomes W = kic-1(Aeks
Be--kz)[CJ„(kp)
DY„(kp)] sin (744 4- an)
(1)
The vector potential derived from this by 7.04 (1) is
A p= —(Aekz
Be-kz)[C.1,.(kp)
A4, = (Aekz Be—kz )[C4(kP)
DYn(kp)1n(kp)-1cos (nd, ± on) (2) DY (kp)] sin (nd, +on)
This is the orthogonal surface vector function defined by 5.296 (8). If, for a given z-value, either component of k x A vanishes when p = a then at this z-value k x A may be expressed as a sum of such functions. A rarely used form containing ¢ is obtained by setting 5„ = 0, letting n --> 0 and keeping the products of nA and nB finite. In order to obtain suitable solutions for expressing the tangential components of A over the curved surfaces, we use LLie forms of 5.292 (3)
§7.06
VECTOR POTENTIAL IN SPHERICAL COORDINATES
which are orthogonal in z and 4. This gives in place of (2) yk) cos (ncl, A„ = -[aIn(kP) DKn(kP)1n(kp)-1cos (kz cos 7h) sin (kz (n4i A4, = [CI:,(kP) DIV,,(kp)]
287 5.) (3)
The z-component satisfies the scalar Laplace equation and is written Aa = [C'I,,(kp) D'Kn(kp)] cos (kz 71) cos (ncb (5:,) (4) The above solutions are inadequate when k = 0 and n 0 0. For the curved surface, solutions corresponding to (3) and (4) are A,, = (Apn-1 Bp-n-9 (Cz D) sin (n4 + on) A A pn-i + Bp n 1) (UZ D) cos (n0 + an) A. = (A' pn D') cos (nci, B'p-n)(C'z (5) 0, there are no solutions orthogonal in both p and 4) suitable for When k = the end surfaces. When both k and n are zero, some forms of interest can be found from 5.291 (9) by taking Cz D) In p W = k[(Az4, (Ez F)4,] rGz (6) A,, = Ap-'z In p Bp-' In p E p-1Z F -
-
A4, = -Ap-14 - Bp-'0 A, = (GO H) In p 141
- Dp-1 -1-Gp (7)
Other solutions not often needed are found by using i, i, or r in W. 7.06. Vector Potential in Spherical Coordinates.—In Chap. V the general solution of Laplace's equation in polar spherical coordinates was built up of a sum of terms involving spherical harmonics. We need analogous solutions for the vector potential which possess orthogonal properties on the surface of a sphere so that we can express the tangential components of the vector potential thereon as a sum of such solutions and thus determine its value at any interior point. Choosing the spherical harmonic solution of V2W1= 0 given in 5.23 and setting u = T, the W and A of 7.04 become W = T(Arn Br-n-9[CP;n,(cos 0) + DQ„ 'n(cos 0)] sin (m4, 8„,) (1) A B = (Arn Br-n-1)[CP„m(cos 0) + DQ"'„ (cos 0)]m csc 0 cos (m4, + 8,n) = (Arn+Br-n-9[CP"'„' (cos 0)-F INT' (cos 0)] sin 0 sin (m4,+ am) (2) This A is the orthogonal surface vector function mentioned in 5.231. When m = 0 and 8„, = -br, this equation becomes r Br-71-1)LCPVCOS A4, = (Arn 0) + DQVCOS 0)] (3) It is often useful to have B in terms of W. If u = r we have
B = V x (V x rW) = -V x (r x VW) = (VW) + 2vW Writing out the components of this equation gives a2 (rW) a2(rW) 1 a 2(rW) B - are Bo B,,, - . r ar ae' r sin 0 ar act.
(4)
288
MAGNETIC INTERACTION OF CURRENTS
§7.08
7.07. Vector Potential in Terms of Magnetic Induction on Axis.— Magnetic lenses for focusing electron beams are usually constructed from coaxial current coils or equivalent magnets so that the vector potential has only a 0-component. To calculate the focusing properties of such lenses, it is convenient to have the vector potential expressed in terms of the magnetic induction and its derivatives along the axis. If W = kW, where W is the solution of Laplace's equation given in 3.15 (1), 7.04 (1) yields, if •If is written for 43', A = —
aw ap
f
=
2a
\If (z jp sin 0) sin 0 d0
27r r,
(1)
where xlf(z) is a real function of z. The Taylor expansion is, by Dw 39, a2*(z) (jp sin 0)2 + • • axli(z) jp sin (2) jp sin 0) = xlf (z) xlf (z 2! az2 1! az Substitution of (2) in (1) and integration from 0 to 27r give a 2n+14/ ( z)(pyn+1
— A—
n!(n + 1)!
a z,-+i
2)
(3)
n=0
The magnetic induction V x A on the axis where p = 0 is Bo(z) = [ 1 a(pA4,)
p ap
1 =0
_ .34/(z) az
(4)
This, when put in (3), gives Ao(p, z) in terms of Bo(z) and its derivatives. ea —
n!(n
—
a2nBo(Z)
1)! az2n
fp \ 2n+1
(5)
n=0
The current elements that generate fields of this type are coaxial circular current loops so the range of p and z over which (5) holds are the same as for such loops. To find the range insert Bo(z) from 7.10 (8) into (5) and compare the result with that obtained from problem 27, which holds for all values of p and z, when J1(kp) is expanded by 5.293 (3), the summation and integration are interchanged, and the integrals are replaced by derivatives of 5.298 (6). The identity of the results shows that (5) gives a unique vector potential at all values of p and z which can be reached from the axis without crossing a current sheet. The restriction is needed because the external fields of any currents inside a closed surface are unaffected if these currents are removed, provided suitable currents are set up in the surface. 7.08. Equation of Axially Symmetrical Tubes of Induction.—A magnetic field is most easily visualized by mapping the stream lines or lines of magnetic induction. When we have axial symmetry so that all
§7.09
VECTOR POTENTIAL AND FIELD OF BIFILAR CIRCUIT
289
sections of the field made by planes passing through the axis look the same and all currents are normal to these planes, we can obtain the equation of these lines very simply from the vector potential which, in this case, has only the component Ad,. When rotated about the axis, each stream line generates the surface of a tube of induction. Any particular tube may be specified by stating the flux N through it. This is obtained by integrating the normal component of the induction over any cross section S of the tube. Thus the equation of the surface of such a tube is given by N = fsB• n dS f(p, z)
(1)
By means of Stokes's theorem [3.01 (1)], this becomes N= fsV x A • n dS = A • ds
(2)
If the section is taken by a plane normal to the z-axis, then the path s is a circle on which Aois constant and its length is 27rp so that the equation of the tube of force becomes N = 27rpAo (p, z) (3) In an axially symmetrical system of orthogonal curvilinear coordinates ui, u2, and 47, the distance p will be a function of u1and u2 so that the equation of the tubes of force is N = 2lrp(ui, u2)A4,(ui, u2) (4) 7.09. Vector Potential and Field of Bifilar Circuit.—We shall first apply 7.02 (5) to finding the magnetic field due to a long straight wire with
a parallel return at a distance a from it. Since all elements of the wire lie in the x-direction in Fig. 7.09, the vector A has only the component A. Thus, we have, from 7.02 (5), A. = pl I+
— dx = — _ ri 2rfo r2 a/ In ri x , a2 — =—n i — 2r r2 x0 27r al
[(a?
+
x2 ) — i
—
(4
+
x 2) -1] dx
(1)
290
§7.10
MAGNETIC INTERACTION OF CURRENTS
This shows the surfaces of constant vector potential to be circular cylinders coinciding exactly with the equipotentials in the electrostatic case where V = —q(2.2-€)-1In (a2/a1) if the two wires carry charges +per and —pd. For p/ = 27, the equation of these cylinders is given by 4.13 (3), where A. is substituted for U and z for x. The magnetic field which is given by V x A is at right angles to the electric field given by V V but has the same numerical value. From 3.04 (1), (2), and (3), the components of the magnetic induction, are, in vacua, since h1= h2 = h3 = 1,
aAz
aA„
aAz ay
B. = — — = 0, By = az ay az Since a1= [(y — ia)2 z 2]1and az = [(y By pIzt
1 W V4
B. = — 2,.
B. = --
ia)2
z94, this gives
1) 2 az
(3) y— 1a) a2
(4)
7.10. Vector Potential and Field of Circular Loop.—Let us compute the vector potential at the point P shown in Fig. 7.10. From symmetry, we know that in spherical polar coordinates the magnitude of A is independent of 4,. Therefore, for simplicity, we choose the point P in the xz-plane where q5 = 0. We x notice that when equidistant elements of length ds at + and —95 are paired, the resultant is normal to pz. Thus, A has only the single F IG. 7.10. component A. Let dsobe the component of ds in this direction, and 7.02 (5) becomes a cos 4,c/4, 2rio (a2 p 2 +Z2 — 2ap cos 40)1 z2)i >> a, If our loop is very small, this becomes, since r. = (p2 = pI redS5
40) r
A=
era cosq5t1 apcos ¢1do_ a2pI p _ a2A/ sin 0 4r3 r?, I 4r2
27rjo ( r.
(1)
By taking, from 7.00 the magnetic moment M of the loop equal to ra2/ and directed upward, this may be written
A = p(M x r) (4z-r3)-1
(2)
If this approximation does not hold, let 4, = r + 20 so that d4, = 2 dO
§7.10
VECTOR POTENTIAL AND FIELD OF CIRCULAR LOOP
291
and cos 0 = 2 sin2 0 — 1, then this becomes (2 sin2 — 1) d0 gaIf iT 414, = — z2 — 4ap sin2 7 0 [(a + p)2 Rearrange and let z2]-1, p)2 k2 = 4apRa m = [1 — (1 — k2)i][1 -F (1 — k2)11-1 l k3( 1 43k2 7258k4 240 = 14 p [(1 — 21—10)K — E]= 32W +
(11
= LI L-1(--a—Y [K(m) — E(m)1 = mp 4
+ -83-m2 + 6145 m4 +
p
where K and E are complete elliptic integrals of the first and second kind. When z = 0, the m modulus is simply p/a or a/ p. To determine the magnetic induction, we must write, from 3.04 (1), (2), and (3), the components of the curl in cylindrical coordinates. From 3.05, this gives h i = 1, h2 = p, and h 3 = 1 so that
a 1 a 8A4, Bp= ---(pAgs) + --(A = — az paz pact. )
B# =-aa--i(A„) — 4(Az) = 0
( 5)
1a a 1 a B. = —(Ap) +— —(PA9s) = — — (P214,) P ad) P ap P aP For the derivatives of K and we use the formulas Dw 789.1 and 789.2
K OK_ E — k 2) T ak
and
aE _E K al T Tc
We also have, from (3), that
ak_ — = -az
4ap
ak = k k 3
and
Op 2p 4p
4a
Carrying out the differentiation, collecting terms, and substituting for k give
Bp = 27 °' pRa + p)z 2
E + z2P
K + a 2 ± p2 ± z2] 2E (a — p) 2 +z
2 2 22 E E K+(a a— p 4_ zz 2 .B. = 27 PI 4_ pr ± z z i[ _ p)2 [(a
(6)
—
]
(7)
Numerical values of B, and B. can be computed for any values of p and z by finding k from (3) and looking up the corresponding values of K and E in a table (Dw pp. 208-210 or Pc p. 121). On the axis p = 0 we have 0
Bp = — —> 0 o
and
B.— ba2i
(a2 + z2)1
(8)
292
MAGNETIC INTERACTION OF CURRENTS
§7.11
7.11. Field of Currents in Spherical Shell.—We define the value of the stream functionI,G, at any point P on a thin spherical shell of radius a, to be the total current flowing across any line drawn on the surface of the shell between P and a point where IP is zero. The components of the current density are therefore related to IP by the equation —1 a,p a‘p io a sin (1) 0 a cp = ae We wish to find the vector potential and magnetic field due to these currents. Since any possible i/i can be expressed as a sum of surface harmonics, it will suffice to calculate the field of the distribution i'> = 4>), for we can then, by superimposing, obtain that due to any 1,G. We shall designate the induction outside the shell by Bo and that inside by B. Let us apply 7.01 (2) to a small circuit around an element of shell of length AO, lying in a constant (IS plane. We obtain pica De = [(No — (130),]a AO Using (1) and introducing the scalar function W, of 7.06 (4), we have
a‘p a2(rW 0) ae = ar ae
a2(rWi)
ar ae
Similarly, taking a circuit in the q5 direction gives a,p = —1 0 2(rWo) 1 a2(rWi) sin 0 ar sin e ao sin 0 ar 3 4 If we multiply the first equation through by d0 and the second by sin 0 dcp and subtract, both sides are total differentials so that, if we integrate and remember that Wo = Wi when r = a, we obtain
aaw 0
1.4
—(rW — rWi) =
aw)
(2) ar — ar To give 4> the harmonic form S:(0, 4>), to make W0 = W, at r = a, and to have Wo finite at infinity and Wifinite at the origin, we must choose, from 7.06 (1), the forms n-1-1 pc (a — W0 2n ± 1 r) Sn (0' 4)), — 2n + 1(a) 877(8' (k) (3) Hence, from 7.04 (1), the vector potential has the form ia\"+1a in = 0, (2n 1) sin 0?-) . p[A3'1(°' C6)] m aNn+1 /.1 (0, 4)] "6°' — 2n + —
ar
—
("0
(A,.); = 0, '
—
i = (2n ("
sin t9(arY46[S:(°' 4>)]
— AO ae 2n + 1 a nl[S709,
(4)
ZONAL CURRENTS IN SPHERICAL SHELL
§7.12
293
The 0 terms in A gs and A Bhave exactly the form of the first and second terms of 5.231 (7), so A is an orthogonal surface vector function on the spherical surface. 7.12. Zonal Currents in Spherical Shell.—When the currents in a spherical shell flow along parallels of latitude, we have the most important practical case. In this case, axial symmetry makes all quantities independent of cp. The stream function may then be written CaP„(cos 0)
=
(1)
n=1
The current density having only a 4, component, becomes
1 84, 4= a ae —
c'inaPn(u) a ae
Ca sin0 aPn(u) a au
n=1
enPl(u) (2) — —— a
n=1
n=1
where u = cos O. The function W, if all space has the permeability it, takes, from 7.11 (3), the form . . wo = --1 —AC7, ia\n+ip r Wj = (3) W° 2n + 1 a 22n + 1W n(u)' —ACC _ n=1
n=l
The vector potential, from 7.11 (4), may be written ra
A = (141
n=1
(4)
211 C+Y 1(a) P(U)
(arr
A=
2—± Cni iP(U) (5) n=1 n We may note here that (43)x = — sin q5 and (+)„ = cos 4' so that Az and Av satisfy 7.02 (3) in free space where i = 0. We also see that A has the form given in 7.06 (3). The components of the magnetic induction may be obtained from (3) or (4) and (5). Thus, by 7.06 (4), r a, we obtain
F = ---„II, r sine ,3
1 ay-",,i(cos ,),(cos 3) n + 1(b
(4)
n=1
7.20. Vector Potential and Magnetization.—The definition of the vector potential given in 7.02 (4.1) has proved adequate for all the cases so far treated in which the whole region is occupied by a medium of a single constant permeability. If p is variable or discontinuous, it is necessary, in order to define a magnetic vector potential uniquely, to consider the nature of magnetization. In 12.06, we shall give experimental evidence that magnetization is due to circulating currents or spinning electric charges within the body. We define a vector M to be the magnetic moment per unit volume of such currents or spins and call it the magnetization. With the aid of 7.10 (2), setting A = At, and measuring r from the field point the vector potential of M becomes
Am = — Pv iM xr dv = 14 f M 4r v 4r v r3
r
(1)
We can transform this equation by using the vector relations v x pq = Vp x q + pV x q
(2)
s fvVxMdv = fnxMdS
(3)
and
where n is the unit outward normal vector to the surface S which encloses v. We prove (3) by putting A= Mxa in Gauss's theorem [3.00 (2)] where a is a constant vector so that V • (M x a) = a • V x M, giving a•f VxMdv = fMxa• ndS = a• fnxMdS
v
s
s
Since this is true for all values of a, Eq. (3) follows. Applying (2) and (3) to (1) gives 4ir A --tam— ilv
VX VX Mxn M dv I Vx (1 dv = Iv r M dv ± si r dS (4) r fv r v
When the magnetization is uniform in the interior of a body, (4) is more
§7.21
MAGNETIC BOUNDARY CONDITIONS
303
useful than (1) since the volume integral is zero. If i is the actual current density, then the total vector potential is now given by
n
A=14 f i+V xM dv-F Av 471- fs Mr dS 4ir v
(5)
In order to apply this formula in most cases where the data include only the current distributions and the permeabilities, we must express M in terms of g and B. To do this, we note that, at great distances from closed circuits of finite extent, the induction and, hence, the magnetization are inversely proportional to the square or higher powers of r so that the surface integral in (4) vanishes. If A is constant in this region, and M is due entirely to i the vector potential must also be given by 7.02 (4.1) so that, comparing it with (5), we have = Ar(i + V M)
(6)
In a medium that is isotropic but not homogeneous we defineµ so that this equation is still valid. With it we eliminate Ai from 7.01 (3), and because the resultant equation must reduce to 7.01 (3) when g is constant and it is known experimentally that M is proportional to B and in the same direction, we have
vx
= = vx
\. 11 M=
(1 1.)B Au
A
(7) (8)
Just as we can get 7.01 (2) from 7.01 (3), we get from (7) the relation /
• ds =I A
(9)
7.21. Magnetic Boundary Conditions.—In the last article, we defined the vector potential A for regions where the magnetization is not uniform and on surfaces where it is discontinuous. To find the boundary conditions that A must satisfy, we observe that each of the three components of A is defined by a scalar equation analytically identical with 1.06 (6), which may be written in vacuo f p dv vr
r dS js r
(1.06) (6)
This defines the electrostatic potential in free space due to a volume distribution of electric charge of density p and a surface distribution of density a. From electrostatic considerations, we know the value V, of these integrals just inside the surface S is the same as the value V. just outside. Furthermore, by applying Gauss's electric flux theorem
304
MAGNETIC INTERACTION OF CURRENTS
§7.21
to a small disk-shaped volume fitting closely an area dS of the surface and so thin that dv is negligible compared with dS, we have, after canceling out dS, the relation
i av ay° — =
an
an
Ev
Thus, we know the boundary conditions that apply to integrals of the form of 1.06 (6) and hence to each component of 7.20 (5). Adding the components and writing 1.4 for 1/e„, we have Ao = Ai
(1)
and with the aid of 7.20 (8),
aA0
3A, Av(M x n)
Av[(V x Ai) x n]
=
(2)
If there is a magnetization M' on one side of the boundary and M" on the other, we may find the boundary conditions by imagining a thin layer of permeability between the boundaries, writing down (1) and (2) for each boundary, referring them to the same normal, and eliminating Ao and 0Ao/an, giving A' = A" (3) and
aA" WI,
an
liv[(111/
(4)
3/1") x
To get the second boundary condition in terms of the permeability, we write n • V for a/an and use the relation n • V (A' — A") = V[n • (A' — A")] — n x [V x (A' — A")] — (A' — A") • Vn (A' — A") x (V x n) —
The last two terms drop out because of (3), so that, writing B' for V x (A' A"), by 7.02 (1), and rearranging, we have
—
B"
—
V[n • (A'
—
A")] = n x (B'
—
,uoM'
—
B"
vM")
Substituting from 7.20 (8), we have V (A',
—
AZ) = Avn
B' B")
—
From (3) A'n — AZ is equal to zero over all the boundary so that its gradient along the boundary is zero. Thus V(A'n — A'„') is a vector normal to the boundary. But the right side is a vector directed along the boundary surface so that writing out the tangential components of this equation gives B' B" nx — = 0 (5) A
§7.22
EXAMPLE OF THE USE OF
A
AND A
305
We may now write down the boundary conditions on the components of A in the orthogonal curvilinear coordinate system u„ us, and u,, treated in 3.03, 3.04, and 3.05. Let us take ur to be constant on the boundary; then, from (3), we have
AT =
,
A's = A's',
and
A; =
(6) Writing V x A for B in the left side of (5), we have, by 3.04 (1), 3.04 (2), and 3.04 (3), the two equations
Fa (hr, L aut,,
a (hag u t,,)1 = 1[a(hr. sA"r ,$) rA r,:t,c) j aut, r These equations (6) and (7) give the required boundary conditions. To obtain the boundary conditions on B, we note that (3) shows the differences in the vector potential between two points on the interface to be the same in either medium. Hence, the derivatives of the vector potential on the two sides of the interface, in the same direction parallel to it, are equal. The vector n x A lies in the interface, and so V • n x A involves only such derivatives. We have the well-known vector equation for the derivative of a cross product s)
(ht
att„,
V •nxA= When we substitute A'
—
—
n•VxA+A•Vxn
(8)
A" for A, the last term vanishes by (3) giving
n•VxA' = n•VxA"
(9)
Writing B for V x A gives n•B' = n • B"
(10)
Thus, the normal components of the induction are continuous. From (5), we have for the tangential components of the induction 1 -7
1 (n x B') = ,7 (n x B") —
(11)
It is often possible to choose two vectors A' and A", different from and simpler than A' and A", whose curl gives the same induction at all points but which, as will be shown in 8.04, satisfy, instead of (3), the boundary conditions As' and (12) A',.' A '„t = It is evident in the example considered in the next article that these vectors may be more convenient for computation than A' and A". They are not uniquely defined, however, as are A' and A", by integrals of the type of 7.20 (5). We shall refer to them as quasi-vector potentials. 7.22. Example of the Use of A and A. As an instructive example of the use of the boundary conditions in the last article, let us consider an infinite wire carrying a current I in the z-direction, positive and nega—
306
§7.23
MAGNETIC INTERACTION OF CURRENTS
tive values of z having permeabilities A' and A", respectively. Clearly, from 7.02 (5), the simplest vectors A' and A" that we can write which satisfy 7.21 (7) but not 7.21 (3) and whose curls give us the correct values of B' and B" are A' = A" = — kG-A--144" In p) A' In p) and (1) The vector defined by 7.20 (5) requires that A' = A" at the boundary. The part of the volume integral involving I in 7.20 (5) is given by 7.02 (5). We must superimpose on this a potential to meet the new boundary condition. Inspection of the solutions available in cylindrical coordinates in 7.05 (7) shows that the suitable form is cz/p. Putting this in and evaluating the constants, we obtain A' = Fr-111 - kA,(In p (2) ILOZP-1 + C211 +e A" = -kA,,(1n p C1) eRA" — C2]) (3) The constants CI and C2 are arbitrary. The second terms represent the contribution from the integral of 7.20 (1). We get the same value of B' and B" from (2) and (3) as from (1). 7.23. Current Images in Plane Face.—The similarity of the boundary conditions for B in 7.21 to those for D in 1.17 suggests that the image method of 5.05 might be used in getting the magnetic field when an electric circuit is near the plane face of a semi-infinite block of material of permeability A. Let the plane face be at z = 0, the circuit lie in the region of positive z, and the material of permeability A occupy the entire region of negative z. Let A be the vector potential due to the circuit alone. The image law of 5.05 suggests that the quasi-vector potential above the interface can be given by A + A' and that below by A", where y, z) kf 3(x, y, z) A= y, z) (1) A' = iCY1(x, y, -z) .iCLf 2(x, y, -z) kCLI3(x, y, -z) (2) A" = kCI73(x, y, z) y, z) iC'2'f2(x, y, z) (3) A'; and A.,, A'z'„ = A'x',,„ or From 7.21 (12) at z = 0 Az + A; (4) 1 + CI = 1 + = 1+ {(
—
The boundary conditions at z = 0 on the derivatives given by 7.21 (7) are satisfied if we have 14(1+ CD = A„C3 (5) u(1- CD = Pven ti(1- CD = have the same form, Solving, we obtain, since A" and A
Ci = C2 =
-
-
-r Av
-
-
2A A
,
uo
(6)
Thus, the magnetic induction outside the permeable medium appears to be caused by two circuits, the original carrying a current I, and an
u image circuit carrying a current I' -
7 u„I.
AL 'I” Ao
The direction of I' is
§7.25
TWO-DIMENSIONAL MAGNETIC FIELDS
307
such that the projections of I and I' on the interface coincide in position and direction. In the permeable medium, the magnetic induction appears to be due to the original circuit alone but carrying a current I" = 2Iµ/(µ ± Ay) instead of I. 7.24. Magnetic Induction and Permeability in Crystals.—The similarity of the boundary conditions derived in 7.21 for B to those derived in 1.17 for D suggests that the method of measuring D and E in 1.18 may be applicable also to B and µ. Let us excavate a small disk-shaped cavity in the solid, whose thickness is very small compared with its radius so that the induction in the cavity, far from the edges, will be determined entirely by the boundary conditions over the flat surfaces. If we orient the cavity so that the induction inside is normal to the flat faces, then from 7.21 (10) this B equals that in the medium. Let us now excavate a long thin cylindrical cavity so that the induction inside, far from the ends, is determined entirely by the boundary conditions over the curved walls and orient it so that this induction is parallel to the axis. We now find, from 7.21 (11), that BIliz inside equals B/A in the solid. These two measurements give us both B and A. If we perform this experiment in a magnetically anisotropic medium we shall find, in general, that B and B' are in different directions. There will be at least three orientations of the field with respect to the crystal for which B and B' have the same direction. These are the magnetic axes of the crystal, along each of whichµ may have a different value. In the more common case, B and B' have the same direction for all orientations of the field in some plane, and for the field normal to it. This subject is further discussed in 9.02 and 9.03. 7.25. Two-dimensional Magnetic Fields.—In rectangular coordinates, the boundary conditions for the tangential component of the magnetostatic vector potential are identical with those for the electrostatic scalar potential when e is replaced by 1/A. In two-dimensional problems, involving only magnetic fields parallel to the xy-plane, the currents must flow in the z-direction and so the vector potential can have only one component Az, which is, necessarily, tangential to all surfaces. When WA, becomes very large, we see, from 7.21 (7), that the surfaces on which Az is constant, i.e., the equivector potential surfaces, are normal to the boundary. Thus, at this point, Az behaves as the electrostatic stream function does at an electric equipotential boundary. We also obtain B from A. in the same way as we obtain the electric field from the stream function; for, from 3.04 (2) and (3), we see that, in rectangular coordinates, where h1= h2 = h3 = 1 we have
B. =
OA
ay
These are the same as 4.11 (1).
By =
OA.
ax
(1)
308
§7.26
MAGNETIC INTERACTION OF CURRENTS
"Where there is no current, 7.02 (3) becomes v2Az = 0
(2)
so that all the methods of Chap. IV are available for its solution. 7.26. Magnetic Shielding of Bifilar Circuit.—To illustrate the application of the results of the last article, let us compute the magnetic induction outside a cylindrical shield of permeability 1.1, with internal and external radii a and b which surrounds, symmetrically, two parallel conducting wires carrying currents in opposite directions. Evidently, the problem calls for the use of the circular harmonics of 4.01. In 7.09, the vector potential due to two wires carrying currents in opposite directions was found to be
AZ = 14"T In r2
(1) r1 By 4.02 (1), setting Oo = 0 and 00 = r a==d po = c, we have the expansions, when p > 2r
1 c ' -(- cos nO + In p n p
In r1 = — n
_1(cy( — 1)n cos nO + In p n p
In r2 n =1
When we take the difference, even terms go out so that, writing 2n + 1 for n, we get 2..-1-1 cos (2n + 1)0 (2) In = ri 2n + c\ n=o Let us now superimpose a potential inside the shell, due to its influence, which is finite at the origin, giving in this region /2n-F1
Al z = iiii
A2.+1.p2M-1
.[
+2n 1+ 1
o In the region 2, of permeability 11, have n=
A2. =
—
7i n=0
p
cos (2n + 1)0
(3)
is neither zero nor infinity so we may
_L
2n-1\)
cos (2n
1)0
(4)
309
CURRENT IMAGES IN TWO DIMENSIONS
§7.27
Outside the potential must vanish at infinity so that cos (2n + 1)0
A3z =
(5)
n=0
These equations must be satisfied for all values of 0 so that each term separately must meet the boundary conditions 7.21 (6) and (7), giving after multiplying through by a 2n-E.' or b2"-", when p = a A2.+1.64n+2
2n+1 = B2,04a41.-1-2 _L
r
(6)
2n + 1c
and p2=12„.4_1a4n+2
C2n+1
2n + 1
AvB2n+
ia4n-1-2
(7)
when p = b B 2,,-Fib4n+2
(8)
C2214-1 = D2z1-1-1
and — ev-2n-F1
(9)
= PD2n+1
Solving for D2„.4_1, we have [
A 3= = 41//
/a 4n+2 1-1 1 (K„,± 1) 2 — (K„
— 1) 2
V,)
Cyn-l-l
2n ± 1.13
cos (2n
+ 1)0
n=0
(10) The field outside is given by 3.04 (2) and (3) where, from 3.05, h1 = 1 and h2 = p to be 1 .0.A aA z B = 139 = (11) " p 80 op 7.27. Current Images in Two Dimensions. The vector potential due to a straight linear current I and the scalar potential due to a line charge q have exactly the same form. Furthermore, as we have seen, the magnetic vector potential and the electrostatic potential, in two dimensions, satisfy the same form of Laplace's equation and the boundary conditions, if 1/Kn, is substituted for K, are the same. It follows that the results of Art. 4.04 apply to a linear current I parallel to and at a distance b from a circular cylinder of permeability pi and radius a. Thus, the vector potential in the region outside the cylinder due to its presence is the same as if the cylinder were replaced by a parallel image current I' located between I and the axis, at a distance a2/b from the latter, plus a current of strength — I' along the axis. The vector potential inside the cylinder is the same as if I were replaced by a linear current I". I' and I" are given by substituting 1/K„, for K in 4.04 (7). These —
310
MAGNETIC INTERACTION OF CURRENTS
§7.28
images lead to the potentials Azi
p 23. If the spheroid in 21 is prolate instead of oblate, show that, when n < no,
A = — +A(n20 — 1)4Z n(nC+ 1) Q;,(no)P;,(n)P.;,(t) 24. In prolate spheroidal harmonics, the coordinates of a loop are to, no. Show that the vector potential due to such a loop carrying a current I is —431A/[(7120 — 1)(1 — tNiln-2(n + 1)-2(2n + 1)P;i(to)(2;.(no)n(n)P.;,(t) 25. In the last problem, a prolate spheroid of permeability A whose surface is given by n = n1is inserted. Show that the additional vector potential due to its presence is given by
2n+1 AL,•/ n(o)Ono)P1(ni)P.(n) ——(K„-1)(7 2 -1)1(1—EDIz (21(n)P1(t) 2 n2(n+1) 2(2;,(ni)P,,,(no) — K4n(ni)P;,(771) where n > no.
PROBLEMS
323
26. Show that the vector potential due to a plane loop of wire of radius a at z = 0 and carrying a current I is, if p > a, mahr-Ifo I1(ka)Ki(kp) cos kz dk If p < a, interchange p and a under the integral. 27. Show that the vector potential due to a plane loop of wire of radius a, carrying a current I, is A4, = +.a/ f Ji(ka)../i(kp)e-kl'l dk
28. A loop of wire of radius a is coaxial with an infinite cylinder of radius b and permeability A and carries a current I. Show that that part of the vector potential outside the cylinder due to its presence is Avahr-i f 43(k)K,.(ka)Ki(kp) cos kz dk 0 (Km — 1)kbI0 (kb),I 2(kb) ~(k)= (K„, — 1)kbK 0(kb)7 2(kb) + 1
where
Write out the vector potential for the region inside. 29. Show that the vector potential, due to a section of length 2c of a circular cylinder of radius a around which circulates a uniformly distributed current I, is, when p > a A ct, =
AzaI f 1
rc o k
i(ka)KOP)
cos kz sin kc dk
30. The plane of a circular loop of wire of radius a, carrying a current I, is parallel to and at a distance b from the face of an infinite plate of material of permeability At and thickness t. Show that the vector potentials on the loop side of the slab, in the slab, and on the other side of the slab are given, respectively, by A l = itiva/f0 Ji(kp)Ji(ka)[e-41z1
A2 =
f i(kp),12(ka)C 0
— 1)(1 — e-2k0ek('-'20 ] dk 1)e-k= —
—
14 1),.-20+ti] I die
A3 = 2/2„Kmal f Ji(kp)Ji(ka)Ce-la dk 0
where p is measured from the axis of the loop, z from the plane of the loop, and C = [(K„, + 1) 2 — (K„, — 1) 2e-2kti-1
31. An infinite tube of external radius aiand infinite permeability is coaxial with an infinite tube of the same material and internal radius a3. In the z = 0 plane between al and a2 there is a current sheet. The current density ict, in the ranges al < p < b1 and b2 < p < a2 is —I[p In (a2/a1)]-2, and in the range b1 < p < b 2 it is I[p ln (b2 /b1 )]-1— I[p ln (a2/a1)]-2 , so that the total circulating current is zero and = 0 on both walls. Show that the vector potential when al < p < a2 and z > 0 is = ZCI.e-k.zRi(k.p) n=1
324
MAGNETIC INTERACTION OF CURRENTS
where R,„(k„p) = Yo(k„cti)J.(k.P) — Jo(k.cii)17.(k.p) and k„ is chosen so that Ro(kna2) = 0 (see HMF, Table 9.7), and C. =
Air2/1R0(k.b2)— Ro(k.bi)M(k.a2)12 41n (b2/bi) [Jo(k.a1)]2— Vo(k.a2)121
32. The potential A = —Boz/p of 7.05 (6) gives only a radial magnetic field Bo/p. Thus by superimposing the potentials In (a2/a1 —F 1 when z > 0 and zt4/z[p In (a2/a1)] -1when z < 0 on the result of the last problem, wipe out the circulating current except in the strip b1 < p < b2 of the z = 0 plane where i4, becomes I[p In (b2/b1)]-'. From 6.16 (2) this is the current distribution for a steady current
in a conducting ring. 33. The permeability of the walls of the region bounded by p = ai, p = a2, and z = 0 is infinite. At z = c there is a current sheet bounded by p = b2 and p = bi in which the current density ict, is I[p In (b2 /b1)]-1. Show from the last two problems that when ai < p < a2 the values of the vector potential, A:0 when z > c and when c > z > 0, are A'm
—Iz(p In — al
+ 2 ZC.e-kn. cosh k„cRi(k„p) n =1
= —Ic(p In — ai
+ 2 ZC„.e-kn' cosh k„zRi(k,,,p) n=1
34. An infinitely permeable rectangular conduit has its corners at x = ±a, y = 0, and y = b. Wires parallel to its axis carry currents I at c jd and —I at f jg. Show with the aid of 4.29 that the vector potential in the conduit is
+ 2[(xi — ci)2 — yncg + d: [(xi — ci)2 + In 47 [(xi — fi )2 + yn2 + 2[(xi — fi)2 — yng; g: sn mx dn my sn mx dn mx sn my dn my 1 — cn2my + k sn 2 mx sn2 my my + k sn2 sn2 my
A. =
Xi =
Similar formulas hold for xi = cl or fl and x = c or f and for yi = di or gl and y = d or g. The modulus k is found from a/b = K(k)/K[(1 — k 2)+] and ma = K. 35. An infinite wire carrying a current I is situated at x = a, y = b, between two faces of infinite permeability y = 0 and y = c. Show that the vector potential between the faces is U=
4r
iry)2 rb 71-(x — a) ir(x — a) . iry In [ (cos — -- cosh cos + sinh2 sin e — c
which is the real part of W=
ir(z — a)] [ rb In cos — — cosh
Note that the field is uniform and oppositely directed when x >> a and when x a. Show that the force pulling the loop toward the z = 0 plane is raI 2 A( c
Zni,i— nra ,..,.d ra 2n2b is.1— sin n=I
c
c
c
46. Show that the magnetic induction in terms of the flux function N of 7.08 (3) and the differential equation satisfied by N where no current is present are
aN — ap
27rpl3z,
aN
— —27rpBp, az
a2N a 2N aN az'
apz
—
pap
0
Note that the sign difference in the last term rules out the solutions of 5.29 (1) with n = 0 but yields instead 2rpAo as given by 7.05 (2) and 7.05 (3). Show that the boundary conditions on N at the interface between Ai and az are N1
= N2,
aNiaN 1 aN 2 aN2) A2(91- + k— ) = P1(91— ±kaz Oz ap 01, References A.
MAGNETIC
and R. BECKER: "Classical Electricity and Magnetism," Hafner, 1950. Gives simple vector treatment. ATTwoon, S. S.: "Electric and Magnetic Fields," Wiley, 1941. Gives graphical method for plotting fields. BucHHoLz, H.: "Electrische and Magnetische Potentialfelder," Springer, 1957. Gives theory and many solutions of special problems. CORSON, D. R., and P. LORRAIN : "Introduction to Electromagnetic Fields and Waves," Freeman, 1962. Well-illustrated exposition of basic ideas. DURAND, E.: "Electrostatique et Magnetostatique," Masson et Cie, 1953. Gives theory with many examples and field maps. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XV, Berlin, 1927. GRAY, A.: "Absolute Measurements in Electricity and Magnetism," Vol. II, Macmillan, 1888. Extensive calculations of fields of circuits and shells. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extensive treatment without vectors. MASON, M., and W. WEAVER,: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Fine treatment of vector potential and boundary conditions. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives extensive treatment with good field maps and no vector analysis. OWEN, G. E.: "Electromagnetic Theory," Allyn and Bacon, 1963. Well-illustrated exposition of basic ideas with simple examples. ABRAHAM, M.,
328
MAGNETIC INTERACTION OF CURRENTS
K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Excellent treatment of vector potential in Chap. VIII. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Gives complete theory of the material of this chapter. WALKER, M.:. "Schwarz-Christoffel Transformation and Its Applications," Dover, 1964. Works out important applications in detail. WEBER, E.: "Electromagnetic Fields," Vol. I, Wiley, 1950. Gives many examples and literature references. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. PANOFSKY, W.
B. See references of Chap. V.
MATHEMATICAL
CHAPTER VIII ELECTROMAGNETIC INDUCTION 8.00. Faraday's Law of Induction.—A century ago Faraday and Henry discovered, independently, that when the magnetic flux N, defined by 7.29 (1), through a closed conducting circuit changes, a current is generated in the circuit. The direction of this induced current is such as to set up a magnetic flux opposing the change. Thus if the flux through a circuit in a certain direction is increasing, the induced current sets up a flux in the opposite direction, and if the flux is decreasing, this current sets up a flux in the same direction. The induced currents always seek to maintain the status quo of the magnetic field. The induced electromotance in volts equals the negative rate of the change of flux in webers per second so that
dN = — dt
(1)
with possible limitations mentioned later. It is immaterial what means we use to change the flux. We may move the source of the flux with respect to our circuit, we may change its strength, or we may move or change the shape of our circuit. Faraday proved (1) for a closed metallic circuit, but since we know that the tangential component of ye = E is the same on either side of the wire surface, (1) must apply equally well to the electromotance around a path just outside the wire. We may therefore assert that (1) holds for any path whatever and write, from 7.29 (1), for a two-sided surface S
fE • ds =
-dt fs B*
n dS
(2)
Applying Stokes's theorem [3.01 (1)] to the left side, we have
n • V x E dS = — f — • n dS sdt B Since this must hold for any arbitrary surface, we have
dB (3) dt in webers per square meter-second. If E is due entirely to electromagnetic induction, it is a special case of the E" of 6.01 and possesses no V xE=—
329
330
ELECTROMAGNETIC INDUCTION
§8.01
sources or sinks so that its divergence is zero. If A is obtained from 7.02 (5), its divergence is also zero. Let us now substitute V x A f or B in (3) and interchange d/dt and the curl, and we see that the curls of the vectors E and —dA/dt are equal. If two vectors have everywhere the same curl and divergence, they differ at most by a constant, so that
E = dA dt
(4)
which connects the electric intensity or electromotance per meter and the vector potential whose change induces it. Equation (1) seems adequate to deal with all cases involving a change of flux in rigid circuits and also with those cases in which the flux is changed by distorting the shape of the circuit, provided that, during the process, all elements of the circuit initially adjoining, remain adjoining. Experiments can be devised, using sliding contacts, to which the application of (1) is obscure or ambiguous. These cases can be treated by focusing our attention, not upon the area surrounded by the circuit, but upon the elements of the circuit themselves, and then applying (4). Another formulation of the law of induction is worked out in Chap. XIV, where it is shown, from the principle of special relativity, that when an observer moves relative to the fixed circuits producing a magnetic induction B he will observe, in general, an electromotance. If v is the velocity of the motion and E that part of the observed electric intensity which is due to the motion, we have, from 14.13 (12), if E is volts per meter, v meters per second, and B webers per square meter,
E = [v x B]
(5) 8.01. Mutual Energy of Two Circuits.—Let us now consider the work required to bring near each other two circuits carrying steady currents I and I'. Each circuit contains a source of electromotance that supplies or absorbs energy at the proper rate to maintain the currents constant at all times. In circuit 1 we have, from Ohm's law, 6.02 (1), and 8.00 (1),
IR = 6 —dN dt where 6 is the electromotance just described. The energy consumed in circuit 1 while bringing it up from infinity in time t is then, from 6.03 (2),
I 2R f tdt=4'6dt±I dN = f te dt ± NI where N is the flux through the first circuit due to the second circuit in the final position. Evidently, the last term represents the work done in the first circuit due to the magnetic field of the second. At the same
§8.01
331
MUTUAL ENERGY OF TWO CIRCUITS
time, work of amount +NT has been done in the second circuit to maintain its current constant. Thus, the total energy expended by the electromotance sources in the two circuits to maintain the currents constant while they are brought together is
± (IN I'N') Substituting the vector potential in 7.29 (1) and applying Stokes's theorem and 7.02 (5), we have N= fB' •ndS fVxA' • ndS= A'• ds
= xx ds • ds' 4r J' Y
and similarly
Elx peds• ds' LITYY r
(1)
so that the energy expended becomes
We
=
(IN ± I'N') = +1.1II' fr erds• ds'
(2)
2r YY
Faraday's law shows that if the two circuits attract each other the induced electromotance sets up an opposing field which tries to reverse the current in each circuit so that the sources of electromotance must supply energy to maintain I and I'. If they repel each other, the reverse is true. Let us now consider the mechanical work done against magnetic forces in moving the two circuits. From 7.18 (5), the force on circuit 1 in newtons is
f
F = I ds x B'
All'xfds x (ds' x r) r3 4r Y
All'
d '(ds • r) — r(ds • ds')
r3
47r
The first term vanishes when integrated around circuit 1 since the integrand r/r3can be written as the gradient of the scalar 1/r. The mechanical work done is then, for pure translation from r = co to r = r, TY„ =
f rF • dr =
erds • ds'r • dr _
47r
YYJ.
2-3
4a
3
f ds • ds'
(3)
Comparing (2) and (3), we see that half the energy supplied by the battery in the circuit is used in doing mechanical work. Since the only difference between the initial and final states is in the magnetic field surrounding the circuits, the remainder of the energy must be in the magnetic field. Thus, when two constant-current circuits are moved with respect to each other, the mechanical work done by the circuit and the energy in the magnetic field increase or decrease together and at the same rate. This explains the apparent anomaly mentioned in 7.18. It follows that if we know the energy WB in the magnetic fields of
332
ELECTROMAGNETIC INDUCTION
§8.02
two circuits we can get the mechanical force or torque trying to increase any coordinate 0 by taking the positive derivative of this energy with respect to 0, so that
ow',
F = + ao
(4)
8.02. Energy in a Magnetic Field.—We shall now find the energy required to establish the magnetic field of a single circuit, utilizing the results obtained in the last article and assuming all space filled with a homogeneous, isotropic medium of permeability pc. Let us build up the field step by step by bringing together infinitesimal current filaments. If the final current density is nowhere infinite, the denominator in 8.01 (3) causes no difficulty because finite current filaments are a finite distance apart. When 8.01 (3) is applied to a single circuit, we must include the factor to get the correct energy since our integration includes not only the work done in bringing filament a up to b, but also that in bringing b to a. When we write in current densities in place of currents by 6.00 (2), and make this correction, 8.01 (3) becomes TYB =
dv •
dv'
(1)
871-iviv
where r is the distance between the volume elements dv and dv', i and i' are the current densities in these elements, and the integration is performed twice throughout space, where, as in 8.01, we have assumed that is constant in the region of integration. Substituting for i from 7.01 (3) and for fr (e/r) dv' from 7.02 (4), we have Wa = 1 f VxB•A dv
(2)
v
From the formula for the divergence of a crossproduct, we have the relation A•VxB=B•[VxA]—V•[AxB]=B2 —V•[AxB] which gives 1 ,,1 f B2 dv — f V • [A x B] dv v 41.1 v where the integrals are over all space. By Gauss's theorem [3.00 (2)], we may transform the second integral into a surface integral over the sphere at infinity. This integral vanishes, since, from 7.02 (4), A goes to zero as 1/r whereas B = V x A goes to zero as 1/r2, and the surface area increases only by r2, so that fvV • [A x B] dv = fs[A x B] • n dS
0
§8.03
MUTUAL INDUCTANCE
333
Thus, the final expression for the energy becomes WB = r, f B2 dv GA V
(3)
This energy can be looked upon as residing in the space occupied by the magnetic field surrounding a current, the energy density being 1B2/p. This may be compared with 2.08 (2) which gives the energy in an electrostatic field. We see that the magnetic field, just as well as the electrostatic field, can be visualized in terms of a system of stresses. 8.03. Mutual Inductance.—The coefficient of mutual inductance M12 between two circuits is defined as the flux N12 through circuit 1 produced by unit current in circuit 2. The mks mutual inductance unit is the henry. This is expressed mathematically by writing (1),(2) B2 • n d81 = 4'A2 • dsi Js, where A2 is the entire vector potential due to unit current in circuit 2. From 8.01 (1), we may write M12 =
M12 =
XXdS1 • ds2
r
4rYJ
M 21
(3)
The flux through circuit 1 due to a current /2 in circuit 2 is, from (1), N12
=
(4)
M12/2
From 8.00 (1), the electromotance in circuit 1 produced by a fluctuating current in 2 is el
d/2 = 1V1 12-
(5)
dt
From 8.01 (3), the mutual energy of the two circuits is given by W12
=
(6)
M121112
From 8.02 (3), the total energy of the two circuits is = fv(Bi+ B2) - (B1 + B2) dv = (f B? dv + 2f B1 • B2 dv 211 v
dv) B2 The first term is the energy required to establish I l alone and the last term that for /2 alone so that the remaining term gives the energy used in bringing the two into interaction. Hence, from (6), we have 1 M12/112 = -
f
B1 • B2 dv (7) i,r From 8.01 (4), the force or torque tending to increase any coordinate of
334
ELECTROMAGNETIC INDUCTION
position 0 of one circuit relative to the other P' =
is
M12 2a
§8.05
ae
(8)
8.04. Boundary Conditions on A.—In 7.21 appear the conditions that derivatives of the quasi-vector potential A satisfy at the boundary between two regions of different permeability. In 8.03 (2), we have, for the first time, a relation involving an integral of A, and we must therefore consider what boundary conditions to impose on A itself in order that this equation will be valid where such discontinuities occur. Let us take a small rectangular circuit, the long sides of which are very close together but on opposite sides of the boundary between regions of different permeability. As these sides approach each other, the area of the rectangle approaches zero, so that the flux through this circuit, due to a current /2 flowing in a second circuit, must become zero. But by 8.03 (2) and (4), this flux is N = .9642 • ds. The integral is taken around the rectangle, the ends of which are vanishingly small, so that the whole contribution must come from the sides, which are of equal length L and which are short enough so that the vector A is constant along a side. From 7.21 (10) and (11), the tangential component of A has the same direction on both sides of the boundary, so orienting our rectangle along ut and ut we obtain for the corresponding components of A, respectively, — A.,t • ds,,t = =0 Thus, we have the two equations for the tangential components of A. A's,g = A'Zt
(1)
8.05. Mutual Inductance of Simple Circuits.—As a simple example of the computation of a mutual inductance coefficient, consider the two closely wound coils A, B, shown in Fig. 8.05. Coil B contains n turns and is wound on a ring of permeability A. Coil A has m turns. All the flux goes through the ring so that, from 7.29 (4) and 8.03 (4), we have (a2 by] (1) M12 = imm[a — If a >> b, we may factor a out of the radical and expand the remainder by Pc 753 or Dw 5.3, keeping only the square terms. This then becomes FIG. 8.05. i.annb2 (2) M12 — 2a If A = rb2 and ni = n/(27ra) is the number of turns per unit length,
MUTUAL INDUCTANCE OF CIRCULAR LOOPS
§8.06
335
then this gives the mutual inductance between an infinite solenoid wound on a cylinder of area A and permeability /.4 and a coil of m turns encircling it to be M12 = µn1m A (3) This could have been written down directly from 7.15 (7).
8.06. Mutual Inductance of Circular Loops.—The coefficient of mutual inductance for two coaxial loops of wire can be written down by
Ca) FIG. 8.06.
the use of 8.03 (2). The vector potential of A is entirely in the 95-direction and so has the same value for all elements of loop B and is parallel to each element so that, from 8.03 (2), MBA
=
4AA
• dSB = 2rb AA
:
P=b
Thus, from 7.10 (4), we have M12
= 2µk-1(ab)i[(1 — 1k2)K — E]
where, from 7.10 (3), k 2= 4ab[(a
(1)
b) 2 ± c2]-1
(2)
Numerical values of K and E may be found in Pc 121 or Dw 1040 and 1041. Another expression for M12 may be written down in the same way 0 = 0, giving, by 5.157, if b2 c from 7.13 (2), letting a =r 2 < a2, 1—> .o
n
n +1
1.3.5 •
n
n • 2 • 4.6 • •• (n
b2
1)( a2
c2y"
Pl(cos 13)
(3)
odd
n+1 c2)] 2 for c2)/a2]4n. [(b2 We may compute the mutual inductance when the two loops are in any position, provided that their axes intersect, by means of 8.03 (2), 7.13 (3), and 5.24 (5). Thus if, in Fig. 8.06b, b > a, we have, taking the origin at the point of intersection and measuring 0 from the a-loop axis, 21a (4) [NI) sin 0 d0 diri) M12 = a2f o o When b2
c2 > a 2, write [a2/(b2
ELECTROMAGNETIC INDUCTION
336
§8.07
If 0' is the angle measured from the b-loop axis, then substitution of 5.24 (5) for Pa(cos 0') in 7.13 (3), where 7 = 0, a = r, and I = 1, give
[B,lb =
sin 13
a
2a .a.7..j 0 P;,(cos /3) n=1
n (2 — 62,) (91— in) Irn, (cos -y)P7n(cos 0)
cos my6
m).
m=0
When this is inserted in (4), the cb-integration eliminates all but the m = 0 terms. By 5.154 (5) the integration with respect to 0 gives ens a sin' aP'n (cos a) — sin aPli(cos a) i) n(n n(n 1) Pn(cos 0)d(cos 0) = so that the mutual inductance is M12 = rya sin a sin )(3
1 1)(b n(n ± 1)
«)P71(cos 0)P.(cos 7) (5)
n=1
8.07. Variable Mutual Inductance.—Variable mutual inductors are frequently built in the form shown in b Fig. 8.07. Each coil consists of a single . . layer of wire wound on a bobbin with a spherical surface. One coil rotates, .\ I /1/1/// / .'- . relative to the other, about a common \ / axis. Let the angle between the axes / IN I I of the coils be 7, let the coils subtend — --r i 1e angles 2A and 2B at the center, and let I\ / i /;y \\ \ the radii of the two surfaces be a and b. \\ / , \ \/ // Consider each coil as being made up of a1. Assuming that the currents will distribute themselves uniformly on the bands, show that the mutual inductance between them is 2,ualaz Ic-2/1(kai)K1(kaz) sin kci sin kc2 dk cie2 fo 17. Using problem 30 of Chap. VII, show that if the plane of a circular loop of wire of radius a is parallel to, and at a distance b from, the face of an infinite plate of material of permeability and thickness t the self-inductance of the loop is increased by the amount AL11
= 13 1110 - (1 — 1 2)E 132("—')M. [
71-1 2A„a[( k where M„ = 1 — -=)K„ — Ed, — k. 2
A
— +
a2
Ay
1.4 , k 2—
a2 +(nt b)2
and K.
and E. are the complete elliptic integrals of the first and second kinds of modulus k„. 18. An infinite plate of thickness t and permeability At is inserted between the loops of Fig. 8.06a, with its faces parallel to them. Show that the mutual inductance between the loops is now given by 00
M = 2/.1.„(ab)1(1 — #2 ) El (1 — —E 2 n n=0 k' K. and E„ are complete elliptic integrals of modulus k., where —
4ab (a + b)2 +(c + 2nt)2
and
13
-
A
-
µ+
19. Show that the self-inductance of a solenoid of diameter d, length c, N turns, and small pitch is igN2d{ csc a[(tan2 a — 1)E + K] — tan2ce) where d/c = tan a and E and K are complete elliptic integrals of modulus sin a. 20. A wire loop of radius a is coaxial with a solenoid of radius b and n turns per unit length. If the greatest distance from a point on the loop to the near end of the solenoid is c and to the far end it is d, show from 7.10 (4) without further integration that the force between loop and solenoid when they carry currents I and I' is
2Arn//' (ab)4 fkri[(1 —
—
— 41(1 — ikZ)K2 — Ezil
where the modulus of the complete elliptic integrals is ki = 2(ab)1/c and kz = 2 (ab)i/d. 21. Two coaxial solenoids of diameter d and small pitch, with n and m turns per unit length, are placed so that the distance between near ends is b and between far ends is c. If one is of length a, show that their mutual inductance is
PROBLEMS
347
4 iihnmd3Z ( —1). sec an[(1 — tan2 ot„)E(k„) + tan2a„K(kn)]
n=1 where tan al = c/d, tan a2 = (c — a)/d, tan a3 = b/d, tan at = (a + b)/d, and k„ = cos a,,. 22. Show that, if the currents in problem 21 are I and I', the force between the coils is 4
(-1)" sin ct. sect a„[E(k„) — K(k„)] n=i 23. Show by 5.35 (4) that the mutual inductance between two loops of radii a and b with parallel axes a distance c apart and planes a distance d apart is, if c > a -I- b, 2ktd2nmIr
I i(ka)l i(kb)Ko(kc) cos kd dk
21.1abf
0
24. Show that, if a > b + c, the mutual inductance in the last problem is .. 2gabf€i
K i(ka)Ii(kb).10(kc) cos kd dk
25. Show with the aid of 5.298 (7) that the results of problems 23 and 24 may be written irpab f'J i(ka)J1(kb)J0(kc)e-ka dk 0
26. Show from problem 23 that the mutual inductance between two solenoids of radii a and b, length 2A and 2B, and m and n turns per unit length, with parallel axes a distance c apart and central loop planes a distance d apart is, when c > a + b, 2
AB Aabmnfo k-2I 1 (ka)I i(kb)Ko(kc) sin kA sin kB cos kd dk 27. If coil b of the last problem lies inside coil a, show from problem 24 that . 2 M - — Aabmnf k-21Ci(ka)l i(kb)I o(kc) sin kA sin kB cos kd dk AB o 28. From problem 25 write the mutual inductances of problems 26 and 27 in the form 1 ' -2 AB irliabmnf k Ji(ka)..1,(kb)4(kc) sinh kA sinh kB e- I'd dk 0
29. The most general position of a loop of radius a relative to one of radius b may be stated in terms of the shortest distance c between their axes, the distances A and B of the centers of each from c, and the angle 0 between their axes measured from each center toward c. Show that the mutual inductance is /lab 47, 0
22-f 22.
0 [D2 ± a2 +
(cos 95 COS 9/,' cos # ± sin 0. sin 0') dcl, de b2
+ 2ab(cos 4, cos 0' + sin 0 sin 0' cos 0) — 2f(0, 0')]i
where D = (c2 + A2 + B2— 2AB cos i3)i is the distance between their centers and
AO, 4)') = a(c cos 0 + B sin 0 sin 0) + b(c cos 4.' + A sin 0' sin 0) 30. A bifilar lead of two parallel wires with a spacing c is placed symmetrically in the gap of width a between two parallel sheets of infinite permeability with its plane normal to them. Show that the additional self-inductance per unit length due to the presence of the sheets is GOO In ([2a tan (4irc/a)]/(7rc)}.
348
ELECTROMAGNETIC INDUCTION
31. Obtain a series solution of problem 25 by first establishing, from Watson, page 148, and HMF 15.4.14 and 8.2.5, the formula
J,
Abk) =
( -1Na2— `‘.../
m=0
(in
1)!(m
+ 2)!
a' b2 k a 2 — b2 2
Substitute this in the integral of problem 25 and integrate the result by IT I (9), page 182. Thus show that the mutual inductance is 2rAtab m=0
a2 (-1)TM(2m 1)!! (a2 — b2).+Ip1 ( ÷1 a2 (2m 4)!!
b2) b2 P2„,+2 (cos a)
where the angle between r, the line joining loop centers, and either axis is a. 32. If a = b in the preceding problem, show that the mutual inductance becomes Z(2m + 2)!( — 1)m(2m + ip(a)2'n+3 P2m+2(cos r m!(m + 1)!(2m + 4)!!
2rkta
m=0
33. Apply the method of problem 31 to problem 28 and thus show that the mutual inductance is 3 r
8AB
0:1
( -1)m÷'(2m — 1)!! (a2 — b 2)m+1
AcibmnZ a=0 m=0
(2m + 2)(2m
4)!!
r2;'+'
(a2
b2
Pm ' ÷, a2
b2)P2„,(00s a,)
where if po d + A + B, pi = d + A — B, p2 = d — A — B, and p3 = d — A + B, then r: = c2 p:,c4 is the angle between r, and either coil axis, and d > A + B. 34. If a = b in the preceding problem, show that the mutual inductance becomes 3
Z th 7//a 3 nina=0 m=0
(2m + 2)!(-1)m÷s(2m — 1)!!
(2m + 2)m!(m + 1)!(2m + 4)!!
(al 2m-1-1
r,
P2,” (COS
as)
References All the references of Chap. VII contain some of the material of this chapter. In addition there is one very important reference. GROVER, F. W.: "Inductance Calculations," Dover, 1962. This contains the most
extensive tables and instructions for making all kinds of inductance calculations.
CHAPTER IX MAGNETISM 9.00. Paramagnetism and Diamagnetism.—In most substances, with some notable exceptions to be considered in Art. 9.04, the magnetic permeability depends very little on the field strength so that taking it constant, as we have done, introduces no appreciable error. Unlike the relative capacitivity, the relative permeability may be either greater or less than 1. Substances in which it is greater are called paramagnetic, and those in which it is less diamagnetic. Let us consider the forces acting on such bodies when placed in the field of a fixed circuit carrying a constant current. From 8.01 (4) and 8.03 (4) or 8.08 (3), we see that, if any infinitesimal displacement or rotation of the circuit will increase the flux through it, then there is a force or torque trying to produce this motion. If there are any bodies in the field of this circuit whose displacement or rotation will increase the flux through it then, by Newton's law of reaction, there will be corresponding forces or torques acting on these bodies. In 7.29, we saw that the magnetic induction or flux density and the permeability are related in magnetic circuits in exactly the same way as the current density and the electric conductivity in electric circuits. Thus we may state theorems 3 and 4 of 6.05 in a form applicable to magnetic circuits. If the permeability of any element in the magnetic field of an electric current is increased or decreased, the reluctance of the magnetic circuit is decreased or increased, respectively. As we have seen, these are forces acting to increase the flux and hence to decrease the reluctance. Thus, in an inhomogeneous field, there is a tendency for bodies that are more paramagnetic or less diamagnetic than the ambient medium to move toward the more intense parts of the field, and vice versa. If a paramagnetic or diamagnetic body of elongated shape is placed in a uniform field, there is a torque tending to set its axis parallel to the field. This can be seen in the case of spheroids by substituting /./. for e, pi, for e„, and H or B/A for E in problem 84, Chap. V, and observing that the torque formula contains H1 — H2 which has the factor A in it so that the torque depends on (14 — ,u,,)2 and has the same sign for A > Ae, and < µv. Quantum mechanics gives a theoretical basis for the empirical fact that the permeability of diamagnetic bodies is usually independent of temperature. For weakly paramagnetic substances, the permeability is often 349
350
MAGNETISM
§9.02
independent of temperature. In strongly paramagnetic, but not ferromagnetic, substances, the permeability usually depends on temperature, obeying the equation =µa+
Ae,C
(1)
T+ 0
where C and 0 are constants and T is the absolute temperature. This empirical relation is called Curie's law. A theoretical basis for it is found in quantum mechanics. For gases, 0 is usually zero. 9.01. Magnetic Susceptibility.—It is often convenient to use a new quantity magnetic susceptibility defined, in an isotropic medium, in terms of the quantities already treated in 7.20 and 7.28 by the equations
IcH = M =
iho
\ — 1.)
(1)
Thus susceptibility and permeability are related by the equation K=
A;1(kt
—
Fp)
(2)
= Km — 1
When a substance is placed in a magnetic field, its energy is decreased if it is paramagnetic and increased if it is diamagnetic. From 8.02 (3) and (2), the change is given by „ H2 B2 (4 µ)H2 Aw r-v — = = 2
2µ
2
2 KH
(3)
We notice that, for paramagnetic bodies, K is positive; for diamagnetic bodies, it is negative. Curie's law [9.00 (1)] may be expressed more simply in terms of the susceptibility by the empirical equation K
_ C T +
(4)
A theoretical basis for this equation has also been worked out. 9.02. Magnetic Properties of Crystals.—Many substances, especially crystals, possess different magnetic properties in different directions. It is even possible with some materials, such as graphite, to prepare specimens that are paramagnetic in one direction and diamagnetic in another. In such cases, it is found however that for any given orientation the magnetic induction B is proportional to the field intensity H and makes a constant angle a with it. This relation is similar therefore to that connecting the electric displacement D and the electric field intensity E in a crystal and can be formulated in the same way as in 1.19 by writing a set of equations analogous to 1.19 (1) and (2). We find, therefore, from 1.19 (3) that the components of B and H are con-
§9.03
CRYSTALLINE SPHERE IN UNIFORM MAGNETIC FIELD
351
netted by the relations Bz = By =
11211-1vA31tlz A22Hy + 1232Hz
12Hx
B, = 1213H.
j. 2 3H7/
(1)
+ 1133Hz
where /212
=
A21,
/113
=
A31,
1.1 23 = 1.432
(2)
Thus B and H are now connected by a quantity having nine components of which six are different. The permeability, formerly a simple ratio, has become a symmetrical tensor. By a suitable orientation of axes, (1) may be written in a form analogous to 1.19 (7), giving Bz =
By = A2Hy,
Bz = 1.13Hz
My = K2H y,
My = K3Hy
(3) When (3) holds, the coordinate axes are said to lie along the magnetic axes of the crystal. By means of 9.01 (1), we see that the corresponding equation connecting the magnetic susceptibility and intensity of magnetization are Mx = "Ix,
(4) where Arsci. = µl — mu, etc. By rotation of coordinates, we can get a set of equations, connecting M and H in a crystal, analogous to (1) and involving a tensor magnetic susceptibility. 9.03. Crystalline Sphere in Uniform Magnetic Field.—As an illustration of the manipulation of the formulas of the last article, let us find the torque acting on a crystalline sphere when placed in a uniform magnetic field of induction B. Let the angle between B and the x-axis in 9.02 (3) be a, and let 122 = 123. The boundary conditions will evidently be satisfied if we superimpose the case of a field of induction B cos a in the x-direction, acting on an isotropic sphere of permeability and the case of a field B sin a in the y-direction, acting on an isotropic sphere of permeability /12. If we take B in the xy-plane (Fig. 9.03), then problem 18C, Chap. VII, and 7.10 (2) show that the field outside the s2here is exactly the same as if we superimposed on the original field a field due to a magnetic dipole of moment Msin the x-direction and that of one of strength muin the y-direction where — µv
47a3B cos a
and
_
— At,47ra3B
sin a
112 ;Iv The field due to these two dipoles equals that due to a single dipole of -
Zuz
352
MAGNETISM
§9.04
strength M making an angle 13 with the axis, where tan = my = (A2 — Av)(At ± 2/4) tana (1) Ar. (Ai — µv)(µ2 + 2/4) = M = of! + 471-Ba3[(ili — A.) 2(12 + 214) 2 (1 — sin a) + (µ2 — 14) 2(mi + 2;4) 2 sin ali — iL.)(A2 + 2/4) (2) The angle that Al makes with the field is, from Dw 405.02 or Pc 593, tan (fi — a) =
3(A2 — ai)u„ tan
— µv) (µ2
2/4)
(g2 —
a + 214) tang a
(3)
The torque acting is then, from 7.18,
T = 31B sin 63 — a)
(4)
This torque tends to rotate the sphere so that the axis of Al lies in the direction of the field if AI > A2 and so that it lies at right angles to the field if µ l< 112. 9.04. Ferromagnetism.—There is an important group of materials whose permeability varies with the magnetizing field, depends on the previous treatment of the specimen, and is much larger than that of
ordinary substances. These materials are said to be ferromagnetic and include iron, cobalt, nickel, and the Heusler alloys and, at low temperatures, some of the rare earth metals. Let us construct, of ferromagnetic material, a simple magnetic circuit such as the anchor ring considered in 7.29 in which the magnetizing force, or the magnetomotance per meter, can be computed easily. Starting with the zero values of both the magnetic induction B and the magnetizing force H, we find that as H is increased B also increases but the ratio of B to H which is the permeability A first increases and then decreases. Typical curves of B and A as a function of H, for a ferromagnetic material, are shown in Fig. 9.04. In many ferromagnetic substances, if we measure the magnetization
§9.05
HYSTERESIS. PERMANENT MAGNETISM
353
with sufficiently sensitive instruments, we find the steeper part of the magnetization curve to have a step structure. This is called the Barkhausen effect. It suggests that a large region, involving many similarly oriented atomic magnets, changes direction as a unit. Experiments on ferromagnetic single crystals indicate that certain preferred directions are easiest for magnetization. Since most ferromagnetic substances with which we work are polycrystalline, it is believed that the major part of the magnetization is due to the magnetic units in a microcrystal which are partially aligned with the magnetizing field growing at the expense of those which are not. When all units are so oriented, the Barkhausen effect ceases, and further increases in the magnetizing field force the magnetization of these units continuously from their preferred directions toward the direction of the field. This accounts for the absence of step structure in the flatter part of the curve where the magnetization approaches saturation. A theoretical basis for this picture is found in the electron configuration of the iron atom. In the calculation of magnetic fields, inductances, magnetic forces, eddy currents, and such quantities, we have hitherto assumed that the permeability of a given substance is fixed. We see from these curves that if there is much variation in the magnetizing force in a region occupied by ferromagnetic material this assumption is not justified. In such cases, we take a mean value of i.i, which is approximately correct for this substance, for the range of H used. Our knowledge of the magnetization curve of the particular specimen involved usually does not justify higher precision. In case, however, all ferromagnetic material used is very homogeneous and has been carefully annealed, we may be able to determine from this solution and the magnetization curve the value of i.t which should be used in different regions. It is in geometrically simple cases only that this knowledge can be used to find a more rigorous analytical solution. 9.05. Hysteresis. Permanent Magnetism.—Suppose that, having increased H to I/1, thereby increasing B to B1, as described in 9.04, we now decrease H to —H1. We find that the induction B does not retrace any portion of the path shown in Fig. 9.04 but decreases less rapidly, following the upper curve in Fig. 9.05a, and, for a normal specimen, reaches the value —B1 at —H1. If we now increase H again to +H1, we follow the lower curve in Fig. 9.05 and reach the original curve at B1, H1. This lag in the induction is called hysteresis, and the closed curve in Fig. 9.05a is called a hysteresis loop. It is clear that with a given specimen we get a different hysteresis loop when we start with a different point on the magnetization curve, but if we vary H continuously from H1 to — H1 and back we repeatedly retrace the same loop.
354
§9.06
MAGNETISM
The hysteresis loops for different ferromagnetic materials may differ greatly. For magnetically "soft" specimens, the area inside the loop is very small, giving the inner curve in Fig. 9.05b, whereas in magnetically "hard" substances it is very great as shown in curve D. In the latter case, we still have a large residual induction or retentivity 13,, when the magnetizing force is zero and it actually takes a large reverse field H„ called the coercive force, to destroy the induction in the specimen. Here, we observe for the first time the presence of a magnetic field when electric currents are apparently absent. Magnetic phenomena were
(6)
(a)
FIG. 9.05.
originally discovered associated with such "permanent" magnets, and the whole theory of magnetism was developed on the basis of experiments made with them. 9.06. The Nature of Permanent Magnetism.—Hitherto, we have considered the energy in a magnetic field as essentially kinetic, it being associated with the motion of electric charges. As the magnetic fields produced by permanent magnets appear in every respect to be identical with those produced by electric currents, it is natural to seek a similar origin for them. The nature of permanent magnetism is revealed by a group of phenomena known as the gyromagnetic effects. Since no electricity is entering or leaving a permanent magnet, any motion of electricity therein must be circulatory and this circulation or spin must be about axes that are oriented, on the average, in a definite direction to produce a definite external field. If, as we have seen in 1.04, the electrical carriers possess mechanical inertia, then when circulating in closed paths or spinning, they possess angular momentum and therefore are subject to gyroscopic forces. Such forces were predicted by Maxwell but could not be detected with the experimental technique of his day.
UNIFORM MAGNETIZATION
§9.07
355
Two effects immediately suggest themselves. The first of these is magnetization by rotation. A well-known fact in mechanics is that when the supporting system of a gyroscope is rotated and its axis a is free to turn only in the plane common to it and the axis b of rotation of the system, then a tends to set itself parallel to b. Thus, if an unmagnetized body possessed circulating or spinning electricity with axes oriented at random, a rotation of this body should produce an alignment of these axes with the axis of rotation, and the body should become magnetized. Such effects have been detected and measured by Barnett in ferromagnetic substances. The second effect is the converse of the first, rotation by magnetization. From the law of conservation of angular momentum, if the random axes of rotation are aligned by a magnetic field, then the body as a whole must rotate in the reverse direction to keep the resultant angular momentum zero. This effect was first measured by Einstein and De Haas and has since been done with greater precision by other experimenters. Both effects show that there is a rotation of negative electricity in ferromagnetic bodies and that the average magnetic moment of the individual gyroscopes is slightly greater than that of a spinning electron. The excess is supposed to be due to an "orbital" motion of the electrons. Thus, we see that the magnetic fields of permanent magnets are not different from those already studied. 9.07. Uniform Magnetization. Equivalent Current Shell.—In permanent magnets, the magnetization M is, by definition, independent of applied fields. It can be defined, as in 7.20, as the magnetic moment per unit volume of the permanent circulating currents or spins. From 7.20 (1) and(4), the vector potential due to M is given by the equation
,f
pa Mxr gyfVxM iMxn dv = dv dS (1) 47 v r3 s r 47r v r where the volume integrals are throughout the volume of the magnet and the surface integral over its surface. When M is the same in magnitude and direction in every element of a given region, we say this region is uniformly magnetized. For such a magnet, the second volume integral in (1) is zero. Let us examine the remaining surface integral. Suppose that M is in the x-direction so that M = iM. Let 0 be the angle between i and n, and let ds and dsi be orthogonal vectors lying in the magnet surface, ds being normal to i and n. Then, we have M x n dS = M sin 0dsi ds = M dx ds so that
Am —
A
Prof Cm dx ds _ _47, r
(2)
356
§9.08
MAGNETISM
We see that this is identical in form with 7.02 (5) so that the whole magnet can be replaced by a current shell, coinciding with its surface, around which the currents flow in planes normal to the direction of magnetization x. The current density, in terms of x, is uniform and equals the intensity of magnetization M. Such a shell is called an equivalent current shell and is frequently very useful in treating permanent magnets. 9.08. Magnetized Sphere and Cylinder. Magnetic Poles. For a uniformly magnetized sphere, the current flowing in the equivalent current shell between 0 and 0 — dO is —
i dO M dx = Ma sin 0 dO where i is the angular current density. The stream function is then given by e. (1) o x dO = Ma(1 — cos 0) = M a[Po(u) — P1(u)]
=f
where u = cos 0. This is identical in form with 7.12 (1) ; and setting n = 0 and 1 in 7.12 (5) and (3), we see that the vector potential outside the sphere is u,,Ma3 (2) A0 sin 0 3r2 -
X
Comparing with 7.10 (1), we see that the field is identical with that of a small current loop of moment M
-
47rMa3 3
(3)
We saw in 7.00 that the magnitude and configuration of the magnetic field at a distance from such a loop are identical with that of the electric field about an electric dipole having the same moment. For a right circular cylinder magnetized parallel to its axis, the equivalent current shell is seen to be a solenoid with zero pitch, the current per unit length being M. From 7.15 (4), the magnetic induction at any point P on the axis of such a solenoid is given by Bz = tu,M.(cos 132 — cos 0)
(4)
where 132 and 01are the angles subtended by radii of the two ends at P. The assumption that B can be found from the normal magnetization M. on a magnet surface from the inverse-square law /2„M(47r-r 2)-i also gives (4). Take Mzuniform over the flat ends of this magnet, and consider a ring of radius p = c tan 0 and width dp = c sect 0 dO which subtends an angle 0 at an axial point a distance c from its center so r = c sec 0 and .1=L at c is
§9.10 SPHERICAL PERMANENT MAGNET IN UNIFORM FIELD
seen to be A,M.1 9227rp dp cos Bs = r2 47r
Jo
1.1,M
2
fs2 s sin in 0 dO =
2
357
(cos 02 — 1) (5)
Subtraction of a similar term for the opposite end gives (4). For a magnet of length 1, this is just the electric displacement given on the x-axis with electric charges qi = itora2M at x = and qz = —1.4.2ara2M Thus, experiments with magnetic needles lead naturally at x = to the hypothesis of magnetic charges or poles. The region from which the magnetic lines of force emerge is called the north pole, and the region in which they reenter the magnet is called the south pole. As we have seen, the actual existence of magnetic charges is impossible if the divergence of the magnetic induction is to be zero everywhere. 9.09. Boundary Conditions on Permanent Magnets.—As the name implies, we assume that the intensity of magnetization of a "permanent" magnet is unaffected by the fields in which it is placed. This means that if we replace the magnet with an equivalent current shell, the region inside is assumed to have a permeability /.4,. If the magnet is immersed in a region of permeability A and an external field is superimposed, it will be distorted at the surface of the magnet to meet the boundary conditions between a medium of permeability /Ivand one of permeability /./. These boundary conditions have already been stated in 7.21, 7.22, and 7.28. 9.10. Spherical Permanent Magnet in Uniform Field.—As an example, let us compute the torque on a uniformly magnetized sphere immersed in a medium of permeabilityµ in which the magnetic induction B, before the introduction of the sphere, was uniform. Let a be the angle between M and B. Obviously, the current sheet can produce no torque on itself; so we need compute only the induction due to the external field. In spherical coordinates, the vector potential of B is
Ao Orsin 0
(1)
In 7.10 (1), we have a form of vector potential in which 0 enters in the same way but which vanishes at infinity. This is therefore the logical form to choose for the additional term due to the presence of the sphere with permeability 14. Thus, outside the sphere, we have
B A0 = —(r — r2 sin 0 2
(2)
Since Aimust be finite at the origin and must include 0 in the same way, it must have the form = Dr sin 0 (3) To determine C and D, we apply the boundary conditions of 7.21 (6) and
358
MAGNETISM
§9.11
(7) at r = a. These give the equations Bil
= -
and
201
+
= Da
(4)
Solving for C and substituting in (2) give
2L
=— 1 24r
ar ) sin 0
(5)
When r = a, this has the same form as (1) and so, from (3), represents a uniform field in the directio:i of B. The torque on the current ring element lying between 01 and 01 +d01, where 01is the colatitude angle measured from the axis of magnetization, equals the product of the current in the ring by its area, by the magnetic induction, and by sin a, giving dT = ira' sin' 01B[1
2 A — Av i do1sin a 21A + /iv
Substituting for i dOifrom 9.08 and integrating by Dw 854.1 or Pc 483 give an-Ava'MB sin a T 471-1.4a3MBsin a sin' 01d01— T= (6) 2µ + If the field in which the sphere was placed were not homogeneous, we would have a force as well as a torque. This would consist of two parts, viz., the force on a sphere of permeability 14 and the force on the current shell in the field at the surface of this sphere. These forces could be calculated, using 10.06 (13) and 7.18.
9.11. Lifting Power of Horseshoe Magnet.— For lifting ferromagnetic objects, one frequently uses a permanent magnet in the form of a horseshoe. In treating this, we shall assume for simplicity that the "legs" are eliminated, leaving only half a ring as shown in Fig. 9.11. If the legs were present, we could use an approximate method FIG. 9.11 like that in 6.16. We shall assume that the half ring was magnetized by placing it in contact with another similar piece of steel to form a complete ring, winding uniformly with wire and passing a current. To a first approximation, the magnetization M will be the same function of r, the distance from the center of the ring, as the magnetomotance was, so that we have, from 7.29 (3), M=— r
(1)
§9.11
LIFTING POWER OF HORSESHOE MAGNET
359
To find the equivalent current shell, we take a thin layer of thickness dr and radius r in which M may be considered constant. Then, if i is the angular current density, we have for the current flowing between 0 and 0 — do, as in 9.08, idO=Mdx=( C )rdo=CdO
(2)
When the layer current shells are superimposed, adjacent current layers cancel and we have a constant angular current density C around any section of the ring. Now let us suppose this magnet is placed in perfect contact with a block of infinite permeability. The lines of magnetic induction inside the current shell are then semicircles since they enter and leave the block normally by 7.21 (2). The magnetomotance around the circuit is, by 7.28 (3),
St= fove
= 71-C
(3)
The reluctance of the layer dr is, as in 7.29, length — irr Ilya dr !Iv X area since the permeability inside our current shell is 1.1.. The flux in this layer is dR —
Ba dr = dR =A,Ca dr The tension across a section of this layer is, from 8.14 (2), dT =
aB2 u C2 dr — • "a dr 21.1. 2r2
The pull on the block at each contact is T—
aC2 f 14 2
b
r2
ilvaC2(c— b) 2bc
(4)
An upper limit for C can be computed roughly from the magnetization curve of the steel. If the total number of turns in the original magnetizing winding was n and the current used was im then the magnetomotance per meter was nim/(irr) ampere-turns per meter from 7.28 (3). Suppose that the wide loop in Fig. 9.05b corresponds to our specimen and that HD = ni„,Arb). Then, for all values of r greater than b, the magnetizing force will be smaller than HD and the hysteresis loops will lie inside of that shown. We have assumed that the ratio of B. to BD is the same for all these loops (let us call it P) and that the ratio of BD to HD is also a constant (let us call it 12'). Then, from (1), C B. ni.A'P M = r ,
360
MAGNETISM
§9.111
so that we have max
ni,./213
(5)
This is an upper limit for C since any mechanical shock, heat-treatment, or "softness" of the steel will diminish C, especially after the ring has been separated into two magnets. The approximation made in assuming 1.z = 00 for the block is not serious, for in most soft iron specimens A/A2, > 500. 9.111. Field of Cylindrical Magnet.— In order to produce strong steady fields over small areas, permanent magnets are frequently made in the form of rings, with small gaps, which are magnetized, as described in the last article, by winding uniformly with wire and passing a current. To simplify the calculation of the field in this case, let us suppose that the ring is so wide in the direction parallel to its axis that we may take it as a long cylinder of internal and external radii a and b which has been magnetized, before the gap is made, by winding wire uniformly over the J P walls parallel to its axis. As explained in the last article, the resultant M is inversely proportional to r. After magnetizing, a gap with radial walls is created, without disturbing M, by removing part of the metal, leaving only the portions lying be44110 tween el = +a and Oi = — a. If b is FIG. 9.111a. sufficiently small compared with the length of the cylinder, the problem far from the ends becomes a twodimensional one. In this case, the lines of magnetic induction coincide with the lines of constant vector potential as proved in 7.25 and may be most easily calculated by the method of circular harmonics considered in 4.01, 4.02, 4.03, and 7.17. As shown in the last article, the equivalent current shell consists of that portion of the original magnetizing winding not removed with the gap and carries the current I d0 between 0 and 0 + do. Since all elements of the winding are parallel to the axis and the ends are so distant that they do not affect the vector potential A, the latter is also everywhere parallel to the axis and the current elements may be considered infinite in length. Considering the contribution of the inner elements negative and that of the outer ones positive, we find that dA at the point P at r, 0 due to the elements at +01 and —01is given from 7.09 (1), by
dA
R1 — In R2 — In Rs + In R4) des
FIELD OF CYLINDRICAL MAGNET
§9.111
361
FIG. 9.111b.—Lines of magnetic induction and constant vector potential in a permanent magnet formed by magnetizing a long thick cylindrical shell and removing a sector to form the air gap. Calculated from Eqs. (1), (2), and (3) of Art. 9.111 in which C = 1, a =7a/8,a= 1, and b = 2.
Substituting for the logarithms from 4.02 (1) and writing C for Av/hr give
dA = C n =1
ni(an
; b) cos nO1cos nO dOi
Integrating this from 01 = 0 to 01 = a gives 03
r > b,
A = cn12(an r—n b)
sin na cos nO
(1)
n=1 Similarly,
b > r > a,
A = C4celn —b
n2L
•na cos n0} (2) On sin -6
362
MAGNETISM
§9.12
and a > r,
A=
C{a ln a b- M1-2[ ( a)"—
sin na cos nO}
(3)
n=1
The lines of induction when C = 1, a = 77/8, a = 1, and b = 2, calculated from (1), (2), and (3), are accurately drawn in Fig. 9.111b. 9.12. Magnetic Needles.—The most familiar form of permanent magnet is probably the magnetic needle. This consists of a thin cylinder of steel more or less uniformly magnetized lengthwise. As we saw in 9.08 (5), the magnetic induction of such a magnet approximates in form that of the electric displacement about two charges porat2M and — pora2M placed at a distance 1 apart, where a is the radius, 1 the length, and M the intensity of magnetization of the magnet. This analogy fails if the magnet is placed in a medium of permeability /./ Ai,. However, in the case of the magnetic needle whose length-to-diameter ratio is very great, another approximation can be used. Its equivalent current shell, if filled with a medium A, and immersed in a medium behaves by 7.24 and 1.18 like a long slender cavity so that the axial flux through it from an external source is the product of gi,H or i.L,B/p. by ra2 cos 0 where 0 is the angle between its axis and B. From 9.07, the current in a length dl of the equivalent current shell that links this flux is M dl where M is the magnetization. From 8.03 (6) and 7.28 (5), the energy of the shell is W = gv/Ifira2rH cos 0 dl = gyMira2 (e2 —
(1)
This has the form of 1.071 (1) so that a thin needle uniformly magnetized lengthwise acts exactly like a pair of equal and opposite electric charges of size tz,Mra2placed in an electric field of potential V = St, where Si is the magnetomotance or scalar magnetic potential. In the absence of external magnetic field sources, the flux threading the slender current shell is practically independent of the permeability of the medium surrounding the needle for the reluctance of the magnetic circuit involved lies almost entirely inside the shell where the medium does not penetrate and the permeability is p,,. Thus the magnetic flux from a sufficiently thin magnetic needle, like the electric flux from an electric dipole, is independent of the medium surrounding it. The scalar magnetic potential or magnetomotance of such a needle is, therefore, inversely proportional to the permeability of the medium around it. These forces may be computed by the formulas of 1.071 as if the magnets possessed a mutual potential energy. Thus if r is the radius vector from ke, to it4 which make angles 01and 02 with it and a with each other and, if sp is the angle between the planes intersecting in r which contain Arl and h4 then,
PROBLEMS
§9.12
363
from (1), 1.071 (4), and 1.071 (5), this potential energy is w — - (cosa — 3 cos 01 cos 02) 4./1-Ar3 . = 471 .Lr3(sm 0 sin 02cos ¢— 2 cos 01 cos 02)
(2) (3)
All forces and torques exerted by one dipole on the other can be found from these formulas by the usual method of differentiation with respect to the coordinate involved. In particular, the force of repulsion is
F= —
aW = dr
2(sin 01sin 02cos IP — 2 cos 01 cos 02)
47rAr3
(4)
The apparent potential energy of a needle in a field of induction B, from which the forces are found by differentiation, is, from 1.071 (2), —W= be • H= Ise •
B14-1
(5)
The vector potential at a distance r from a magnetic needle is given in spherical polar coordinates by 7.10 to be, when r >> 1. -
M' sin 0 47r2
(6)
which, like its total flux, is independent of the medium's permeability. It should be noted that the classical magnetic moment AT' used in this article differs dimensionally from the loop magnetic moment defined in 7.00 and 7.10 (2). The formulas of this article are rigorous only for infinitely thin magnets. Forces on or energies of permanent magnets of large cross section in mediums where cannot be found by integrating these formulas. Problems Problems in the following group marked C are taken, by permission of the Cambridge University Press, from the Cambridge examinations as reprinted by Jeans. The classical magnetic moment is used throughout these problems. 1. An iron pipe runs east and west in a certain region. To locate the pipe, measurements of the dip at 5-ft intervals, starting from a fixed point, in a northerly direction are made. At distances x ft from the point, the dip deviates from the normal 60° dip by the following amounts: x 185 210 215 220 225 230 235 200 205 190 195 AO —4.5° —5.0° —5.5° —7.5° —6.0° +0.0° +14.5° +0.0° —1.5° —1.2° —1.0° Find x for the center of the pipe and its distance below the surface. 2. A steel pendulum bob of radius a is uniformly magnetized in a vertical direction with an intensity I. The strength of the vertical component of the earth's field at a certain location is V and the acceleration of gravity g. Find the ratio of the period to that when unmagnetized if it swings in an east and west plane, its mass being m and the distance from center of mass to pivot, 1. 3. A sphere of permeability 1000 and radius 10 cm is placed 1 m northeast (mag-
364
MAGNETISM
netic) of a compass needle. Show that, within 1 per cent, the deviation of the needle from the magnetic meridian, neglecting the magnet's own image forces, is 5'. 4. Find the magnetic field outside and in the cavity of a thick spherical shell uniformly magnetized in the x-direction with an intensity M. 5C. Two small magnets float horizontally on the surface of water, one along the direction of the straight line joining their centers, and the other at right angles to it. Prove that the action of each magnet on the other reduces to a single force at right angles to the straight line joining the centers and meeting that line at one-third of its length from the longitudinal magnet. 6C. A small magnet ACB, free to turn about its center C, is acted on by a small fixed magnet PQ. Prove that in equilibrium the axis ACB lies in the plane PQC and that tan 0 = — a tan 0', where 0, 0' are the angles that the two magnets make with the line joining them. 7C. Three small magnets having their centers at the angular points of an equilateral triangle ABC, and being free to move about their centers, can rest in equilibrium with the magnet at A parallel to BC and those at B and C, respectively, at right angles to AB and AC. Prove that the magnetic moments are in the ratios 31:4: 4. 8C. The axis of a small magnet makes an angle 0 with the normal to a plane. Prove that the line from the magnet to the point in the plane where the number of lines of force crossing it per unit area is a maximum makes an angle 0 with the axis of the magnet, such that 2 tan 8 = 3 tan 2(0 — 0). 9C. Two small magnets lie in the same plane and make angles 0, 0' with the line joining their centers. Show that the line of act pri of the resultant force between them divides the line of centers in the ratio tan 0' + 2 tan 0: tan B + 2 tan 0'. 10C. Two small magnets having their centers at distances r apart make angles 8, 0' with the line joining them and an angle e with each other. Show that the longitudinal force on the first magnet is 3mm' (47rAtr4)--'(5 cos2 0 cos 0' — cos 0' — 2 cos e cos 0). Show that the couple about the line r which the magnets exert on one another is mm'(471-Ar4)--id sin e, where d is the shortest distance between their axes produced. 11C. Two magnetic needles of moments M, M' are soldered together so that their directions include an angle a. Show that when they are suspended so as to swing freely in a uniform horizontal magnetic field, their directions will make angles 0, 0' with the lines of force given by sin 0
sin 0' M
(M2 M'2 + 2MM' cos a)-i sin a
12C. Prove that if there are two magnetic molecules, of moments M and M', with their centers fixed at A and B, where AB = r, and one of the molecules swings freely, whereas the other is acted on by a given couple, so that when the system is in equilibrium this molecule makes an angle 0 with AB, the moment of the couple is 3MM' sin 20 87µr3(3 cos2 B + 1)1
where there is no external field. 13C. Two small equal magnets have their centers fixed and can turn about them in a magnetic field of uniform intensity H, whose direction is perpendicular to the line r joining the centers. Show that the position in which the magnets both point in the direction of the lines of force of the uniform field is stable only if H > 3M (4n-pr3)-i. 14C. Two magnetic particles of equal moment are fixed with their axes parallel to the axis of z, and in the same direction, and with their centers at the points + a, 0, 0. Show that if another magnetic molecule is free to turn about its center, which is fixed
365
PROBLEMS
at the point (0, y, z), its axis will rest in the plane x = 0 and will make with the axis of z the angle tan-1[3yz/(2z2 — a2 — y 2 . Examine which of the two positions of equilibrium is stable. 15C. Prove that there are four positions in which a given bar magnet may be placed so as to destroy the earth's control of a compass needle so that the needle can point indifferently in any direction (and experiences no translational force). If the bar is short compared with its distance from the needle, show that one pair of these positions is about 1/ times more distant than the other pair. 16C. Three small magnets, each of magnetic moment M, are fixed at the angular points of an equilateral triangle ABC, so that their north poles lie in the directions AC, AB, BC, respectively. Another small magnet, moment M', is placed at the center of the triangle and is free to move about its center. Prove that the period of a small oscillation is the same as that of a pendulum of length 4774t/b3g/(351)1mM', where b is the length of a side of the triangle and I the moment of inertia of the movable magnet about its center. 17C. Three magnetic particles of equal moments are placed at the corners of an equilateral triangle and can turn about those points so as to point in any direction in the plane of the triangle. Prove that there are four and only four positions of equilibrium such that the angles, measured in the same sense of rotation, between the axes of the magnets and the bisectors of the corresponding angles of the triangle are equal. Also prove that the two symmetrical positions are unstable. 18C. Four small equal magnets are placed at the corners of a square and oscillate under the actions they exert on each other. Prove that the times of vibration of the principal oscillations are )]
[ 47rAnik2 d3 27r m23(2 + /2-2)
{47-Amk2 ds 27r-
2119 —
y,
27r
[ 47rumk2 d32 (2)i
3M2
J
where M is the magnetic moment, and mkt the moment of inertia, of a magnet, and d is a side of the square. 19C. A system of magnets lies entirely in one plane, and it is found that when the axis of a small needle travels round a contour in the plane that contains no magnetic poles, the needle turns completely round. Prove that the contour contains at least one equilibrium point. 20C. Prove that the scalar potential of a body uniformly magnetized with intensity I is, at any external point, the same as that due to a complex magnetic shell coinciding with the surface of the body and of strength Ix, where x is a coordinate measured parallel to the direction of magnetization. 21C. A sphere of hard steel is magnetized uniformly in a constant direction, and a magnetic particle is held at an external point with the axis of the particle parallel to the direction of magnetization of the sphere. Find the couples acting on the sphere and on the particle. 22C. A spherical magnetic shell of radius a is normally magnetized so that its strength at any point is Si, where Si is a spherical surface harmonic of positive order i. Show that the scalar potential at a distance r from the center is [
+ 1) is,0' 2i + 1 a
.*4-1 [2i4-Fri113'0
when r < a when r > a
MAGNETISM
366
23C. If the earth was a uniformly magnetized sphere, show that the tangent of the dip at any point would be equal to twice the tangent of the magnetic latitude. 24C. Prove that if the horizontal component, in the direction of the meridian, of the earth's magnetic force was known all over its surface, all the other elements of its magnetic force might be theoretically deduced. 26C. From the principle that the line integral of the magnetic force round any circuit ordinarily vanishes, show that the horizontal components of the magnetic force at any station may be deduced approximately from the known values for three other stations which lie around it. Show that these six known elements are not independent but must satisfy one equation of condition. 26C. If the earth was a sphere and its magnetism due to two small straight bar magnets of the same strength situated at the poles, with their axes in the same direction along the earth's axis, prove that the dip S in latitude X would be given by 8 cot (5 +
= cot iX — 6 tan 4X — 3 tan3iX
27C. Assuming that the earth is a sphere of radius a and that the magnetic potential 7 is represented by ft = Si(r/a) S2(r/a)2 Si(a/r)2 S2(a/r)3, show that is completely determined by observations of horizontal intensity, declination, and dip at four stations and of dip at four more. 28C. Assuming that in the expansion of the earth's magnetic potential the fifth and higher harmonics may be neglected, show that observations of the resultant magnetic force at eight points are sufficient to determine the potential everywhere. 29C. Assuming that the earth's magnetism is entirely due to internal causes and that in latitude X the northerly component of the horizontal force is A cos X B cos 3X, prove that in this latitude the vertical component reckoned downward is 4B
6B
2(A + --) sin X — — sin3X 3 5
30C. A magnetic particle of moment M lies at a distance a in front of an infinite block of soft iron bounded by a plane face, to which the axis of the particle is perpendicular. Find the force acting on the magnet, and show that the potential energy of the system is M2G4 3271-Att,a3(A
+
31. A small magnet of moment M is held in the presence of a very large fixed mass of soft iron of permeability At with a very large plane face, the magnet is at a distance a from the plane face and makes an angle B with the shortest distance from it to the plane. Show that a certain force and a couple — A.0 M2sin B cos 32ri.tv(Ai
/12,)a3
are required to keep the magnet in position. References
S. S.: "Electric and Magnetic Fields," Wiley, 1941. Clear practical treatment of ferromagnetism and permanent magnets with diagrams and tables. BITTER, F. T.: "Introduction to Ferromagnetism," McGraw-Hill, 1937. Excellent modern treatment.
ATTWOOD,
REFERENCES
367
R. M.: "Ferromagnetism," Van Nostrand, 1951. A classic in modern magnetic experiments. EwING, J. A.: "Magnetic Induction in Iron and Other Substances," Van Nostrand, 1900. The standard work of its time. FLUGGE, S.: "Handbuch der Physik," Vol. XVIII, Springer, 1966. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XII, Springer, 1927. GRAY, D. E.: "American Institute of Physics Handbook," 2d ed., 5-164, McGrawHill, 1963. Valuable tables and bibliography. MATTIS, D. C.: "The Theory of Magnetism," Harper and Row, 1965. A very modern and advanced treatment of magnetism and cooperative phenomena. RADO, G. T., and N. Sum.: "Magnetism," Vol. III, Academic, 1963. First of three volumes on magnetic theory and materials by many contributors. SPOONER, T.: "Properties and Testing of Magnetic Materials," McGraw-Hill, 1931. Gives detailed experimental information. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. WILLIAMS, S. R.: "Magnetic Phenomena," McGraw-Hill, 1931. Treats experimental f acts. BOZORTH,
CHAPTER X
EDDY CURRENTS 10.00. Induced Currents in Extended Conductors.—This chapter covers the laws of magnetic interaction of currents and the induction laws of Faraday discussed in Chap. VIII as they apply to extended conductors, including skin effect, induction heating, drag on moving magnets, and the magnetic shielding by thin sheets. We should note that this treatment, and all that have preceded it, involves one approximation, viz., the assumption that electric and magnetic fields are instantaneously propagated. This is equivalent to saying that Maxwell's "displacement current" is neglected. The error so introduced is perfectly negligible if the frequencies are such that the wave length is large compared with the dimensions of the apparatus. If this condition is not met, we must resort to the complete Maxwell equations treated in Chap. XI. . Faraday's law of induction states that, if the magnetic induction B in a conductor, is changing, an electric field E is produced which is given, in magnitude and direction by 8.00 (3), to be VxE=—
dB d
In terms of the magnetic vector potential A, 8.00 (4) gives A E=—d dt
(1)
(2)
Since this electric field is produced in a conductor, a current will flow according to Ohm's law. If T is the resistivity and i the current density, we may, with the aid of 6.02 (3), write (1) and (2) in the form
(v x i) = —ITBI
r
(3)
dA (4) : = — dt These currents, flowing in the conductor of permeability A, will produce magnetic fields which, by 7.01 (3) and 7.02 (2), are given by V x B = Ai (5) v2A = — pd (6) The equations that must be satisfied by i, B, and A in a conductor when a changing field is present derive from (3), (4), (5), and (6). Elimination Ti
368
§10.01 SOLUTION FOR VECTOR POTENTIAL OF EDDY CURRENTS 369
of B by (3) from the time derivative of (5) gives, by 6.00 (3),
1.2 di = _ [v x (V x i)]= v2i — v (vi = • ) dt
V2i
T
(7)
Elimination of i from (4) and (6) gives
dA = V2A
(8)
r dt Elimination of i from (3) and (5) gives, with the aid of 7.01 (1),
II. dB = _ [V x (V x B)] = V 2B — V (V • B) = V2B dt
T
(9)
Equations (7), (8), and (9) all have the form of the well-known equation of the conduction of heat, but the dependent variable is a vector instead of a scalar. In rectangular coordinates, each component, regarded as a scalar, satisfies the same equation. This is not true for the components in any other system of coordinates except in special cases. Eddy current problems fall into two classes, transient and steady state. In the latter case the current density is often written as the real part of a complex Fourier series with terms of the form A„e3n't where n is an integer, w is the angular frequency, and A nis a complex constant called a phasor. The latter is indicated by a small flat v or inverted circumflex over the symbol, and its conjugate complex by a circumflex over the symbol. Vectors, shown by boldface type, may also be phasors. If i, the current density with amplitude i0 has its maximum value when t is zero, then the instantaneous rate of dissipation of energy in an element dv is, from 6.03 (2), ,
dP = Ti2dv = rig cos' wt dv = Tii cost wt dv The power averaged over a cycle is (10)
dP = Tii dv
10.01. Solution for Vector Potential of Eddy Currents.—We can obtain solutions of 10.00 (8) when T co in just the same way we did when T = 00 in 7.04. As in 7.04 (1) and (2), let us write A in the form
A = V x (uF rid-uxVW 2)
(1)
where u = 4i, k, or r. From 7.04 (3) and (4), we have
x v(V2 W2)] V2A = V x [uV2W1 u Substituting from these two equations in 10.00 (8) gives V x [u(V2W2
d W 2) T dt
u x V(V2W2
T
dW)] =0 t
370
EDDY CURRENTS
Thus, if W1and
W2
§10.02
are solutions of the heat conduction equation, viz., dW =r dt
(2)
then, with the aid of (1), we can obtain from them solutions for the vector potential of eddy currents. We calculate B in terms of W1and W2 by making use of 7.04 (5) and (6). Thus, B=VxA=Vx(VxuWi)+Vx(uV 2W2) (3) -dd-t(uW2) — u x vWi
Vx
(4)
J
where u = t, Is k, or r. Since B and A satisfy 'deistical equations, 10.00 (8) and (9), the similar form of (1) and (4) is to be expected. We notice that now both W1 and W2 contribute to B. We can simplify this expression somewhat further by using 7.04 (6), which gives u = i, k,
B=
u=
B=
T,
d — (u x VW 2 uW 1) u • V(VW1)
T dt
ud r dt
(5)
rWi) r • V(VW1) + 2V Wi (6)
When u lies along ul, a solution of (2) of the form U(ui)F(u2, u3, t) makes B normal to A as in 7.04. 10.02. Steady-state Skin Effect.—A simple solution of 10.00 (7) applies to a medium of permeability A and resistivity T filling all positive z-space. If the phasor current density Ix is uniform, x-directed, and of angular frequency w on the surface, then in the interior 10.00 (7) becomes jWilx = j C447IX
a rz 2 = aZ 2
=
d 21x dz2
(1)
since L is a function of z only. The solution of this equation is Ix = ae-(i.wr)iz
fle(iwwy)l,
If Ix is finite when z = 00, then 13 = 0; so writing 1
(2)
j for (2j)1gives
Lem = 2oe--(.i7)izoka-(1.4)iz1
(3)
where iois the value of Ix at the surface. Taking the real part gives
ix = ioe-(iw")iz cos [cot — (icy/17)/z]
(4)
Thus the ix amplitude decrease is exponential and the phase change uniform. The net current I cos (wt + ,p) per meter width, given by expanding the cosine and using Dw 863.1 and 863.2 or Pc 506 and 507, is Jo m
ix dz = io(2wiry)-i(cos wt + sin wt) = io (co,u7)-i cos (wt —
(5)
SKIN EFFECT OF TUBULAR CONDUCTOR
§10.03
371
This would be increased by removing all conducting matter below a certain depth since, for certain values of z, the current is reversed. The power absorbed as heat, per square meter of surface, is, from 10.00 (10), with the aid of Dw 440.20 and 565.1 or Pc 363 and 401, = O.Yr
2ir /7;
„42
e-2(1'1'7)1z dz 2 0 (6) = 7/1(10447)1 = 12(76)-1 =-icokibg = where 1„ = (2)--4/ is the effective value of I. Thus the resistance equals the d-c resistance of a skin of thickness b = (coify)-1. A uniform y-directed magnetic induction Bo cos (wt - fir) at the surface would induce the eddy currents just considered. From 10.00 (7) and (9) the same differential equation describes the behavior of i and B in the conductor so that both die out exponentially. To find the power dissipated, take the line integral of B around a rectangular path normal to x whose sides lie in the z-direction and whose meter long ends lie, one just outside the surface where the induction is Bo and the other far inside the conductor where it is zero. Only the Bo end contributes to the integral so that from 7.01 (2) (or from Maxwell's first equation) P1
2arfo o
i2dt dz =
Bo = kc„/ = 2iµ„/, (7) From (6), the power dissipated by eddy currents in a surface area S in terms of the magnetic induction amplitude Bo just outside the surface is
P = f Pi dS = (2/4,76)-1f
Bg dS
(8)
The inductance per square vicontributed by the magnetic field inside the conductor is found from (4), (6), 8.08 (1.1), and 10.00 (3).
T
27r
57)
L:I! w P Ti! - , i • A dt dz dt dz = — = (9) o s 2 26) 2w& 7rJ. f = 47r f. Comparison with the resistance R: per square gives the relation 1 R” L: = — = = (10) wb way co At high frequencies g is completely negligible. 10.03. Skin Effect on Tubular Conductor.—If the frequency is so high that the depth of penetration is small compared with the curvature of the conducting surface, then the results of the last article can be applied to the usual cylindrical wire or tubular conductor. If this is not true, we must use cylindrical coordinates. In this case, if we have symmetry about the axis so that iz is a function of p only, 10.00 (7) becomes aqz , 1 al. icokir • • (1) = 3coirrtz = 37). 0 = ap2 p
ap
EDDY CURRENTS
372
§10.04
Let v = (jp)i p, and this becomes
a2E, , 1 81„ (2) 5172. This is exactly Bessel's modified Eq. 5.292 (1) of order zero, the solutions of which were found in 5.32 (6) to be "
IZ = C/o(v)
liKo(v) = 0/4(./P)IP] /51CoRiP)iPi (3) From 7.16 (2), we know that there can be no magnetic field in the cavity in this tube because of the symmetry. If the external and internal radii of the tube are a and b, respectively, we see that the boundary condition at p = b is B = 0. From 10.00 (3), this gives x i)„_b = 0
or
\op
=0
(4)
p-b
At p = a, the boundary condition is = io From (3) and (4) using 5.33 (4), we have 0 = 0/1[(jp)ib] —
(5) Ki[(ip)ib]
and from (3) and (5), io = 0/0[(jp)ia] + 13K 0[(jp)i a] From these, we obtain
10[(ip)4a]Ki[(jP)4 b] + iRip)41911(0[0p)ial I iRip)iblio .75 — . I 0[(.7P)ia]K iRi P)Ibi -I- I ai P)INK oR j p) a]
(6)
(7)
10.04. Skin Effect on Solid Cylindrical Conductor.—If we have a solid wire instead of a tube, we must set D = 0 in 10.03 (3) since Ko(x) is infinite when x = 0 by 5.33. We would then have for fa /0[(iP)1P] . (1) • 2, I 4(3P)ial instead of the result given by 10.03 (3), (6), and (7). For numerical computation, it is necessary to split /0[(j)1x] and Ko[(j)ix] into real and imaginary parts. This is done by means of the ber, bei, ker, and kei functions of Lord Kelvin, for which numerous tables exist; see, for example, Dw 1050 or HMF, pages 430 to 433. Thus /JUN] = berox jbeiox (2) Ko[(j)lx] = kerox jkeiox (3) Series for these functions are easily obtained by substituting in 5.33 (1) and (4). Using ber and bei functions in (1) multiplied by eiwt and taking
=
§10.04
SKIN EFFECT ON SOLID CYLINDRICAL CONDUCTOR
373
the real part, we have for the current density in a solid wire of radius a beiO[(P)1Plyto cos kun, , a) {berORP)1Pi beiff(p)ia] berg[(p)ia] , bero[(p)ia]beioRp)ip] — bero[(p)ip]beioRp)ia] a = tan— bero[(p)ia]bero[(p)ip] + beio[(p)ip]beio[(p)*a]
(4)
is =
(5)
We can obtain the total current in the wire at any instant from the magnetic field just inside the surface, since by 8.09 (1), they are connected by the relation B = µI/(27a). From (1), 10.00 (3), and 5.33 (4), we have, after dividing out ei't,
= — T(N7 x i). = 448ap.1 a— 6 (jP)1/O[ciP)aiTio I 0[(3p)ia]
jc0(11)0 or
= 2rab _ 2irr(jp)iala( jp)iali jco,u10[(jp)ia] °
(6)
The average power dissipated per unit length in a ring of radius p and thickness dp is, from 10.00 (10),
dPav= iTIL122vp dp = xTIZi Z p dp where L is the conjugate phasor of L. We should note here that for electrical purposes the conjugate complex of (-Fj)i = 2—i(1 + j) is ( —j)i = 2-4(1 — j) = —j(j)i. We may write
= LA( —iP)iPl = fo[ —i(iP)iP] The total power consumed per unit length in the wire is then '2 7720 Pay =
0
dPa„, =
. I0P)ictlio[ R3 — j(iP)/a]f
dp
This integral is a special case of 5.323 (1) obtained by setting n = 0, and the result written in terms of berox and beiox is
pa
Tyra
v
(P)1
beroRp)ialbeia(p)ial — bera(p)ia]beio[(p)ial, to berff(p)ict] beiff(p)ial
(7)
From (6), the rms current I, is _ 24r2T2Pa2 fa(iP)iarla(iP)Ial .2 112‘1.12 Io[(jp)ia]Io[(jp)ia]Zo 2ir2a2 bera(p)ial}2 + beia (p)ial } 2.2 p {bero[(p)ia]) 2 + beio[(p)ia]) 2°
(8)
If R = r/(ra2) is the resistance per unit length for direct currents, then the high-frequency resistance R' is R,
_ — Pavra2R _ ct(p)iber[(p)ia]beil(p)ial — ber'[(p)iajbei[(p)ra]R (9) (p)ia] } 2 71,2 — 2 ber'[(p)ia] I 2 +
374
EDDY CURRENTS
§10.05
where, as in many tables, the zero subscript is omitted and from 10.03 (1) p = r'1.10) = ypco
(10)
and a is the radius of the cylinder. One gets the magnetic field energy inside the wire as in (6), thus drrio
jo.)13 = 441 =
49.10[(jp)ia]
as I o[UP)Icti The average energy inside the wire is, from 8.02 (3),
Op .
a a ir 7.2jap ,;I) dp f13- •ap dp — 2u(o2 /oRiP)IalioR —iP)lailo 2A 0 where we have written /0 for /0[(jp)ip] and .1- , for a/o(x)/ax. Since /“x) equals /1(x) by 5.33 (4), we see that this integral is identical with 5.323 (1) when n = 1, so that its value in terms of berix and beia is
la
1-
ap-i[beri(pia)bee,(pia) — berapia)beii(pia)] Using Dw 828.1, 828.2, 829.3, and 829.4 to reduce this to zero order and remembering that this average energy equals IL.,/! where Liis the internal self-inductance per unit length and Ie is given by (8), we obtain Li =
r(p)-1ber[(p)ia]berl(p)ia] + bei[(p)iedbeil(p)ia]
fber'[(p)icd} 2 + {bei'[(p)ict) 2
2irwa
(11)
10.05. Solution in Spherical Coordinates for Axial Symmetry.—Let us assume that the magnetic field producing the eddy currents is independent of ci5 and has no 0-component. Then the vector potential has only a 0-component and may be written A = 4)A 4,(r , 0, t)
(1)
where 4is a unit vector in the .-direction given by . = — i sin . + i cos .
(2)
Apply Laplace's operator 3.05 (1) to A. = —A 4, sin ck and Au = A4, cos 4) separately. After recombining we see that 10.00 (8) becomes il
a,aA4,
T 1-
at
= V2A = 4,[v2A, — 7
sn -AA,
i e0 r2 r sin2
cl3V 2A4, + Act,V2(13
(3)
Writing out V2 in polar coordinates by 3.05 (1) and dividing out 4give AaA0 1 a (r 2aA4, 1 a (sn o i 0m o = T at r2(3r + sin o aen2 are r2 ae ) r 2A sin 0
(1_ .2)ia2R1 = 1 a (r 2aA 9!. ± (1 2 r2 ar
ar
r
u2pAoi
—
au2
(4)
where u = cos 0. We shall now consider the steady-state eddy currents when the magnetic field oscillates with an angular frequency w. As in 5.12, we shall
§10.06
CONDUCTING SPHERE IN ALTERNATING FIELD
375
seek a solution that is the product of a function of 0 by a function of r. Then we write A0 = er-iReiwt (real part) (5) Substituting in (4), multiplying through by r2, and dividing through by Or-iRem give r2 d2R R dr'
r di? h dr
1 4
jpr2 +
(1 — U2)1 d2
[(1 — U2)103] = 0
du2
(6)
where, as in 10.04 (10), we have written P = 7112W = i" (7) Proceeding as we did in solving 5.12 (1), we set the terms in (6) involving 0 equal to —n(n 1) and those involving r equal to n(n 1), thus satisfying (6) and giving, after expanding the derivatives in (6), (1—
u2)cf — 2u du
on
(8) (9)
2
The first of these is identical with the differential equation for Legendre's associated functions with m = 1 as written in 5.23 (4.1), and the second is the modified Bessel equation in x where x = (jp)ir as written in 5.32 (1). Thus, from 5.23 (5) and 5.32, Am is the real part of r--4[A„P(u)
Btia(u)l{anIn+1[(.7P)Irl
1.5.K.+1[(jP)irll el' (10)
If n is an integer, as it must be for Pl(u) and Qgu) unless conical boundaries are involved, we can use /_(,,+i) instead of Kn+1 for the second solution, by 5.37. In a region where the conductivity is zero, the left side of (4) is zero, and if we let A0em = E'Oeic", we obtain (8) as before but instead of (9) we get die\ n(n 1)R' = 0 -c/--7 r, dr ) whose solution is given by 5.12 (3) to be
R' = Arn ± ijr-n-1
(12)
and in a nonconducting region this will replace terms involving r in (10). 10.06. Conducting Sphere in Alternating Field.—Consider now the specific example of a sphere of resistivity T, permeability j4, and radius a placed in a uniform alternating z-directed magnetic field Beiwi . The phasor vector potential of this field is, when ei‘" is divided out,
A = +Or sin 0 = (10rP1(cos 0)
(1)
376
EDDY CURRENTS
§10.06
as can be verified easily by taking its curl by 3.04 (2) and (3). Thus, n = 1 in 10.05 (10) and (12); and, since the eddy current vector potential must vanish at infinity, we have, outside the sphere, a < r < co,
= 0/3(r +
1)-7.-2) sin 0
(2)
At r = 0, is finite, so 5.37 (4) and (5) show that only Ii[(jp)ir] can occur inside the sphere. Thus, setting n = 1 in 10.05 (10), we have
= obor-iIi[(jp)ir] sin 0
0 < r < a,
(3)
From 7.21 (6) and (7), the boundary conditions when r = a are
A. = At
a
-
a
-
au,,-&.(r - sin MO = ii— (r sin 0A 0) ar
and
(4)
Putting r = a in (2) and (3), we obtain with the aid of 5.321 (1), (2), and (3), after writing In for /n[(jp)ia] and v for (jp)ia, a3 + 13 = alO/I = aio[/_i — (2a3— D)/A = Ance[iIi vI Solving for C and
= Anal[(v
— I-110
15 gives 31.wal
—
(5)
— 14)2)1_1+ [An( 1 + v2) — []i* _ (21A — [1.1„(1 + v2) +21.1]Ii.a3 0.4 — [An( 1 + v2) —
(6)
This may be expressed in terms of hyperbolic functions by 5.37 or Dw 808.1 and 808.3. From 10.00 (4), the current density anywhere inside can be obtained from (3) by the equation i = —juryAi = — jpg.-1Ai
(7)
The magnetic field at any point outside is obtained from (2), 3.04 (3), and 3.04 (2) to be Boe
= —b(1 — 13) sin 0
= —r1 1
a (,s.
r sin 0 (30
0Ao) = '41 ±
-)
r3
(8)
cos 0
(9)
By a similar method (3) yields bio and Eir. Comparison of (8) or (9) with 7.10 (2) shows the eddy current field to be like that of a magnetic dipole loop of radius a carrying a current where 1.4a27 = 21315. If 0 from 11.05 the magnetic field is not alternating, then w = 0 so that p (7). From 5.37 and Dw 657.1 or 657.2, we see that Ii(x)
(-2 ) 4(x + 6 rx
and
2 (1 /_4(x) ---> (7rX ) 1
2)
377
POWER ABSORBED BY SPHERE
§10.07
Thus (5) and (6) simplify so that (2) and (3) become
A. = (1)i r
. 2(Ifn, - a31 1)2 sin 0 (K m +2)
(10)
3Kmb
(11)
- 412(Km 2) r sin
These are the correct static fields. From (7), we obtain as a first approximation for slowly alternating fields
3jco.K„,yr3 14, = — 2 ( ICm + 2)r sin 0
(12)
The same result is obtained if the resistivity is made infinite. When the frequency is made very high, we see that A.-> 0
4,-> (14/3(r - a 3r-2) sin 0
(13)
because, from 5.37, Ii(x)
( 2 yez
/3(x) --> Li(x) ot,
71-X
2.
thus, there is no magnetic field inside, and the eddy currents are confined to the surface as we would expect. To visualize the magnitudes of the numbers involved, we shall compute p by 10.05 (7) for a field alternating at 60 cycles per second (so that w = 1207r) for several substances. If r and A are in mks units, we find = 4r X 10-7henry per for copper r P-11.7 X 10-8ohm-meter, a meter, and p 28,000. For a typical specimen of iron, r 10-7, 1.1 480r X 10-7for a magnetic field intensity of 1.5 X 105/(47r) ampere8 X 10-8, turns per meter, and p = 570,000. For graphite, r A r-=, 47 X 10-7, and p = 60. Thus for this frequency, over distances of a few centimeters, (12) would apply to graphite but not to iron or copper. The initial assumption that A, is zero and Bo tangential to the surface greatly simplifies the calculation. All the results of this section, being written in phasors, give both the amplitude and phase of the quantities involved. The same forms for the electromotances and the same types of boundary conditions apply equally well to any number of thick concentric spherical shells and could be used, for example, to calculate their screening effect. The results would be much more complicated than those given here. The distribution of a given amount of material in several separated shells will increase the screening effect. There are optimum thicknesses and spacings. 10.07. Power Absorbed by Sphere in Alternating Magnetic Field.— We shall now compute the power absorbed by the sphere in the last article. From 10.00 (10), the power absorbed in the volume element dv is
dP
dv = irriir2 sin 0 dr d0
378
EDDY CURRENTS
§10.08
Substitution for i and I from 10.06 (7) and (3) and integration with respect to 0 from 0 = 0 to 0 = 7 give
P - irw32B2aef aIiRjp)irlIg —jp)irk dr
(1)
Integrate by the last form of 5.323 (1) and note from 5.37 that .4[( ±jp)ia] = Fir( ±jp)iai—i sinh [i(2p)i(1 ± j)a] I4(±jp)ia] = [lir( ±jp)ia]-1cosh [1(2p)' (1 ± j)a]
(2)
Application of Dw 651.06 to 651.09 or Pc 669 to 672 gives for the integral (wpia)-14(2pa2)1[sinh (2pa2)1+ sin (2pa2)4] — cosh (2pa2)i cos (2pa2)1 Writing out CC by (2), 10.06 (5), and 11.05 (7) gives for P, if U = µ U2[(pa2
3,a 5w2B2A2T-1[_u(s
s) — C
-
c],
+ 1)C (pat— 1)c — u(S s)] + Upzi,pa2u(S — s) mv2p2a4(c —
where u = (2p)fa, C = cosh u, c = cos u, S = sinh u, and s = sin u. 10.08. Transients in Conducting Sphere.—In the last two articles we solved the steady-state problem of a sphere of permeability pi and resistivity T in a uniform alternating magnetic field. Let us now solve a transient problem by calculating the effects when the same sphere is placed in a uniform magnetic field B which is suddenly removed at the time t = 0. Clearly, in this case, as in the last one, A will have a 0-component only. At t = 0, surface eddy currents prevent the interior field from changing, and since the vector potential must be continuous across the boundary, we have, from 10.06 (11), at t = 0,
A, = CI)
3K,J3 2)rPi(u)' 2(Km
A,, =
3„,,13 '2(K„,
a
131.(u)
2) r2
(1)
Since T = 00 outside, the subsequent behavior of A is determined from 10.00 (8) by solving the equations
v2 Ai = T dt
V2A. = 0
(2)
In 10.05, we found a solution of the first of these equations when the time entered exponentially. Let us see if we can fit the boundary conditions in the present case by using a sum of such solutions. Clearly in the present case there will be no oscillations, so we shall need to substitute a negative exponent —gat for jest. Let us write lc! = r--41.tqa =
(3) We must then substitute —k for jp throughout 10.05. This gives us jk, instead of (jp)i in 10.05 (10) and leads to ordinary Bessel functions. From 10.05 (10), writing sin 0 for Pi(u) and noting that Aimust be finite
§10.08
TRANSIENTS IN CONDUCTING SPHERE
379
at r = 0 and contain 0 in the same form as (1), we obtain Ai = 4•ZA.r—ifi(ka r) sin Be—ot
(4)
8
A. is finite at infinity and equals Ai at r = a for all values of 1. From 10.05 (12), it therefore has the form = 42/38r-2 sin 0 —g.t
(5)
a
In addition to satisfying (1) when t = 0, we must satisfy 10.06 (4). This gives at r = a, dividing out sin 0, a(rA0)
1.,° a(rA,)
A° Ai,
a
=
Therefore, at r = a for all values of t, it is necessary that
AsaiJi (k.a) = B,,
p„Asa
d [aiJi(ksa)] = da
(6)
—
Differentiating the product in the second equation, multiplying the first equation by pi, and adding and dividing out As, we obtain a)] k8 + ilvada[ji(
1
—1.1,)T(k.a) = 0 2
(7)
To meet the boundary conditions, we must choose values of k, that satisfy this equation. This determines, by (3), the values of q, that appear in (4) and (5). The values of k, can be found with the aid of 5.31 (3) and a table of trigonometric functions. From (1), (4), and (5), multiplying through by ri and setting t = 0, we have 3KmB
2(K., ± 2)
r =
(8)
This is identical with 5.297 (1), last case, where n = I and, from (7), I-. From 5.294 (7), the B of that article equals (/.0.4) a 3K Bat , VJI(k v) dv — ' vf(v)J3(k,v) dv — 3K,„B 2k, Km± 2)J o(kea) 2(Ka. ± 2)1 With the aid of (7) and 5.294 (2), we obtain ±2 Ji (kaa) = —Ji (k,a) gsa ji(kaa) = keft ji(k8a)
f
a
So that, from 5.297 (6) we have
A. —
[k!a2
31C,0Bal (IC„, — 1)(1C„,
2)]11(k.a)
(9)
380
EDDY CURRENTS
§10.09
Substituting this in (4) and (5) gives Ai and A,. With the aid of 10.00 (4), we obtain for the current density inside
411(ksr)
3Bal sin ONI
i = +
Ayr*
2)1,11(1c,a)e
L-1[14a2+ (Km— 1)(K„,
—k 1174-1t
(10)
8
10.09. Eddy Currents in Plane Sheets.—We shall now calculate the vector potential A(x, y, z, t) of the eddy currents of area density i induced in a very thin plane sheet of area resistivity s at z = 0 by a fluctuating magnetic field whose vector potential is A'(x, y, z, t). In the interior of the sheet, the electric field —a (A' + A,)/at drives to the surface charges whose electrostatic field exactly neutralizes it. The effect of the associated current is negligible, so that inside the sheet we need consider only the tangential components A, and A. Thus using 10.00 (4), we may write
d(A: A,) = si (1) dt Let the eddy currents be confined to a finite region of the sheet which may or may not extend to infinity, and let us define the stream function c13(x, y) at any point P in the sheet to be the current flowing through any cross section of the sheet extending from P to its edge. From 7.01 (2), we have, for the closed path bounding this cross section, but not including the surface, since B is symmetrical with respect to the sheet, =
(fB • ds = 2fX C Bs dx = 2f dy
(2)
where the path of integration is in the positive x- or y-direction when z is positive. Differentiating this equation, we have 2 aA _ M = a4) = 2B„ = j'y = act. = = as
ay
/.1,
ax
az
ktv az
Adding the components gives, inside the sheet
i=
2 Ay
aA, az
(3)
Substituting this in (1) gives, inside the sheet,
d(A: + As) _ 25 9A, dt Az az
(4)
Outside the sheet, A consists of two parts, a magnetic part arising from the eddy currents and an electric part, the gradient of a scalar, arising from the electric double layer produced by Ai and Az. Since A:, the tangential component of the exciting potential, is known and since A, is continuous across each face of the double layer, (4) gives the boundary condition on A, just outside either surface of the sheet. This, combined
§10.10
EDDY CURRENTS IN INFINITE PLANE SHEET
381
with the equation V 2A = 0, determines A everywhere outside the sheet. When the sheet is finite, the boundary condition in its plane, but beyond its edge, becomes i = 0 or dasIdt = 0. The right side of (4) is finite at all times, which means that if of —4 0, then 5(A; + A,) —> 0. Thus an abrupt change in A' instantaneously induces eddy currents such as will maintain A + A' and B B' unchanged in the sheet. Therefore, for a specified change in A', the initial value of A is known and, if no further changes occur, its subsequent values, as the eddy currents decay, can be determined by putting dA'/dt = 0 in (4) and solving. A second abrupt change in A' produces a second set of eddy currents, and so forth. At any instant, the actual field of the eddy currents is a superposition of these. As the magnitudes of the discontinuous changes in the external field become smaller and the intervals between them shorter, we approach, as a limit, a continuously changing magnetic field. 10.10. Eddy Currents in Infinite Plane Sheet by Image Method.— Suppose that a thin infinite plane conducting sheet at z = 0 lies in a magnetic field produced by sources in the region z > 0. At t = 0 the field is changed, the vector potential being given by A'1 = fl(x, y, z) when t < 0 and by A', = f 2(x, y, z) when 0 < t. From the last article, the eddy currents produced at the instant t = 0 keep the vector potential over the surface of the sheet the same so that, initially, the whole field on the negative side is unchanged. Thus we have, on the negative side of the sheet, from the eddy currents alone = (1) = fl(X1 y, z) — f2(x, y, z) —
This field could be produced by reversing the sign of the new source and replacing the old source. These hypothetical sources which can replace the actual eddy currents are images as in electrostatics. Since A2 is not a function of t, 10.09 (4) reduces to
dA = 2s aA µv az dt A general solution of this equation satisfying (1) at t = 0 is
(2)
A = f i(x, y, —Izi — 2stc1t) — f2(x, y, Hz' — 241,710 (3) A identical at ±z The sign of z was chosen to make as required by symmetry and to make it die out with time. Thus, the equation shows that, added to the field ig which would exist if no sheet were present, there is a decaying field due to the eddy currents which appears, from either side of the sheet, to come from a pair of images on the opposite side which recedes with a uniform velocity 2s/A,,. Maxwell gives a formula for this law of images which applies to any type of field variation. Suppose the inducing field vector potential is A' = f(t, x, y, z)
(4)
382
EDDY CURRENTS
§10.11
The change in this field in an infinitesimal interval of time dr is
s a
dT
=
a
x, y, z) dr
The initial field of the eddy currents formed in that interval must be equal and opposite to this. This field dies out, as we have seen, as if it were due to an image on the opposite side of the sheet moving away with a uniform velocity 2s/1.4,. Thus, the vector potential of the eddy currents at the present time t due to images formed in the interval dr at a positive time 7 before the present is given by
dA =
a f(t —
at
2s y, —1z1 — — r) dr
T, x,
At the time t, the total vector potential due to eddy currents is
a 2S A= —f —f(t — x, y, —1z1 — —r) dr (5) o at When z < 0, z replaces — jzi and, by Pc 863, war = —avat — (2s/g )af/az so that substitution for of/at in (5) and integration give, since f is zero at T = 00 and equals A' at T = 0, for the total resultant field 2s a f x, 2s A A' = -F -, f(t — T, y, z — r) dr (6) az o This is often simpler to integrate than (5). 10.11. Torque on Small Rotating Current Loop or Magnetic Dipole.— If a magnetic system is rotated about an axis normal to a conducting M --- sheet, the field of induced eddy currents will, in general, set up a retarding torque on the system which is, as we shall see, proportional to its = angular velocity if this is not too great. By measuring this torque, the angular velocity can be found, and this is the basic principle of some o automobile speedometers. The simplest magnetic system is a dipole or small current loop, P. 1 which rotates about its center keeping its axis 2sr parallel to the sheet. In order to visualize the /L. images, we shall not use 10.10 (5) but shall sup-. )— 2 2sr pose that at regular intervals of time T the magp. net is instantaneously rotated through an angle 3 WT. From this result, by letting the intervals approach zero, we shall obtain the result for FIG. 10.11. continuous motion. Figure 10.11 shows the images formed by the last four jumps at the instant the dipole, of moment as, arrives in the position indicated, as viewed from the upper side of the z < 0,
TORQUE ON SMALL ROTATING CURRENT LOOP
§10.11
383
sheet. The actual torque on M will be the sum of torques due to all the images. Since all image dipoles are perpendicular to the axis of rotation, ei= 82 = ir in 1.071 (4) and, from 1.071 (7) and 7.00, the torque due to a single image making an angle 1p with M is
-fry Af 2 sin ip aip a = 47— r 3
T=
(1)
The torque due to all the images is then, if p = co/4/s, et1
T=
i.c.„m2
47r
sin nwl- — sin (n 1)wr [2c + (2ns//12,)rJ3
(2)
n=0
1.4,„p 3m2
sin non- — sin nun- cos wr — cos ncor sin wr
(pc + ncor) 3
— 327r -LJ n=0
As the motion becomes continuous, r dt, nr —> t, and sin wr so that (2) passes over into the integral
T—
— cos cot
ici,cop3m2 7r 32
0
(pc + wt)3
(or —> w dt
dt
(3)
Let x = pc + wt so that cos wt = cos (x — pc), and this becomes „v,n3m2 cos x 'sin x (cos pc dx + sin pc —T— X 3 oZT fp, x 3 dx fpc Applying Dw 441.13 and 431.13 or Pc 345 and 344 gives
—T—
647r
p2c2 COS
pc
dx — sin pc
• x sin
dx
(4)
These integrals may be evaluated by the series Pc 346 and 347 or Dw 431.11 and 441.11, or they may be put in the form of the standard cosine integral Ci and sine integral Si. Thus 7r 'cos x I sin x dx = — — Si(pc) = —si(pc) (5) 2 , x dx =Ci(pc), x
fp
Numerical tables and graphs of Ci and Si are available in Jahnke and Emde and other mathematical handbooks, especially HMF, Chap. V. If pc is much greater than unity, we integrate (3) thrice by parts and obtain for — T,
1.4,P 3m1 327r
(pc + 0) 3
co sin wt 3 coswt + 12 sin wt — 60 dt (pc + (00 50 fo (pc + cot)6 (pc + (4 4
Neglecting 20w/(pc)2 compared with unity, we have 3A04.2
Pc >> 1
T=
32irpc4
(6)
384
§10.12
EDDY CURRENTS
Neglecting w(pc)2compared with unity (4) gives
T — AvP 312 6471-c2
pc in Eq. 6, Art. 10.12, are shown when coA„M/(47rs) = 1.
1 cm distance, on the scale shown, above the plane of the paper and is parallel to the c1 = 0 line. We see at once that the currents shown will give a field at right angles to the magnet. This will produce the torque calculated in the last article.
10.13. Shielding of Circular Coil by Thin Conducting Sheet.—Let us use 10.10 (6) to find the effect of the insertion of an infinite thin conducting sheet on the flux linkage between two coaxial loops of radii a and b at
386
EDDY CURRENTS
§10.14
a distance c apart. Without a sheet the vector potential at one loop due to the other is, by 7.10 and problem 27, Chap. VII, the real part of A'4,0" = Arir—iemfoir(u2
c2)--i cos 4c/4) = Ne" Ji(ka)Ji(kb)e—kc dk (1)
where u2 = a 2 b2 tab cos 4 and N = 11.4„aI. Then writing X for 2s/A, and c Xr for — Izi — Xr in 10.10 (6) gives at the second loop A4.
= — NXf: kJ i(ka)J i(kb)e—kc f e—(xkl-mr dr dk (2)
= —N f: Xk(jco Xk)-1J i(ka)J i(kb)e—kc dk The general solution is in problem 33, but if Xk/ co < 1, this becomes CO
CO
t42a/N(2js f '
-.J\1.4,,co
n=1
2j5 \nanA'c°
knJ i(ka)cl i(kb)e—kc dk =
0
n=1
Avw)
acn
(3)
If b and c replace p and z, the k series of 7.10 (4) is a good form for 4 From 8.06 the ratio of the new to the old flux linkage is Ro —
271-b(A0 A;). 27rb(A:Orroo
(4)
If 2s/(wuv) is small, only the first term in the series is needed so that 2s B 2s 1 aA,6' Ro = — — — ac
C01.4 A0
(5)
where B„ and Ao are given by 7.10 (6) and (4), respectively, and are evaluated at p = b, z = c. If, in addition, either a or b is small compared with c, then from 7.10 (1) we get R0 = wi,„(a2
6sc b2
c2)
(6)
As we saw in 10.12 for a sheet of copper 0.1 mm thick and a frequency of a million cycles per second, couv/s is 46,700 which justifies the approximation in (5). Hence, if a = 1 cm, b = 10 cm, and c = 10 cm, then, from (6), Ro 0.0006 so that the linkage is greatly reduced by the shield. 10.14. Rotating Sheet in Magnetic Gap.—The reasoning of 10.10 applies to the scalar as well as to the vector magnetic potential and the magnetic induction. As an example consider an infinite thin horizontal sheet of area resistivity s parallel to the horizontal plane faces of a gap of height 2h in an infinite vertical circular cylinder of radius a having a uniform permanent magnetization M parallel to its axis. The sheet
ROTATING SHEET IN MAGNETIC GAP
§10.14
387
rotates with a constant angular velocity w about an axis parallel to and at a distance c from the cylinder axis, or the magnet may rotate with the sheet stationary. In any case, the system of eddy currents is fixed relative to the magnet. The magnetization is assumed to be absolutely rigid, so that the magnet acts as a cylindrical current sheet and does not perturb the magnetic field of the eddy currents as a ferromagnetic magnet would. The integrations in this article will be carried out exactly, partly to show what can be done with the mathematical references now available, especially IITF, IT, and HMF, and partly because the resultant formulas which involve many repetitive operations are easily set up and evaluated on a digital computer. From 5.16, 7.28, 9.08, and 5.298 (6) the potential on the axis of a ring of radius p' and magnetization M is, when z < h, dJ0 = Marp' dp' [4.7r[p' 2
dp' f Jo(kp')e-"h-z) dk (1)
(h2 — z2)]1 }-1 =
Here the loop definition for M is used so pi, is omitted. Integration by 5.294 (7) from zero to a gives the potential on the axis of one pole face:
Sta = f dS2a = aM f k-lJ i(ka)e-h(h-z) dk
(2)
Off the axis, in the plane z = 0 where the sheet lies, this becomes = ictMf' k-lJ o(kpi)J i(ka)e-kh dk
(3)
it, for cp to express rotation, so that Expand this by 5.298 (7), write cot when t = 0, c lies on the 4, = 0 axis (see Fig. 10.14), and obtain = aM (2 — 32).1. 0%-1J3(kp)J8(kc)Ji(ka)e—kh dk cos [s(wt 8=o
4))] (4)
This replaces 10.10 (4). Similar changes in 10.05 (5) give the eddy current Si, so write t — r for t and h (2s/A„)r for h in (4) and put 1 swilvh P. = 2 s
2s Alv
swh
(5)
ps
Now take 8/cat, set t = 0, and integrate with respect to r, so that = am
7_),,f°°,1-0(kP)J.(kc).11(ka)e_
h
k[k2
kh(k sin so
pah—' cos so) dk
(6)
8 =1
The s = 0 term drops out since Pscontains an s factor. From 10.12 (1)
388
§10.14
EDDY CURRENTS
the stream function 4 is, adding that of the lower pole,
= 4f 1-1„dp = —4f Gdp =
(7)
Now let k = p8t/h, 2R = ps p/h,2c1 = pac/h, and 2a1 = pact/h so all symbols after the summation are dimensionless, then 4) = 4aM
8(2Rt), 8(2cit)J i(2ait)e_p. t(t sin sq5 + cos sO) dt (8) t(1 t 2)
fo =1
Watson gives on page 148 a power series in t for J 8 (2tci)J 1(2tai). (-1)mcIm+3Cms(ai,c1)0'n+8+1 J8(2tc1)J1(2ta1) = a 1 m= o s)!]-1 Cms(ai,c1) = 2F1( —m, —m — s;2;aler2)[m!(m
(9)
— 4)-1] (10)
= (4c1.2— 1)".[(m + 1) !(m + s)!)--11V)[(ci
by HMF 22.5.44, where P;,!, 8)(x) is a Jacobi polynomial whose recurrence relations and coefficients appear in HMF, pages 782 and 793. In the integral 8 (2tc1)J i(2tai) is therefore replaced by t2m+3+1, so the integrals in the sine or cosine terms are .t2m+a-1-1j 8 (2tR)
1
1 + t2
.
dt
or
r Om+8,18(2tR)e_" dt 1 + t2 jo
Expand J.(2tR) in series by 5.293 (3) and write N for m
1YR2r1f t2N+10—P'td, r=0
r!(r
s)!
1 ± t2
or
s
(11)
r, so (10) is
r .t2Ne- p,t dt J o 1 + t2
(12)
If N were zero, the integrals would be HMF 5.2.13 or 5.2.12. Since N is never zero because s j 1, divide the numerator by the denominator until the remainders have this form and integrate the quotient terms by Dw 860.07, so the integrals of (12) are
(2N — 2i ± 1) ! (— 1l
—F N (P0 = i=1
(-1)N[Ci(ps) cos p8
si(p8) sin ps] (13)
N
( -1)i(2N — 2i)!
—GN(p8) =
( 1)N[Ci(p8) sin ps— si(p8) cos p8]
i=1 (14)
389
ROTATING SHEET IN MAGNETIC GAP
§10.14
See 10.11 (5) for Ci and si and see HMF, pages 238 to 244, for tables.
Thus the final value of the stream function 4) is CO
Z
—4aM ( — 1)maicrn-"Cms(ai,ci) Z ( —1)rR2r+8 r0 8=1 m= o X [r!(r s)!]-1[F7 (pz) sin scp G, (p8) cos sq5]
(15)
The torque about the axis of rotation will be calculated only in the case where h is so small compared with the radius a of the pole pieces that the fringing field is negligible. With this narrow gap the magnetic induction under the poles has the constant value µ 2 M, which is that in the equivalent current sheet solenoid, from 9.07. The torque equals the product of the normal induction Bz by the radial current density i,, and by the lever arm p integrated over the pole area. T = f pBzi, d8
where i,, = —
Ocl) P
(16)
a 4)
For the sine term in (8), i is even about 4) = 0, and for the cosine term it is odd. Thus for the latter the torque on one-half will cancel that ,
x
(a)
(b) FIG. 10.14a, b.
on the other so it need not be considered. From Fig. 10.14, the torque will be given by (8) if the J8(2tR)t sin s4, factor is replaced by af M AvSt f0 —1r cos .34) J ,(2tR) pi dpi do Let 2/i1 = poi/h.
(17)
Then from HMF 9.1.79 we have
J,(2tR) cos s4, =
J.+8(2tc1), n(2tRi) cos nO
(18)
It=
The limits — 7 to -Fir leave only n = 0. The R integral is 5.302 so (17) is —27r/4,8J8(2t,c1)4112 73:2f0U 0(2tR OR' dB]. = —27M psahm,--1,1„(2tci)J1(2tai)
390
EDDY CURRENTS
§10.15
Insert the factor 27/1/4„sa in (8), omit the 4) terms, and replace .18(2tR) in (11) by hp,7't-iJs(2tci)J1(2tai). Then use of (9) gives ‘.2 T = 271.4„a3M2 Z Zsai(-1)m+eciNC..(ai,ci)Cr.(al,ci)FN(p.) (19) 8=1
m=0 r= 0
10.15. Rotating Disk in Magnetic Gap.—Suppose that the infinite rotating sheet of the last article is replaced by a disk of radius A rotating in a magnetic gap. The primary change is that the circulation of the eddy currents is restricted to the region inside p = A. A secondary result is that the magnetic fields of the eddy currents can now pass freely around the edge of the disk without penetrating the sheet. The main problem is therefore to find a means of confining the eddy currents to the region p < A. One method to look into is that of images. In 10.14 (3) if p = a, then pibecomes (c2 A 2— 2cA cos 6)1, and if A2/c replaces c, this becomes (A/c)(c2 +A2 -2cA cos Of. Thus if a and h are replaced by Aa/c and Ah/c and the integration variable is changed from k to Ak/c, then the value of the integral in 10.14 (3) is unchanged and the new potential or magnetomotance 0' is also the same at p = A if aM is the same. In passing from 10.14 (3) to 10.14 (6), the substitution h (2s/ µ„)T was made for h in 10.14 (3), so for the new magnet (A/c)[h' (2s/ A„).7] must replace h' and thus in 10.14 (5) p.' = (c/A)ps replaces ps. Therefore if the original magnet is replaced by one centered at c' with a pole face radius a', a gap 2h', a magnetization M', and a new p87 then the value of the stream function at p = A will be unchanged provided that , A' c=
,
a
=
Aa
—,
h h' = c,
M' = A' '
p' = s A
(1)
Thus if both magnets are present, with M' reversed in the second, the circle p = A becomes the stream line 1 = 0. The stream function for a disk is then given by adding to the 43 of 10.14 (15) and (12) a function calculated by the formulas using c'1, ai h', —M', and p' instead of ci, al, h, M, and p. in 10.14 (14). If s/(A„(.0) is large compared with a2 and c2, a simpler procedure than that of 10.14 can be used. From Fig. 10.14, in the z = 0 plane (note that there is no because the loop definition is used for M) ,
mfarr =
4a 0
p' dp' da
(h2
d2
-
2
pip cos a)}
(2)
where pi is [c2 p2— 2cp cos (4) + wt)]} for a rotating field. The 10.10 (5) substitutions are now made, but because w p' and 0 if p1 < p', so the upper limit for the p' integration is a when p1> a and p'when p1 < a. Multiplication by 4, because of 10.14 (9) and the lower pole, gives for the stream function for an infinite plane P1 > a pi < a
M yvwcpa2sin cp 2 spi MA,,cocp sin 0 4' = 2s
(1,—
(outside pole piece)
(4)
(under pole piece)
(5)
These formulas replace 10.14 (11) and (12) when Avo)/ s is small. For the disk of radius A, the stream function of the second pole found by making the substitutions for c, a, h, M, and p, indicated in (1) must be subtracted from (4) and (5). Thus clpu under the pole piece and between the pole piece and p = A become A 2a2 MA,,cocp sin y5[ (6) 1 2s c2p2 + A 4 — 2pcA2cos 4) A2 1 MiLvWCPa2 sin ck [ c2p2 ± A 4 2cpA 2cos 4) p2 ± c2 2s 2cp cos 4) 4). —
410
(7)
Figure 10.15 shows 4,plotted from the above equations for a single pole (a) and, by superposition, for two poles of opposite polarity (b), so that the currents from both poles flow in the same direction along the line between centers. In Fig. 10.15a and b, MAcos-1 is 10, c/a is 0.7, and A is 0.1 meter. The stream function in amperes has the value zero on the boundary and increases by steps of 0.1. The torque about p = 0 is calculated by 10.14 (13). The use of polar coordinates r, 0, coaxial with the pole applying the torque, simplifies integration. The needed relations between r, 0 and p, 4) appear in Fig.
392
§10.15
EDDY CURRENTS
(a)
10.15c and d.
(b)
For the single magnet these are
A' cf = — c R2 = p2 + (c')2 — 2pc' cos0 .,___ (c '— c)2 + r2 — 2r(c' — c) cos 0 p cos cp = c ± r cos 0,
p sin cp = r sin 0,
(8)
From (6) and 10.14 (16), the torque with one magnet is T = _i_m21,20.,a2s-i f Rca-2— c' li-2)p cos 4) + 2(e)2R-4p2 sine 0] dS s = _ MWwa2s lf a.1.1.[(ca-2 — CR-2)(C + r cos 0) oo + 2(c')2R-4r2 sine 0]r dr dO (9)
Integration of the R-4term by parts changes it to an R-2 term, so the 0 integrals become Dw 859.121, 859.122. The simple r integration then gives the torque for a single magnet to be T = im21,2wc2a27rs9 — A2a2(A2 — c2)-21
(10)
The two magnets of Fig. 10.15b give approximately double the torque of one. The additional AT due to the eddy currents from one pole flowing
§10.16
ZONAL EDDY CURRENTS IN SPHERICAL SHELL
393
under the other may be set up and calculated with the aid of Fig. 10.15d. In the given by (7) and 10.14 (16) for one gap, 4) is replaced by it — cto to put it in the coordinates of the other gap. The integral is put in terms of r, 0 by the geometrical relations of Fig. 10.15d. The resultant integrals, evaluated as before, give AT = IM2/.12wc2a2rs-1[1a2c-2— A2a2(A 2
c2)-2]
(11)
The total torque for Fig. 10.15b where magnets have opposite polarity is T' = 2(T + AT)
(12)
When the poles have the same polarity it is T" = 2(T — AT)
(13)
If the magnet gaps 2h and 2h' of the M and M' magnets are small compared with the distance of their edges from the boundary p = A, the primary eddy currents induced by each will be unaffected by the field of the other. There is, however, a slight distortion of the eddy currents on one side of the boundary by the magnetic fields of the eddy currents on the other side. This should have little effect on 43. However, the magnetic field of the disk eddy currents, except near its surface, cannot be found from the scalar potentials 10.14 (6) and 10.15 (3) as modified to include the image magnets. They may be found by 7.02 (5) or 7.14 (1), where the integrals are taken over the regions p < A and the ip and i4, used are derived from the stream function 10.14 (14) and its image or from 10.15 (6) and (7). Note that Eqs. (2) to (13) apply to small rotation speeds or highresistivity disks. This means one made of material of poor conductivity or one made very thin of good conductivity. Since the resistivity T of copper is 1.75 X 10_8 ohm-meters and Ay = 1.25 X 10-6, for a sheet 0.1 mm in thickness, 2644,, is 280 and w must be considerably less than this for (2.1) to hold accurately. The formulas of the last two articles should be applied with caution to electromagnets if the eddy currents are strong. These cannot be replaced by an equivalent current sheet as a permanent magnet can. The permeable core tends to short-circuit the magnetic fields of the eddy currents, which then produce a demagnetizing force that in extreme cases can greatly reduce the torque. 10.16. Zonal Eddy Currents in Spherical Shell.—Let us now consider the eddy currents in a thin spherical conducting shell when we have axial symmetry so that all the eddy currents flow in coaxial circles. Let the total vector potential be A' + A, that part due to the eddy currents being A. The electromotance e producing the current in the ring of width a de at 0 is induced by the change in the total flux through the ring. By 8.00 (1) and 7.08, we can write E in terms of A' + A and then relate it to
394
EDDY CURRENTS
§10.16
the current by Ohm's law 6.02 (1), giving = —
dN = dt
d [2ra sin 8(A',, ± A0)] — dt
lirassin a d0
Zia d0
so that, if the vector potentials are zero at 0 = 0, (1)
i4, -7t(A +114') = s
where i4, is the current density and s the area resistivity. Let the eddy currents in the shell be expressed as a series of zonal harmonics, the nth term being in. In 7.12 (2) and (5), we found a simple relation between in and the nth term in the expansion of the vector potential produced by these currents. The relation is
mua
(2)
(A 's")' = 2n + .
If A'4, is also expanded in spherical harmonics, we see by substituting (2) in (1) that the expansions of A'4, and A95 are related at the surface of the shell r = a by the equation
d IA, dt‘ n
+ An)_ '12n ± 1Ans
(3)
n
n
If, after the time t = 0, the inducing field A'd, is constant and if, at the time t = 0, the field of the eddy currents is known to be A4,
=
(4)
2CnAn
then, clearly, a solution of (3), showing how the eddy currents decay is, 2n+ lst
Ad, = ZGAne µ
(5)
n
Suppose that the vector potential of the inducing field can be written, on the surface of the sphere, in the form
Aye = ZCn An.(t)
(6)
The change in this field in the infinitesimal interval of time dr, at a time t, is given by
r before the present time
0A0, dr
at
n
aAns(t — r)
at
dr
§10.16
ZONAL EDDY CURRENTS IN SPHERICAL SHELL
395
The eddy currents set up in that interval dr exactly neutralized the change at the surface of the sphere at that time but have been dying out according to (5) since, so that the present effect is 2n-I-1,7 CnaAns( 1-)e — A" dr at
dA 08 = — n
The total vector potential of the eddy currents at the surface of the sphere at the present time due to all these past changes is, then, A08 = —
C11.1 0 n
(t
"
r) e 2n-I-1,
—
at
dr
(7)
Replacement of aAna/at by —.9Ana/ar and integration by parts with the aid of (6) give, for the total vector potential from all sources at r = a, Ark. + A',58 = +---
(2n
1)Cni Ans(t — r)e
_ 2 n +1sT
Ava CiT
(8)
n
In the steady-state a-c case A'4„ which may result from both external and internal sources, is given at r = a by (6) where An, = Pni (cos 0) cos wt
(9)
Integration of (7), with N substituted for (2n + 1) s/(i.coa), by Dw 860.80 and 860.90, gives for the eddy currents alone at r = a CO
A le = ZCn/1,(cos 0)(01/42
N 2)-1(N sin cot — w cos cot)
n=1 ao
= — ZC.P;;,(cos 0) cos On cos (cot + On)
(10)
n=1
where tan tkn is N/co. Thus the present eddy currents are in phase with those formed at a time 1//n/co before the present. The phase lag en is The value of Ad, that vanishes at r = 00 and gives (10) at r = a is the vector potential outside the sphere due to eddy currents alone. Thus
Ao =
—
a 11+1 Cn(—) Pnl(cos 0) cos en cos (cot — r
En)
(11)
n=1
tan en —
(2n
1)s
tiv aw
(12)
Inside the shell, the eddy current vector potential As must be everywhere finite, so (r/a)n replaces (a/r)n+1. For an external source, r enters A' as
396
EDDY CURRENTS
§10.17
(r/a)n. From (6), (9), and (10), the total internal field is, by Dw 401.03,
Ai+ A' = (1:•
— C.(OnP;,(cos 0) sin en sin (cot — en)
(13)
=1
This could also have been found directly from (6), (8), and (9). The ratio of the nth term in the harmonic expansion of the new amplitude to the corresponding term in the expansion of the old amplitude is
R. — IA'
± A/I =sin en
(14)
From (12), if s is very large, this is unity and the field is unchanged, but if the shield is a good conductor, i.e., if s is small compared with izyco, Ron is very small and the shielding is high. The eddy current density in the shell as given by (2) and (10) is 2n + 1 CnPnl(cos 0) cos en cos (wt — en) /Iva
i = (1:•
(15)
n
The instantaneous energy dissipation rate in an element of area dS of the shell is its dS from 6.03 (2). In the whole shell it is, writing u for cos 0, Sf i2 dS
2 = 2sraTi2 sin 0 dO = —2sra2 f +12, 1 du '
When (15) is squared and integrated from u = —1 to u = 1, all crossproduct terms vanish by 5.13 (2) leaving only a sum of integrals of squares. Thus each harmonic component in cos (cot — en) of i behaves as an independent circuit so that the mean rate of energy dissipation from 6.03 (2) is Sf i2 dS
+1 .2 +1 2 sra2f [Zin ] du = srct'V f 1 du
Integration by 5.231 (3) gives for the energy dissipation rate in the sphere P = 271-sA,72Zn(n + 1)(2n + 1)C,2, cost en
(16)
n=1
10.17. Spherical Shell in Alternating Field Solenoid Gap.—As an application of the formulas of the last article consider a spherical shell of radius a centered in a gap of width 2d in an infinite solenoid of radius c having a current density io cos wt amperes per meter, as shown in Fig. 10.17. The potential d444, due to a ring element of width dzo at z = zo may be written down by 7.13 (2) where I is replaced by icb dzo. Superimposing the results
§10.18
GENERAL EDDY CURRENTS IN SPHERICAL SHELL
397
FIG. 10.17.
for a ring at r = a, 0 = a and one at r = a, 0 = r — a eliminates even values of n and doubles the odd values so that, writing 2n + 1 for n, sin a (2n + 1)(2n + 2) a)
dA.0 = 1.44 dzo
a)PL±i(cos 0) cos cot
n =- 0
(1)
The n = 0 integration gives
fa
a-1sine a dzo = f
C 2(C 2 4)-1
2)-1 = 1 — cos dzo = 1 — d(c2 d (2)
For s > 1 a useful formula, verifiable by differentiation, is a—. sin
a 2 = c2
zZ ,
aP!(cos a) = [(s — 1)!]-1(-1)8+1
aa+ia
az80+11 s>1 a sin a = c,
a cos a = zo,
(3) (4)
The zo integration of (3) merely reduces the derivative order by one, and (3) expresses the rtsult in terms of a, so that = Ecvnir2n+ipi n+,(cos 0) cos cot,
Co = 4ihic6(1 — cos ()
n=0 = I.liob-2n[2a(2n + 1)(2n + 2)]-' sini3PL(cos0)
Cn
(5) (6)
The applied vector potential is then 10.16 (6) where r = R, so that A„B = PL+1(cos 8) cos cat, Cl = pvioR sine 40 C 2n+i = AvioR2n-Flb-2n sini3PL(cos 0)[2n(2n + 1)(2n + 2)]-1
(7)
(8)
Thus the eddy current density i and the energy dissipation are given by 10.16 (15) and (16) provided n is replaced by 2n + 1. 10.18. General Eddy Currents in Spherical Shell.—It was shown in 7.04 and 7.06 that the vector potential whose curl gives a magnetostatic B in spherical coordinates has only 0- and 0-components given by A„,,„ = V x r(Arn
Br—n—')S:(0 45)
(1)
398
EDDY CURRENTS
§10.18
In 7.11 the vector potential of an arbitrary current distribution in a spherical shell was expressed in terms of these components. These relations provide a basis for the solution of the general eddy current problem when a thin shell is placed in a time-varying field whose wave length is much longer than the shell diameter 2a and when the skin depth is much greater than the shell thickness. First the orthogonality relations of these solutions must be found. From 7.11 their form is m [ sin ml A,,,,, = 01 + 4:111 sin OFV(cos B) PZ(cos 0) sin m 4) sin 0 cos mcb
ml
—
(2)
The 4) integration of Anin • Apq over the unit sphere gives integrals of sin m4, sin p4, or cos m(11 cos p4, from 0 to 2r. From Dw 858.516 and 858.517 these vanish unless m = p in which case the result is 7. Thus
r
+1 _i r 2,r Am?, • Am, sin 0 d0 d4, = r f PZ(cos 0) • PQ (cos 0)sin 0 de 1. J i J.
(3)
From 5.232 (7) and 5.232 (8) this is zero if n / q, and if n = q, it is +1
J-1
2n
o
Amn • IL, sin 0 de (14) =
2rn(n + 1)(n + m)!
(2n + 1)(n — m)!
(4)
Let the total vector potential be A' + A, the eddy current part being A. Then from 10.00 (4), if i is surface current density and s is area resistivity, —d(A' + A) _ si dt
(5)
The relation between the current i and the vector potential it produces at r = a is found by a procedure like that in 7.11 and 7.12 to be
A.. =
Ala
im,. 2n + 1
(6)
The reasoning of Art. 10.16 starting with 10.16 (1) and (2) applies without change to the present case, if C,nn, Am, and A replace Cn, An, and A0, so that if the applied vector potential at the surface is
ZC,,,n A,,,,,,s(t)
A8 = 71.
(7)
m
then the total vector potential at the surface of the sphere at the present time due to all past fluctuations is
,
1-, mn
- a Amns(t — T)e_ [ ( 2n + 1) / (m,,a) 1 sr dT at
(8)
TORQUE ON SPINNING SPHERICAL SHELL
§10.19
399
10.19. Torque on Spinning Spherical Shell between Magnet Poles.— The theory of the last article yields the solution for the retarding torque on a spinning thin spherical conducting shell of radius a centered in a gap between the coaxial halves of an infinite cylindrical bar magnet having a uniform permanent magnetization M parallel to its axis. The edges of the plane magnet faces are at a distance b from the sphere center, where any face radius subtends an angle 13 as shown in Fig. 10.19. The vector potential for the equivalent current shell (see 9.07) is given by 10.17 (5) if M replaces Operations are simplified by having the harmonic axis and the rotation axis coincide. To accomplish this, observe from 7.06 (1) and (2) that the A' of 10.17 is given by
cb
A' =VxrW =VxrZC;r 2n-"P2n+i(COS
(1)
n=0
where the prime indicates the 10.17 coordinate. By 5.24 (5), taking 0 to be br, the expression for W referred to the new axis is 00
n
2(2n — 2m — 1)!Ir2n+17,9,--1 C'n (2n ± 2m ± 2) !! (cos 0) cos (2m + 1)4) (2) n=0 m=0
Now taking V x rW by 7.06 (1) and (2) gives Eqs. (4) and (5) for the vector potential referred to the rotation axis, in terms of C. = 2(2n — 2m — 1)!![(2n + 2m + 2)!!]-1C',
A; =
(2m + 1)C.r2n+1csc OPrn nV(cos 0) sin (2m + 1)cp
(3) (4)
n=0 m=0 n
A.; = —
c.r2.+1 sin 0(fninV)'(cos 0) cos (2m + 1)0
(5)
n=0 m=0
Substitution of O. + wt for puts in rotation. The Aos of the eddy currents is now found from 10.18 (8). Substitute t — T for r, differentiate with respect to t, put t = 0, and perform the T integration. Writing S for sin (2m + 1)0 and C for cos (2m + 1)0, the last factors in (4) and (5) are (2m
+ 1)cofo lsCcos (2m + 1)cor
S .
—c
sin (2m + l)cor i-[(2.-1-1) /AA= dr (6)
400
EDDY CURRENTS
§10.20
where the upper symbols refer to A 9and the lower to Ao. Integration by Dw 860.90 and Dw 860.80 gives the results for (A.v.8)0 and (A.,,z)0: 1)95 + 1//mn] (7) emn cos ikm.nr2n-"(2m + 1) csc OPM-N(cos 0) sin [(2m _Gan Cos4,mnr2n1-1 sin 0(P2V)'(cos 0) cos [(2m + 1)0 + 1,Gznn] (8)
where tan tkinn is (2n + 1)s/[(2m 1)Anaco)]. Clearly Am., with these components has the same orthogonal properties as 10.18 (2) since from 0 to 27 the integration of the ck factor products is independent of'/,nn. The torque on the sphere is found by using the i of 10.18 (6). The rate of energy dissipation is constant, as in 6.03 (2), so that T = Pco—' = co—'si • i dS = 27rws n=-0 m=0 a4n-I-4(2n
1)(2n + 2)(4n + 3)(2m + 1)2(2n + 2m + 2) Cl,„ [(2n 1)2s2 + (2m + 1)2/4o/201(2n — 2m) !
(9)
10.20. Eddy Currents in Thin Cylindrical Shell. When all the eddy currents in a thin infinite conducting cylinder of radius a parallel the axis, the vector potential of the field producing them also parallels the axis and the magnetic field is a two-dimensional one. In this case, as we saw in 7.25, the vector potential behaves as an electrostatic stream function so its value at any point represents the flux between this point and some fixed point. Let the total vector potential he A; + A,, that part due to the eddy currents being A,. Then the electromotance per unit length induced in a strip of width a dO at a, 0 by the change in total flux N linking unit length of this strip gives, by Ohm's law, 6.02 (1), —
dN E = -- = dt
d(A;
A,) dt
=
s . zza dB a dB
(1)
where iz is the current density and s the area resistivity. If i is written as a series of circular harmonics, then from 7.17 (1) and (4), there is a simple relation between the nth term of this expansion and the nth term of the vector potential produced by it. At the surface of the shell this relation is (2) (A n) p=a = 2µva1Z 2n If both A; and A, are expanded in circular harmonics, we see, by putting (2) in (1), that these expansions satisfy the relation
dt n
2ns An Aza
An ) = n
(3)
§10.20
EDDY CURRENTS IN THIN CYLINDRICAL SHELL
401
Except for a constant factor, this equation is identical with 10.16 (3), so that if the inducing field at p = a is given, as in 10.16 (6), by AL = ZCnAns(t) n=
(4)
then the eddy current potential is given, as in 10.16 (7), by CO
Azs =
a Ans(t — at
n
r)
e Ava
(5)
n=1
Replacement of
aAns/at
by — aAnda, and integration by parts give 4-
flC nf Ans(t — r)e
A;s = (92c
A=s
– 2n — A'a7dr
(6)
n=1
For a sinusoidal inducing field we have, as in 10.16 (9), A ns = cos (nO + on) cos cot
(7)
As in 10.16 (11) and 10.16 (12), the eddy currents alone give, when p > a, A0 =
—
kEenanp—n cos (nO + an) cos encos (wt — En)
(8)
n=1
tan
En = — 2nS(LJA,a)-1
(9)
When p < a, (p/a)n replaces (a/p)n and for external sources (see 10.14), Ai + A' = —kZen pna-n cos (nO
bn) sin En sin (wt
—
En)
(10)
n =1
The ratio of the nth harmonic with the shell to that without the shell is
±AEI
IA' I
= sin En
The same remarks apply to this as to 10.16 (14). For internal excitation the new and old amplitude ratio outside the shell is still given by (11). The eddy current density in the shell is, from (2) and (8), i = — k2(i.1„a)-1
nCn cos (nO + an)
n=1 power
cos En cos (wt
—
En)
(12)
EDDY CURRENTS
402
§10.21
From 10.02 (6), the average power consumption per unit length is
=
2r
o
si 2 do
When (12) is squared, the integration from 0 to 2r eliminates all the cross-product terms and leaves only the sum of integrals of cos' (nO (5„) which equal 7r by Dw 854.1. Thus each n term behaves as an independent circuit, and the rate of energy dissipation per unit length is P = 27r s(1.4,0- 1n2C7, cost
(13)
n=1
10.21. Eddy Currents in Rotating Finite Cylindrical Shell.—A device often used for the damping of torsional oscillations consists of a thin z
conducting cylinder of area resistivity s suspended in the gap between one or more transverse bar magnets and a soft iron external yoke which serves as a return path for the magnetic flux between poles as shown in Fig. 10.21. The treatment is rigorous, subject to the conditions of 10.00 and 10.09, for an infinite thin cylinder with a periodic arrangement of transverse uniformly magnetized bars so that the circles at z = 0, ± b, ± 2b, etc., are eddy current flow lines. It is nearly rigorous for a single section of length b as shown if the magnet poles are not too near the edge. It is pointed out in the next article that the damping can be worked out as soon as the solution for the torque on a uniformly rotating cylinder is known. It is convenient to use the scalar potential Sl throughout, and the procedure involves the same steps used in the preceding articles.
ROTATING FINITE CYLINDRICAL SHELL
§10.21
403
Suppose the stream function is expanded in the harmonic form =
C,n„ sin n— Irz sin (m4 n
6„,)
(1)
m
so that 43 = 0 at z = 0, z = b. From 7.01 (2), the line integral of H around a path from (00)p on the outer face around the edge to the point (0i)p on the inner face is 00 — 0, which equals 43, the total current flowing between P and the edge. One can see by inspection that the potentials gi and 0° are (see 5.292 and 5.324), Sti = rab-1ZZnC„,n./.(nrpb-1)K'„(nrab-1) sin (nrzb-1) sin (m4) +
nm (2) 120 = rab-1ZZnC„,„1:„(nrab-1)K„.(nrpb-1) sin (nrzb-1) sin (nick + Sm) nm (3) because if the difference Sti— 00 at p = a is taken and the modified functions eliminated by the Wronskian 5.32 (7), the result is (1). The rate of decay of these eddy currents is found from 10.00 (3), applied at p = a.
= -pv
ao
ap
_L H) ‘1)az
r a 24,
a COO'
Vx
=6'1 a' ao2
024,
al)
aZ2
k a az
(4)
= -p,,n272ab-2/:„(nrab-1)K:„(nrab-9C...
sin (nrzb-1) sin (m4,
3,„) (5)
As noted at the start of 10.09, B, has negligible effect on a thin shell, so substitution of (5) for B and (4) for V x i in 10.00 (3) gives
_aBrnn at
[1 + )2 ][i n (n7rb a) K4nrb a)]-'13nin Ilya nr a
(6)
The same law of decay applies to B and 0, so that [2.]t = Pm,, = [1 + m2b2(nra) ][p„a1:„(nrab-1)K:„(nrab-91-1
(7 )
(8)
Suppose the inducing potential at the cylinder surface is =
(9) n
m
Then the change in this potential in an interval dr at a time T before the present was compensated by the eddy currents induced at that time. The present set of eddy currents is the result of all past changes, each change having died out according to (7) so that the present potential of
404
EDDY CURRENTS
§10.21
the eddy currents is ''' aD.,nn(P,d),z,t— T) e_,,,,,,,,
at
dr
(10)
The form of Dm„(p,ck,z,t) for Fig. 10.21 must now be found. The bar is uniformly magnetized with intensity M (loop definition), but the ends are rounded so the normal M is M cos cpo where —a < 44 < +a. The pole strength in an area dS is then cM cos 4)0 dzo d4)0. The magnetic field is normal to the permeable boundary at p = d and to the plane z = b so St' vanishes there. Thus the problem is exactly that of a point charge in an earthed box which was solved in 5.324 (1) and (10). The following substitutions must be made: cM cos 4o for —q/E and zo, b, c, d for c, L, b , a. There is another positive element at —44, which by Dw 401.06 changes cos m(4) — q50) to 2 cos mq5 cos m4 0. There are negative elements at it — rpo and 7 + 00, which cancel the positive element terms when m is even and double them when m is odd. For the four elements the last term is then replaced by 4 cos (2m + 1)4) cos (2m + 1)45 o and the integration range is 0 to a. Thus from 5.324 (10), writing s for 2m + 1, c/12' —
8cM cos cti o dzod4)0 rb
n = 1 m =o
/3(nircb-1)M(mirb-1,d,p) I $(Tordb—')
sin (nrzob-1) sin (rorzb-1) cos sck cos s4)0 (11) li2(nrb—',d,p) = K3(nirdb-913(nrpb-1) — I s(nirdb-1)Ks(nirpb—') (12) Integrations with respect to zo and 4)0 give
foacos s4)0 cos 4)0 d40 = (4m)-1sin 2ma + (4m + 4)-1sin (2m + 2)a(13) f: sin (rorzob-1) dzo = b(mr)-1[cos (nirkb-1) — cos (nirhb-1)]
(14)
Thus, if s = 2m + 1, (9) may be written St' = 2cMir-2ZZF,,,,,R2(nrb-1,d,p) sin (nirzb-1) cos s(4) + wt)
(15)
n m
Fm„ --- I s
rock (nirc)( mrh\ cos — cos b b b) (m + 1) sin 2ma + m sin (2m + 2)a nm(m + 1)/s(nrdb-1)
(16)
The potential of the eddy currents is now given by (10). Substitute t — 7 for t, differentiate with respect to t, and put t = 0, so that the eddy
§10.23
TRANSIENT SHIELDING BY A THICK CYLINDRICAL SHELL 405
currents are given at the instant that cp = 0 splits the pole piece. The integral involved is then, if tan is written for P„,„ss-'co-1, scof sin (scb — swr)e -Pm•sIdT = — cos %Gm cos (s0
11.„0
(17)
This replaces cos 8(:15 + wt) in (15) for p > a to give S20 at t = 0. Note that for a perfectly conducting cylinder or for a high frequency, 1,1/„, is zero, so that S2 completely neutralizes St' outside the cylinder. At p = a on the inside surface C2, = —S20, so 4,is 2C20 at the surface and i is given by (4). The rate of energy dissipation is constant and equals
P= sfi•idS=
b 4safj oo():=‘ [t
as2 \2 dz dI
(18)
The integrations are straightforward and the torque is given by P/co. Thus writing out cos' &,,, gives finally
T
8c2.1112cos ir'ab
.EL,,,[14m+i(nirb-1,d,a)]
2[n2,2a2+ (2m + 1)2b2]
P;„,,s2 ± (2m + 1)2(.02
(19)
n=1 ,a=0
where F„,„ is given by (16), RL+1by (12), and P„,„ by (8). 10.22. Eddy Current Damping. Numerous calculations of eddy current torques and forces have been made in this chapter, especially in the case of conducting shells whose thickness is small compared with the skin depth at the frequencies involved. In every case where the linear velocity v or the angular velocity w is small, the forces and torques are proportional to v and co. There is nearly always, however, as in the last two articles, a term involving ii,v2w2a2 plus a s2 member in the denominator somewhere which, if i.c,coa is large enough compared with s, will eventually take charge. Only the case where co is small or s large gives a simple damped solution. Then the differential equation is that of an R, L, C circuit where, in the angular case, R is the coefficient of co in the torque expression such as 10.19 (10) or 10.20 (19), L is the moment of inertia I, and C-1is the torque constant of the supporting wire or spring. Formulas for the various types of damping can be found in any circuit theory text. 10.23. Transient Shielding by a Thick Cylindrical Shell. Sinusoidal longitudinal eddy currents in a thick cylindrical shell and arbitrarily varying longitudinal currents in a thin shell were treated in Arts. 10.03 and 10.20, respectively. As a relatively simple example involving both cross and longitudinal components, consider an infinite cylinder of permeability u, conductivity y, internal radius a, and external radius b which makes an angle 47r — a with a uniform field of induction B. Find the field inside at a time t after the external field is removed or estab—
—
406
EDDY CURRENTS
§10.23
lished. Examination of 7.05 (5) shows that suitable solutions for the static vector potential when p > b, b > p > a, and a > p are, respectively, (1) E[4:14 sin a(p — Cop') + k cos a(p — qp-') sin q5] (2) sin sin 0] k cos a(D' o p-1) o p — E' a(Dop — EoP-1) (A2)0 = B[4 B[4 sin k cos aF/ aF o p + (243)o = 013 sin ct] (3)
(A1)0
=
These give the prescribed field at p = 00 , and the field is finite at p = 0. The values of the constants needed to match Al and (B x p)/ p. at p = a and at p = b are, if 13 is written for (I.L — p,,)/(p ki„) and G for b2/(b2 — 02a2), Co = µT 1(4 — ilv)(b2 — a2), Do = Eo = — A0b2, Fo = 1 (4) )G, E', 3(a2 — b2)G, D', = (1 + = 3 = 0(1 + 13)a2G, = (1 — 13 2)G (5)
Ca
After t = 0, the same solutions of 10.00 (8) are valid where y = 0 except that in A lthe p-terms disappear because B = 0 at p = co . For a < p < b, solutions are needed where Az has the factor sin 0 and A4, in independent of 0. As in 10.08 we try sums of exponentially damped solutions so that A2 = BZ[(1)1?(p)e—P.'
kP(p) sin 0e-m]
(6)
For A z10.00 (8) is a scalar equation from which sin 0e-g.' divides out and which, upon substitution of v2 = iryq 8 p2, yields 5.291 (3) with n = 1. For A4, 10.00 (8) becomes 10.05 (3) which, upon inserting — 4/3-2for V24 from 10.05 (5), dividing out e-P.' and writing v2= wyp8p2 gives again 5.291 (3) with n = 1. Thus, when t > 0, (1), (2) and (3) are replaced by Al = BZ[4 sin aC,p-ie-P•g
k
cos «C;p-i sin 0e-m]
A2 = BZ{(1) sin aD3R1[(WYP8)1 p]e-P•' A3 = B[(1) sin aF,pe-P.'
k
k cos agP1Rryq8)1p] sin
(7) 0e-g.g) (8)
cos aF's p sin 4,e-9•1]
(9)
8
where R1 and P1are Bessel functions. At p = b, equate Al to A2, write k, for (ryps)i, 1, for (iryq,)i, and cancel sin 0 out of Az. Thus we get (1)C3
kC; = 4thD5R1(ksb)
kbgP 1(18 b)
(10)
Equating tangential components of A-1/3 = p 1v x A at p = b, canceling sin 4,out of the 0-component, and multiplying through by b2give —
4)11C", = kilybD,[kablVi(k.b)
Ri(k.b)] — 4)14-18b2D:P',(15b)
(11)
§10.23
TRANSIENT SHIELDING BY A THICK CYLINDRICAL SHELL 407
The k-term is zero so, from 5.294, Ro(ksb) = 0 and, from 5.293 (9), R1(k3b) = 47r(kab)-1,
Ri(ksp) = o(ksb)Ji(k.p) — o(ksb) Y i(ksp) (12)
Similarly the boundary conditions at p = a give, writing out Pi(lsa), 43Fsa
kF;a = 43D3R1(ksa)
I7].(/3a)]
kg[E'sJ i(lsa)
kW's — 444F; = kiinksDsRo(ksa) — itiiinIsD;P;(13a)
(13) (14)
Eliminating Fs, Ds, F;, and D; from these four component equations gives ksaRo(ksa) — 2 (PL,T1A)Ri(k.a) = ksaRi(kaa) + (1 — 2A,71A)R1(ksa) = 0 (15) i(lsa) AP1(/3b ) = 0 = p„lsal'i (lsa) — (16) iA3/3bPi(i3b)
Only values of p3 = 14(1.1-y) which satisfy (15) can be used. Insertion of Yo(k8b) from (15) into Ri(ksa) and application of 5.293 (9) give Ri(ksa) = 2113J0(ksb)/1--1[A3k3aJo(k3a) — 2AJI(k3a)]-1 (17) If Pi and P iare expressed in terms of Po and P2 by 5.294 (2) and (3) and if 0 replaces (p. — p,,)/(µ + A„), (16) may be rearranged to give — Es
0Yo(i3a)
Y2(/sa)
Yo(Lb)
0,10(18a)
J2(13a)
Jo(lsb)
0Y2(/,,b) 0J2(ta b)
(18)
Only values of qs = l;/(µy) satisfying the second equality can be used. Insertion of E; into Pi(lsa) and Pi(tsb) and application of the recurrence formulas of 5.294 and of 5.293 (9) give Piqsa) —
2(1 — 0) —2(1 — 0) Pi(isb) = irtaa[0Jo(La) ± .12(4a)] 71.b[Jo(lsb) 0,12(18b)]
(19)
It remains to determine Ds and D; so that, when t = 0, (8) agrees with (2) when a < p < b or so that, dividing common factors out of each component, 41)( D oP — Eop— + k (D'op — girl) = Zl(I)DaR i(ksp) kD;P ( 4)1
(20)
3 Take the scalar product of both sides by P[43Ri(knP) kPi(lnp)] dp and integrate from a to b or subtract the 0 to a from the 0 to b integral. By (11), (15), (16), and 5.296 (3) only the nth term on the right survives, and for A4, its value is given by 5.297 (6) with B = 1 — 21A;Ii.I. Integrate the left side by 5.294 (7) and (8), solve for Dn, and from (13) F„ is Fn, —
/.1. 471.13Ri(kna) k!a2g)72[Ri(kna)]2 )
kna{41.1,2, — (41.42—
(21)
408
EDDY CURRENTS
§10.23
For Az 5.297 (6) with B = kcip. gives the right side. Integrate the left side by 5.294 (7) and 5.294 (8), solve for and from (5) F',, becomes Fn
41.4.4,1)(1 [Pl(inb)]2
(11,2,1V)2+ 142—
n
13)Pi(lnb) (11:0ba 2 + A2 — AMPI(/.02
—
(22)
The total field inside at any time is given by (9), (21), (22), (17), and (19) where k,, and ln are determined by (15) and (18). If there is no field when t < 0 but at t = 0 the field B is established, then it is clear from Eqs. (3) and (9) that the vector potential A; inside is
p < a,
A', =
(A3)0 — A3
(23)
These solutions are obtained by a different method in Phil. Mag., Vol. 29, page 18, 1940. The field, when B varies arbitrarily with time, can be found from these results by methods similar to those used in preceding articles. Problems Problems marked C in the following group are taken from the Cambridge examinations as reprinted by Jeans, with the permission of the Cambridge University Press. 1C. An infinite iron plate is bounded by the parallel planes x = h, x = —h; wire is wound uniformly round the plate, the layers of wire being parallel to the axis of y. If an alternating current is sent through the wire, producing outside the plate a magnetic force Ho cos pt parallel to z, prove that H, the magnetic force in the plate at a distance x from the center, will be given by H = Ho tan —
(cosh 2mx + cos 2mx)i
cosh 2mh + cos 2mh
cos (pt /3)
— sinh m(h x) sin m(h — x) — sinh m(h — x) sin m(h x) cosh m(h x) cos m(h — x) + cosh m(h — x) cos m(h x)
where m' = hip/r. Discuss the special cases of mh small and mh large. 2. All points of a circular loop of wire are at a distance c from the center of a ball of radius a, permeability 12, and resistivity T, and any radius of the loop subtends an angle a at the center. A current I cos wt flows in the loop. Show that the vector potential inside the sphere is the real part of 2n + 1) (a y Anh,+1[(ip)ir]Pni (cos 0)eiwi n(n+1 C
aaociiI sin a 274 ti =
where An
=
—
ao(jp)ictIo_i[(jp)ia]}-iP1 (cos a) and p = aw/T
3. A sphere of radius a, permeability and resistivity r lies in a region where, at t = 0, a field is established that, but for the sphere, would be of uniform induction B.
PROBLEMS Show in the notation of 10.08 that, when t Ao =
0, the potentials are
- m120a7.z 3 _ _ -Br + B Bd. 2e q.t] sin 0 2 (AL + 20 2
[ 3/2Br 2(.1. 2bt„)
Ai =
409
ZAJ-1,11(k,,r)e-g•i] sin 0
4. An infinite circular cylinder of radius a, permeabil ty it, and resistivity 7 is wound with wire carrying an alternating current. If H0 at the surface is uniform and of magnitude Ho cos wt, show that the magnetic field at any point inside is given by 10.04 (4) and (5) with B, and Bo substituted for i„ and io. Find the density and direction of the eddy currents by 10.00 (5). 5. A circular loop of wire is located at r = a, 0 = a and a spherical conducting shell of area resistivity s at r = b. A current is built up in the loop according to the equation i = 1(1 - e-" L). If the sphere and loop do not move appreciably before the eddy currents have decayed, show that the momentum imparted is 71-4b/ 2 x.■ sin' a by-,(2n +1)01, + 2m„bR i P ( cos a)P41(cos a) 2s %-/ (n + 1)(2n + 1)(a (2n + 1) sL
n=1
6. A small loop of wire of moment M cos wt lies above an infinite thin plane sheet of area resistivity s, with its axis perpendicular to the sheet. Show that, if 1 >> orro where sro >> Atr, the magnetic induction on the far side of the sheet is
Br =
AvM[ 277.3
cos 0 cos ca ±
Avcor
4s
sin sin 0.4 ,
Be =
479.3
sin 0 cos cot
7. Show, in the preceding case, that the current density igs at a distance p from the axis and the average rate of dissipation of power are given, approximately, by
g,copM sin wt 47s(p2
c2)1
Atm 2,, 2
and
64r SC 2
8. The magnetic moment of a small loop carrying current falls linearly from the value M at t = 0 to 0 at t = T. If the loop is at a distance c above an infinite thin sheet of area resistivity s with its axis normal to it, show that the additional B on the axis at a distance z above the sheet due to its presence is AB -
/42„Mt[A,,(c z) st] 271-T(c 2st]2 z) z)2[th,o(c
when 0 < t < T and that, when t > T, it is
-
s(2t - T)] z) 4M[A,,(c z) 27[µ,,(c 2s(t - T)Pkiv(c z)
2 s tp
9. A magnetic loop dipole of moment M is parallel to and at a distance c from an infinite thin sheet of area resistivity s and is moved parallel to the sheet and normal to its axis with a uniform velocity v. Show that in a steady state the eddy current retarding force is br-lA2,,M2vs(2c)-4[(4s2 2s]-2. /.42„v 2)1 10. A loop dipole of moment M moves with a uniform velocity v in a straight line at a distance c from an infinite plane sheet of area resistivity s. If M parallels the
410
EDDY CURRENTS
sheet and makes an angle 4, with v, show that the retarding force is F = [1 +
4s cost c6
(140
4s 2)}
F
where Fo is the retarding force in the last problem. 11C. A slowly alternating current I cos wt is traversing a small circular coil whose magnetic moment for unit current is M. A thin spherical shell, of radius a and specific resistance s, has its center on the axis of the coil at a distance f from the center of the coil. Show that the currents in the shell form circles round the axis of the coil and that the strength of the current in any circle whose radius subtends an angle a at the center is, if tan c„ = — (2n + 1)5'/(i.i,,coa), M/( ztirPa )-1Z (2n + 1) (t'a).P;,(cos a) cos en cos (wt — en)
12. A thin spherical shell of area resistivity s and radius a rotates with a uniform angular velocity w about an axis normal to a uniform field of induction B. Show that the retarding torque is 67B 2,,,a4 (9,2 + ga 2,42) -1. 13. Show that the average power dissipation in the shell of the last problem when placed in the field B cos wt is 37sB2w2a4(9s2 +1,2va1/4,2)-1. 14. Show that the field inside the shell in the last problem is, if tan E = —35'/(A„coa), 3B0(902
A2va2c4,2)—i
sin (wt — e)
15. A thin circular cylinder of area resistivity s and radius a is placed in a field E = —2s / (Anwa), show that the field inside is
B cos wt normal to its axis. If tan 2sB(4s2
w2ga2)-i sin (wt — e)
16. Show that the retarding torque per unit length, when the cylinder of the last problem is spinning on its axis with an angular velocity w in a uniform magnetic field B normal to its axis, is 471-cosa3B2(4s2 17. The cylindrical coordinates of the wires of a bifilar circuit which carry the outgoing and incoming currents are, respectively, b, 0 and b, Ir. When the current in the circuit is I cos 0.4, show that the vector potential at the point p, 0 outside a thin tube p = a of area resistivity s is 2i.t,,shr-1Z[4(2n
(p- lb ) 2n+1 cos
1)25'2
(2n + 1)0 sin (cot — €,,)
where tan €,, = —2(2n + 1)s(tt,,coa)-1and the summation is from 0 to co, a > b. 18. A thin oblate spheroidal shell g- = ro has a variable thickness such that we may consider the area resistivity to be s = h2so• At the time t = 0, a uniform magnetic field of induction B is suddenly established parallel to the axis. Show that if N = 2[(1 a )NQI(jg-0)]-1the eddy current density is given by 24,
BN (1 + 31)(1 — 2 (e + 30 )1
_Ariot
e
19. A long solid cylinder of radius a = (x2 + y2)i, permeability A, and resistivity is placed in an x-directed alternating magnetic field Bei•'t (real part). Show that the z-directed vector potentials A i and A 0 inside and outside the cylinder are
T
Ai = 0/ i[(iP)1P]ei"' sin 41 (real part),
A. = B(p
n p-i)ei40,sin (real part)
PROBLEMS
411
where, writing f o and /2 for Io[(jp)1a] and /2[(iP)Ia], =
4/.4B
n - (L
and
(ip)11(m, + AO/0 - (12 -
-(A
+120/0 -
+ A')/2,2
- Arv)/2
20. Show that the mean power dissipated per unit length in the last problem is Llirulia 2B2[bero(Pla)bei2(pla) — beio(p1a)ber2(p1a)1
P— [(m
12,)bero
(IA — p.,)ber212+ [(A + ti,,)beto -I-(A — ti,)beiz12
where — (ber2x jbei2x) is written for /2(j1x). The arguments in numerator and denominator are the same. If p is small, show that this reduces to P = 1,m 2w 2a4B27( A Av)-2
21. A long solid cylinder of radius a, permeability 12, and conductivity y is rotated about its axis with a uniform angular velocity w in a magnetic field B normal to this axis. Show that the retarding torque per unit length, due to eddy currents, is 871.,a 213 2(bero(P1a)bei2(pla) — beio(p1a)ber2(0a)]
T—
+
[Cu
— I.1,,)ber212
[Cu
/4)beio
— 1.1.)bet212
The arguments in numerator and denominator are the same, and p = 7µw. 22. A loop of radius a is coaxial with and at a distance d from the bottom of a cylindrical box of radius b and height c. If the skin effect is so large that B is nearly tangent to the walls, show that the decrease in self-inductance is AL = j
M 1„
47rma2 Z[Ii(rorac-1)]2[c/i(n7rbc-1)]-1K1(nirbc-1) sin 2 (iordc-1) n=0
n= —
M = 2µa [k,7'[(1— 114)K1 — E1] — (1 — 3,!)k-'[(1 — ik 2)K —
n2c2)-I. where the modulus k1 is a[a2 (nc — d)2]-1and k is a(a2 23. In problem 22 the skin depth is .5, the conductivity y, and the current I. Show from 10.02 (8) that if k„, is chosen so that Ji(k,,,b) is zero, the power dissipation is P=
[ /1(nira/c) sin (nrd/c)]2
27/ 2{a2 -yo bcn _ i
I i(nirb/c) -2 b
=1
[ Ji(k,„a)]2[sinh2 k„,(c — d) sinh2(k„,c1)1 } [Jo(k mb) sinh (k„,c)] 2
24. A thin disk of radius a and area resistivity s is placed in a region where, before its insertion, there was a uniform field of induction Bo cos wt normal to it. Find the secondary currents from the vector potential of the primary eddy currents induced by Bo cos wt. Thus show that the total eddy current density is WB 0 {
— p sin (0t
2s
+ 6n p s[(a
p) (2 p 2 a 2)K
+
(a
+
p)(2p2 — a2 )E] cos wt + • • •
p)-and terms in (6.4La/s) 2are neglected. where the modulus of K and E is 2 (ap)1(a Note that at 60 cycles this is good for 0.1 mm copper sheet. 25. If in the last problem the frequency is so high that the skin depth 3 is much less than the disk thickness so that B can be taken tangent to its surface, show that the
412
EDDY CURRENTS
eddy current density, including both faces, at a distance p from the center is -
srBp2 ma(a2 - 1)2)1
This icb is completely rigorous for a perfectly conducting disk. 26. The eddy current density in an infinitely long cylindrical shell of radius a, area resistivity s, and thickness much less than skin depth is (i0)0 at t = 0 and i, at t = T. If (4)0 can be written as a Fourier integral, show that /95 =
(4)0 = f0 *(k) cos kz dk,
f
4/(k) cos kze-DIal,(WICI(ka)] 4r dk
27. The vector potential of the inducing field at the shell's surface in problem 26 is (A;),-a = -
1.(k, a, t) cos kz dk
Show by the reasoning of 10.16 and the last problem that, due to eddy currents, f .0 a -4,(k, a, t - r)e-Ip.ar ickoic,(kanlsr cos kz dk dr
(A0 )5=. = -
at
28. A loop of radius b carries a current I cos cot and is coaxial with a very long shell at p = a of area resistivity s and thickness much less than the skin depth. Show that the shell increases the loop's resistance and inductance by amounts
AR = 214b2wf
sin cos 0[Ii(kb)] 21(1(ka)[Ii(ka)]-1dk
0
AL = -2i.Lb 2 f
0
sine 0[Ii(kb)PKi(ka)[Ii(ka)]-' dk
if b < a. If b > a, K and I are interchanged and tan 0 = 5-1cogaIi(ka)Ki(ka). 29. From problem 40, Chap. VII, and 7.04 show that the external self-inductance of the torus formed by rotating a circle of radius b about a coplanar line c meters from its center at high frequencies where there is a strong skin effect is
llrr
L = 2pc cos a COSla[(1 - 1- k)E -
2
2
-2 n =1
(-1)nC_4(csc a)
(4n - 1)(4n + 1)(ti(cse
a)
where sin a = b/c and the modulus k of K and E is given by k 2 =cos a sect 30. A wire loop of radius a is coaxial with and outside an infinite conducting cylinder of radius b and carries an alternating current I of such high frequency that the eddy current density in the cylinder is confined to its skin. Show with the aid of problems 26 and 28 of Chap. VII that this density is f 'ICI(ka)
= bar
Ki(kb)
cos kz dk
31. In the foregoing problem show with the aid of 8.06 (1) that the presence of the cylinder decreases the self-inductance of the loop by an amount = 2pc12.10.* Ii(kb)K1(ka)Ki-1(kb) dk
PROBLEMS
413
Show that the eddy current losses increase the resistance of the loop by an amount 4a2f Ki(ka)] 2 R=— -yab 0 K1(kb) dk where -y is the conductivity and B the skin depth for the cylinder material. 32. Show from HMF 9.6.25, 12.2.3, and 12.2.4 that the integrals of 10.12 (5) are X -K i(Xp)
cos Au du
and
f (P 2 u 2)1
7rX [I i(Xp) 2p
Li(XP)!
A
csin Au du
f 0 (p 2
P
2l 2)1
Note that Ki(z) and /1(z) - Li(z) are both tabulated in HMF. If Ac < 1, then cos Au and sin Au may be expanded in series and the integrals done. If (Ac) 2 is neglected compared with 1, show that in 10.12 (5) becomes, if II. is 0 - Ac. CrlilvM {[
47rs
C
XKI(Xp)
p(p2
C2)i
sin
7rX (—[/1 (XP ) 2
LI(Xp)]
Xp (p 2
c2)1
cos Vr}
33. Solve the shielding problem of 10.13 exactly as follows: From Watson, page 148, and HMF 15.4.14 and 8.2.5 establish the formula = Ji (az)Ji (bz)
as + b2)( z)2.+2 (-1)m(a 2- b2),.+1pi ( = Lj C ,,z 2m+2 (m 1)!(m 2)! m+1 a2 - b2 2 m=0 m=0
Substitute for Ji(az)Ji(bz) in the integral of 10.13 (2), integrate the result by IT 1(11), page 135, and evaluate Ei( -jwc/X) by HTF II (4), page 145. Replace wc/X = luvwc/s by p and thus show that
Aof _
_Avai
Cm (7 p12m.+3{
2
2m+3
en' (Ci p
(s -
j si p) s =1
For Ci and si see 10.11 (5). 34. It is desired to calculate the induction heating of a solid sphere of radius a, permeability pc, and resistivity r at the center of a cylindrical current sheet of radius c and length 2d whose current density is io cos cat amperes per meter. From 7.15 (7) and 7.02 (8) the vector potential inside an infinite solenoid with opposite current density is - luior/31(cos 8) cos wt. Superimpose on this the potential of 10.17 (6) so as to wipe out the current sheet except for a length 2d. Thus show that the potential inside the sheet is the real part of
Co = - 111,,i0 cos 0,
A.95 = C,t7-2n+2P1,, +1(cos 6)0c" n =0 C. = Atsicsd-2n[2n(2n + 1)(2n + 2)1-1sin OPL, (nos 13)
where tan 0 is c/d. Now apply the theory as in 10.06, and show that the vector potential inside the sphere is, writing P for (jp) ,
A ILI 2n (15r)P2n +1 (cos 6)
A 56i = n=0
A. = Kmai(4n
3)a2n+2C4[(K„, - 1)(2n
1)/2n+4 (Pa)
(Pa/2ct+I(Pa)1-1
414
EDDY CURRENTS
3 Now show from 10.07 and 5.231 (1) that the power absorbed by the sphere is
P = irrpay-2
(2n + 1)(2n + 2)(4n + 3) -4,,An(Pi2n÷112n+1 Pl2n+112.+I)
n=0
where 1„, and pm are written for 1„,(15a) and I,,, (Pa) and K,,, is ii/tc„. 35. A solid sphere of radius a, permeability A, and resistivity r is spinning between the magnet poles as shown in Fig. 10.19. Show that the interior and exterior vector potentials are given by 10.19 (4) and (5) provided that +,,t1 and
Amnr
(C,,,,,r 2'+'
15.nr-2')02m+ + .t]
respectively, replace C„,„r2"1cos [(2m + 1)cis + cd]. Show by using the boundary conditions of 10.06 and writing K„, for pip, that ii,„„ is given by Amn =
(4n +3)K,,,a2n÷1C,,,„ (2n + 1)(K,,, — 1).12,,+1
(jP)4(1/2,,+1
where 12+1is written for 72,,+1[(jP)la]and C,,,,, is defined in 10.19 (3). Find the power dissipation and show that the retarding torque is n 7(.°Ct
n=0 m=0
(2n + 1)(2n + 2)(2n + 2m 2)! [( (4n + 3)(2n — 2m)!
.
if
,
jp),./ 2n-Fli 21,44 (7P)112n+i12n+1]
References H.: "Electrical and Optical Wave Motion," Cambridge, 1915. Discusses boundary conditions, gives solutions of wave equations and an extensive list of references. FRENKEL, J.: "Lehrbuch der Elektrodynamik," Vol II, Springer, Berlin, 1928. Gives vector potential treatment of eddy currents. GEIGER-SCHEEL: "Handbuch der Physik," Vol. XV, Berlin, 1927. Gives surprisingly meager treatment. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Follows Maxwell. MACDONALD, H. M.: "Electromagnetism," Bell, 1934. Uses long notation but gives rigorous treatment of eddy currents in solids. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Treats current sheets. MOULLIN, E. B.: "Principles of Electromagnetism," Oxford, 1932. Gives simple approximate treatment of eddy currents. OLLENDORF, F.: "Die Grundlagen der Hochfrequenztechnik," Springer, Berlin, 1926. Treats a-c resistance of wires and coils. RUSSELL, A.: "Alternating Currents," Cambridge, 1914. Treats manipulation of ber and bei functions. SCHELKUNOFF, S. A.: "Electromagnetic Waves," Van Nostrand, 1943. Treats chiefly very high frequency eddy currents. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Treats eddy currents associated with electromagnetic waves. BATEMAN,
CHAPTER XI PLANE ELECTROMAGNETIC WAVES 11.00. Maxwell's Field Equations.—In a region whereµ and e are continuous and in which there may be an electric charge density together with electric convection or conduction currents, we shall assume that the electric and magnetic quantities are connected by the equations B . OD + pv VxH=Vx— =1-1OB VxE = — — at V • D= p V • B= 0
(1) (2)
(3) (4)
Without the last terms the first of these equations expresses Ampere's law 7.01 (3), the second generalizes Faraday's law of induction 8.00 (3), and the third and fourth are 3.02 (1) and 7.01 (1), respectively. The D term in the first equation, introduced by Maxwell, is new and, when currents and charges are absent, completes the symmetry of the group and provides a companion equation for (2). It asserts that a change in the electric flux through a closed curve in space produces a magnetomotance around the curve; just as Faraday's law of induction states that a change in the magnetic flux through a closed curve produces an electromotance around it. Maxwell apparently thought of the magnetic field as being caused by an actual displacement of electric charge, but such a picture is not needed to justify (1). These equations are justified by the far-reaching validity of the conclusions based on them. They may be only approximately true, but for the present precision of measurement we may consider them exact. If charge is conserved, the equation of continuity, 6.00 (3), holds so that
ap v•i+— =o at
(5)
In isotropic bodies, i and E are connected by Ohm's law, 6.02 (3), which is E =ri
or
i = ^yE
(6)
In anisotropic bodies, with a proper choice of axes, 6.23 (2) gives 6.17 i=
kl'sEs
415
(7 )
416
PLANE ELECTROMAGNETIC WAVES
§11.01
The relations between D and E for isotropic bodies [1.13 (1)] or for anisotropic bodies with proper choice of axes [1.19 (7)] are or
D = eE
D = iel Ez je 2E,
le€3Es
(8)
In (6), (7), and (8) µ was taken independent of field strength so these equations cannot be used for ferromagnetic materials at some frequencies. 11.01. Propagation Equation. Dynamic Potentials. Gauges. Hertz Vector.—If, in the vector formula V x V x V = V(V • V) — 02 V, B or E replaces V, 11.00 (1) or (2) replaces V x B or V x E, zero replaces V • B or V • E, and then 11.00 (2) or (1) replaces the remaining V x E or V x B and yE from 11.00 (6) replaces i, the resultant propagation equations are
art v2B =µy at +
a2B at,
or
a2E aE v2E = wY at +12' at,
(1)
These hold in charge-free space. The first term on the right gives energy dissipation by heat and a wave damping that is absent when 7 is zero. The static scalar and vector potentials may be redefined to include rapidly fluctuating fields. The vector potential whose curl is B must reduce for constant fields to the magnetostatic potential. Replacing B in 11.00 (2) by V x A and changing the differentiation order gives B= VxA
V x E = —V x
aA
at
(2)
Integration of V x E removes the curl but adds an integration constant with zero curl which must be the gradient of a scalar. Thus
aA
E = -at
— v,fr
(3)
The potentials for a specified E and B are not unique. If A is solenoidal, then V • E along with 11.00 (3) and (5) yields Poisson's equation: '7211. =
—
p/e =
Pc' • i dt/€
(4)
This depends only on the lamellar part of i. Also, when V • A is zero, writing V x A for B in 11.00 (1) and using the above formula yield V xVx
A -= +1.4,3E v2A
at =
Let E, and is be the solenoidal parts of (3) and of the source i, and let 7E8 be the current induced in the medium by E8. Then from (3)
v2A —
aA cat
—
ate = mi„
(5)
§11.01
PROPAGATION EQUATION
417
Clearly for static fields, this becomes 7.02 (2). The potential choice in which A derives from the solenoidal part of i and ,k from the lamellar part is called the Coulomb gauge. When time-dependent, only A propagates and 1G gives a localized field in the same time phase everywhere. This A comes from a scalar solution W of the propagation equation by using V x uW or V x (u x VW), a method that appeared in 10.01 and is applied later in Chaps. XII and XIII to wave guides and cavity calculations. Another and more usual choice of potential is the Lorentz gauge, which requires that both A and 1t. have the same propagation equations as E and B. In a region of conductivity y with no sources, these equations are (3A a2A v2A =µy at + PE ate
alp °2% =µY at + PE W
(6)
This gives another relation between A and ik for, if p is zero, taking the divergence of both sides of (3) gives zero for V • E by 11.00 (3), so that
av • A at
v24, —
Comparison of this with (6) shows that for V21,1/ to agree requires that
aip v • A— — A-4 — E 797
(7)
To prove this also gives VA in (6), take the gradient of both sides of (7), expand VV • A by the initial vector relation of this article, write V x B in terms of E by 11.00 (1), (6), and (8), and eliminate V1,t by (3). Cancellation of the E terms gives v2A as in (6). The whole electromagnetic field may be represented by a single vector Z, the Hertz vector, from which A and 1,1. can be obtained by the equations
az t,
A = A-yZ + ALE 6- ---
= —v • z
(8)
These equations satisfy (7) and also (3) if we take E = V (V • Z) — V 2Z = V x (V x Z)
az a2z v2z = 1.4-y— at ± E a te
The magnetic induction is given by (2) and (8) in terms of Z. B = iryV x Z
1.4€
a(v x z) at
(9) (10)
418
PLANE ELECTROMAGNETIC WAVES
§11.02
Equation (10) contains all the properties of the electromagnetic wave. In a lossless medium the fields of (9) and (11) may be written
E=VxVxZe and
H=
Ea (v
x ze) at
(12)
This Ze is the electric Hertz vector. If 11.00 (1) and (2), in a medium without charges or currents, are put in terms of H and E, then the same two equations result if E and H as well as mu and E are interchanged. Making the same interchanges in (12) gives a new Hertz vector Z„„ thus
H=VxVxZ,„
and
E—
/la (v
x Z,, )
at
(13)
This is often called the magnetic Hertz vector and is sometimes more convenient than the Coulomb gauge potentials in solving wave-guide problems. 11.02. Poynting's Vector.—Multiplication of 11.00 (1) by —E and of 11.00 (2) by B/µ and addition of the results give
B B ap B a s — • (V x E) — E • (V x— = —i • E — E • w — • — a 1.4
(1)
Integration of this over any volume v gives by Gauss's theorem, 3.00 (2),
fv
[i. •(vx
E) — E • (v x
dv = fvV • (E x — au ) dv = isn • (E x — 12 ) dS
for the left side. The right side remains a\\volumeintegral so that
— f n • (E x — B)dS = [i • E
a (D • E B)1 dv at 2 2i.L
(2)
By 6.03 (2) and Ohm's law, the first term on the right represents the electric power absorbed as heat in v. By 2.08 (2) and 8.02 (3), the second and third term represent the rate of change of the electric and magnetic field energies in v. If all these terms are positive and if the law of the conservation of energy holds, then v absorbs power from outside, and the rate of flow of energy through its surface must be given by the left side of (2). Thus the outward normal component of the vector H
=
ExB
(3)
when integrated over a closed surface, represents the rate of flow of energy outward through that surface. The vector H is called Poynting's vector We should note that a meaning has been given only to the surface integral of this vector over a closed surface. We have not shown that H can represent the rate of flow of energy through any element of surface.
§11.03
PLANE WAVES IN DIELECTRIC INSULATOR
419
11.03. Plane Waves in Homogeneous Uncharged Dielectric Insulator. An electromagnetic disturbance is a plane wave when the instantaneous values of B, E,1., A, and Z are constant in phase over any plane parallel to a fixed plane. These planes are called the wave fronts and their normal, given by the unit vector n, is the wave normal. In the dielectric where y is zero so that the first term on the right of 11.01 (11) is missing, differentiation shows that, if v = (AE)-1, a general solution is Z = f i(n • r — vt) ± f 2(n • r + vt) = fiRn • r + ye) — v(t + t')] + f2[(n • r — vt') + v(t + t')]
(1)
so that fl has the same value at the point n • r + vt' at the time t + t' as at the point n • r at the time t. Thus it represents a wave moving in the n-direction with a velocity v. Similarly, 12represents a wave of the same velocity in the — n-direction. The vacuum velocity of an electromagnetic wave is c = (1.1„E,,)-1and equals 3 X 108meters per second. The ratio of c to the velocity v in a medium is its index of refraction n. c
(2)
n= -
v
Since Eqs. (1), (2), (6), (7), and (10) of 11.01 are identical, the simplest solutions for a wave in the positive n-direction have the form
D = Do f(n • r — vt) E = Eo f(n • r — vt) B = Bo fi(n • r — vt)
(3) (4) (5)
where Do, Eo, and Bo are the vector amplitudes of D, E, and B, respectively, and f(n • r — vt) is a scalar. When p is zero, (3) and 11.00 (3) give V • D = n • DoRn • r — vt) = 0 Thus, either f' is zero, giving a static field of no interest here, or n • Do = 0
(6)
A similar substitution of (5) in 11.00 (4) gives the similar result n • Bo = 0
(7)
Thus D and B lie in the wave front. From (4), (5), and 11.00 (2) we have
V x E = n x Eo f'(n • r — vt) = — y aB t = vilo f“n • r — vt) This holds at all times so that fiis proportional to f, and we take n x Bo = vB0
or
n x E = vB
(8)
420
PLANE ELECTROMAGNETIC WAVES
Similarly, substitution of (3) and (5) in 11.00
where i is zero gives
n x B = 1.4vD
or
n x Bo = --AvDo
(1)
§11.04
(9)
The scalar product of (8) by (9) gives, since n • Bo is zero by (7), (n x E 0) • (n x B = — 1.10 Do • Bo = (n
(E0 . Bo) — (n • B0)(12 • E o) = Eo . Bo
By (8) the right side of this equation is zero, so that E0 • Bo = 0
and
Do . Bo = 0
(10)
Thus both Eo and Do are normal to Bo. From 11.02 (3), Poynting's vector is (11) H= 1 E0 x Bo[An • r — vt)] 2 = lIo[f(n • r — vt)] 2 whereas from (6) and (7) the normal to the wave front has the direction of Do x Bo. Thus the energy propagation direction H makes the same angle with the wave normal n that E makes with D. When all B vectors parallel a fixed line, then from (10) all E vectors parallel a line normal to it and the wave is plane polarized. The polarization plane is defined as that parallel to H and B in optics and as that parallel to H and E in radio. 11.04. Plane Wave Velocity in Anisotropic Mediums. We now consider the propagation of electromagnetic effects in a homogeneous anisotropic medium of zero conductivity and permeability 1.4. By 1.19 (7), in such a medium there are axes for which D and E are connected by the equations = ciEz, Dy = € 2E„, D, = € 3E, (1) —
These coordinate axes are chosen to coincide with the electric axes of the medium. By 2.08, the electrical energy density is given by
aw 1D .E = _1(elE! av — 2
2
E2.q
+ !1:
€3ED = 21
(2) The last relation is obtained by using 11.03 (9) for D, interchanging dot and cross, and substituting 11.03 (8) for n x E. It shows the equality of electric and magnetic energy FIG. 11.04. densities whose sum is twice (2). Let the direction of ray or energy propagation be n', then B, D, E, n, and n' are related as shown in Fig. 11.04. Evidently the vectors D, E, n, and n' lie in the same plane since all are normal to B. We also notice that sin a =
n• E
——
n' • D
(3)
RAY SURFACE AND POLARIZATION
§11.05
421
To find the wave front velocity v along n, we start with 11.03 (8) and (9). m 4v2D =
—
n x (n x E) = E
—
n(n • E)
(4)
We now introduce the crystal constants v1, v2, and v3defined by 1,
AvE221 =
'4603 = 1
1,
(5)
Elimination of E from the components of (4) by (1) and (5) gives mo(vi
v2)Dx = i(n • E), 14(v2 — v 2)D, = m(n ,u4(v3 — v2)Dz = n(n • E)
—
(6)
where t, nz, and n are the components or direction cosines of n. From 11.03 (6) n • D = iDx .D z = 0. Insertion of Dx, D0, and Dz from (6) gives 12 V2
—
V
n2 m2 ± V 2 — V22± V2 —
—0
(7)
This is Fresnel's equation and gives the velocity of the wave front propagation in terms of the direction cosines of its normal. In general, for every particular direction of the wave normal, there are two different values of this velocity. We can easily verify by substitution in (7) that, if vl > v2 > v3, the only two directions in which v has a single value are to =
+ vi — 4)3/4 24 — mo = 0, no = + — vg)i(v1 — v3)—i (8) (
(
and that the velocity in these directions is v2. The directions defined by 10, ?no, and no are called the optic axes of this biaxial crystal. When v1 = v2 v 3, v is single-valued in the z-direction only, for w—.c— bi 1-1 no = 1 and t o = mo = 0. If vi v2 = v3, v is single-valued in the x-direction only, for which to = 1 and mo = no = 0. In such cases there is only one optic axis and the crystal is said to be uniaxial. 11.05. Ray Surface and Polarization in Anisotropic Mediums.—The energy propagation velocity v' in any direction in a crystal is found by dividing 11.02 (3) by E2/(Av) and multiplying through by n • E giving by 11.03 (8) n• E
E2
n = vn • E
Ex B — n(n • E) E2
(n E2E)2 •
(1)
We can then solve 11.04 (4) for n(n • E), write out components, eliminate Ds, D0, and Dz by 11.04 (6), and insert in the right side of (1), giving Av
n•E rix • = [1 E2
,
—
( n • E)21 Ex,,, E2
(2)
422
PLANE ELECTROMAGNETIC WAVES
§11.05
Figure 11.04 and 11.04 (3) show that the relation between v and v' is („)2 v'
sin2 a = 1 — (n E)2 E2
= COS2 a = 1 —
(3)
Also IIz,y,z = llon'z,y.zso that we can rearrange (2) in the form E2 AvIIo(n • E)
n'.,y,z
(v/2)1,2,0 2 — (v/v')2
Let c', m', and n' be the direction cosines or components of n' and notice that, from 11.03 (11), n' • E = l'Ez m'Ey n'Ez. Then multiplying these equations through by m', and n' and adding, we obtain /
,24.
v'2 —
/2 2
in V2
n"Vi
V"— Vi
—
=0 (4)
This equation gives the ray velocity v' in the direction /', m', and n' as a function of the crystal constants v1, v2, and v3. In general for every particular direction there are two ray velocities. If, at some instant, an electromagnetic disturbance starts from the origin, then in 1 sec the wave will be given by r = v'. When v1 > v2 > v3, the form of this wave in one octant is shown in Fig. 11.05. In the direction of OP and of its image in the yz-plane, both rays have the same velocity. Substitution in (4) shows the directions of the ray and optic axes to be related by lova 2
= m 0 = 0,
nr
noVi V2
(5)
The optic axis is shown by 00' in Fig. 11.05. In each of the coordinate planes, the section of one of the sheets of the double surface is circular and of radius v1, v2, or v3as shown. If two of the quantities v1, v2, and v3 are equal, one sheet of the wave surface is a prolate or oblate spheroid whose axis is a diameter of the other sheet which is spherical. The rays that produce the spherical surface are said to be ordinary rays, and those producing the spheroidal surface are called extraordinary rays. Let us designate the two solutions of 11.04 (7) by v. and vb and the components of the electric displacement corresponding to these two normal velocities by DP),,,z and Dr,;,z. Multiply each equation of 11.04
§11.06
ENERGY, PRESSURE, AND MOMENTUM OF A PLANE WAVE 423
(6) with v = v¢by the same equation with v = vb and rearrange. This gives „ 2 j)(a) D (b)
--x,y,z--x,y,z
(n • E)2
- v!)
- vt)
(6)
Split the right side by partial fractions and add the three equations: 2D(a) D(b)
(n • E)2
n2
1
v?, -
vx,y,z
- v?,
x,y,z
n!,,,,z - v/
By 11.04 (7), each term in the sum on the right is zero so that we have Do) Do) = 0
(7)
Thus the v¢and vb rays are plane polarized at right angles to each other. 11.06. Energy, Pressure, and Momentum of a Plane Wave.—Imagine a plane wave in an isotropic medium striking a thin infinite plane absorbing sheet normally, and consider a right prism which encloses a unit area of the sheet and whose axis is normal to it. An integration of Poynting's vector H over its surface gives the radiant energy entering it. But we have already seen, from 11.03 (6) and (7), that on the sides of the prism vBn = =0 so that II,, is zero. It is also zero on that end which is on the opposite side of the absorbing sheet from the impinging wave. Thus, the only contribution to the integral is from the base on which the wave impinges. Therefore, from 11.03 (11) and (8), the instantaneous rate of energy absorption in the sheet, in watts per square meter, is n =
E x B nE2 - (n • E)E p.v
In isotropic mediums n • E = 0 and
B = (.4,€)tE from 11.03 (8), so that
EB2 B2 1 (B2 + E D) n,, - no,€)i = niloic)i - n(p,e)i 2
(1)
But, by 2.08 (2) and 8.02 (3), the energy density in the wave field is
aw = av
E.E2 B 2 —
2
21.4
(2)
The propagation velocity is (µ6)-i so that the energy per square meter of wave front equals that in a cylinder of unit area and length (AE)--i. From 1.14 (6) and 8.14 (2) there appear to be in electric and magnetic fields pressures normal to the lines of force of iEE2 and 02/A, respectively. Since in the plane wave case, such fields parallel the face of the absorbing sheet on one side but not on the other, we should expect a total pressure,
424
PLANE ELECTROMAGNETIC WAVES
§11.07
in newtons per square meter, against the sheet, of amount
+ 412-1132
P=
(3)
From (2) this equals the energy density at the surface of the sheet. Consider the absorbing sheet free to move but so heavy that P gives it a negligible velocity. Then the relation between P and the momentum p is dp P= (4) dt The mechanical law of conservation of momentum, if applied here, requires that the momentum acquired by the sheet must have been carried by the electromagnetic wave which impinged on it. A comparison of (4), (3), and (1) shows that this implies that the wave possesses a momentum gn per unit volume, in the direction of propagation n, equal to gn
= m€11„
(5)
11.07. Refraction and Reflection of a Plane Wave. Let us now consider a plane wave in a medium of permeability A and capacitivity E, impinging due on the infinite plane boundary of a second 0 medium whose constants are A" and e". This wave may give rise to two waves: one, known as the reflected wave, returning into the first medium, and the other, known as the refracted wave, entering the second medium. Let the unit vec0 tors in the direction of propagation of the incident, reflected, and refracted waves be d, d', and d", respectively. Let d and d" make angles El and 0" with the normal to the plane surface drawn into the second medium, and d' make an angle 0' with the opposite normal. The plane of incidence is perpendicular to the FIG. 11.07. interface and contains d• Let M and N be any two points in space, the vector from M to N being p. If, as before, we take n to be a unit vector normal to the wave front, then n • p/v seconds after passing through M this wave, which has a velocity v, will pass through N. In the case being considered, the same wave front passes through a given point P in the first medium twice, once before and once after reflection. A wave front which passes through a point 0 on the reflecting surface at the time t = 0 will pass through P before reflection at t = d • r/v and after reflection at t = d' • r/v, where —
§11.08
INTENSITY OF REFLECTED AND REFRACTED WAVES
425
r is the radius vector from 0 to P. If the same law of reflection holds for all parts of the surface, the interval of time between these two passages must be the same for any other point Q at the same distance from the surface as P. The vector from 0 to Q will be r s where s is a vector lying in the surface. Equating these time intervals and multiplying through by the velocity give — df • (r
d • T — d' • T = d • (T
s)
or
d • s = d' • s
(1)
Therefore d and d' make the same angle with any vector drawn in the interface which is possible only if d is the mirror image of d' in the surface. Thus both reflected and incident wave normals lie in the plane of incidence on opposite sides of the normal to the interface and make the same (acute) angle with it. To locate the refracted beam, suppose a wave front requires a certain time to pass from P in the first medium to its image point P" in the second. Let the radius vector from 0 to P be r and from 0 to P" be r". If two other points Q and its image point Q" are at the same distances from the interface as P and P", the time for the wave front to pass from Q to Q" will be the same. The radius vectors to Q and Q" are r s and r" s, where s lies in the interface. Equating the time intervals in the same way as for the reflected wave normal, we have (11"E").}(d" • r") — (1.1E)+(d • r) = (.i."E")i[d" • (r"
s)] — (p,E)i[d • (r
(m"E")1(d" • s) = (p.E)1(d • s)
s)] (2)
If s is taken perpendicular to d, which means normal to the plane of incidence, then the right side of (2) is zero, so that d" is also perpendicular to s. Thus both reflected and refracted wave normals lie in the plane of incidence. If s is taken in the plane of incidence then, if v and v" are the wave velocities in the first and second mediums, respectively, v
cos ds cos d"S
sin 6 =n sin 0"
(3)
This is Snell's law of refraction. The ratio n of sin 0 to sin 0" is called the index of refraction. The preceding reasoning is equally valid for an anisotropic medium, although v and v" may be different for each angle of incidence so that n depends on O. 11.08. Intensity of Reflected and Refracted Waves. — The law of conservation of energy requires that the energy passing into a medium through a square meter of surface must equal the difference between the incident and the reflected energies. From Fig. 11.08 and 11.03 (11), this gives (1) (II — II') cos 0 = II" cos 8"
426
PLANE ELECTROMAGNETIC WAVES
§11.08
We shall identify all vectors associated with a beam whose magnetic vector lies in the plane of incidence by the subscript 1, and those associated with a beam whose electric vector lies in this plane with the subscript 2. In finding the intensity of reflected and refracted beams, it is necessary to treat these cases separately.
FIG. 11.08.
For the beam whose magnetic vector lies in the plane of incidence, E1 is parallel to the surface; so, from 1.17 (3), we have
= E','
Ei
In this case, from 11.06 (1), setting µ = A" €1(E?
— _OM cos
(2) =
(1)
becomes
0 = ENVI' 2 cos 0"
Dividing this by (2) and using 11.07 (3), we get
E, — E' — sin ti cos 61"E" 1 sin 0" cos 0 1
(3)
Solving (2) and (3) for E', and E',' gives E;. = E',' —
sin (0— 0") Ei sin (0 + 0") 2 sin 0" cos 0 s in (0 + 0")
El
It follows from 11.03 (8) and 11.07 (3) that sin (0— 0") B' — Bi 1 sin (0 + 6") sin 20 B1 131' = sin (0 + 0") Using 11.06 (1), we get for the reflected intensity sin2 (0 e")ll sin2 (0 + 0") 1 —
(8)
Substituting this in (1), we get for the refracted intensity — 2 sin 0" cos 0 sin 20 sin2 (0 + 8")
(9)
§11.08
INTENSITY OF REFLECTED AND REFRACTED WAVES
427
For the beam whose electric vector lies in the plane of incidence, B2 is parallel to the surface so, setting 1.1 = /I" = A„, we have B2 + B2 = B2
From 11.06 (1), setting µ =
(10)
= my, (1) becomes
(B2 — B22) cos 8 = B22e"--1cos 0,,
E1
Dividing this by (10) and using 11.07 (3), we get B2 - B' — 2
sin 20" B„ sin 20 2
(11)
Solving (10) and (11) for R2and B2 gives tan (0 — 0")B B' _ B22 2 tan(0 + 0") sin 20 B/2' B2 sin (0 + 0") cos (0 — 0")
(13)
It follows, from 11.03 (8) and 11.07 (3), that E,
= tan (0 — 0") E2 tan (0 + 0") 2 sin 0" cos 0 E'2' = E2 sin (0 + 0") cos (0 — 0") 2
(15)
Using 11.06 (1), we get for the reflected intensity tan2 (0 — 0"), tan2 (0 + 0")112
11'
(16)
Substituting this in (1), we get for the refracted intensity 11'2' —
2 sin ;" cos 0 sin 20 sine (0 ± 8") cos2(0 — 0")112
(17)
At normal incidence, the cosines in (1) are unity, so that in place of (3) 6" )1 „ v (— E = -Ti E”
E — E'
V
Combining this with (2) gives E' E"=
v
v — v" E v v" 2v" E v"
(18) (19)
428
PLANE ELECTROMAGNETIC WAVES
§11.09
From 11.06 (1), the reflected intensity is H
,
tv — v"\211 + v"
(20)
4vv" (v v")211
(21)
Combining this with (1) gives _
It should be noted that the intensity of radiation given in (8), (9), (16), (17), (20), and (21) is the energy passing in 1 sec through 1 sq m of area parallel to the wave front. To get the actual radiation energy in 1 cu m divide II by the wave velocity in the medium. Evidently from (12), if 0 + 0" = ir , B2 = 0, which means that at this incident angle radiation whose electric vector lies in the plane of incidence is not reflected. This is called the polarizing angle because unpolarized radiation incident at this angle is reflected with its magnetic field in the plane of incidence. From 11.07 (3), this "Brewster's angle" is sin
sin Op —
(fir
0p)
= tan Op =
=n
(22)
11.09. Frequency, Wave Length, Elliptic Polarization.—So far we have used the general solution for a plane wave, f(n • r — vt), in deriving the laws of reflection and refraction. For discussing circular and elliptic polarization it is convenient to take a regularly periodic function which can be built up by Fourier's series from simple sinusoidal terms. Thus
f(n • r — vt) = D cos [w(t — v—'n • r)
(1)
where n is a unit vector in the propagation direction. The mathematical manipulation is considerably simplified by writing
D cos [co(t — v—'n • r)
= Dejoemt—"''o) (real part)
(2)
The term DO' is usually written as a phasor D. The angular frequency w and the cyclical frequency I, are related by =
2irp
(3)
The wave length X is the shortest distance in the direction of propagation along a "frozen" wave in which the electrical conditions repeat. It is related to the frequencies and the wave velocity v by the equations v
A = =
277-v
=
1
211-
P (AO
W (AO
2a
(4)
§11.09
FREQUENCY, WAVE LENGTH, ELLIPTIC POLARIZATION
429
Now consider the superposition of two plane electromagnetic waves of the same frequencies traveling in the z-direction. Let one be plane polarized with a y-directed magnetic vector and the other with an x-directed one so that the equations for the electrical intensities are Ex = El cos [w(t — v—'z)] Ey = E2 cos [0)(t V-1Z) + 6]
(5) (6)
Ex and Ey are shown as a function of z in Fig. 11.09. To find the locus
in the xy-plane of the end of the vector E whose components are Ex and Ey, set z = 0 and eliminate t from (5) and (6). This gives 2ExE, cos 3 = sin' S El + E2 E 2E,
(7)
This is the equation of an ellipse, shown at the left in Fig. 11.09. Such waves are called elliptically polarized. If S = mr, (7) becomes E2Ez ±
=0
(8)
which represents a straight line, so the resultant wave is plane polarized. If S = -(2n + 1)7 and, in addition, E1 = E2 = E, then (7) becomes = E2 This locus is a circle, and the resultant wave is circularly polarized.
(9)
430
§11.10
PLANE ELECTROMAGNETIC WAVES
11.10. Total Reflection.—Our first example of elliptically polarized radiation occurs in connection with the phenomenon of total reflection. If e" < E in 11.07 (3), then 0" > 0 so that 0" may equal 47r when 0 < tea. The value of 0 which gives 0" = a is called the critical angle Oc. The agreement between experimental facts and the results of the following interpretation of this phenomenon justifies its use. When 0 > 0c, let us write 11.07 (3) in the following form, when A = A", cos 0" = (1 — sin2 0")1 = 1 (—
e
E'
sin2 0
i
E
= j —,- sin2 ei
Equation 11.08 (3) must then be written El _ E1 i[sin2 0 — (e"/OPE--1, cos 0
(1)
which, combined with 11.08 (2), gives = cos 0 — j[sin2 0 — (E'/E)]1E = . ev' El El cos j[sin2 0 — (e"/e)P 1 2 cos 0[sin2 0 — (E'/E)]1 tan ik1 = cost 0 — sin20 + (E" /6)
(2) (3)
Corresponding to 11.08 (11), we have B2
sM2 0 — 1])i I3 cos 0
j{(E/e")[(E/E")
P2
which, combined with 11.08 (10), gives _ cos 0 2
tan
ik 2
—
{(e/e")[(e/e") sin2
—
(e/e") [ (E/e") sin2 — 2 cos 0{ (e/e")[(e/e") sin2 — 1]}i COS2 0 — (e/e")2 sin2 0 (6 / E" )
cos 0 +
{
_
P2—
e
2
(4)
(5)
It is evident from (2) and (4) that 1E11 = 1E11 and 1B21 = 1B21 so that the reflected and incident rays have the same intensity. Thus the law of the conservation of energy permits no energy in the refracted beam. This can be verified by computing E"," and B12' and from this II". A comparison of (2) and (4) with 11.09 (2) shows that there is a phase change for the beam whose magnetic vector lies in the plane of incidence and 4/2 for the beam whose magnetic vector lies normal to it. Thus, for plane polarized incident radiation, whose two electric vector components are E1 and E2, the reflected radiation is elliptically polarized. The phase difference S can be obtained from the ratio of (2) to (4), since B2 = (µE)lE2 and
PLANE WAVES IN ISOTROPIC CONDUCTORS
§11.12
431
B2 = (1.1,e)1E2by 11.03 (8), giving j cos 0[sin2 0 — (€"/e)]1 sine 0 — j cos 0[sin2 0 — (e" /O]1
e,5 = ejoh_4,0 = sine
Taking the ratio of the imaginary to the real part of this gives us tan 3, which, with the aid of Pc 573 or Dw 406.02, may be reduced to tan
— (66)]i 7 3 cos 0[sin2 = 2 sine 0
(6)
11.11. Electromagnetic Waves in Homogeneous Conductors.—In a medium whose conductivity is not zero, we must use all the terms in the propagation equations 11.01 (1) and (2). In deriving 11.01 (2), we set p = 0. To justify this, take the divergence of 11.00 (1) which gives Ile— (V • E) at
iryV • E = 0
Write p for eV • E by 11.00 (3) and integrate from 0 and
pt, to t' and p'. (1)
= Ploe
Thus, an electrical distribution p'0is dissipated independently of electromagnetic disturbances so that, if p; is zero initially, p' is always zero, which justifies setting p = 0. We define the relaxation time T to be T=
TE
=
(2)
-rie
Take a simple periodic disturbance so that both B and E have the form B = beic." (real part) = Boelo"-E0) (real part) Now
a/at may be replaced by jco, and 11.01 (1) and (2) become 02B = gjoyy — u.,203 — 0,20E V 2E µ ( jwy
(3) (4)
These are identical in form with 11.01 (1) for a periodic disturbance in an insulator. The equations will have identical solutions if we write E - j-y/co for E. The complex quantity (ile)1which occurs frequently may be separated into real and imaginary parts by Dw 58.2. [A(E — ic0-17)]/ = n n={
2(1+ 0)72622)1 +1 J}1'
k
ik
(5)
0)1 — 1 ]}1 (6) 2(1 ±—;
11.12. Plane Waves in Homogeneous Isotropic Conductors.—Certain properties of plane waves in conductors can now be obtained, by the
432
PLANE ELECTROMAGNETIC WAVES
§11.12
substitution suggested above, from the corresponding properties in a dielectric derived in 11.03. In isotropic mediums the nature of e does not affect (6), (7), or (8) of 11.03 so that in a homogeneous conductor B is normal to E and both lie in the wave front. From 11.03 (9) k2)+Eoeitan-1(kIn) (1) Eo = (n 4- jk)E o = (n2 Thus, the electric and magnetic vectors differ in phase by tan—' (k/n). Writing E in the form of 11.09 (2) and substituting in (1) give &awe = ,Alecolco•reaco(t—no•r)
(n2
&jot = (, x
k2)IA lecuka•re j[w(t—no•T)-1-tan -l(k/n)]
(2) (3)
where E is a unit vector along E, and 6 is a unit vector normal to the wave front. We note that both E and B decrease exponentially as the wave advances, indicating a rapid absorption if k is large. We call n the index of refraction of the conductor and k the coefficient of extinction. It is useful to know the magnitude of these quantities in nonferromagnetic conductors whose permeability is nearly equal to lie. In conductors the capacitivity is not accurately known but seems to be of the same order of magnitude as in insulators. As a typical example take copper, whose conductivity is about 5.8 X 107 mhos/m. Designating wave length in vacuo by Xo and using 11.09 (4), we have, for copper, writing K for E/Ev, 7
Xo7
WE
2rcE
Xo -„ X 3.48 X 109
Except for very short waves, this quantity is certainly large compared with unity so that, from 11.11 (6), n and k are well approximated by n ~ — k ~ (110■ 07)1 (21-c)-1
(4)
The distance an electromagnetic wave must travel in order that E and /3fall to e--2' = 0.002 times their initial values is, from (3), 2r
Xo (47X oy
wk
kc \FLIT
meters
(5)
For air wave lengths of 1 cm, 1 m, 100 m, and 10 km, this gives d the values, in copper, of 0.0025, 0.024, 0.24, and 2.4 mm, respectively. It is also of interest to compare the average electric and magnetic energies in the wave. From (3) and (4) and 11.06 (2), we have We _ E2 _
Wm
2rEc
n2 ± k2 Xo-y
(6)
On the assumption that E = 2e,, this gives, for the waves just considered, 5.7 X 10-8, 5.7 X 10-18, 5.7 X 10-12, and 5.7 X 10-14, respectively, so that nearly the entire energy is in the magnetic field.
§11.13
REFLECTION FROM CONDUCTING SURFACE
433
11.13. Reflection from Conducting Surface.—We saw in 11.11 that in a conducting medium (A"e")i is replaced by
=n
ii1"(E" —
ik
Thus, sin 0" and cos 0" become, by 11.07 (3) with the aid of 11.11 (5), sin 0" =
LE ) 4 sin n(+ jk
cos 6" = [1
At si'0 (n + j k)2
(1)
We saw in the last article that, for electromagnetic waves over 1 cm long. —k
n > 1.4 X 104
So that, from (1), 0" is a complex and extremely small angle. Thus, from 11.08 (8), when the magnetic vector lies in the plane of incidence, , sin'(0 — 8")11.1 _> II 111 = sin' (8 + 8") 0,,,o
(2)
When the electric vector lies in the plane of incidence, we get, from 11.08 (16), provided 0 is not too close to 4r, ,tan' (8 — 0")ll 112 = tan' (0 + 6") 2 v,-,o
112
(3)
Thus, for all angles of incidence and for all states of polarization, a sufficiently long electromagnetic wave is completely reflected from a conducting surface. The same results apply to longer waves at lower conductivities. From the last article, the extremely small fraction not reflected is very rapidly absorbed so that quite thin sheets of metal can be very opaque for short waves. To study the state of polarization of the reflected beam, we divide 11.08 (14) by 11.08 (4) and obtain _
cos (0 + 0")E2 _ cos (0 — 0")E1
(cos0" — sin 0" tan 0)E2 (cos 0" + sin 0" tan 0)E1
(4)
For the wave lengths just considered, we have seen that n + ,9k = n(1 — j) If µe/fin jkl 2 b. When used as a transmission line, show that, from 4.30 (2), its characteristic impedance is 2k = I nK(b2 a- 2){1f1(1 - bet-1)N -1 where n is the intrinsic impedance and K(k) is an elliptic integral. 21. A strip of width b with one edge on the axis of a cylinder of radius r, where r > b, forms a transmission line. Show from 4.30 that 2k = 4nK[(1 - b2r-2)4][K(br-9]-1
22. A transmission line is formed of two unequal coplanar strips, the wider having a width w. The gap between the strips is q and the distance between their outer edges h. Show from 4.30 that the characteristic impedance is 2k = inK(k){K[(1
-
k 2 = hq(w q)-1(h -
where n is the intrinsic impedance of 11.15 (15) and K(k) is an elliptic integral of modulus k. 23. A strip lies on a diameter of a cylinder of radius r, one edge being at a distance c1from its center and the other at a distance c2. Show from 4.30 that it - \ 2k = 4nKR1 - k2)11[K(k)]-1,
k = r(ci
e2)(r2
cic2)'
24. An infinite cylinder of radius c lies midway between two infinite parallel planes a distance 2b apart. Show from 4.31 that the characteristic impedance of this line is 2k = 4nK(k)(KR1 -
k = cos[Zircb-,(1 + X)]
where X is given by 4.28 (6). This is accurate to less than 1 per cent if c < lb. The attenuation may also be worked out by the method of 11.18 with the aid of HMF formulas. 25. An infinite strip of width 2a lies parallel to and halfway between two infinite parallel planes at a distance 2b apart. Show from problem 61 of Chap. IV that the characteristic impedance of this "strip line" is
2k =
inK(e-T.b-1) K[(1 - e-"rab-1)4] } -1
where K(k) is a complete elliptic integral of modulus k.
446
PLANE ELECTROMAGNETIC WAVES
26. The two halves of an infinite plane are separated by a gap of width 2h. An infinite strip of width 2d lies in the gap normal to the plane so that its center line bisects the gap. Show that the characteristic impedance of this transmission line is 2k = nK[h(d,
h2)-iliK[d(c12
10)-1]1-1
where n is the intrinsic impedance of the medium defined in 11.15 (15). 27. An infinite strip of width 2c lies midway between and normal to two infinite parallel planes a distance 2d apart. Show from problem 63 of Chap. IV that the characteristic impedance of this line is, if a =
2k = inK(cos a)[K(sin cO] —' 28. The center line of an infinite strip of width 2h is coplanar with, and at a distance
b from, the edge of a semi-infinite plane. The strip is normal to the plane. Show from problem 64 of Chap. IV that the characteristic impedance of this line is 2k
- 7/11((k)1-1-KI(1
k)2)11,
-
k = b-1[(h2
b 2)1— h]
29. A transmission line consists of two infinite conductors, a cross section of which forms two separate portions of the circumference of an infinite circular cylinder as specified in problem 65 of Chap. IV. Show that for this line
2k = inK(k)(K[(1 — k 2)4]}-1 where the modulus k is defined in the problem referred to. 30. A transmission line consists of two parallel strips of width 2h in two parallel planes a distance 2d apart. Their center lines lie in a plane normal to the parallel planes. Show from problem 66 of Chap. IV that
2k = nK(ba--1)(K[(1 — b2a-2)+] where ba-' is found as directed in the problem referred to. 31. One side of a transmission line is a circular cylinder of radius c at the origin. The other consists of four hyperbolic cylinders given by ±xy b2, Show that the character-
istic impedance of this line is
2k =
inK(k)(KR1 — k 2)111 -1
where the modulus k is found as directed in problem 79 of Chap. IV. 32. One side of a transmission line is a circular cylinder of radius a. The other is a parabolic cylinder y2 = 4p(x p) whose focus is the cylinder axis. Show that the characteristic impedance is
2k = inK(k){K[(1 — k2)4]1-1 where the modulus is found as directed in problem 80 of Chap. IV. 33. One side of a transmission line is a conductor of a rectangular cross section bounded by x' = ±b, y' = ±c. The other side is two conducting planes at y = ±a. Show that the characteristic impedance is 2k 7171K
[ MI] {K[(1 — A-IB)I ] I -1
where A and B are determined as directed in problem 82 of Chap. IV. 34. A plane wave in a medium of permeabilityµ and capacitivity E strikes normally a plane perfectly conducting mirror. Show that reflection is prevented by placing a
REFERENCES
447
thin layer of material of thickness d, capacitivity e', and conductivity y' one-quarter wave length in front of it if y'd ~ (e/ it)i and y' >> References R. BECKER: "Classical Electricity and Magnetism," Blackie, 1932. Uses vector notation and treats waves on wires. BATEMAN, H.: "Electrical and Optical Wave Motion," Cambridge, 1915, and Dover. Discusses boundaries and wave solutions and gives extensive references. BORN, MAX: "Optik," Springer, Berlin, 1933. A complete, illustrated treatment. COLLIN, R. E.: "Field Theory of Guided Waves," McGraw-Hill, 1960. Chapters III and IV cover subject matter of this chapter in detail. DRUDE, P.: "Theory of Optics," Longmans, 1920, and Dover. This is the classical book on this subject and uses the long notation. FLUGGE, S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. Several of the articles in this volume contain material on plane waves. FORSTERLING, K.: "Lehrbuch der Optik," Hirzel, Leipzig, 1928. A complete and elegant treatment of the subject using vector notation. FRENKEL, J.: "Lehrbuch der Elektrodynamik," 2 vols., Springer, Berlin, 1926, 1928. GEIGER-SCHEEL: "Handbuch der Physik," Vols. XII, XV, XX, Berlin, 1927, 1928. Gives elegant treatment of electromagnetic theory in Vol. XII. HARRINGTON, R. F.: "Time-harmonic Electromagnetic Fields," McGraw-Hill, 1961. Chapter II covers much of this material with circuit analysis. HERTZ, H. R.: "Electric Waves," Macmillan, 1893. Drawings of radiation fields. JACKSON, J. D.: "Classical Electrodynamics," Wiley, 1962. Chapters VI and VII are pertinent. Uses Gaussian units. JEANS, J. H.: "The Mathematical Theory of Electricity and Magnetism," Cambridge, 1925. Gives extensive treatment using long notation. KING, R. W. P., H. R. MIMNO, AND A. H. WING: "Transmission Lines, Antennas, and Wave Guides," McGraw-Hill, 1945. Practical detailed treatment of transmission lines, with some field theory. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. Gives original treatment using long notation. PANOFSKY, W. K. H., AND MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Chapter XI is pertinent to this chapter. PAPAS, C. H.: "Theory of Electromagnetic Wave Propagation," McGraw-Hill, 1965. Contains a very lucid and detailed treatment of Maxwell's equations. RAMO, S., AND J. R. WHINNERY: "Fields and Waves in Modern Radio," Wiley, 1944. Simple treatment of waves on wires. SCHELKUNOFF, S. A.: "Electromagnetic Waves," Van Nostrand, 1943. Whole theory treated by means of the impedance concept. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Extensive and rigorous mathematical treatment of the whole subject. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Chapters IX, X, and XI cover this chapter with additional material. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. ABRAHAM, M., AND
CHAPTER XII
ELECTROMAGNETIC RADIATION 12.00. The Radiation Problem.—Any system generating electromagnetic waves and incompletely enclosed by conductors loses energy because waves escape into space. If the system is not supposed to radiate, this represents a leakage loss and the only information usually required of a calculation is the power radiated. If the function of the system is to radiate, as in the case of antennas or horns, the additional information needed may include the polarization and distribution in space of the radiant energy, the nonradiative losses, the current and charge distribution in the system, and its input impedance. These will depend on the impressed frequency and electromotance, on the system's dimensions and geometrical configuration, and on the materials of which it and its surroundings are composed. When the antenna lies in empty space, a complete solution of the problem must give fields which satisfy Maxwell's equations for empty space outside the antenna and for the antenna medium inside it and which satisfy its surface boundary conditions. These fields must also lead to electric intensity and current distributions that match those of the transmission line over its junction surface with the antenna. For existing materials and configurations, such perfect solutions appear impossible. There are, however, methods of calculating one or more of the desired quantities with considerable accuracy. One can always find, by means of retarded potentials, the fields of a given charge or current distribution. Often the distant radiation is sensitive only to the major features of this distribution, of which a rough estimate is then adequate. This also gives the ohmic losses, if they are small, with sufficient accuracy so that with the aid of Poynting's theorem the real part, but not the imaginary part, of the input impedance can be found. Information as to the actual current distribution in perfectly conducting antennas of simple geometrical forms like spheres and spheroids can be obtained by a rigorous solution of the boundary value problem in terms of orthogonal functions similar to, but more complicated than, those used in electrostatics. Skin effect formulas give the ohmic losses in terms of the magnetic fields just outside the surface. With the exception of the biconical antenna, such solutions give little information on the input reactance. The radiation from orifices is often found by assuming initial field values which are then corrected by trial to fit the boundary conditions. In the articles that follow examples of most of these methods are given. 448
§12.01
449
SPHERICAL ELECTROMAGNETIC WAVES
12.01. Spherical Electromagnetic Waves. Dipole and Quadrupole Radiation.—When the strength of a multipole that appears in 1.07 or 7.10 varies in time, it generates a spherical electromagnetic wave. The electrostatic potential of the electric multipole of order n and moment q(a) may be written 1 an(r-9 4ren! axi ax; • • • axk
(1)
) n
In rectangular coordinates xi = x, x2 = y, and xs = z. (1) 4,1 = qz x 4rer3
qy(1) y
q P)z
4rer 3
4rer 3
When n = 1, (2)
This is the NI,for a dipole of moment [(qp)2 + (412 + (c))9+ with direction cosines qPq-i When n = 2, qa), ey) and qL2) are linear quadrupoles on the x-, y-, and z-axes and Q,I2y) = g1,2x),ql2z) = q L2) and q,22 = 42) are square quadrupoles in the xy-, xz-, and yz-planes. Thus the moment is a symmetrical tensor with three different square moments and two different linear moments. One linear moment was eliminated by ,32 ay2 (3 2/az2)r—i = 0. the relation (a2/3x2 When q()oscillates, the currents transferring the charges determine the moments just as well as the charges do. Equation 7.02 (4) shows that the magnetostatic Azof an infinitesimal current element io dzo is a function of r only. If the current io f(t) depends on time and the field travels with the same velocity in all directions, then the radiation Az should be a function of r only. Thus the propagation equation and its solution are
_ a2A, r-2 a (219 Az) = ea2Az V2Az = tz at 2 v2 at 2 ar ar dzo (r1 cos 0 - 0 sin 0)f(t ± v-'r) A _ µiodzo i(t ± v-lr) 4irr 4rr
(3) (4)
The plus sign gives a converging wave and the minus sign a diverging one. From 1.07 (1), if the potentials of the two charges qofi(t) of opposite sign at a distance dzo apart are superimposed, then since a/ azo= — a/az,
_
dzo alfi(t ± v-lr) - q(i) cos or 4r ad_
47rEr2 Lf1(t
1r) + rti(t
v-11 e (5)
If the current element in (4) feeds the charges in (5), then io = ago/at and q(1) f1(t) = io dzo ff(t) dt. Thus (4) and (5) satisfy 11.01 (8), and Az and NI/ are the potentials of an electric dipole in the Lorentz gauge of 11.01.
§12.01
ELECTROMAGNETIC RADIATION
450
If f(t) is cos cot, then 11.01 (1) gives the phasor fields
_n
io dzo(1 _a Az _ \r2 Ora cos Oe—or car at _L 1 = nic = aA, sin 0e— jor 3 ) Ora r at r ae aAr = 1) sin 0e-20" B _ a(rA0) 4n- \ r ' r ar r ae Er
=
(6) (7) (8)
Replacement of io dzoby jwq(1)in (6), (7), and (8) gives the fields in terms of the moment q( i) cos wt based on the charges. Note that f3 = w(i.te)i =
= were = wmn 1
=
(9
)
Far from the dipole the 7-1terms alone survive and form the radiation field, while near it the r-3terms dominate and produce the quasi-static field. The r-2terms give the induction field. In the Lorentz gauge the Hertz vector gives more compact formulas. From 11.01 (8), using jovo) for io dzo and phasor notation, (4) becomes (1)
Z = Ze"'t =
ej(wt—sr) 41-Er
(10)
The linear quadrupole, le in (1), is found from pi') as in the static case by differentiation of ZP) with respect to z and division by 2!. In the wave equation a/ az and V2 may be interchanged only when applied to scalars or rectangular components of vectors. The formulas needed for putting Z into spherical components are, for the unit vectors, j
=
i = r1sin 0 cos ck ± 0 cos 0 cos 0 — 43 sin 0 r1sin 0 sin 0 + 0 cos 0 sin 0 + 4) cos 0, k = r1cos 0
—
(11) 0 sin 0 (12)
The result in phasor form with cos 0 written for z/r is 8wer 2
= q2) (/3r — j)(ri cos 0 — 0 sin 0) cos 0e—or
(13)
Similar procedures for Zxx and i„ give 8,ffer2isz = iqP2(f3r — j) sin 0 cos ¢e— jor 8rer2E„ = jqa)(13r — j) sin 0 sin 0e—or
(14) (15)
For a square quadrupole centered at the origin in the xy-plane with sides parallel to the x- and y-axes, Z is found by superposition of z aZ11)/ay and 28Z1,1)/ax where (10) is Z 1). This gives 167rer2112y) = qg) (Or — j)(i sin 0
j cos ck) sin 0e —or
(16)
§12.01
451
SPHERICAL ELECTROMAGNETIC WAVES
Likewise for square quadrupoles in the yz- and zx-planes, we get k sin 0 sin (p)e-of k sin 0 cos 0) e—or
167E742 = ez)((r - j)(j cos 1671-er2ZT = qg) (Or - j)(i cos 0
(17) (18)
From the above tensor components the most general electric quadrupole field can be built up. From 11.01 (1) E and B satisfy the same propagation equation, so for a magnetic dipole field, from 11.03 (8) as r 00, vB should replace the E of the electric dipole field of (6), (7), and (8), and from 11.03 (11) to propagate outward, -E should replace the B. Thus writing jcoM for iodzo gives Br
= I.LM (ji3
1 ± 775) cos 0e—or
241.
B9 = 47rI — ( T d r + -j4= wi/ MC - 47r r
7,1
(19)
sin 0e—i15r
(20)
1-
(21)
j ) sin 0e-jor r2
Note that as r -> 0 the r-3terms give exactly the curl of 7.10 (2). The mean energy radiated from a multipole source is found by integration of the mean Poynting vector, 11.17 (3), over the sphere of large radius r. This may be put in terms of the Hertz vector by noting that in its curl the only derivative which does not disturb the r-1factor is r-1 a(re-or)/ar, and when r is very large this simply replaces co' by - jOe-o r. It acts only on the 0- and 0-components of Z, so that if Zt„ denotes the tangential component of Z as r 00, 11.01 (12) gives, since jo.q.Le • -ji3 is w 2v
3,
kw = co2v-3(-02,,, + 420.,)
(22)
Now from 11.03 (8), ri x Et = vBt, so that Et. = co2v-2(020. + 4,2o.) H = 1,-4 x a= ,_i0,4v_5,1(z9,028. + zoca,w)
(23) (24)
For the electric dipole, from (10) and (12), 28,0 -
II = rico4 in 0[q(I)p = 2fl
-jq(1) sin 0e-ifir, 47rEr
P = _MT dS -
co4[q(1)] 2
4,3v[q (1)12
127rev3 =
30.4
1671-2E2i.w5r2 0)2 (io dzo) 2 r(io dzo) 2 r-----127rev3 30X2
(25) (26)
For the magnetic dipole, from (19), (20), and (21), the radiated power is
0,4m2 _ 47.3m2 w4/2A 2
P = f E13 ° dS = 127rev5 2A
3euX4
127rev5 —
473/2A 2 3euX4
(27)
ELECTROMAGNETIC RADIATION
452
§12.02
In 7.10 (2) M was defined as IA for a current / in a loop of area A. For a linear electric quadrupole, from (13) and (24), w 6r,(212
R 2w 4r n(2)12 f
P Jr dS =
2,5,[q (2)12
sin3 cos2 d0 — " " 480rEv5 12872E211v' 0
(28)
1500
For a square quadrupole, from (16), (11), and (12), then (24), = Oa° cos 0 sin 20 + 4) cos 2¢] sin Oe-or
f 2'S
P = ft dS — wqq1.2'12
25672 €0 o o
(29)
cos2 20) sin30 d0 d¢
(cost 0 sins 2
= w 6r,,(2)12 L'ixy J
3207-05
,52,ral2y)12 iv
500
(30)
12.02. Retarded Potentials.—A useful method of finding solutions of Maxwell's equations given in 11.00 is by means of retarded potentials. From 11.01 (6) and (7), the vector and scalar potentials propagate in a homogeneous isotropic dielectric with a velocity (1.46)-1. The object is to set up solutions of these equations analogous to 7.02 (5) and 3.09 (1) that give the potentials A and NI,at the point P at the time t. Both the electric current i and the electric charge density p are assumed fixed in position but with magnitudes varying with time. The treatment of moving isolated charges appears in Chap. XIV. The contribution of an element dv at the point xiyizi, having traveled a distance r from dv to P with a velocity (.1E)--1, must have left dv at the time t — (i./.6)1r and thus must represent the conditions at dv at that time. Summing the effects of all elements gives the P potential, from 7.02 (5) and 3.09 (1), at time t, A(x,y,z,t) =
‘1(x,y,z,t) =
iff
i[xi,
yi, zi, t
— (i.‘€)1r]
4.irj 4re
f fP
[xi yi
zi t r
"
(1)
dx i dyi dzi
— (1./e)i r]
(2)
dxidyi dzi
The fields derive from these retarded potentials as in 11.01. Note that the integrands for Az, A,,, Az, and ‘If in (1) and (2) have exactly the form f[t — (1.1E)1]/r of the propagation equation solutions found in 11.03 (1). In connection with 12.01 (4) and (5), it was shown that if the charges in the element dv in (2) are fed by the current in dv in (1), then 11.01 (8) is satisfied. Thus A and \If for the element dv satisfy 11.01 (8), and A is i.LE 8Z/dt. From this one writes down by inspection an expression for the Hertz vector that combines (1) and (2). Thus 1rcifffi[xi, Z(x,y,z,t) = iT
yi,zi, t r
(E)irl dxi dyi dzi
(µ6)1r]
dt
(3)
§12.03
453
RADIATION FROM LINEAR ANTENNA
The usefulness of (1), (2), and (3) depends on the accuracy of the estimate of the current and charge distribution in the source. If the source is a very thin perfectly conducting wire lying on a curve s, then no component of the electric field given by 11.01 (3) can lie along s and the equation of continuity must be satisfied. Thus
E$ =
at as
aA at
,= o
ap — at
— as
0
(4)
But, since the wire is infinitely thin, we may go so close to it that the wave length and the radius of curvature of s is infinite compared with our distance r. Then, if radiation is neglected, A and ir are given by the formulas for a long straight wire •If = — 27 ln r
A=
27r
In r
(5)
Insert these values into (4), differentiate one with respect to t and the other with respect to s, and eliminate A, or NI,from the result. Thus 321 as2
321 wat2
32 p as2
320 Aeat2
(6)
If the current is harmonic in time, the solution of (6) is
/(s) = I iej('t —es)
Leic.t+so
P(s) = biel((u —m
15 . 20(''+56)
(7)
The distribution of current and charge are therefore sinusoidal. This is the current form nearly always used in antenna theory but, as will be evident in the next article, the assumption of no radiation is not the only approximation involved. 12.03. Radiation from Linear Antenna. Retarded potentials may be used to calculate the radiation field surrounding a linear antenna. We use the cylindrical coordinate system p, cP, z and write z' for the coordinate of a point on the axis of which the antenna occupies that portion between z', and 4. Its diameter to length ratio is so small that the current forms a sinusoidal standing wave. The origin lies inside or outside the antenna at a point where there is, or would be, a current node so that —
I = I o sin Oz' cos wt
(1)
We can use either the potentials or the Hertz vector to get the fields. The physical nature of the terminal conditions is somewhat clearer in terms of the potentials. Whether or not the current is zero at the end depends on the terminal connection. The linear charge density is, by 11.00 (5), (2) a- = — (11E)1I 0 cos $z' sin cot
454
ELECTROMAGNETIC RADIATION
§12.03
Note that a is everywhere finite so the solution applies only when the lumped terminal charges do not radiate. It is a very simple matter to add to the solutions, when necessary, the additional terms arising from such sources. From 12.02 (1) and (2), the retarded potentials are Az =
2'sin 0z' cos (wt — 3r) dz' A/0 z
(3)
jzi'
= 0/0 f z2 _ cos (3z' sin (wt — 3r) dz'
(4)
4s-cot zi,
u
au au — = --, az'
_sin sin (wt— Or) OIL =
at
To get Ez from 11.01 (3), we substitute
p2.
where r2 = (z' — z)2
04410 fz2' 40
z
av
v = cos Oz',
UV,
wijo f 22f v du
az
,,
= —0 sin Oz'
470
z,,
Thus from 11.01 (3), we have for Ez Ez = —
coll.! 0
uv
"' Zi
cos 04 sin (cot — Or') r1 47r0
W it
cos 04 sin (cot — On)] r2
(5)
To find E, and I I 4, without the integration of (3) and (4), we substitute u = r z' — z and v = r — z' z so that r du = u dz' and r dv = —v dz' in the integrals of (3) and (4). With the aid of Dw 401.01 or Pc 591, these may then be written, using the upper sign for (3) and the lower for (4), Su'
(wt— 0z — /3u) du 2u
f v2sin (cot ± Oz — Ov)dv 2v
(6)
Note that p appears only in the limits of the integrals and that au1,2 =
497)1,2
P
op
r1,2
Op
(7)
so that differentiation of the integrals with respect to p gives p[ 2
sin (wt — — Oz') r(r z' — z)
sin (wt — Or + 13z1"' r(r — z' z)
(8)
Putting over a common denominator, substituting p2 for r2— (z' — z)2, and combining the sines of the sum and difference of cot — Or and 13z' give coil/ cos (wt — 0r2) sin 04 — cos (wt —girl) sin 134 470p 4 z 4 z sin (wt — 13ri) cos 04] (9) sin (wt — 0r2) cos 3z2 + r1 r2
Ep = = OP
RADIATION FROM LINEAR ANTENNA
§12.03
455
aAz 0[ — sin (cot — Or 2) cos 04 — sin (cot — Ori) cos 04 = — ap = 47rp
4
z
7.2
cos (cot — /3r2) sin )34
z
z
cos (cot — Ori) sin Ozd (10)
ri
Equations (5), (9), and (10) apply to antennas whose terminal loads are not zero, provided the loads charged by them are prevented from radiating by earthed shields like the coaxial termination shown in Fig. 12.03a. The cases shown in Fig. 12.03b and c can be treated by adding to the scalar potential of (4) the terminal charge retarded potential. If the load is replaced by an antenna section extending to the nearest current node, the charge on this section equals that on the load since the same
(a)
(c)
(6) FIG. 12.03.
charging current flows into it. Thus the integrals of o from the nearest node to zi and from 4 to the nearest node give, respectively, the charge Qi on zi and Q2 on z2 so that, from (2), Qi = — co-1/0 sin 04 sin cot,
Q2 = co-1/0 sin 04 sin cot
If the terminal loads are small enough to be treated as point charges, the corresponding retarded scalar potential to be added to (4) is
I
0 sin 04 sin (cot — Or2) ANY = / 47r-weL r2
sin 0.4 sin(cot — 13r1)] r1
(11)
The contributions to Ez and E, are, respectively, A.E'z—
402
— sin 04
a [sin az
(0.4—
07.2)]
r2 ± sin 0.z i
a [sin (cot —
az
a sin (cot — 0r2)]
AE, — ``)/L1°{— sin Oz'2ap 4702
r2 ± sin Oz'la
01
r1
[sin (cot — r1
ap
(12)
}
B.
Ori)
(13)
456
ELECTROMAGNETIC RADIATION
§12.04
The magnetic field of (10) is unchanged. It is interesting to note that Bechmann's derivation of the linear oscillator field by use of the Hertz vector, which is quoted by Stratton, gives this load type at both ends. If the antenna terminals are current nodes, then from (1), Oz',. and Oz', must be mor and mor, respectively, where m1 and m2 are positive or negative integers so that the cosine become (-1)m,and (-1)T42. The fields given by (5), (9), and (10) then take the simplified form (-1)m2 sin (cot — ,3r2 )
Ez _
wad-(-1)"4 sin (wt — (3r1) ri — 470 L z)sin (wt— On)
E, —
BQ =
r2
(4 —
-
(-1)miri
47rOpl_ 1.410
z) sin (wt— 13r2) ]
(-1)m2r 2
1Y41 sin (wt — On) — (-1)m2 sin (wt — 13r 2)]
(14) (15) (16)
Here m2 — mi = n is the number of current loops in the antenna. Suppose an antenna of length 21 has end a at z' = — 1, end b at z' = 1, and is driven from the center at an arbitrary frequency so that there may or may not be a loop there, but there are nodes at the ends. Find the fields for each half separately by (5), (9), and (10) and superimpose the results. The substitutions for ends a and b are, respectively,
zi = 0, 4 = 1, ri = ra, r2 = r, z — zi = Ti cos 0“, z — z2 = r 2 cos 0 zi = 0, z2 = —1, r 1= rb, r2 = r, z — zi = r1 cos Ob, Z = r2 cos 0 The angles 0, 0z, and Ob are measured from the positive z-axis. Substitution in (5), (9), and (10) gives, writing co = cO, Ez = cido r sin (wt — (3ra)
4/r F0 =
CA I iri:[2
4
sin
(wt
ra
—
Orb)
2
rb
cos131sin(wt
—
Or)]
(17)
cos 0 cos 01 sin (wt — (3r) — cos Oa sin (wt — Or.)
Bo = ,
[2 cos 01 sin (wt — (3r) — sill (wt — ziirp
— cos Ob sin (wt — (3rb)]
(18)
ra) — sin (wt — Orb)]
(19)
(
Note that /0 is the maximum current amplitude and, if it lies on the antenna, may occur at any point except at the ends. 12.04. Distant Radiation from Linear Antenna. We shall now obtain the radiation pattern at a great distance from a linear antenna of length 21 center driven at an arbitrary frequency with nodes at the ends. In l cos 0 for ra, r — l cos 0 for rb, and r sin 0 for p, 12.03 (19), write r whence —
B, = ( AolE e —
o[eos 131
— cos
(01 cos 0)]
27i-r sin 0
sin (wt — (3r)
(1)
DISTANT RADIATION FROM LINEAR ANTENNA
§12.04
457
From 11.17 (3), the mean radiation intensity in the r-direction is II = liEei
(2)
= 4/1-1E-11Boi 2
If the antenna is driven at resonance with a loop at the center so that 201 = nlr, it is clear from these equations that, when II is plotted as the radius vector in polar coordinates as a function of 0, the resultant curve has maxima and minima corresponding in number to the standing wave loops and nodes. The dashed curves in Fig. 12.04 show such a plot for n = 3. The total power radiated is found by the integration of tI over a large sphere centered at the origin. We shall calculate only the
FIG. 12.04.
resonance case. When we substitute -inir for f31 and write the square of the numerator in (1) in terms of twice the angle by Pc 577 or Dw 404.22, we get /Ica f'1 -(-1)'cos (nor cos0) P= 0 , sin 0 d0 1 - cos' 0 or Jo -
(3)
Substitution of u for cos 0 and resolution into two integrals by partial fractions show that the integrals are equal because one changes into the other when -u replaces u as the variable of integration. Thus,
r +1. ± cos ram +1 1+cos min/ 1 ± cos nr u du du du + 1+u j _1 2(1 - u) ) 2(1 + u r+1
5-1
Let v = nif(1 + u) and (3) becomes P=
mcio2 r2-1 — cos v
Sr Jo
dv
AcI8 [C + In (2n7r) - Ci(2n7r)] 87r
(4)
458
ELECTROMAGNETIC RADIATION
§12.05
where Ci(2nir) is the cosine integral tabulated in Jahnke and Emde and C is 0.5772. Insertion of numerical values gives for the radiation resistance Rr =
P
2P
=
= 72.4 + 30 In n - 30 Ci(6.28n) (5)
When n = 1, this becomes 73.13 ohms. The radiation resistance when there is not a connected loop at the driving point is given in 12.07. The assumption of an infinitely thin antenna is only one of the defects in the foregoing treatment. To maintain the postulated standing wave, one must supply to each element the energy lost by ohmic heating and radiation. The former is greatest at the current maximum near the driving point, and the latter is greatest near the ends of the antenna as can be shown with little difficulty from 12.03 (14) and (16). Thus it is evident not only that it is impossible to maintain the assumed current from any single driving point, but also that, to supply the losses, a damped progressive wave must be present which has no nodes and radiates in quite a different fashion from the standing wave. A rough qualitative idea of the effect on the radiation pattern of the weakening of the current near the ends of the antenna is obtained, for example, by superimposing the fields of two antennas, one oscillating in the n = 1 mode and one in the n = 3 mode. Assuming equal strengths and frequencies, this gives the radiation from a three-loop oscillator when the inner loop has twice the amplitude of the outer ones. Using (1) with 131 = +wir and simplifying by Dw 403.23 give 0.401.E0 =
,u/0 cos (iir cos0)cos Or cos 0) . sin (cot - Or) irr sin 0
(6)
When substituted in (2), this gives the radiation pattern shown by the solid lines in Fig. 12.04. We note that, although the outer loops have the same amplitude as before, the outer lobes are actually cut down in magnitude, but the central lobe is quadrupled in magnitude. 12.05. Radiation from Progressive Waves.—Consider a wave of uniform amplitude that travels along a wire from z' = -41 to z' = +41 where it is absorbed without reflection, giving a current /0 cos (cot - Oz'). From 12.02 (1), the field at x, y, z is, when r = (p2 z 2)4 is large,
aA,
= (A )+E9 - -
ap
f +1a f cos [cot —
4r -Op t
+ z') E} dz ]
r'
where r' = [(z - z') 2 p2I1 r - z' cos 0 and p = r sin 0. If r' >> 1, we may neglect terms in r-2so that, using Dw 401.11 or Pc 597,
Bm -
ao sin 0 sin(cot -(3r) sin [041 - cos 61 )] 2irr 1 - cos 0
(1)
CONICAL TRANSMISSION LINES
§12.06
459
The mean power radiated then becomes, by integration of 12.04 (2),
P
C"sin2 sine[01(1— cos 0)] sin B do (1 — cos 0)2
zlirEljo
(2)
Substitution of u for (31(1 — cos 0) gives, using Dw 403.4 or Pc 576, P
=
f 21311 — COS u du -
47rEI[ j 0
1
r 2f1/
2,31j (1 -
cos u) du]
The first is given on page 3 of Jahnke and Emde in terms of Ci 201 so that the substitution of numerical values and of 271-/X for 3 give / 47/ in — — Ci
P = 30I o[
sin (47//X) 4 /X
(3)
12.06. Conical Transmission Lines.—The radiation fields of a linear antenna driven at the center are given by 12.03 (17) to (19). In 12.04 (5), we found the radiation resistance of such an antenna with a current loop at the driving point. No antenna reactance information is obtained by considering such infinitely thin wires because their inductance per unit length is infinite. If, to avoid this difficulty, we choose a radius not zero, then the infinite capacitance appearing across the infinitely narrow gap between input terminals again prevents reactance calculations. Schelkunoff eliminated both these difficulties by considering a biconical antenna composed of two cones whose apexes meet at the driving point. To simplify matters we should first discuss the biconical transmission line. The propagation equation in spherical polar coordinates for a wave of angular frequency co is, from 11.01, 3.05 (1), and 11.15 (3), 1 sin
a(sin
ao
a r2 aw a¢22 + ar ar
1 a2 W
NI
+ sin'
r 2.2W = 0 (1)
To break this into two equations, we equate the first and second pair of terms separately to zero and obtain 5.12 (6) and 12.01 (3) with P2 in place of —v2/w2. Thus from 5.12 (7) and 12.01 (4) for an expanding wave, W = r —q7(0, 4))e—'r cos (wt — Or)
(2)
where a is the attenuation constant, O the wave number, and U(0, 0) + jV(0,
= F(eio tan 20)
(3)
The vector potential for a transverse electromagnetic wave is A = V x rW r--— 71x V2Ve—ar cos (0.4 — Or) = V2 Ue—ar cos (Cot — Or) (4) where V2 has 0- and 0-components only. This is identical in form with 11.14 (5) and shows that, when an alternating potential is applied between the apexes of two or more perfectly conducting cones whose surfaces are
ELECTROMAGNETIC RADIATION
460
§12.06
generated by radius vectors, a spherical wave is transmitted that is just equivalent to a plane wave on a cylindrical transmission line for which x = cos 4) tan 20,
y = sin 4,tan IO,
z
(5)
Let us now consider the special case of two circular cones of half angle X1 and x2 whose axes intersect at an angle 21//, as shown in Fig. 12.06. Clearly, if the trace of these cones on a sphere of radius r is projected
/
/
/
..,
/ ..•••'"*. ••• / ..,..••• / I ..,--"""
14, - _ ..,_ I--- , , " I
-..
\\ \
\
\ \ ,„,_, \
.47/
...."1
\ 1
-
-'---,r-.------1'-{ -.... S . I'
I
\■
.... ....-
/
2R2
/
FIG. 12.06.
stereographically on the tangent plane, as in 6.20, the equivalent transmission line is the two cylinders of radii R1 and R2 at an axial distance D whose sections are shown in Fig. 4.13b. From Fig. 12.06,
D+
R1, 2 =
2r tan -i(//
xi,2),
4./) —
R1, 2 =
2r tan 4Ni — xi,2)
Take the ratio of these equations, apply Pc 602, and solve for 4D/R1,2.
D
_ sin 1,/, sin X1,2
2R1,2
(6)
The capacitance per unit length of the equivalent cylindrical line is, from 4.14 (2) or 4.13 (5), according to whether x =xi = X2 or xi xi,
e CY
1 sin 1,G 1 4 sine >L — sin2 — sin2 = — cosh-1 or 2.n.cosh-1 (+ or sin x 2 sin xi sin x2
(7)
The negative sign applies if one cone encloses the other, otherwise the positive sign is used. By 11.15 (10), the characteristic impedance is 2k = e
C -y
jwilE y= (a ± ji3)C jcoe
jcoli
(8)
§12.07
THE BICONICAL ANTENNA
461
where 12 is the permeability, E the capacitivity, and y the conductivity of the medium surrounding the perfectly conducting cones. When I/. = lr, simpler formulas are given by taking 0 = 0 as the axis of one of the cones so that U is independent of 4). For this we choose F(u) to be C In u in (3) so that from 5.212 (4) U voln 0 1 n ttaann : x
v 0, W/cos 0) vo cos x)
(9)
where the potential between the cones and the capacitance per unit length are, respectively, V = 2Voe— 'r cos (wt — Or)
C = rc(ln cot +x)-1
(10)
12.07. The Biconical Antenna. When both biconical transmission line cones terminate, they form a biconical antenna. If open at the ends or if closed by spherical caps, they fit into the spherical polar coordinate system to be discussed in 12.11, and the field calculation is a boundary value problem in this system. The method works best for cones whose half angles are nearly zero or 90°. In the small angle case, numerical work is simplified by using a different method. Take perfectly conducting coaxial cones of equal length whose half angle x is so small that by 12.02 (4) to (7) the current distribution is —
i = /0 cos cut sin (0/ — 13r)
(1)
The fields are then given by 12.03 (17) to (19) and the radiated power by an integration of Poynting's vector over the conical surfaces. Let xa, x, and xb be the angles subtended at z = —1, z = 0, and z = 1, respectively, by a radius p of the cone ending on its surface at a distance r from its apex. By Poynting's vector 11.02 (3) and the magnetomotance law 7.01 (2), the total instantaneous power radiation from both cones is P = 2.10.Erli-021rp dr = 2.1 0 (Ez i cos x
E,, sin x) dr
(2)
Inserting 12.03 (17) for E„ 12.03 (18) for E„, and (1) for i gives
P=
CAP cos
2ir
wtfiRcos x cos
x°)S. (COS X
rb
COS Xb
)8b1 sin 01 — )3r) dr
where 8„ and Sb have been written for sin (cot — )3r.) and sin (wt — Orb), respectively. The terms containing cos /31 cancel out because p = r sin x. r and 1 — r replace ra and The half angle x is chosen so small that 1 rb and cos xa, cos x, and cos xb equal one. If 8„ and Sb are split up so that sin wt and cos cot factor out, then with the aid of Dw 401.05 to 401.07
462
ELECTROMAGNETIC RADIATION
§12.07
or Pc 595 to 597, we obtain for 27rP(c/2/2)-1 1 sin 20.4f 1 1—) sin (L — R)]dr (sin L — sin R) ( 1 1 —r r 4 6,[ (1 r l — 1 cos 2cot f [1 — cos (L — R)]} dr (cos R cos L) r(1 — r) 2 6 { r(1 r) where R is written for 20r and L for 201. The quantities S and 5' are to be set equal to zero except where so doing makes a term infinite. In Jahnke and Emde we find the cosine- and sine-integral formulas asin Ax dx = Si(Aa), jo x
— cxos Ax d
x = C + In (A a) — Ci(A a)
where C = 0.5772. With these formulas, the power expression integrates easily and In 5 and In 5' cancel out so that all terms are finite. Thus
P
cµ/8 sin 2cot
.
{2S1(20/) — Si(40l) cos (200
In (131)] sin (MI [Ci(400 C c/2I cos2cot [C Ci(401) — 2Ci(2(3/) {2C + 2 In (200 — 2Ci(201) 47r [Si(401) — 2Si(2(3/)] sin (2001 (3) + In (131)] cos (200 —
—
Comparison with 11.15 (18), where 62 is replaced by /822, shows that the coefficient of ig sin 2wt is 22, sin 1,G or 2X,, where X, is the radiation reactance and that of Ig cos2 wt is 2, cos 1,/, or R, the radiation resistance. If there is a current loop at the driving point so that 23l = n7r where n is odd, the value of R, is identical with that of 12.04 (5). With a node at the driving point, n is even so the result is different because in (1) and (2) we assumed the current similarly directed in the two halves, whereas in 12.04 (4) they are taken in opposite directions. Although we have now established the power loss from the antenna, we still do not know where to put it in the line to give the correct input impedance. Examination of (1) shows that Ho and hence the power loss are zero at the current nodes so it cannot be put at the ends. A simple way to discover its location is to consider the case in which the perfectly conducting antenna is tuned so that the only power loss is by radiation. There is then a current loop at the driving point so that 131 = (2n + 1)r and the input current amplitude is /0. Insertion of this value in 11.15 (13) after setting the attenuation constant equal to zero gives for the input impedance and the mean power expended
2, =
P = 422'
(real part)
(4)
ANTENNA ARRAYS
§12.08
463
where 2t is the terminal impedance. But (3) gives 12,n for the mean power so for a tuned antenna Z,Z, = Zk. We try this value in 11.15 (13) for the input impedance of an untuned antenna and, to justify it, show that for thin antennas the resistive power loss agrees with (3). Thus 2, =
2k cos 0/ ± j2, sin SI Z, cos 0/ + Z k sin (3/
(5)
Now let Zk become very large, which means a very thin cone, so that the input current becomes /0 sin 0/ and the power input is, from (4),
Pi = 12,g sin' 0/ = 1(2, — ij2k sin 200/g (6) This gives a resistive power loss of 2/?,n as it should. We note that (5) is the same expression that would have been obtained from 11.15 (13) 350 300 250 v3200 E o 150
100 50 0
it
cm
IITEMB1111 III MILIMIEWAI■ 11111M111 EFIL MOM■ NII / NI !A Eli 11011111 2 3
4
5
2rrl/a
6
1
8
9
10
FIG. 12.07.—Real (solid) and imaginary (dotted) parts of ik2 /Z plotted against 271-1/X.
if we had written 0/ — 1r for pl and 2, for 2L. This is equivalent to cutting 4Xoff the line and using Z, as the terminal impedance. Thus the effective position of the radiation impedance is one-quarter wave length from the end of the line. The inverse of the radiation impedance given by Schelkunoff in Proc. I.R.E., September, 1941, pages 493-521 appears in Fig. 12.07. Clearly from (5) the resonance wave length at which Z. is real is affected by the radiation reactance, but it is still near 20/ = na if Zk »X,. 12.08. Antenna Arrays.—The radiation pattern of a linear antenna is symmetrical about its axis. To concentrate the radiation in a single direction, a "directional array" is needed in which several radiators of length 2/, usually identical in type and parallel to each other, are set in some regular pattern. These may differ in amplitude and phase of
464
ELECTROMAGNETIC RADIATION
§12.08
excitation. Let us consider radiators parallel to the z-axis and lying in the positive octant of a rectangular lattice structure, the spacing between and z-direction, respectively. adjacent ones being a, b, and c in the x-, With integral u, v, and w the distance of radiator uvw from the origin is
ruvw = iva
.ivb
(1)
kwc
Let the radius vector from the origin 0 to the very distant field point P be rr1so that the difference in travel distance of a signal from 0 to P and from radiator uvw to P is r1 • ru„.. Assume the dimensions of the array small compared with r so that all radiators can be considered as equidistant from P for field amplitude calculations. If oscillator uvw lags behind oscillator 000 by a phase angle iku,no and has a current ./uvw, then from 12.04 (1) its contribution to Bo at P is the real part of ,,[cos j31— cos (01 cos 0)] . 27rr sin 0
u
Writing
Fu. for the first factor gives for the whole array 14 0 = (11f)1E0 =
(
U
v
Or
(2)
W
For identical radiators F is constant, and this formula simplifies if, when going in the positive x-, y-, or z-direction, each radiator shows a constant phase lag n, or respectively, behind the preceding one. The triple summations in (2) then breaks down into a triple product of summations.
FZZ
ei[u (Orr ja– E) +v (Pr rjb
-Fw woe, _0]
u v w FZeiu(firi.ja-0 Zeiv(Orrjb—n)Zeiw(Prpitc—r)
(3)
Each factor is a geometrical progression which may be summed by Dw 26 and gives, when mxis the number of oscillators in the x-direction, &my n=o
sin = 1 — ejm-# 'Pe—A. (nix-1)# .— sin Ilk 1 — 04'
(4 )
From 11.17 (4) the mean Poynting vector is 1(A3E)--ito • Bo so that F2 sin2[1m.(1'r1 is — )] sin' [Inty(Rri •ib n)] sin2 [ilrl(Ori • kC 2 mitt S1112 [El3r 1 • is E)] sin2[1(Or1 • ib — 0)] sin2 [1(07.1. kc — —
(5)
465
ANTENNA ARRAYS
§12.08
As already stated, an array is used to concentrate radiation in special directions. The directivity or gain G is defined as the ratio of the maximum intensity (Im to the average intensity .130on a large sphere concentric with the array. In decibels we write it Gd. Thus G
=
c13m
G d = 10 logio
30
cI3m
(6)
—
The gain function G(0, 95) in any direction is the ratio of 4)(0, 0) to Thus for a half wave antenna, from 12.04 (2) and (4), the ratio of maxiC — Ci(27)J-1which gives a mum to average intensity is 4[1n numerical value of 1.64 for G or a gain of 2.15 decibels in the equatorial plane. The transmitting pattern is the surface. r=
G(0, 4)
)
c,(0,
)
(7)
.4)3,1
Gm
Now consider the special case of m identical, in phase, n-loop antennas, lying a half wave length apart along the x-axis. Take my = m2 = 1,
FIG. 12.08.
= 0, i3a = 113X = 7, and 3l = inir, so which becomes
- img 8,2r2,1
On • is
is r cos 4) sin 0 in (5)
cos (lnir cos 0)121-sin (Inor cos .45 sin 0) sin sin (fr cos 4) sin 0)
_IL
12
(8)
This is called a broadside array because the second factor maximum is at right angles to the plane of the array when 4) = fir. Figure 12.08 shows the relative values of II in the plane 0 = 0 when subsidiary maxima are omitted. Near this maximum the second factor sines are small so it becomes m2. We cannot assume from this that, for the same total power input, the radiation in this direction is m times that of a single oscillator because the oscillators interact. To get the actual gain, we must calculatect,o. We note from (4) that the last factor in (8) may be written, if a is 7 cos cb sin 0, —
1
sin Imaeii(m-1). 2 sin la
2
m-1
= + 2 (m — p) cos pa p =0
P=1
466
§12.09
ELECTROMAGNETIC RADIATION
Expansion of the cosine by Pc 773 or Dw 415.02, integration over 4) by Pc 483 or Dw 854.1, and combination of the terms independent of B give rn-1
2M71"
('2r (m — p)f cos (pr. cos 4) sin 0) d4)
p =1 M—1
a.
= 4m,
021
(m — P)CiPir sin 6
228-1(s!) 2 p =1 s =1
To get the power radiated, multiply this by the remaining factors in (8) and by r2 sin 0 de and integrate from 0 = 0 to 0 = 7r. The first term was integrated in 14.05. The other integrals have the form cos2 (4717r cos 0) sines-10 de = j [1 + cos (nir cos 0)] sin28-10 de
We may integrate this by Dw 854.1 or Pc 483 and 5.302 (5) which gives
—
[(s
1) U2228-2 (2s — 1) !
1) ! 2 8—i 221-8-1 n 3-1\mr1
(s —
The s-summation of the first term by Dw 442.11 or Pc 347 gives a result expressible as a cosine integral by page 3 of Jahnke and Emde. Thus pal
2 s =1
(-1)8(137)28 = —2f 2s(2s)! c■
— cos xdx = —2C — 2 In (pr) x
2Ci(pr)
Collecting terms gives for the total power radiated p
A/ C{m 2C + m 2 In (2nr) — 8r —
1
— p =1
rr
m2Ci(2nir) co
Ci(pr) — C — In (pr.) ' s =1
(-1)878±1p28 r 4ss!(2n)8-1 ''s-1(nir) (9)
From (6) and (7) the directivity in decibels is 10 log [11.4cI 2m2/ (irP)]. For n = 1, m = 2, the ratio 43mj4 0 is 3.81 which is more than twice that of a single half wave antenna. The directivity is 5.81 db. 12.09. Earth Effects.—Dielectrics or conductors near an antenna react on it or distort its field or both. The commonest such object is the earth's surface. To a nearly spherical wave one may apply the laws of reflection and refraction at a conducting boundary derived in 11.13. If, on the other hand, the antenna is so near the surface that the angle of incidence is a complicated function, this becomes laborious. Often
§12.11
467
SOLUTIONS OF THE WAVE EQUATION
the surface may be taken as flat and perfectly conducting so that the fields above it are unchanged if it is replaced by a second identical or "image" antenna located and oriented so as to make the resultant electric field normal to the former earth's surface. If all the original antennas were either normal to or parallel to the earth's surface, then evidently the resultant fields may be calculated by the formulas of the last article. 12.10. Uniqueness of Solution. Before deriving the wave equation solutions used in boundary value problems, one should find what data are needed for a unique solution. Consider a region, bounded internally by the surfaces 51to 5„ and externally by So, wherein and y are functions of position but not of field strength, and which contains no sources. Let E, B and E', B' be two solutions of Maxwell's equations which are identical throughout the region when t = 0. Poynting's theorem 11.02, and Ohm's law give, writing AE for E E', AB for B B', etc., —
E,
—
f,
(Ai) 2
{
7
a[E(AE) 2 2
at
—
1} dv =
(AB) 2 2/.4
is,AEx AB
n dSi
o
In order that the surface integral vanish, it is necessary that AE x AB • n = AB x n • AE = n x AE • AB = 0 Thus, if n x AB or n x AE is zero when I > 0, the surface integrals are zero. The energy term in brackets is zero or positive and was zero at t = 0 so, if it changes at all, it must become positive. But the first term is zero or positive, and the whole integrand is zero so that AE and AB are zero. Thus the E, B and E', B' solutions are identical and are determined by the initial values of the fields in the region together with the tangential components of either E or B over its surface when t > 0. In practice we are usually concerned with steady-state solutions of the problem, in which case the values at any time determine all previous values. 12.11. Solutions of the Wave Equation in Spherical Coordinates. In an isotropic insulating medium all waves are spherical at distances from the source great compared with its dimensions. In radiation problems, therefore, the most useful solutions of the propagation equation are in polar coordinates and have the form of a sum of products of orthogonal functions with coefficients which can be determined from given boundary conditions. We saw in 11.01 that in such cases the entire radiation field is derivable from a vector potential with zero divergence. Expressions for such a potential can be obtained from two solutions of the scalar propagation equation as indicated in 10.01. Thus from 10.01 (2), if Wte and Wem are solutions of the last two of the equations, V2A = _ mtc02A, v2w,, = v2wtm =— µew 2Wtm (1) —
468
ELECTROMAGNETIC RADIATION
§12.12
then a solution of the first equation is obtained from the formula r x VWt”.)
A= v x (r1,17t
(2)
Note that the solution derived from Wte gives a vector potential, and hence an electric field, normal to r. This is called a transverse electric wave as the subscript suggests. From 10.01 (4), the magnetic induction is r x vWte) B = —v x (Acce2rW,„, (3) Thus the magnetic field derived from W,„, is normal to r, and the wave is called transverse magnetic. We solve (1) as in 10.05 except that for more generality the factor 4)(0) is added. We therefore write ITV = r-IE(r)0(0),I3(0)
(4)
Substitution of (4) in (1) gives the differential equation of 10.05 (9) for f?, that of 5.14 (2) for 0, and d24,/d 02 = _m 2c13 for (D. The solution for W then becomes, writing ,3 for w(ME)1and u for cos 0, W = [APB(u)
BQZ(u)][0j„(13r)
bk.(j3r)] cos (met. + 3,„) (5)
This spherical Bessel function notation is that of 5.31 and 5.37 where j, ( 3r) = (2(r)-IJ„±1(Or),
k„(jOr) = (1j7r0r)-1k„±1 (jOr)
(6)
The reason for using these forms is that the first, combined with em, represents a standing wave and the second, from 5.37 (9), an expanding or contracting traveling wave according as v is positive or negative. 12.12. Polynomial Expansion for a Plane Wave.—A plane wave expansion in the form of 12.11 (5) is often needed to fit spherical boundary conditions. For simplicity consider a wave parallel to the 0 = 0 axis. It can then be referred to any other axis by 5.24. We saw in 11.15 (4) and (5) that z = r cos 0 enters plane wave formulas only in the exponent and multiplied by P. Thus we need the coefficientof P,,(cos 0) in the Legendre polynomial series expansion of err cos O. From 5.156 (3), +1
an =
2n + 1(pr)ni 2. +in ! • , 1
errna
u2\n du
(1)
Expansion of the exponential by Pc 759 or Dw 550 gives 2n + 1
an=
"K1 (fr)8 (P-r)"2 j s.
s =0
r 1
u(1 - u2)„ du
(2)
The integral vanishes when s is odd, so write 2m for s and take twice the value of the integral from 0 to 1. For the (2m)! in the denominator we may write 22mm!ir-IF(m + 4-) by Dw 850.7 and 855.1 and obtain with the
§12.13
RADIATION FROM UNIFORM CURRENT LOOP
469
aid of 5.32 (2), after writing 1'(n + 1) for n!, CO
a"'
r(nt + i)F(n + 1)
(Fr)n+,-
(2n + 1)71-1
2"-T(n + 1) L122"'rn,!r(m in=o
(m n
(3)
CO
.= (2n + 1) / -
irt!r(m
(2n + 1)G, ) in±.}(fr)
n
m=O
—
When there is no attenuation, we replace is by j0 and obtain by 5.32 (2) r P n+I(Or) = jn(2fl an = (2n + 1)jn6
1):7„03r)
(4)
The expansion is, therefore, \\ 09' ens 9 = Zjn(2n
1)jn(3r)Pn(cos 0)
(5)
n=o
For e—jo' replace /3 by —13 or cos 0 by cos (ir — 0) which is equivalent, by 5.293 (3) or 5.157, to inserting the factor (-1)" on the right. Thus
e--Jos == e--,fic cos o =
( 3) (2n + 1)jn(3r)P„(cos 0)
(6)
n=0
12.13. Radiation from Uniform Current Loop. Magnetic Dipole.— The orthogonal solutions of 12.11 usually occur in boundary value problems but may also be preferable for sources with given current distributions. For example, the radiation from a uniform current loop of radius a found, when X >> a and r >> a, by retarded potentials and given in 12.01 (19), (20), and (21) is more difficult by that method if r < a. From symmetry this source gives a transverse electric field with no free charge and hence, from 12.02 (2), no scalar potential and a vector potential with only a ct,-component. Therefore when r > a we have, from 12.11 (5), Wee =
ZA0 P0(cos 0)j.((3a)k.(jor)
(1)
n=0
If r < a, interchange r and a. By 12.11 (2), the vector potentials are r>a
r,
(6)
Therefore those parts We'„, and Ceof W,,„ and W,e that represent the plane incident wave are found by multiplying the nth term of (4) and (5) by —r[n(n + 1)]-1. Those parts that represent the diffracted wave must be diverging waves and so contain kn(jfir). Thus we have
= E cos oN 2n + 1
[(
(03 .Jn(n + 1)
j)nj.(0r)
Ankn(j,3r)]Pl(cos
(7)
n=1
E sin ON 2n + 1 . . bnicn(jOr)]P!,(cos 0) . Jit(n ± 1) [ ( 3)ni n(00 + n=1
(8)
§12.15 OSCILLATIONS OF DIELECTRIC OR CONDUCTING SPHERE
473
Inside the sphere, fields must be finite at the origin so that E cos ON 2n ± 1 . . ( 3)'2=1„,,n(O'r)P4(cos 0) 6.0 G,J n(n 1) n =1
E
W ei
(/)
3u)
2n ± 1
j„(f3/r)P;i(cos 0)
1)`
n =1
From (6) and (2), equating normal displacements gives e,An,j,t(tVa) = e[j„(0a)
jnilnicn(j0a)]
From (6) and (2), equating normal magnetic inductions gives —bnon(13/a) — in(0a)
jntinkn(j/3a)
From (6) and 12.11 (2), the tangential components of A are
1 a2(rtivt,.)] ae ar
A, = i f 1 a-rk, sin
e ac5
r
1 a2(r ar ae + r sin 0
Equating tangential components of E or A involves only the r-derivatives of Wtin because the Wte terms are already equal by (12). The same relation is obtained from the 0 or 0-component. -
A a[ajn(O'a)]
as
_ a[ain(0a)] as
.„A- a[arcn(i0a)] 3 as
(14)
By comparison of 12.11 (2) and (3), the contribution of Wt„, to A, is identical with that of WIe to B,. Equating either 0 or 4) components of 1.4,-'B at r = a involves only the r-derivatives of W,e because from (11), (13), and 12.11 (3) the Tkt„, terms are already equal. Thus -
1 a[aj.(0/a)] — B„, as -
lia[ajn(i3a)] p as
-
• 3nBn a[ak n(i0a)1} aa
(15)
We now equate values of A,, from (11) and (14) and solve for An. Use of 5.37 (13), 5.31 (11), and 5.31 (12) gives (1 — jN)--' for In where N is Oae1in(tTa)nn-1((a) — O'ctenn.(0a)in-1(O'a) — n(Ei — e)in(O'C)nn(0a) (16) Oafijn(3'a).j.-1(0a) — OictEjn(0a)in-1(13/ a) — n(Ei e)in(0/ a)in( 3a) —
The same formula is obtained for E„ by solving (12) and (15) except that Ai andµ replace Ei and e. When the sphere is perfectly conducting, An and Bn are obtained by equating the left sides of (12) and (14) to zero. Thus An 11
nnn(0a)ri J i3ajn_1(13a) — nj.(13a) —
_
jn(3a)
in(Oa) — jn.n((3a)
(17)
474
ELECTROMAGNETIC RADIATION
§12.16
The energy scattered by the sphere in a specified direction is, from 11.17 (3), the real part of the complex Poynting vector 411-it x
a=
x (ri x E) = -iriA— Y(PeRe + to R0)
(18)
From 5.37 (12) at a great distance, neglecting terms in r-P when p > 1, k„(jOr) =
a[ric.000] =
e-or,
1 ;flr
(19)
r ar
Substitution in (7), (8), and (13) gives for the tangential component of the scattered electrical intensity at a great distance from the sphere
= _i043= jEe_or
2n + 1
{0[Ansin OP;,/(cos 0)
L-J7i(n ± 1)
n
B„P
(cos 0)] cos 95 sin 0
4[
P,' (cos 0) ssin in 0
in sin 0P.;,' (cos 0)1 sin cb} (20)
To get the total energy scattered, (18) is multiplied by r2 sin 0 de chi) and integrated over the ranges 0 < < 27r and 0 < 0 < 7r. Substitution of (20) in (18) and integration with respect to 4) bring in a factor 7r and leave, omitting the argument in the Legendre polynomials, (2n ± 1)(2m + 1) [ (A.A”, 2/.41(327.2 LJ LJn(n 1)m(m ± 1) fiirE2
ju n (sin2 )
sPinP21 0)
n =1 m
— (A.13. + nA„,)(P!' P;7,
P.,1131')]
(21)
When multiplied by r2 sin 0 c10 and integrated from 0 to 1r, the first group of Legendre polynomials yields the integral following 5.231 (7) which is zero if m n and 5.231 (8) if m = n. Integration of the first term of the second group gives the negative of the second term and cancels it. The result is real so that the total scattered power is 7rekE2-1 P = 0/37 (2n + 1)(IA.12
IBni 2)
(22)
n -= 1
Interesting cases are discussed in the books by MacDonald and by Stratton listed at the end of the chapter. 12.16. Solution of Propagation Equation in Cylindrical Coordinates.— In 11.14 we considered the special type of cylindrical wave moving in the z-direction obtained by setting those terms in the scalar wave equation containing z and t separately equal to zero. If instead we equate the first group to +0;„„ and the second to -Ti3,„2„ and assume a sinusoidal time
§12.17
EXPANSION IN CYLINDRICAL HARMONICS
475
dependence so that 132 = co2ue, we obtain the equations
± 0;„„0 = 0,
c 22 dz2
032 T- 131n) 2 = 0,
W=
02 (1)
Comparison with 5.291 (1) to (5) and 5.293 (6) shows that the form of W in the p, 4 z system is, writing k,2„„ for 32 — 131„, and k:,& for 0 2 + ,
W = (Aelkm-z = (Aepe,nnz
jje-7e„,„.)[eim (F3 nnp)
.5,) Y.(13.np)] cos (mcb 15K„,(0„,n p)] cos (mt. + om)
+
(2) (3)
When 0, /5, knin, and k„, ' „ are real, both these equations give waves propagated only in the z-direction. If 01„ > 02 so that km„ is imaginary, (2) gives a wave exponentially damped in the z-direction. If C is real and 15 is complex, we have a radial propagation component in (2). If z is absent, k,n„ and Kin are zero, and we have cylindrical wave fronts in (2) and (3). From 5.293 (4) and 5.295 (8), D = —jC gives p-directed waves. If transverse electric and transverse magnetic waves are defined as those whose electric and magnetic fields, respectively, are normal to the z-axis then, putting u = k in 10.01 (1) and (5), we obtain A = piativte B
ap ,t, a2tivte + aqkte = az p ack az
a t7 ,m)]
3214-7„,„ [ ank ap az + p ao az +k 132W
az2
(4)
8214-7)
k (02'W te
az2
13 2 014-7 el
P
,.
+ 4)02
-wfm ap
(5)
12.17. Expansion in Cylindrical Harmonics for a Plane Wave.—From 11.15, the formula for a plane sinusoidal wave traveling in the direction n is greiwe
Atti, u2)0(‘"-13n.r) = f
u2)eme--10Pco. (a-0)
(1)
where n is normal to the z-axis and makes an angle a with the plane yt. = 0 and u1and u2 are coordinates in the plane normal to n. A plane wave may be expressed in cylindrical harmonics by expanding the last exponential factor in (1) in a complex Fourier series. In order to utilize formulas already derived, it is simpler to expand real and imaginary parts separately. Let us first expand cos (x sin 4') in a Fourier series. a„ cos n4',
cos (x sin 4/) = n=0
2 —
—
27r
so
n
f —„
cos (x sin +,&) cos nik dik
(2)
476
ELECTROMAGNETIC RADIATION
§12.18
Application of Dw 401.06 or Pc 592 gives two even integrands whose integral from n to it may be replaced by twice that from 0 to r. Thus — -
an =
2
—
2a
6° n[i cos (4 — x sin IP) d¢ +
Joy cos (4
x sin1,G) did
(3)
From 5.302 (4), the integrals are Jn(x) and J„( — x) or (-1)V„(x) so that an is zero when n is odd, and we may write 2m for n which gives cos (x sin 1,t) =
J„(x) cos 4 (4)
(2 — 5,,)J2m(x) cos 2no,G = m=o
n=—
We can expand sin (x sin lk) in a sine series in just the same way and get the difference of J„(x) and J.( — x) instead of the sum so that sin (x sin ik) =
2J2,, 1(x) sin (2m + m=o
=
J,(x) sin ?AG
(5)
n=—
If we multiply this by j and add to (4), we get the exponential form. If we then substitute Op for x and fir + — a for tp, we get ejaP c°s ('—o) =
jn,/,(13p)ein(.-0) = n=—
0 (3p) + 2 Z j"J,( F6p) cos n(a —
CP )
n=1
(6) 12.18. Radiation from Apertures in Plane Conducting Screens. The rigorous calculation of the radiation passing through an aperture is a very difficult problem. The fields must satisfy not only the propagation equations in the space outside the aperture and specified conditions at its boundaries, but also they must fit smoothly onto the fields inside the aperture. These fields are usually changed by the back radiation from the screen surrounding the aperture which practically restricts rigorous solutions to the very limited number of cases where it is mathematically feasible to treat the entire radiation space as a single volume. It was proved in 12.10 that, if the initial field values throughout an empty volume are given together with the tangential components of either the electric or the magnetic field over its surface, then the fields at any subsequent time are uniquely determined. The second condition alone is sufficient in a steady state. Evidently with conducting screens one should use the electric field because its tangential component is known to be zero on the screen. The best first approximation to the unknown electric field over the aperture seems to be that its value is the same as with no screen. The treatment that follows is particularly applicable to plane conducting screens with no restrictions on the number or shape of the apertures nor on the form of the incident wave. —
§12.18
RADIATION FROM APERTURES
477
We desire a source that will give a tangential electric field E over an area S of an infinite plane and zero over the remainder. Consider a thin double current sheet with a very small distance between layers, one of which has a current density equal and opposite to the other as shown in cross section in Fig. 12.18a. For a sheet uniform in the direction of flow, all the current passes around the edge, but for one stronger in the center, like that shown in Fig. 12.18b, part will turn back before reaching the edge. If the sheet is very thin, the external magnetic field is negligible compared with that between layers so that when we apply the magnetomotance .E law to the rectangle abcd, which is normal to i and fits closely a section of the upper layer, we find B, to be mi. As the flux N = B,6 dl through the rectangle a'b'c'd' changes, the electromotance — dN/dt around the loop equals from symmetry 2E dl as 0 so that the electric field strength E just above the sheet is — 4-jco/.46i. Clearly the double current sheet can be FIG. 12.18. built up out of infinitesimal solenoids of cross-sectional area 6 dl, length dc, and magnetic moment n xi6 dl dc which, in terms of E is —2 (jm.1) —in x E dS. Evidently these solenoidal elements can be combined in such a way as to produce any desired variation in E. It is also evident from symmetry that E is normal to the plane of the sheet outside its boundaries. From 12.13 (5) we see that the vector potential at P of the radiation from a small oscillating loop is normal to the loop axis and proportional to the sine of the angle between this axis and r the radius vector from the loop to P. Thus substitution of the moment just found for ra2/ gives for the diffracted vector potential
A = AO" = 1e coi 271-co
f
s
(j
—
13r)(n xE) x ri e — jor dS 2
r
(1)
where r1is a unit vector along r. At a great distance the j-term in the integrand may be neglected compared with Or. In the most general case, E will vary in magnitude, direction, and phase over the aperture and H will not be parallel to n, the normal to the aperture plane. Let x', y', z' refer to coordinates in the aperture and x, y, z to those of the field point. Then, since r 2 is (x — x') 2 (y— y') 2 (z — z')2, _ ju,(i — (3r)r E = —jwA =
r__ v,(r—ie—asr) = _ x ti x V'cl) dS
= _ vcp (real part)
(2)
478
ELECTROMAGNETIC RADIATION
§12.18
where E' is a function of x', y', z' only. Thus writing —v4, for v'el), observing that then V x (ciln x E') is zero, and using the usual formula for V x 43v give (2) the form E = 47--'eiwtV x
fs(n x E')r —ie— jor dS
(real part)
(3)
Formulas (2) and (3) are derived differently by Jackson. From the first Maxwell equation 11.00 (1), juweg is V x B so, except for the gradient of a scalar, canceling the curl gives
Bt = j13(271-co)—ifs(n x 2)r—'e— jor dS
(4)
This integral equation gives the solenoidal part of the tangential E in the aperture in terms of the tangential B which is known, as will soon be shown. The diffracted field far from the aperture depends only on this part of n x E since B is given by the curl of the A in (1) or (2) and E and B are related by 11.03 (8). A system of sources of electromagnetic wave lies on one side of an infinite plane perfectly conducting sheet having apertures. The eddy currents in this sheet generate additional waves which combine with the original ones to produce the reflected and refracted waves. These currents, being coplanar with the apertures, can produce therein only magnetic fields normal to their plane. Thus the tangential magnetic fields in the apertures must be entirely due to the original sources unperturbed by the eddy currents. When the incident magnetic field parallels the screen and the wave length is long compared with the relevant aperture dimension, it is often possible to calculate by potential theory the exact ratio of the normal to the tangential components of the magnetic induction in the openings and from this the tangential electric field. In Fig. 12.18 a displacement of the rectangle a'b'c'd' along B. decreases 2E dl by an amount equal to the decrease in dN/dt. This in turn equals the rate of change of the flux that escapes from both faces of the shell between the two positions. Thus, for an aperture in the xy-plane when E has only a y-component, we have
—
aEi, = ax
jcoBz
(5)
Integration of this expression gives the tangential component of E to be used in (1). All static potential solutions will give B. in terms of the induction tangential to the sheet far from the apertures. This standing wave value is twice the incident wave value that appears in the aperture. Examples at the end of the chapter include Bethe's small hole results.
§12.19
DIFFRACTION FROM RECTANGULAR APERTURE
479
12.19. Diffraction from Rectangular Aperture in Conducting Plane.— The formula of the last article will now be used to find the diffracted field from a rectangular aperture in a perfectly conducting sheet lying in the xy-plane. The magnetic field parallels the x-axis, and Poynting's vector makes an angle a with the z-axis as shown in Fig. 12.19a. If x1 and yi are the coordinates of dS, r the radius vector from dS to P, and R that from 0 to P, then we have approximately, if R >> a and R >> b, r
R — xicos ck sin 0 — yi sin ck sin 0
(1)
The tangential component Et varies in phase at z = 0 so that Etel't = E cos a
(2)
e3(`"--5711 sin a)
Because n x E parallels the x-axis, (n x E) x R lies parallel to the yz-plane, normal to DP and hence to R and is proportional to sin 0'. Thus A, = — cos 0 csc 0'A, A, =- Ity sin 4) cos 0 — Az sin 0,
A, = sin 61 sin 4) csc O'A A- 4, A., cos 0, A, = 0
When 12.18 (1) is substituted in (3), csc 0' cancels the sin 0'.
(3) (4)
Neglect of
xi and yi compared with R2leaves x1 and ylterms only in the exponent of A. A,
Writing in Et from (2) and using (4), we have la rib OE cos a sin (1:1e-JaR . e jiqx, cos 4, sin 0-1-y i (sin 4, sin 0—sin a)] dx, dyi 2ircoR — lb 2E cos a sin 4) sin (Oa cos sin0)sin [pb(sin 4) sin 0 — sin a)] .0 e-3 E (5) irgo.,R cos 4) sin 0(sin 4) sin 0 — sin a) A = A9 cot 4) cos 0 (6)
f_
The only approximation involved in these formulas is the assumption of an unperturbed electric field over the aperture. Stratton and Chu (Phys. Rev., Vol. 56, page 106) derived them by a superposition or "reflection" of two solutions of Maxwell's equations, which assume unperturbed electric and magnetic fields over the opening. This superposition was necessary to eliminate the tangential electric field over the screen and, as we see from the solution just obtained and from the uniqueness theorem, this is just equivalent to discarding all terms derived from the magnetic field. These authors check (5) and (6) at various b and a values by comparing the yz-plane intensity Ei Egi with the rigorous two-dimensional solution for a slit (a = 00) of Morse and Rubenstein (Phys. Rev., Vol. 54,
480
ELECTROMAGNETIC RADIATION
§12.20
page 895). The results appear in Fig. 12.19b, c, and d. A similar comparison of the xz-plane field with that of a slit (b = co ) for a = 0 is shown in Fig. 12.19e. This fairly good agreement shows the magnitude of the errors made by assuming an unperturbed electric field over the opening. Clearly the Kirchhoff scalar diffraction theory used in optics, which requires Fig. 12.19b and e to be identical, is completely wrong when applied to apertures of wave length dimensions in conducting screens. Dashed lines show rigorous values. 12.20. Orthogonal Functions in Diffraction Problems. Coaxial Line.— The method of the last two articles does not apply to curved screens, and the integral of 12.18 (1) is frequently difficult to evaluate close to the
(6)
(d)
(e) 12.19b, c, d, e.
aperture in a plane screen. In such cases we may start with solutions of the scalar propagation equation in the form ei.tZ ZC,n. Umn(u1) V.(u2) W„.(u3),
ffun„,-vnu„-v, dui du2 = (37,7D,,,n
n m
If u3 is normal to the screen, then u1 and u2 are orthogonal curvilinear coordinates in its surface. The integration is over the screen and aperture surface, and 677 is zero unless p = n and q = m. Such solutions appear in 12.11 (5), 12.16 (2), and 12.16 (3) and others occur in waveguide problems. For a conducting screen, we evaluate Cmn from the tangential component of the electric field, which is given over the aperture. The radiation from the open end of a coaxial line can be calculated by this method. The distant field is found more easily by 12.18 (1), but the method to be used here is better near the opening. The propagation
§12.20
ORTHOGONAL FUNCTIONS IN DIFFRACTION PROBLEMS
481
space in the line is bounded internally and externally by the cylinders p = a and p = b, respectively, and terminates in the plane z = 0 which, except when a < p < b, is perfectly conducting. To find the radiation into the region of positive z, we shall assume that b b,
V(p) = 0,
The
n (p/b) ln (a/b)' V(p) = Vo (2) a > p > 0,
b > p > a,
V(p) = VoI
When these values are inserted in (1), the result may be integrated by Pc 427 or Dw 610.9 and gives b2n+2 _ a 2"+2 (2n+ 1)!! 2)!!r2n±1(cos 0) ( 1)n (2n + 2)r2'.+2 (2n
Vo V
In (b/a)
(3)
n=0
To get the radiation fields, one puts A = 1 and B=o=m= 6=0 in 12.11 (5) so that it gives an expanding wave and evaluates D„ by setting r =b and equating the coefficient of -P2n-I-1(COS 0) in the Ee derived from Wt.m to that in the Et, given by (3). From 12.11 (5) and (2), we have CO
CEO,. = dr [ jwra2(r.W`m)1 80
r=b
±1(cos 0) j[bk2 ±i isb) n,
n db
b
(
]
n=0
Since Ob is small, we may write — (2n + 1)(4n + 1)!!(j0b)-(2n+2) for the derivative in this expression by 5.37 (11) and (12). To get E5 at r = b, we write —b-1d[P2n+I(cos 0)]/d0 = b-'11,,+1 (cos 0) for P2.+1(cos 0) in (3)
ELECTROMAGNETIC RADIATION
482
§12.20
and b2r.+2 for r2 n+2. Equating coefficients and solving for Dn give
_
(2n — 1)!![(00 2n+2— (Oarn+21 jVo ( 4) 1)!! 2)(2n + 2)!!(4n 0.) In (b/ a) (2n The electric and magnetic fields when r is greater than b are then given by 12.11 (2), 12.11 (5), 12.15 (6), and 12.11 (3) to be 00
Et?
=
c [riC2n+I(jOr) ] PL+1(COS
15 -7
0)
(5)
r
n =0
+ 1)(2n
Er =
2)/c2.+1000P2.+1(cos 0)
(6)
n =0
13,15 =
02
nnk2n±l(j3r)P2n+1(COS
n
0)
(7)
=0
These equations hold if r > b, and if b « X they satisfy the boundary conditions (2) rigorously. If we neglect higher powers of Ob and Oa, only the first term of the series survives and, if the case is further simplified by considering the field only at a great distance so that by 5.37 (12) (jOr) —'e--2fir replaces ic2n+ i(jOr), then the fields become 133(b2 — a2)Vo sin 0 cos (wt — Or) (8) = (p 41E0 = 40)r In (b/ a) From 12.01 (8), this is identical with the distant radiation from a current element iodzo cos wt where io dzo is 7ro.)E(b2 — a2) Vo/ln (b/a) so that the power radiated into the upper hemisphere is half that of 12.01 (26).
p
211-5CE(b2
a2) 2 Vg
3[ln (b/a)]2X4 V22 — 3[ln (b/a)]2X4 R, — —± 47r5ce(b2 — a2)2 2p
(9) (10)
The same method applies to other than plane screens, such, for example, as an infinite conducting cone coaxial with the line. In this case we take the same steps but start with 5.261. Other applications will be found in the problems at the end of the chapter. Problems 1. A linear quadrupole consists of charges q, —2q, q at z = —a, 0, +a. Its moment q2) is a2q sin wt. Show from 12.01 (13) that if a
E0 = jwa
It.
ap
ei'i
117u = (4)-1EZ .13.[J„(0p) — j17„(0p)] cos ncil n=0 ( — j)"(2— ON(P/ENn(0a)4(13'a) — (1.LE'N'„(13a)Jn(O'a)] —
(ile')1 in(o'a)[J;(ga) — :717 (0a)1 — (A'E)1J".03'0[Jn(3a) — il7n(Qa)1
16. A plane electromagnetic wave, whose z-directed E-vector is E cos (wt — 13x), impinges on a perfectly conducting cylinder of radius a whose axis is the z-axis. Show
485
PROBLEMS that the total fields are given by the real parts of 0
E, = j,„02gitmeicot,
B, =
, wgme2cot,
P
= a ik
ac
ap
( — j)'(2 — (5,,)[Jn(0a)17.(13P) — J.(RP)Yn(Ra)] = E cos n4, 0.,0 2 J„( pa) — j Yn Oa) n=0 17. In a medium ILE a plane electromagnetic wave whose z-directed electric vector is E cos (wt — fix) impinges on a cylinder µ'E of radius a whose axis is the z-axis. Show that the scattered radiation is given by the real part of B,, =
E, = jc00 2Tkimeic",
02 a We,. P
14
„n
a
ei 2/,,,2 and oike 2pe > v„2 . Show that the velocity of the nth wave is coa(coWe' — u!)-1and that the fields in '€' are
A', = k,,.(cop'e)-1.134, = jOcik.'.11(unP/a),
knaE, = uX„e-ik.2J0(tt.p/a)
where o., 2a2p'e' — = co2a2p,e — 14,2 = Ic:a2 and a < b. 26. If there is no reflector at p = b in the last problem, show that with the same materials and frequencies a plane wave is impossible but that otherwise the results of the last problem hold if the Hankel functions Jo— jYo and J1 — jY1 are substituted for Ro and R1, respectively. 27. An infinite medium 11E surrounds an infinite dielectric cylinder /./'€' of radius a. Show that a plane transverse magnetic wave can be propagated along the cylinder for each value of u„ that satisfies the equations v„E'Jl(u„)Ko(vn.)
unfJo(u„)Ki(v„) = 0,
w2a2AY — u;, = w2a 2/4e
v: = k!a2
provided that w2a2A1 is greater than Show that the velocity of the nth wave is eoa(0, 2a2p'e' — u„2 )-1and that the fields outside the cylinder are
= k„(cope)-113-4, =
k,,at = jvnene-ik,='Ko(vna-1P)
PROBLEMS
487
28. If Yo(v.b/a) is zero, verify that the conditions of problem 25 are satisfied approximately by E = 2E = 2€„, A = = .85 X 100, u, = 1, v, = 3, and = show that the phase velocity in the z-direction is approximately 3.21 X 10' m/sec. 29. Verify that the conditions of problem 27 are satisfied approximately by taking = 1.81€ = 1.81G A = = coa = 0.99 X 10', u. = 2.8, v. = 1 and show that the phase velocity is approximately 2.88 X 10' m/sec. 30. An infinite plane perfectly conducting surface is coated with a dielectric layer 2f2 of thickness a and underlies an infinite dielectric medium Ale'. By 12.16, show a diverging surface TM wave is possible whose vector potentials are the real parts of A l= C1[p1(0r2 A2 = C2 {ei (s:
— —
k/3' HV )(13' p)le-(0" fli0.+iwt 13 2 )111V )' (0' p') 3' 2)1 sin [( 3 — 0'2)z]1-1?)'(13'p) — k/3' cos[(/3s — 0")iz]lle)(0'p)}eiwt -
—
1
where ,(3! = (.0 2141€2, 114 = co 2A2€2, and R2 > (3' > /3,. Show that its velocity is the same as that of the plane wave of problem 7, Chap. XI. 31. An infinite plane perfectly conducting surface is coated with a dielectric layer /12€2 of thickness a and underlies an infinite dielectric medium wei. By 12.16, show a diverging surface TE wave is possible whose vector potentials are the real parts of Al = (1)Cle —(0"-012)1q/i2)(0'p)04'4,
A2 = 8C2 sin [(I3— 0'2]iz]li(12)(0'p)eic"
where Of. = = w2i.42€2, and 02 > > 01. Show that its velocity is the same as that of the plane wave of problem 8, Chap. XI. 32. An infinite perfectly conducting cylinder of radius a which is coated with a dielectric layer A2e2 of outer radius b passes through an infinite medium Atiei. Show by 12.16 that possible vector potentials for a surface TM wave are the real parts of
where 13 = co2121€1,
Al = Ci[—piji3/Ki(piP)
kpiKo(pip)]ei (C0t -P'')
A2 = C2[ — E01,j/3'Rl(P2P)
kP2R0(p2p)]ei (We-9'z)
= W2P2€2, 02 > 0' > 01, p! = 13'2 —
= Oi —13' 2, and
Ro(p2p) = Yo(p2a)Jo(p2P) — Jo(p2a)Fo(p2P) Ri(p2p) = Yo(P2a)J1(P2P) — Jo(P2a)Yi(P2P)
Show that the velocity may be found from the equation —e2piKo(plb)R1(p2b) = eip2Ki(plb)Ro(p2b) If a = 1, b = 2, gl = Al, €2 = 4€1, and /31b = 1.174, show that 01:0'432 = 1:1.048:2 and that, when p = 4.350b, then Bohas one-tenth of its value at p = b. 33. Obtain the result in 12.20 (8) by using 12.18 (1). 34. The plane of polarization of the incident wave in Fig. 12.19a is rotated so that E parallels the x-axis. Show that the diffracted vector potential at a great distance is, if Ae' is given by 12.19 (5), A9
—
— cos ¢' A0 ,
cos a sin 0
-
cos 0
..
21° _ cos a A8'
35. The opening of 12.19 is not rectangular but annular of internal and external radii a and b. Show that the diffracted vector potential at a great distance is -tan cb
Ag =
cos 8
—
A4, =
E cos a sin 4, oiRQ
[a.11(0Qa) — bJ1(13Qb)]c)9 R
where Q = (sin2 a + sin2 e — 2 sin a sin 0 sin 0)i.
488
ELECTROMAGNETIC RADIATION
36. The polarization plane of the incident beam in the last problem is rotated so that E parallels the x-axis. Show that the vector potential of the diffracted field at a great distance is
cot
A0 =
cos
— E cos
=
43.
caRQ
[b‘II(8Qb) — 11(0Qa)lci19R
37. Show that, if in 12.19 or in the last three problems, there are not one but two identical openings centered at y = c and y = — c then, if R >> c the single opening potential must be multiplied by 2 cos [0c( — sin a + y/R)]. 38. If in the last problem the openings are at x = c and x = — c, show that the multiplication factor is 2 cos ((3cx/R). 39. A spherical shell of radius a is perfectly absorbing inside and perfectly conducting outside. An electric dipole of moment M cos wt is placed at the origin pointed in the 0 = 0 direction, and the part of the shell between 0 = a' and 0 = a" is removed. Show that 12.11 gives the external diffracted vector potential where Wt,„ = M(1
—
13 2a2j0a)ZA„P„(cos 0)ic„(j,3r) n=
= e of (2n + 1)Pu(u) — (n 2)uP„_1(u) — (n — 1)uPn+1(u) u = cos a" 87riWee(??,
— 1)(n + 2)a[aL(0a)]/aa
u = cos a'
40. The dipole in the shell of the last problem is replaced by a small coaxial loop of wire of radius b carrying a current /e3.'. Show that 12.11 (2) gives the vector potential of the diffracted field outside where Ikee = ,u1b2(1
jf3a)ZA„Pn(cos 0)k.(07.) n
A„ = e
..[(n LL
2)uP„_1(u) (n — 1)uPn÷i(u) — (2n + 1)P.,,(u)t = Cos a" 8a2kn(j13a)(n — 1)(n + 2) u = cos a'
41. The boundary of a circular hole in an infinite thin plane conducting sheet is p = a. The potential of an impinging wave at the hole surface is Ao = f(p). If the wave in the hole is unperturbed by the boundary, show that the vector potential of the radiated field at a great distance where R >> X is the real part of Ad,
a = 13R-1cos Oei(4" —fi R) f pf(p)J 1(0p sin 0) dp 0
If the source is a very small loop which carries a current leiwt and is coaxial with, and at a distance c from the hole and if a > a, and Op is unrestricted. 45. A plane wave whose magnetic vector is kBelm-0') impinges normally on the slotted plane of problem 42. This creates an electric field across the slit whose form, for slowly varying fields, is shown by the ellipses in Fig. 4.22. Find Cfi(yo) in problem 42 from 4.22 (3) and so, assuming constant phase over the slit, obtain Act, for the radiated field. Find C approximately by equating the curl of A,, when p is zero to the incident B. Thus show that, neglecting (0a) 2 terms,
—
T-Boe ,‘" HO2)(OP) 2[a — j ln (170a )]
where In y = —.11593, p is measured from the z-axis, p >> a, and Op is unrestricted. 46. A plane polarized electromagnetic wave whose wave front makes an angle a with the yz-plane impinges on the slotted plane of problem 42. If its magnetic vector parallels the slit, show from problems 44 and 45 that the diffracted intensity at a great distance R from the slit is, if Oa is so small that phase differences over the slit are negligible and —1n(Oa) >> 1, 1-1 =
{ cos a
0 2a2 sin a sin 412 _
2aR In (Oa)
Iio
4
47. The wave of the last problem has its electric vector parallel to the slit. Show that, if Oa is very small so that phase differences over the slit may be neglected, then from problem 43 the diffracted intensity is _
=
70 3a 4 cost
8R
cosg a_ Ho
48. A hole of radius a is cut in an infinite plane perfectly conducting sheet which bounds a uniform electric standing wave field Eoeiwt. By neglecting phase differences over the hole, show from 5.272 (6) and 12.18 (1) that the vector potential at a great distance from the center of the hole is 49
— jEo [sin (Oa sin 0) — Oa sin 0 cos (Oa sin 0)]e-iPR ro)(3R sing
49. Show by the methods of 5.272 that, if a uniform standing wave magnetic field of induction Boeiwt exists in the x-direction above and parallel to an infinite thin plane perfectly conducting sheet at z = 0 having a hole of radius a in it, the z-component of
ELECTROMAGNETIC RADIATION
490
B in the hole is 2B(x/ir) (a2— X2 — y2)-1. Hence by 12.18, neglecting phase differences, show that at a great distance from the center of the hole below the sheet, where the azimuth angle from the x-axis is 47 and the colatitude angle from the hole axis is 0, the vector potential is AB =
2jBo sin [sin (Oa sin 0) — 13a sin 0 cos (Oa sin 0)] e-i0R, r(32R sins 0
A-0= A8 cos 0 cot 43
50. A plane polarized wave strikes a thin plane perfectly conducting sheet with a hole of radius a in it at an angle a from the normal. If E0 parallels the sheet, show, when Oa is very small in problem 49, that the intensity of the diffracted radiation is -
=
404a6 cos2 a(1 — sin2B 7r2R2
cos2
Note that the standing wave B0 of problem 49 is twice the incident running wave induction. 51. Show that the average power radiated from the hole in the last problem is 16134a6 cost
37r
°
52. In problem 50 let the electric vector lie in the plane of incidence and show, when pa is very small in problems 48 and 49, that the intensity of the diffracted radiation is 134ct6 _2_,2_14 sin2 7r R
cos2 0 + (2 cos 4) — sin a)21110
Note that the standing wave amplitudes in problems 48 and 49 are twice those in 110. 63. Show that the average power radiated from the hole in the last problem is
p
404a6(4 + sin2a)
3r
c II
64. An electromotance e is maintained across the center of a very narrow slot bounded by x = +16, z = ±l in the infinite plane conducting face y = 0. Assuming that the field in the slot in this plane is E0 = ia-le sin [5(l - z)], show by 12.18 (1) that the electrical intensity outside the slot, where y > 0, is E=
1(1±1) (iy — ix)Z f R-3(1 ± jOR) sin [13(/ i(i 1)
where R2 = x 2 + y2 that
(z — z1)2.
zi)leim-OR) dzi
Show by comparison with the curl of 12.03 (3)
[ELia = — 2e(a/0)-1[B].„u,„„„ so that the fields are identical in form with those of an antenna driven at the center but with B and E interchanged so that from 12.03 (18)
= —1e(irp)-'[2 cos 131 sin (cot — Or) — sin (wt — Or.) — sin (wt — Orb)] where ra and rb are the distances between the ends of the slot and the field point. 55. Apply the results of the last problem to the resonant slot 1 = IX and thus show by 12.04 (4) that the radiation resistance of the slot is 363 ohms if it radiates from both faces.
REFERENCES
491
NOTE: The results of the following problems are based on the Kirchhoff theory of diffraction. They are useful in optics but should be used with caution elsewhere. 56. Let U = y, z)el"O be a component of the Hertz vector in an insulating medium so that it satisfies 11.01 (10) when r = 00. Let
V = r-lei(44-13r) = cp(x, y, z)eiwg be a similar component for a spherical wave originating at a point P. Insert V, and (/, in Green's theorem, 3.06 (4), and, taking the volume of integration to lie between a very small sphere surrounding P and some larger surface 8, enclosing the sphere, show that 470, = fs [r-,e-it3'170 — 1,t.V(r-'e-119')] • n dS where Op is the value of at P. This formula, the basis of Kirchhoff's diffraction theory, gives the effect at P in terms of its integral over a surface surrounding P. 57. Let S have apertures, and let U be a spherical wave originating at Q outside S. Assuming that U has the same value over the apertures as if S were absent and is zero over the remainder of S, show that Gr =
—
1(n•r rri r
n• ri)
r1
e-7 00'+'1) dS'
where r1and r are the radius vectors from Q and P to the apertures and both are much larger than 0-1. S' is the area of the openings. 58. Let there be a single aperture, let R and R1be the mean distances from its center 0 to P and Q, and let the coordinates of any point in it, referred to 0, be x' and y'. Expand r and r1in powers of x' and y' and, since /3 = 271-X-1, show that —
j n• R 2XRRI R
27i(R ± RI)
n• R1 R1
2rjF(x' ,y' ,1 / R,1 / RI)
f
dS'
59. Consider a plane wave of intensity /0 incident normally on a circular aperture of radius a so that RI = 00, and take R >> a and P at x = x, y = 0 so that F(x', y',
Er')
= p'
cos 0' sin a
where a is the angle subtended by x at 0. Evaluate the integral by 5.302 (3) and (2), and show that the diffracted intensity is Id = 1/0R-2cote (1a)(Ji(27raX-1sin a)]2a 2 60. Take Q and P on the axis of a circular aperture, of radius a and take R1 and R much larger than a. Show that, if the incident intensity is /0, that at P is 4.1410(R1
R)-2sine [-n-a 2(2X)-'R-1
RT1)]
References Most recent references give extensive bibliographies. S.: "Handbuch der Physik," Vol. XVI, Springer, 1958. "Handbuch der Physik," Vols. XII, XV, XX, Berlin, 1927, 1928. Gives elegant treatment of electromagnetic theory in Vol. XII. HARRINGTON, R. F.: "Time-harmonic Electromagnetic Fields," McGraw-Hill, 1963. Discusses multipoles, Babinet's principle, and pertinent topics. FLUGGE,
GEIGER-SCHEEL:
492
ELECTROMAGNETIC RADIATION
H. R.: "Electric Waves," Macmillan, 1893. Radiation field drawings. J. D.: "Classical Electrodynamics," Wiley, 1962. Good treatment of multipoles, Babinet's principle and diffraction, and cgs units. KING, R. W. P.: "Theory of Linear Antennas," Harvard, 1956. A complete treatment with extensive bibliography. MACDONALD, H. M.: "Electromagnetism," Bell, 1934. Solves diffraction problems. MASON, M., and W. WEAVER: "The Electromagnetic Field," University of Chicago Press, 1929, and Dover. Gives excellent treatment of retarded potentials. MAXWELL, J. C.: "Electricity and Magnetism," 3d ed. (1891), Dover, 1954. The original treatment of the subject. PANOFSKY, W. K. H., and MELBA PHILLIPS: "Classical Electricity and Magnetism," Addison-Wesley, 1962. Covers multipoles and related topics. PAPAS, C. H.: "Theory of Electromagnetic Wave Propagation," McGraw-Hill, 1965. Covers multipoles and antennas with radiation patterns and applications. SCHELKUNOFF, S. A.: "Advanced Antenna Theory," Wiley, 1952. SCHELKUNOFF, S. A., and T. FRIIS: "Antennas; Theory and Practice," Wiley, 1952. Formulas for multipoles and antennas with specific cases in problems. SILVER, S.: "Microwave Antenna Theory and Design," McGraw-Hill, 1949. Treats general antenna theory and diffraction extensively. STRATTON, J. A.: "Electromagnetic Theory," McGraw-Hill, 1941. Extensive and rigorous mathematical treatment of the whole subject. TRALLI, N.: "Classical Electromagnetic Theory," McGraw-Hill, 1963. Treats concisely much of the material of this chapter. VAN BLADEL, J.: "Electromagnetic Fields," McGraw-Hill, 1964. Expands topics of this chapter to include stratified mediums, pulses, and other cases. WIEN-HARMS: "Handbuch der Experimentalphysik," Vol. XI, Leipzig, 1932. HERTZ,
JACKSON,
CHAPTER XIII WAVE GUIDES AND CAVITY RESONATORS 13.00. Waves in Hollow Cylindrical Tubes.—That plane waves may be guided by two or more mutually external perfectly conducting cylinders was shown in 11.14 to 11.18. If the propagation space is enclosed by a perfectly conducting cylinder, other unattenuated wave types will propagate at very high frequencies, and if this is the only boundary, a simple plane wave is no longer possible. Each external and internal (if any) boundary of a cylindrical wave guide is generated by moving a straight line transversely keeping it parallel to the z-axis. It is often considered closed by the z = 0 plane and extends from z = 0 to z = co In the text following 11.01 (5) it was pointed out that in the Coulomb gauge the vector potential A is derivable from a solution W of the scalar propagation equation in two ways, one giving transverse electric and one transverse magnetic waves. The differential equations are
v 2A = am - at 2
vwt. = 1/6
aw-te at2
r• 2 W
t77)
„
a2UT
(1)
We are interested in a steady-state solution of angular frequency w. In this case from 10.01 a vector potential satisfying the first equation and its curl may be obtained from solutions of the scalar equations by t„,), = —V x (o2tivt„, = v x (kikte k x kxVWte) (2) where i32 = co2AE. These equations are written out in cylindrical polar coordinates in 12.16 (4) and (5). As in 12.16, we may write W as a product of a function of z by a function of the transverse coordinates so that either scalar equation in (1) splits into two. Thus
W = u2,
22 dz2 + (o2T- 01.)2 = 0
vw ± 131„u = 0,
(3)
Here U depends only on the transverse coordinates u1and u2, and boundary and symmetry conditions fix 0.„. If n is a unit vector which like. A is normal to the curved surface, then from (2) the conditions there are 0 = n x Age
=
—n • V(kWie)
or
-aU . nte = 0
0= n x [17 (k vtiVt„,)] = n x [kV2Tikt„ — = —132n x
= soln„Wt,
t„, — (n x
k • V (VW t,n)]
tm) = Si[132Tktm
k' a2W as az 493
(4)
a2aw z2t'n]
a vw,m k
as az
494
WAVE GUIDES AND CAVITY RESONATORS
§13.00
where s is a coordinate measured around a cylindrical surface along its curve of intersection with a plane normal to the z-axis so that s1 = k x n. The conditions imposed on U at the boundary are therefore
Ut. = 0
or
Ann =0
and
—o
as
(5)
The second pair of conditions in (5) are those met by the principal waves described in 11.14 to 11.18 whose velocity is independent of frequency. These are possible only with two or more conducting cylinders. The differential equations of (3) are of the second order so each has two solutions, and the general solution is of the form W=
bef'—)
u2) r3U2(ui, u2)](ae-i'--z Ann = :702 — 13104 = amn +
(6) (7)
When > temp, the propagation constant P.„ is a pure imaginary so a = 0 and the first and second exponential terms in (6) indicate unattenuated waves in the positive and negative z-direction, respectively. If a < gm., we put C = 0 when z is positive or D = 0 when z is negative since the fields must remain finite. For the remaining term, P.m is real ' „ = 0 which gives an exponentially damped term. The cutoff so a„, frequency v.n and the cutoff wave length for the mnth mode are P.n =
gm
COmn
n
2r (kie)-1 =
= 24r gmn
(8)
To extend the concept of characteristic impedance to wave guides, the definition of 11.15 (15) is put in terms of B a id B by 11.15 (6). Thus
a
k Wee'
te
12 klte =
k x
to
12kI tm — + Alk-Et —
=+
±jCOAIV x k Weel
jCLIA
= la(V x kW te) 1k x [V x (V x kfk teAl x [V x(kxVW 00]1 T ila(V x kWim)/a21 I _ „
l Btml
(.4EIV
kfkt,„1
(9)
coeiV x kjTVbni
jPmn
(10)
WE
In terms of the cutoff frequency v.., these are
12,, J. = [1 —-±(0 lim n/
E-4
12ki t. =
— (v,../v)9*
(11)
From 11.15 (6), the upper or lower sign is to be used according as the wave is in the negative or positive z-direction, respectively. Note that both wave types give a real characteristic impedance, indicating power propagation above the cutoff frequency and a reactive one below it. Observe that this differs from the transmission line definition in 11.16.
ATTENUATION IN HOLLOW WAVE GUIDES
§13.01
495
From (6) and (7), when # > ft.., waves propagate with a phase velocity co
— /.. 032 — gm)/ [1— (P./ v)2[1 [1— (X/X..)9/ (12) where v is the velocity of a free wave in the medium filling the tube. The signal velocity is given by 11.19 (2) and (12) to be Vmn
=
ac.)
1 NI #'„,„ Ace)i ag„, ' „= . # (pt)i
1 v2 (13) AEv. = v.. Thus v.. > v > (v.)... At sufficiently high frequencies, both phase and signal velocities approach the velocity of the free wave. From (2) and (3), the electric and magnetic fields in terms of U are (v )
—
T,
(
Et. = —jcoAt. = jog, x V Utee—o'—z
ht. = v At. = + koinu„)e-w--. Et. --= —joa. = (wo„„v2ut. + ki,407„„ut„,)e-o'—. ht. = v x At. = 02k x V Utme-7t3 Ims ' ,
(14) (15) (16) (17)
where Ut, and Ut. satisfy (3) and from (2) have different dimensions. 13.01. Attenuation in Hollow Wave Guides.—Wave guides are used to transmit high-frequency power from one point to another so transmission losses are important. If an imperfect dielectric fills the tube, energy will be dissipated there, but in the common air-filled guide such losses are seldom important and their calculation will be left to the problems at the end of the chapter. The eddy current energy loss in the walls is, however, unavoidable and must be considered. In all practical cases, the wall conductivity is so high that field solutions for perfectly conducting tubes are excellent approximations and can be used to calculate P. In any case where losses in a quantity are proportional to the quantity itself, there is exponential attenuation. Equation 10.02 (8) shows that the power loss in a conducting surface is proportional to the square of the tangential magnetic field at the surface. Therefore the fields are damped exponentially as they pass down the tube and, if their attenuation factor is a, that of the Poynting vector is 2a. The periodic factors disappear in the time average P of the transmitted power leaving Pe--2az. The z-derivative of P evaluated by 10.02 (8) and divided by .1-1 is ,
2« =
1 OP = 2122Pt5 P az
B•
h ds
(1)
where aP/3z is the average power loss per unit length, h the magnetic induction next the wall, T the wall resistivity, and 8 the skin thickness. At the wall V2Ut. = siaute/as so that, from 13.00 (15) and (17), /A u \ 2
ht. • hi. = 032
—
+ NmnU20f
h.• ht. = 134v2utn, • v2um (2)
WAVE GUIDES AND CAVITY RESONATORS
496
§13.02
By 11.17 (3), the average z-directed energy passing through unit area per second is the real part of IAL-lt x E so that 13.00 (14) to (17) give =
witni(k x
Ur,,) x V2Utel, = ii.c1c0(02 - 131n)Ir2uce
• v2ute
(3) (k vu,„Olz = -1-A-10'„,,,132[V2U 43202 - 01„)}V2Ut. • V2 Ulm
(4)
To simplify the surface integral of the scalar product in (3) and (4), we expand and note that the first term of the surface integral vanishes when transformed into a line integral around the boundary because, by 13.00 (4) or (5), either U or n • V2U is zero there. Thus, using 13.00 (3), fsV2 U • V2 U dS = fs[V 2• (UV 2U) - UqU] dS = i3,2nnis U 2 dS
(5)
From (1), (2), and (3), the attenuation of transverse electric waves is ate -
T
134[1 — (v„,„/ v) 2]se (aU 3,3) 2 ds
2,avo
(vmn/ v) 2
ds
(6)
[1 - (vm„/ y) 2]iisUL dS
Here S is the skin depth (ic0/27)-i, AI the permeability of the wall, v the frequency, v„,„ the cutoff frequency, v the free wave velocity in /le, and flmn is 221-v-1vmn. From (1), (2), and (4), since V2U is normal to the wall, 3C
T
(aut,,/ an)2 ds
(7)
ahn - 4w 5 oln[i - (vm./ 09i.fsuln dS
This equation shows that the attenuation of transverse magnetic waves increases with frequency in the same way for all forms of cross section. From Eq. (6) the attenuation of transverse electric waves depends on the form of the tube cross section and increases with frequency unless the first numerator term vanishes, as in some circular pipe modes. 13.02. The Rectangular Wave Guide.—Consider a tube bounded by the planes x = 0, x = a, y = 0, y = b. By inspection, solutions U (x,y) of 13.00 (3), which satisfy boundary conditions 13.00 (4) and (5), are mirx nay mr sin (1) U t„ = Cmn cos — cos Ulm = C„,„ sin b a b a m2 n2 m2 n2) v2 2 1 2( I or (2) le n = ir2(a2 ±b2) Pnin =TV — a2 m — b2 =-X,,, 2n if m and n are integers. From 13.00 (14) and (15) the TE fields are th, = jcore(i cos
mrx a
sin
nry m b
-i- sin a
mrx nry) . a cos -v- e-lx--z
(3)
max nryl a jo'm„rr in-aisin max cosnay ± n cos a sin b b a b
t. = {
max nay 20 + kit„ cos a cos .- e- '-^2 (4) }
.
§13.02
THE RECTANGULAR WAVE GUIDE
497
From (1), (2), and 13.01 (6), the attenuation of these waves is a 7 a ± 2b(vmo/v) 2 /Iva ab[l — (v,e0/v)]i Ite,""' latei n2a){1 — (v„,„/v)211 = 27- (m2b (a ± b)2 m2b2 n 2a 2 ab[l — (pm„/v)211 I I mn toot
(5) (6)
From (2), the longest cutoff wave length for the TE waves in a guide for which a = 2 cm and b = 4 cm occurs when m = 0, n = 1, for then Xoi = 8 cm. This is independent of a. The next longest at m = 1, n = 0 is Xio= 4 cm and is independent of b. As we shall see later, thq
TEi0 E lines. AB section
NIN I=1
A
B
C
D TE B-/fines. Top surface
11111 1111 -"I ■ % TE/i Elines.Ab' section
TB B-/fines.Top surface
TM„ B-/ines. AB section
TME B-/h7es. Center secbbr.
FIG. 13.02a.
guide coupling can be arranged to excite some modes and not others. For a 5-cm wave transmitted only in the TEta mode, b 8.5 X 10-7 m in copper. Thus, from (5), the attenuation is ate 0.005 so by Pc 757 or DID 550.2, the field strength drops about -- per cent and the energy 1 per cent in 1 m of guide. From 13.00 (12), the X10 attenuation is au) = 0i° = 0.37r = 0.94 or about 200 times that of X. From 13.00 (12) and (13), the signal velocity is 78 per cent that of a free wave. Figure 13.02a shows field maps of the TE N and TE11waves. From 13.00 (16) and (17), the TM
498
WAVE GUIDES AND CAVITY RESONATORS
§13.03
wave fields are `gym = 0{ cor 0' E
i— in cos MirX m ax n a
Coswiry 1 1sin mrx " Y + 11a b b b max nag + kiwgn sin sin a b nry
m
mrx
Wiry
sin — sin mrx cos + i cos km = 11-020 (-- 1n a b a b a b
(7) (8)
From (1), (2), and 13.01 (7), the attenuation of these waves is atm. =
n2a3 m2b3 2T )[1 ab(m2b2 mv3 n2a2 — (vmn/p)2it
(9)
From (2), the longest TM cutoff wave length in a guide for which a = 2 cm and b = 4 cm occurs when m = n = 1 at Xii = 3.58 cm, and the next is A
B B-lines. R4171 face E-/ines. AB sechhn TE10-TE0,
I
1
B E-lees.l&hCa/ seckbn B-/i nes . AB section TM *TM12 13.02b.
at
X12 = 2.83 cm. A field map of the TM11 wave is shown in Fig. 13.02a. Superposition of two square wave-guide fields gives that of the isosceles right-triangular guide of Fig. 13.02b with the same cutoff frequency. 13.03. Green's Function for Rectangular Guide.—A wave guide is often excited by wire loops or stubs carrying sinusoidal currents. The resultant fields can always be found by integration if the Green's function
GREEN'S FUNCTION FOR RECTANGULAR GUIDE
§13.03
499
for a current element is known. The most general current element is 1(x,y,z) ds = if ,(x,y,z) dx
jiy(x,y,z) dy
kr ,(x,y,z) dz
(1)
It is convenient to work in the Coulomb gauge of 11.01 (4) and 11.01 (5) in which the vector potential comes entirely from the solenoidal part of and satisfies the wave equation and NI,comes from the lamellar part of and satisfies Poisson's equation being everywhere in phase. If 111 is not a function of x, y, z as in a uniform current loop, then only A is needed, but if it depends on x, y, z as in stub excitation, then both A and 11,are needed for the whole field. In all cases the propagated field as well as the magnetic field comes entirely from A. The Green's function must satisfy the boundary conditions that the electric field is normal to the guide walls and the magnetic field tangential to them. The procedure will be to find Green's function for I x, Iv, and rz and form the general case by (1). The calculations are made for elements at z = zo in a guide running from z = — co to z = 00. The function for a guide closed at z = 0 is then obtained by superimposing an image Green's function of suitable sign and orientation in which —zoreplaces zo. First consider a y-directed element. It is evident that the guide walls could be replaced by an infinite set of images in the z = zo plane, each image being parallel or antiparallel to the element and so having a magnetic field normal to the y-axis. Such a field is given by replacement of k by j in 13.00 (2), so that
A = v xi xviTT, B = v=jp2tiV W = — emn sin in 7x cos n71a
(2) (3)
Ampere's circuital law, 7.01 (2), is used to get On0n for which needed. From (2) and (3),
B
x
nay 0' frm Z Zol Om ' „CY...co sin max cos n -
= j02
E. is
(4)
a
m=1 n=0
Now integrate E across the guide at constant y in the z = zo plane except at x = xo, which the path avoids by an infinitesimal detour as shown in Fig. 13.03a. On this path, from symmetry, B, is zero except between Yo and yo dyo where it passes around the current element and equals i.t/ from 7.01 (2). Multiplication of (4) by sin (nom/ a) cos (niry/b) dx dy and integration over the ranges 0 < x < a and 0 < y < b yield ab cos nrY° = jitn021ab(1 dyo sin in7x°
(3(.1)'7,0,n(y)
(5)
500
WAVE GUIDES AND CAVITY RESONATORS
§13.03
This determines emn(y)for A (,,) which takes the form
A(y) =
.1einn(Y) [
mnir 2 maxx sin "Y b cos a ab
m=1 n=0
j(02
n 2r2) b2
max
sin
k ji3„%nnir
b
a
cos
nay
max
sinsin a
n
vi
2 ' n)2 = ,021.te _ Mir 2 _ n'Ir Om = "32 - inn T (
(6) (7)
There remains the local field due to the lamellar part of iy dyo. This is found from problem 113, Chap. V, by interchanging m and n, using the
(a)
FIG. 13.03a, b.
(b)
13.7, notation of (7), writing f„/(jw) for q, differentiation with respect to yo as in 1.07 (1), and replacement of (aV/ayo) dyo by if. Thus .0 dyo N mirx n . mrxo nay „ wry° sin cos sin sin b (8) jo.)Eab2 L-J a a Pmn m=1 n=1 The potentials of the x-component in (1) can be formed from (5), (6), and (8) by interchanging xo and yo, a and b, m and n, and x and y as well as permuting i, j, and k in Am. Thus A
m 2,2
(x) =
m=o
n=1
.Mn7r 2
ab -
(d32 -a2 ) sin "Ycos 7717x -iC ,,n(z)[i a
nry] iz-zoi cos "Y sin max a izjit,nmir sin sin MirX si --b-- e_.0,a
21z dxo N joma2b =
m mrxo nry max n' sin cos sin sin rye a a Nmn — b
n=1
(9)
(10)
GREEN'S FUNCTION FOR RECTANGULAR GUIDE
§13.03
501
As there are no orthogonal functions in the z-direction, a description of the dipole moment in terms of x and y, such as that worked out following 12.21 (8), is needed. This suggests producing a dipole field by the arrangement of Fig. 13.03b in which the z = zo plane is earthed except for the 62area at x = xo, y = yo, which is raised to potential V 00". To get Vo in terms of Izdzo, start with 12.21 (8). IZ dzo =
jcore( 2 — a2)Vo
In (b/a)
a—> b
2j COE V ora2 = 2jcoE Vo dS
(11)
Clearly the magnetic fields of iz dzo and all its images are normal to z and given by 13.02 (7) and 13.02 (8). Thus the tangential component of E in the plane z = zo is .nt = 0)71-
rt,t(z)0,rzn[
a
m=1 n-=1
cos
max .n max sin " — sin Y j cos " bY a b a (12)
Denote the bracket by [ ], take the scalar product of (12) by [ ] dx dy, and integrate over the zo-plane. The integral of [ ] • [ ] dx dy on the right gives 4abfi,n2„. On the left side the tangential E is zero except on the boundaries of the Vo square, and its integral across the boundary is Vo or — Vo depending on the direction of crossing. Therefore the integral of the left side over the vertical edges of the square is 1 1 V o 6— {cos [m —7(xo + — 3)1 — cos [ m —7-(xo— — 5)]} sin "Y° 2 2 a a a The y-integral is similar with m, n, yo, xo, a, b, replacing n, m, xo, yo, b, a. Add the two integrals, combine terms and note that 5 is small, and obtain for the left side VoPrO,„ 2 „ sin (mrxo/a) sin (nryo/b). Thus -2./z dzo amn ( z) = n
otEablini t 7,
sin
norxo
a
sin
n 71-Yo
b
(13)
If 0 in 13.02 (7) is replaced by 0,„„w and it is summed over m and n and multiplied by j/(.0, it becomes A(2) . max miry .n max nay „[i— m a cos asin b + 3 sin a cos
_= m=1
n=1
— kgn sin '71-x 7ria sin nay e—isc-11-41 (14) The scalar potential for the z-directed dipole can also be written down from problem 113, Chap. V, by calculating (a V/azo) dzo and making the
502
WAVE GUIDES AND CAVITY RESONATORS
§13.04
same substitutions as for (8). = 2.Iz dzoN / coma b
sin
max nary n yo mirxo sin sin sin e-0--1z–zol a a
(15)
m = 0 n=0
The integral of the Green's function over a wave-guide antenna plays the same role as the retarded potential integrals in free-space antennas, and both require a knowledge of the current distribution, which is difficult to calculate. A plausible sinusoidal distribution is usually assumed, and the transmitted wave is rarely very sensitive to errors in this assumption. If the wire in loop or stub is not too thick, then the fields outside it are essentially the same as for an infinitely thin antenna along its axis. In input impedance calculations the fields between axis and surface must be excluded to avoid infinite field energies. Each mode acts as an independent circuit shunted across the antenna element. A given current element excites every mode that has an electric field along it. The antenna impedance is minus the sum of the line integrals of the mode electric fields along the wire surface, in the direction of the current. Add to this the capacitive contribution of the scalar potential (if any) field integrated over the surface of the wire. It will be noted in the rigorous inductance formulas of Chap. VIII that the per unit length inductance of a wire becomes infinite if the radius becomes zero. If the input reactance of a thin wire-guide antenna is found by integration of the mode fields along its surface, no single term approaches infinity, and the convergence of the series becomes slower and vanishes at zero radius. Thus in the rectangular-guide case where the walls can be replaced by image antennas, a much better method is to add to the free-space antenna reactance, which can go to infinity and is often in closed form, the mutual reactances of the image antennas which are usually independent of the wire radius. The resultant series is then rapidly convergent. Examples of such calculations appear among the problems.
13.04. Aperture Excitation of Wave Guides. Coaxial Opening.Wave-guide fields set up by waves entering through orifices in the sides or end give many of the same calculation problems encountered with waves diffracted through apertures and discussed in 12.18, 12.19, and 12.20. Few exact solutions are possible but in many cases the assumption that the incident wave in the opening is unperturbed as in 12.19 gives useful results. If the aperture dimensions are sufficiently small compared with a wave length, quasi-static methods give correct tangential fields. Thus the double current sheet method of 12.18 yields accurate solutions.
§13.04
APERTURE EXCITATION OF WAVE GUIDES
503
An example of aperture excitation which uses the double current sheet of 12.18 and the Green's function of 13.03 is the coaxial opening shown in Fig. 13.04a centered at x = d, y = 0, z = c in the bottom of a rectangular guide closed at z = 0. The frequency is such that only the TEiomode is transmitted, and it is absorbed at some positive value of z. The value of the output resistance of the coaxial line is desired. To make E, zero at the closed end z = 0 requires two double current sheets of opposite sign, one at z = c and one at z = — c as shown in Fig. 13.04b. The electric field of the TE10 mode everywhere parallels the y-axis, so the infinitesimal thickness dyo of the toroidal double layer is exaggerated in Fig. 13.04c. The exact value of E over the opening is unknown, but the
z
7
dyo
(a) (c) Fin. 13.04a, b, c.
field of the principal TEM mode in the coaxial line which transports the power and is the only field far from the opening is inversely proportional to r. If the potential of the central conductor is V and the current density in the double layer in amperes per radian is i, then from the text below Fig. 12.18, dy o _
E r
V r In (r2/r i)
(1)
The currents encircle the entire section of the toroidal current sheet of Fig. 13.04c. Thus they are solenoidal and none of the scalar potentials in 13.03 can appear. The amplitude of the TE10 mode excited by a vertical element of width de may be written down from 13.03 (6) by placing the torus in the z = 0 plane in a guide running from z = — 00 to z = co and bounded by x = 0, x = a, y = — b, and y = b. From symmetry the y = 0 plane is then an equipotential except for the coaxial
504
WAVE GUIDES AND CAVITY RESONATORS
§13.04
aperture. Putting m = 1, n = 0, 1 = i d0; y = 0, and 2b for b in 13.03 (5) gives def lo—
dyo sin(71-x0/a) d0 2 jabOi00 2
— V sin (irx0/ a) d0 coabOVio In (r2/r1)
(2)
For the total amplitude this must be integrated over the vertical walls. It is simpler to combine the four elements at d ± r2 sin 0 and c ± r2 cos 0 shown by crosses in Fig. 13.04b so that for the outer wall the integral is 1 irr2 . — 5 cos (— sin 0) cos (310r 2 cos 0) d0 a 20
(3)
In the notation of 13.03 (7) with tan q5 for 7/(13/10a) this may be written Ifol'1cos [Or2 cos (0 — (1))] + cos [Or2 cos (0 + O)]) dB
(4)
so that the value of the Both terms in the integrand have a period of integral is independent of 0. With 4 = 0 it becomes 5.302 (3) so 7rd or2) 2
2f wcos ((3r2 cos 0) d0
2 2A jX
(5)
Addition of similar expressions for the current at the inner edge of the r2 < z, torus and for the image torus in Fig. 13.04b gives, with c A lo = io sin Clo =
ei(c"—,3'ioz)
T-7- sin rd J 0(2/7-r2/X) — Jo(2/rr IA)
jcod(3 0ab In (r2/r1)
(6) (7)
The TE10 electric field and the magnetic induction are B. =
-aA10
az
- j/3i 0A10,
ku =
(8)
The power transmitted is the mean Poynting vector
1EP
ax jiwOlo sm2 — I Ci01 2. a
(9)
Integration of this over the range 0 < x < a, 0 < y < b and setting the result equal to 1V 2/R give for the radiation resistance R=
Ac013'10ab lng (r2/ri) sing (rd/a) sing (13'10c)[Jo(airr2/X) — J0(271-ri/X)P 871-2 sing
The same result is obtained by the integration of the Poynting vector over the coaxial opening, using r1 V[r In (r2/r i)]—' for E and v x A10 for B, but the integration is more difficult.
§13.05
RECTANGULAR GUIDE FILLED BY TWO MEDIUMS
505
13.05. Rectangular Guide Filled by Two Mediums.—Suppose the guide is filled from x = 0 to x = c with a lossless medium Aici and from c to a with /22E2, where a — c = d. Tha wave velocity or wave number and the cutoff frequency are desired. To satisfy the boundary conditions at all times and all y values, the waves in both mediums must have the same velocity and y dependence. This is impossible for most waves purely TM or TE with respect to z but works out well for waves TE or Tit/with respect to x. Such waves derive from 13.00 (2) if i and Wter replace k and Wte. The (irm/a)2 term in 13.02 (2) is still valid but the (irn/b)2 term no longer holds and will be replaced by pi in AiEiand pi in 112E2. Thus the W,„ forms in mediums 1 or 2 that satisfy the boundary conditions at the guide walls are A l sin pix cos
miry .,,„pz
or
A2 sin p2(a — x) cos nwye-oc,,z
From the relations in 13.00 (12), when 13i = (42µ1E1 and i3=
(42122E2,
— (7)2— pi = 4 — (12—
(13:.p)2 =
(2)
For continuity of E, and H, at x = c, (E0 1= (Ez)2 and A2(1301 = Thus awte. since Ez = A2 sin plc = A2 sin p2d ay 1.1.2pii12
cos pic = —mip2A2 cos p2d
since Bz =
(1)
a wtex
(ax (3z)
2•
(3) (4)
The ratio of (3) to (4) gives the transcendental equation ii1232 tan plc = —A2p1 tan p2d
(5)
The same equation results from (Ey) 1/(By) 1= (Ey)2/(By)2 at z = c, so all tangential components of E and H are continuous if (5) holds. Equating the normal components of B gives (3). For a given frequency and medium, piand p2 as functions of Om% from (2) may be inserted in (5), the left and right sides plotted as a function of 13„' p, and its correct value found by their intersection. For the cutoff frequency is zero, so from (2) 2 kFlic = WcAlE1
/MIA 2
(A) c = 04A2€2
97)2 (+
(6)
Insertion of these values in (5) and solution of the resultant transcendental equation as before yield coc. Note that if p i is real in (6), then p2 is imaginary and vice versa. The cutoff frequencies will lie between
506
WAVE GUIDES AND CAVITY RESONATORS
§13.06
those of the equivalent modes in aif i and a2E2. There are an infinite number of values of pi, p2, and Eoc. The boundary conditions for this guide can also be satisfied by waves TM to x by a perfectly analogous procedure. The chief results appear in problem 18 at the end of this chapter. 13.06. Thin Iris in Rectangular Guide.—The junction of two rectangular wave guides, which are affixed normally to opposite sides of the perfectly conducting z = d plane in which holes have been cut to connect their interiors, forms a plane discontinuity. If in both guides only the TEiowave is propagated, then standing waves can be set up so that there is an electric node at z = d. There is then no transverse electric field za
zs Z=0
Z=d (a)
NP
Z=2d Z=0
Z=d
Z =2d
(b)
(c)
Yi
7
X NF.
—x— X t --x-—b —1 +1 +b (d)
(e)
(f)
FIG. 13.06a, b, c, d, e, 1.
there, and no magnetic field links any element of z = d. In a transmission line only a pure shunt element Zsmay be inserted at an electric node without upsetting the pattern. Now consider the special case in which a thin conducting iris at z = d in a rectangular guide partially blocks TEio standing waves set up with nearest nodes at z = 0 and z = 2d. From 13.02 (3), the TED)electric field when 0 < z < d is
( frx Ey = E0 sin T1 )sin Olaz cos wt
(1)
Higher mode fields must be present to cancel this field where it is tangent to the iris. By 13.02 (3), these contain the factor e-1 0'--(z-0which, when X10 is large, will confine them to a z-interval so short that all local fields are in phase and may be found by static field methods. This is called the "quasi-static" approach.
§13.06
THIN IRIS IN RECTANGULAR GUIDE
507
The wave-guide section 0 < z < 2d is equivalent to the short-circuited T section of Fig. 13.06a in which the currents are directed as shown. It is usual in treating wave guides to use "normalized" impedances which we designate by a superior °. These are found by dividing each impedance by the characteristic guide impedance which normalizes to 1. For a guide transmitting only the TE10 mode with no attenuation, 11.15 (13) is zo _ 2i _ 2% cos C/ + j sin /3of z 2, cos o'io/ jZ% sin f3 of
(2)
For the short-circuited T section, 2°,, = 0 and 1 = d, so for resonance = —42,? =
tan iTiod
(3)
Now consider the special case where all sections parallel to the x = 0 plane, and hence to the electric field, look alike. The electric fields parallel the x = 0 plane so that, in the region near z = d where phase differences may be neglected, E at any value of x must obey the equation of continuity in y and z and can be derived from a potential that satisfies Laplace's equation in y and z. Identical currents cross all electric nodes in a guide so that the total charge on top and bottom between nodes is always the same. Let the charge per unit width in the interval 0 < z < 2d at some value xo of x be Qio for the TE10 mode and Qo for all modes. Let the static discontinuity capacitance per unit width be C1and the potential across the guide at x = xo, z = d be Vd = bEd. Then from (1),
C l = Qo —Qc— 2 (fixi'6E, dz bEd bEd 0
f deE, dz)b— oco 2 o
6
cot bo 13'10d
(4)
Insertion of this value in (3) gives
e 1 Z° = . = . ' 3b0i0C1 3WC°
or
C° = —C1 WE
(5)
As a specific example, consider an iris which leaves a window of height c in a guide of height b, as shown in Fig. 13.06b and Fig. 4.22b where the equipotentials are dotted except at the boundaries U = 0 and U = 47. Insertion of a = c, W = jV , and z = jd in 4.22 (6) gives for V, as V and d approach infinity,
V = 11-rd — ln sin 47c— b
b
ln csc — b
(6)
The charge per unit length between 0 and d is E V. The last term times E equals half the additional charge on the guide due to the presence of the iris for a potential difference of fir. Therefore twice the second term multiplied by 26/a gives C1. Insertion of C1into (5) and substitution
508
WAVE GUIDES AND CAVITY RESONATORS
of 27-X9— ' for (310give for the normalized susceptance — In csc B° = wC° = 8b ,
§13.06
B°
-brc
(7)
For the symmetrical iris bounded by U = —ir and U =sir in Fig. 4.22b, the only change is the potential difference, which is now 7r instead of fir, so that (7) must be divided by 2. The least attenuated TE11 mode is now out by symmetry; thus the result is more precise. From Fig. 4.22b by images the center strip behaves exactly the same way as the symmetrical iris. The errors in the above value of B° are about 2 and 9 per cent at b/Xe values 0.1 and 0.2. These can be improved by a variational method given later. Another interesting case is when all sections parallel to the y = 0 plane, and hence to the magnetic field, look alike. Suppose the iris is at z = d and the frequency is such that only the TE10 wave propagates. All fields are independent of y, so n is zero in 13.03 (1) to (4) and the betas are
ai
2r = — X,
=
( 32
01)4)
=
co
Mr
/3 = (40
2r ),
(8)
This case, unlike the previous one, has charges only on the guide walls. Thus in the Coulomb gauge of 11.01 the entire field derives from the vector potential. Take the origin at one end of the short-circuited T section of Fig. 13.06a so that the B component amplitudes at z = d are Bz = A1 f3i cos (,3;4) sin (31x)
mAm1,8'„I cosh It3'„di sin (0mx)
(9)
m= 2 Bz = — Av3i sin (M) cos 0(31 x)
—
mAmQm sinhl0:,,di cos
(0mx) (10)
m=2
Multiply (9) by sin i31x dx and (10) by cos ,31x dx, integrate from x = 0 to x = a, eliminate A1, and substitute from (3) to give tan
Wd) = —
O'i afaBy cos ,31x dx ° 7rf
(11)
B, sin /31x dx
where the shunt impedance is shown in Fig. 13.06a. This requires the static approximation for the higher modes only at z = d. As a specific example, take the symmetrical iris with a gap of width c as shown in Fig. 13.06d. To visualize the equivalent magnetostatic case, let X >> a so the magnetic loops in Fig. 13.02 are much elongated in the z-direction. As the impermeable iris edges push in from the walls, they press the lines together as shown in Fig. 13.06c. No matter where
§13.06
THIN IRIS IN RECTANGULAR GUIDE
509
this occurs in the loop, Bzin the window is positive at one edge, negative at the other, zero at the center, and vanishes on the iris outside the window while Bs is nearly zero in the window and tangent to the iris walls. If the guide walls in Fig. 13.06d are folded outward at the iris junction to the coplanar position, Fig. 13.06e results. These two maps bear the same relation to each other as those in Fig. 4.22b. The field in (b) comes from cutting a slit 2 units wide, shown in Fig. 13.06e in an infinite sheet of permeability zero that separates two equal and opposite uniform magnetic fields. Figure 4.23 shows the result with only one field present. Superposition of a similar field rotated 180° gives Fig. 13.06e, and application of the transformation below Fig. 4.22b, but with the origin at the iris base, is Fig. 13.06d. W = C/(42 — 1)1, Eliminate zi and make z' =
zi = b1sin [:(z' — yt)]
(12)
— c) when z1 = 1 to give
W = C(cos' 7rz — sin' /71 a a
(13)
For the z = d plane of the guide, y' = 0 so this yields, by 4.11 (1), Bz jB, = ±C
a[cos2(lrx/a)— sin'(2rc/a)]1 -
(14)
Thus with respect to the guide axis, B z is an even function that is zero when i(a — c) < x < 4-(a c) and By is an odd function that is zero if c) < x < a. The integration in (4) can 0 < x < i(a — c) and if 4(a be done by parts, letting dv = B, dx and u = sin (Tx/a) in one and dv = Bz dx and u = cos (rx/a) in the other. The uv products vanish, and if sin (71-x/ a) is replaced by t cos (lirc/a) in one and cos (lrx/a) by t sin (irc/ a) in the other, they lead to Dw 350.01. Substitution in (11) gives =— a tan' L-c = jo.)./),? a X,
(15)
Thus the iris is inductive, and the normalized shunt susceptance is B° = = — — 1 =— Yk
X ire) cot' (2a a
(16)
An exact value could be found by solving an integral equation given in 13.08, but this is impractical. A better way is to improve the above solutions by a variational method. This method, also applicable to some static problems, will now be discussed, as in Borgnis and Papas.
510
WAVE GUIDES AND CAVITY RESONATORS
§13.08
13.07. A Variational Method for Improving Approximate Solutions.— Let some physical quantity such as the iris susceptance in 13.06 be expressed in the form of the integral equation
Q=
fsh(r)E(r) dS
(1)
where h(r) is known but E(r) is an unknown function, such as the electric field, of the coordinate r in the surface 8, which can be the window area in 13.06, over which the integration extends. Now consider the integral
h(r) = fs,G(r,r')E(r') dS' where r' relates to area S' and G(r,r') = G(r',r).
(2) Put (2) into (1) so
Q = fs,E(r')G(r,r')E(r) dS dS'
(2)
Division of (3) by the square of (1) yields 1
LLE(7.1)G(r,r')E(r) dS dS'"
Q
[fsh(r)E(r) dS] 2
(4)
This expression has the useful property that, if trial values of E(r) are inserted, the Q value is stationary about the correct value of E(r) so that a first-order variation in E(r) gives a second-order Q variation. To prove this multiply both sides of (4) by the denominators giving
Q fs fs,E(r')G(r,r')E(r) dS dS' = [fsh(r)E(r) dS] 2
(5)
Take the variation, which follows differential rules, thus
Qfs fs,SE(r')G(r,r')E(r) dS dS'
Q fs fs,E(r')G(r,r') SE(r) dS dS'
= — SQ[fsh(r)E(r) dS] 2 2fsh(r) SE(r) dSfsh(r)E(r) dS (6) Since G(r,r') = G(r',r), r and r' may be interchanged in one integral on the left so it adds to the other. In the last right-side term put (2) for h(r) in the first integral and the Q of (1) for the second. Double integrals cancel leaving SQ = 0. The value of (4) is independent of the E(r) units. Borgnis and Papas, Collin, Harrington, Marcuvitz, and others treat the subject fully. Schwinger first used this method in this way. 13.08. Inductive Iris Susceptance by Variational Method.—The technique of the last article, where Q is B° or (B°)-1, may be used to improve
§13.08
INDUCTIVE IRIS SUSCEPTANCE
511
the quasi-static solutions of Art. 13.06. The accuracy of the result depends on how close the trial value of E(r) is to the true value of the tangential E in the window or the tangential B (current density) on the iris. The conformal transformations of 13.06 will provide good trial values in both cases if the static E, and B, have the same directions as the actual E, and B,. To find the latter in the capacitive case, both vector and scalar potentials are needed in the Coulomb gauge because there is charge inside the walls and the divergence of the iris current is not zero. Thus it is easier to use the magnetic Hertz vector of 11.01 (13) for the fields. This shows that E, is y-directed, but B, has both x- and y-components and thus is not compatible with the quasi-static value. This case will therefore be relegated to the problems at the end of the chapter. There is no charge on the inductive iris; therefore only the vector potential of the Coulomb gauge is needed. This gives both E, and B, independent of y in both magnitude and direction and provides an ideal case for using the quasi-static fields as trial values in the variational integrals. From 13.02 (3) and 13.02 (4), after the absorption of some common factors into C., the guide fields become 00
ty(x,z) = /0)
(1)
O. sin i3„,,xe-js'-z m=1
CO
(x,z) =
Cm(—i(3;. sin 0.x + kj0. cos (3.x)e-jtv-z
(2)
m=1
where m is odd since the fields are symmetrical about x = 4a. In the iris plane the electric field is zero on the iris surface and Ea in the window. Multiplication of (1) by sin Ona dx and integration over the guide section give C. which, put into (1), gives CO
2 E(x) = a
sin (3,,,,x f Ea(x') sin 0,ax' dx' m=1
a
(3)
But in the window, as proved in the paragraph following 12.18 (4), .13(x) is that of the incident wave, which is normalized to sin @ix). From 11.00 (1), Bix(x,0) is (.0 -1ag:(x,z)/az. Thus transfer of the m = 1 term, which is the TE10 amplitude, to the left side gives at z = 0 — f Ea(x') sin Oix/c/x'i =
01 sin 4'
a
13',„ sin Ornalata(x') sin fi'mx' dx' m=3
Multiplication of both sides by Ea(x), integration over the aperture, and
512
WAVE GUIDES AND CAVITY RESONATORS
§13.08
division by O'd fEa(x) sin 131x dx]2give a 2f Ea sin Oix dx
1 = NO'„[f Ea(x) sin 0,„x dx] 2 01[ fEa(x) sin 01x dx]2
(4)
m =3
The left side of this equation is expressible in terms of the iris susceptance. Let the total potential and current be V1and I1when z < 0 and V2 and /2 when z > 0 as in Fig. 13.06f. Both the incident wave V, and the reflected wave V, contribute to V 1, so that V1 = Vi Vr,
ZkI 2 = V2
Zkii = V, — Vr,
(5)
Combine with this the relations in (6), valid at z = 0, to get (7). V1 = V2 = 2V, = VI ± ZJI = 2V2
(6)
I 1 = I 2 + IL
jCOL/L,
V2 jc,„'L
so
Zk
j coL
2 (V,
1)
2
(7)
From (3) the ratio of normalized TE10 incident to TE10 transmitted is ri 2
=
Vi V2
2 [-j (
)
„ , ]--1 sin 13 a x
(8)
Substitution of this into (4) and replacement of ji3:„ by (131 — 02)1give Zk
-B Yk
= —Bo = m =3
2031 — (32)4[ fEa(x) sin 0,ax dx] 2 Ea(x) sin Oix dx] 2
(9)
Integration of both integrals in (9) by parts replaces the sines with cosines and Ea(x) by aEa(x)/ax which, by Maxwell's equations, can be replaced by —jcoBz(x) since Ea(x) is y-directed. Writing kBa(x) for jc4E(x) and replacement of some of the betas in (9) by (1) give aB° =
(1 — 4a2m-2X-2)1[
f Ba(x) cos 13,nx dx 2 fBa(x) cos 01x dx
(10)
m =3
The imaginary part of 13.06 (14) gives an excellent trial value of Ba(x). Integration limits are x = 2(a — c) and x = c), but the integrands are even about x = yt and thus the limits za and c) may be used. Since m is odd, replace it by 2n 1 so that the sum goes from n = 1 to n = 03. In 13.06 (14) and in (10), substitute 0 for 131c and -(2n + 1) (7r- + 4)) for 13„,x. The integrals then lead to Mehler's Legendre function integrals found in MO, page 52, WW, page 315, HMF 22.5.36 and
INDUCTIVE IRIS SUSCEPTANCE
§13.08
513
22.10.11, and elsewhere. Thus 'sin 0 sin (n 1)0 d4
g [cos (n — 1 1)0 — cos (n ) 0] d0
1/0 (cos — COS 0)1 — t: 2 0 (cos (1) — cos 0)1 = C3[Pn_3(cos
— P.±1(cos 0)] = C3
2n ± 1 sin OP;,(cos 0) (11) 1)
n(n
When n = 0, the denominator integral is C3(1 — cos 0). Substitution of these values into (10) and application of Dw 406.3 give B° =
a
cot' (7c)[(2n + 1) 2 2a
—
(2111P! cos (irc/a)]2 n(n ± 1) X
(12)
n=1
The convergence of this series can be improved by taking the difference between it and a summable series having identical terms when n is large. Note that if a = 1 and a = 0 in 7.13 (2), then the r integral of A0 from 0 to 00, if r-n-' replaces rn when r > 1, is 2(u/ sin a)-'fo'Ao dr =
(2n + 1)n-2(n + 1)-2[P;,,(cos a)]2 (13) n=1
Here A is the vector potential of a circular loop of radius sin a lying on a unit sphere. Doing the same integration on the first integral in 7.10 with a = sin a, p = r sin a, and z = (r — 1) cos a gives (2n + 1)n-2(n + 1)-2[P,1(cos a)]2 = r-ircos 0 In (1 — cos .0) d¢ = 1 n=1
(14) This might have been foreseen because X = 00 was a tacit assumption used to get 13.06 (16). Thus the series in (12) may be replaced by 2
[(2n
1+
1)2 — 4a2X-91— 2n — 1 I[n-'(n
1)--1P,I(cos --Ca )]
n=1
(15) in which the series converges so rapidly that in the cases checked five terms gave three digits correct in B° and ten terms gave four digits. Jahnke and Emde tabulate dPn/d0 and IP„1 (cos = IdP,i /c181. The integrals in (12) are positive definite and are an absolute minimum for the true fields so (12) provides an upper limit for B°. A short proof of this appears in Borgnis and Papas, page 122. A lower limit is found by use of a variational expression, based on the obstacle current, which
514
WAVE GUIDES AND CAVITY RESONATORS
§13.08
bears a strong resemblance to (9), and reads
fil(131 - 0 2)1 fol(a 4)Bz(x) sin 0„,x dx Yk =
1 = m=3
B°
B
2
}(a-c)
[JO
Bx(x)
sin 131x dx
(16)
]
2
where the obstacle current density iy(x) has been replaced by Bz(x) to which it is proportional when z = 0. The integration is over the iris surface since By is zero in the window. Note that all terms of (9) except the integrals have been inverted. The real part of 13.06 (14) is a good trial value for B5(x). Put this value into the numerator integral of (16), and substitute 7 - (#) for 27x/a, 0 for ac/a, and 2n + 1 for m, which is odd from symmetry. The integral then becomes, disregarding the sign, 'sin 4) cos (n 4)cf, dc/o (cos cos cOi
Cf 0
4)0 - sin(n - cf.] d4 0 - cos 0)1 2n ± 1 sin OP,1(cos 0) (17) = C3[Pn_1(cos 0) - Pn±i(cos 0)] = C3 n(n ± 1)
C2 fe7sin (n
(cos
The second form of Mehler's integral, already referred to, was used here. When n = 0, the denominator integral is C3(1 - cos 0). Substitution of these values into (16) and use of Dw 406.2 give a(7-c (2n + 1) { PRcos(71-c/a)]) 2 a)L-J [1 + 4a2X-2(2n 1)-21in2(n + 1)2
1
(18)
/76 = - )g tan2
n=1
Exactly as with (12) the series may be put in the rapidly convergent form
Jr 1
1+ n=1
4a2 F f Pgcos (rc/a)R2 1}(2n ± 1) (2n + 1)2X2 n(n ± 1)
(19)
This provides a lower limit for the susceptance. The following table gives values of the correction factor by which 13.06 (16) should be multiCORRECTION FACTOR TABLE
X = 0.5
X = 1.0
X = 1.3
c/a 0.2,0.8
0.4,0.6
0.2,0.8
0.4,0.6
0.2,0.8
0.4,0.6
Upper limit
0.9768
0.9582
0.8995
0.8149
0.8102
0.6385
Lower limit
0.9758
0.9558
0.8745
0.7550
0.6760
0.2940
Arithmetic mean
0.9763
0.9570
0.8867
0.7849
0.7434
0.4662
Handbook
0.9765
0.9578
0.8913
0.8045
0.7594
0.5881
§13.09
THE CIRCULAR WAVE GUIDE
A
515 C
B MI mode. E-lines in AB section spaced to give equal intensity increments.
T.Eol mode. Principal section of equiflux tubes of magnetic induction.
A
rTh TEI1 mode. E-lines in AB section spaced to give equal intensity increments on center line.
TEli mode. B-lines in principal section spaced to give equal induction increments on center line.
A
TMol mode. B-lines in AB section spaced to give equal induction increments.
TMol mode. Principal section of equiflux tubes of displacement.
A
C
B
D
T/14-11 mode. E-lines in principal section TM„ mode. B-lines in AB section spaced to give equal induction increments on center spaced to give equal intensity increments on line. center line. Fin. 13.09.
plied according to (12) and (18) and their average compared with the accurate value of the la formula on page 221 of the "Waveguide Handbook" over the ranges plotted on pages 222 and 223. It is clear that the variational treatment has made a great improvement in the quasi-static value and that the upper limit alone does pretty well. 13.09. The Circular Wave Guide.—Consider a wave guide of radius a. From 12.16, a solution of 13.00 (3) which is finite on the axis is 0(p,4) = J „,(0.7,p)(Om, cos m4) +
b„,„ sin m4)
(1)
516
WAVE GUIDES AND CAVITY RESONATORS
§13.10
where, by 13.00 (4) and (5), Omn must be chosen so that I aJm(amna)/aal to = 0
(2)
1,/m(Retteta)lt. = 0
The most important zeros of these functions are J'0(3.832) = J'0 (7.016) = Pi (1.841) = J'1(5.331) = 0 J0(2.405) = J0(5.520) = J1(3.832) = J1(7.016) = 0
(3) (4)
From 13.00 (14) and (15), the fields of the TE waves are [E]te = jco[pimp-I J.(3mo)(Omet sin mct) — /5„,„ cos melo) „,„. ± 43/3m,,Fett(Oette?)(C,,,„ cos m(#1 ± Dm. sin mcp)]e--)0'
(5)
[bite = 1:713:tta — ViOmetP,,,(0emP)(Ctu. cos m4) + /5,,,,t sin md›)
+ (1)mP 1,1.(0..P)(C>r sin mqS — /57.1„ cos mcb)] + kol„Jm(0.p)(Cmn cos MO + Dmn sin m et,) }e 'A'n'ne
(6)
From (1), (2), 13.01 (6), and 5.295 (6), the TE wave attenuation is ate =
T
[1
/Iva
—
Pm, 2
—4
M2 Olna2
—
m2
±
1,„, 2
I
(7)
We notice that the circularly symmetrical waves for which m = 0 have the unusual property of a decreasing attenuation with increasing frequency. The longest TE cutoff wave length A ll is, from (2), (3), and 13.00 (8), at 3.42a and the next, An, is at 1.64a. Take a = 2.34 cm so that A n is 8 cm and compare with the 2- by 4-cm guide of the last article at the same cutoff. The 5-cm free wave length attenuation is 0.0028 in-' which is considerably less than the 0.005 m--' of the rectangular guide. Field maps of some of the TE modes appear in Fig. 13.09. From 13.00 (16) and (17), the fields of the TM waves are ( ./5,„„ sin Ind>) E tm = oge,t[pAtt.T„t(t3mnP)(eett cos mcA — 4mp 1Jett(0m,tp)(C,,,„ sin mg!) — limncos mcb)] kiwnteint(t3ettitP)(Cmn cos mci) + Ann sin m(5)}e--0'—z (8) Btm = [p1o2mp-1J,.(0,..p)(c. sin m(tt — /5„,„ cos m40) 15— sin ntt)]e-30'—' (9) + (13/32t3ttta'n(t3,7t.p)(0.,, cos mcP From (1), (2), 13.01 (7), and 5.295 (6), the TM wave attenuation is )2
at
[1— „, = pava
n
I
(10)
The longest A,en values are A01= 2.61a, A11= 1.64a, and A02 = 1.14a. 13.10. Green's Function for a Circular Guide. The wave patterns set up by a wire loop or stub carrying sinusoidal currents can be found by integration of the Green's function of a current element. Instead of the x , y-, z-components used for the rectangular guide in 13.03 (1), the cur—
-
§13.10
GREEN'S FUNCTION FOR A CIRCULAR GUIDE
rent element will be resolved into p-, I(p,4),z) ds = oil „(p,4.,z) dp
517
z-components, thus
4)i i,k (p,4),z)p d4)
klz(p,4),z) dz
(1)
As in 13.03 the Coulomb guage, in which A arises entirely from the solenoidal part of I(p,4),z) and 4/ entirely from the lamellar part, is convenient. The comments of 13.03 also apply here. The calculations and results for radial p- and longitudinal z-cornponents follow, but the 0-component results, whose calculation resembles that for the p-component, are given in problem 27 at the end of the chapter. Consider a current element of strength pi./0" and of length dpo at po, 0 in the z = 0 plane. If the element is at 0 = 00 instead of at 0 = 0, — oo replaces 4) in the equation of the last article. The complete TB solution is a double sum over m and n of terms like 13.09 (5) and (6). To find the amplitude of the pq-term, one sets t = 0, z = zo, and Cm. = 0 in 13.09 (6), E, being an even function of 0, takes its scalar product by [910pq,P9(07,p) sin pct. + 4:1)Ipp-',/p(Op0 p) cos p4)11) dp d4)
and integrates from 0 = — r to 4) = r and from p = 0 to p = a. The first integration eliminates all but the pth term in the m-summation and puts r in place of sin' p4) and cos' 730. If v1is written for i3 ,,p and v2 for 13„p, the remaining p-integrals are 11-1pq0;ilf:P'a [JVV1),Pp(V2)
,2 dv .-Jp(V1)Jp(V2)11 1
v iv2
(2)
By 5.296 (11), the integral is zero unless n = q when, by 5.296 (12), its value is 1(027,0a2 — p2),Ig0„a). In the zo-plane, 13, is zero everywhere and Bois zero except over the element dpo at po, 0. But by 7.01 (2), at po f Bo p do is —1ALI . Solving for D„ gives PA/ dpo Jp(0„p o)
(3)
[/5„]te ==rj 4 ,a)]2 po0;),(13La2 — p2)[(13
For the TM modes excited by the radial element, set t = 0, z = zo, and D.. = 0 in 13.09 (9) and take its scalar product by [pipp-iJp(i3„p) sin MI + 4303,J;(13p,p) cos p4dp dp d4)
And integrate from 0 = — 7r to ¢ = 7r and from p = 0 to p = a. The integrals are identical with those found before except for a 2 — SP factor, and the integral in (1) is 1[090aJ;(0„a)]2because Jp((3„a) is zero, so
=
dpo J;(f3„p0)(2 — 5;1 ,) 0,
2r)3„[(3aJ;(f3„a)]2
e
"
(4)
518
§13.10
WAVE GUIDES AND CAVITY RESONATORS
For the longitudinal component, use the source of 13.03 (11), set z = Dmn = 0 in 13.03 (8), and take the scalar product of both sides by - (1)A4(014/3) sin PO] dP dcb = Rp0 dP
[000p0Pp(i3p0P) cos
dcb
Integrate over the ranges 0 < p < a and 0 < (42
(12)
The first and third terms have the same form and the third lies closer to Therefore the function Ur") obtained by solving (2) for Ur") gives
§13.15
GUIDES OF ARBITRARY SECTION. ELLIPTIC GUIDES
525
an improved value of M. Any desired precision can be obtained by repetition of the process. The foregoing technique works much better for the dominant mode cutoff frequency of an elliptical guide than that used for the special case of a circular guide in 13.09. Take the origin on the guide axis, so that the equations for the boundary and its slope are
(
X:)2 a
+
(12 =
1,
dy' _ —a 2 y' dx' b2x'
(13)
where boundary coordinates are primed. To visualize a suitable trial function, imagine the ellipse formed by rounding the corners of the rectangular guide in the TEio E-line map in Fig. 13.02a. The E-lines bend to stay normal to the boundary, and the TE10 B-line map is little changed in the y = 0 plane, but the right and left boundary close in as Yi —o b. Thus from symmetry the Bz to which U is proportional by 13.00 (15) is oppositely directed in the right and left halves and so is odd in x. The B-line maps in the +y- and —y-plane are identical if lyI < b so B z is even in y. A simple dimensionless trial function meeting all these conditions is
Uo) = a (1 + Bx2 Cy2)
(14)
From (1) VU(0)must be tangent to the boundary ellipse, and thus from (13)
aU(')/ ox'1 + 3B(x') 2 C(0 2 _ a2y' b2x' auovayi 2Cx'y' —
(15)
Collect all the terms of the last equation on one side, put over a common denominator, factor out x', and write (x')2 in terms of (y')2 by (13). The resultant equation must hold for all y' values so the constant term and the (y')2 terms are separately equal to zero. The constant term gives B, which is then put into the coefficient of (y') 2to give C. Thus (14) is
U
=
a
(1
y2
X2
3a2
2a2
b2
(16)
The integrals of (6) then become I(VU)2 dS _ [irb(5a2 U2 dS [irab(101a4
21)9 (3a2 + b2)]/[6a(2a2 b2)2] b2)2] 80a2b2+ 17b2)]/[144(2a2
(17)
If p is written for b/a, the last inequality in (11) is 13,a
c.
1
z
j-, sin
m=1
mad max a sinh ji3mc sin a
Mr21 . , mirx jmr cos — e-713,0 sin — smh ji3„,c : sin a a a — t3„,a
2A/ z mad
a
m=1
When z < c, interchange z and c in E and for B write
ji = 2µI sinMird max —i1 cosh j0„,z sin sin a lq a a m=1
., max jmr sinh 30„„ cos —ie-i0'.° a (3„,a
k,
1)v 12. In the last problem take the exciting frequency v so that nv < 2av < (n where v is the free wave velocity in the tube dielectric. Show by integration of the Poynting vector across the tube section at large z that the radiation resistance is n
, mrd 2c,nthz , R, = (3„)-1S1112 (3,,c sine — a a m=1
548
WAVE GUIDES AND CAVITY RESONATORS
13. A rectangular wave guide, a = 10 cm, b = 2 cm, and closed at z = 0, transmits waves whose free wave length is 15 cm. It is driven by a 0.5-mm radius wire across the guide at x = 5 cm, z = c. Show that, to make its resistance equal 100 ohms, c should be 2.62 cm, 14.0 cm, etc. Show that the phase and signal velocities are 1.52v and 0.66v, respectively, and that the inductive reactance is 274 ohms. 14. The wave guide of problems 11 and 12 is short-circuited at x = d2, z = c2 by a perfectly conducting wire. The impedances Z1 and Z2 of each wire alone in terms of its position and size may be calculated from 11 and 12. Show that, when both where are present, the impedance 2; of the first wire is 2; = [2i22+ 0,2(10) 2,ji,22
made 2Ab z M° = — (3„,)-1sin —sin a a a m=1
pence '
where 4, is real for transmitted modes. 15. It is desired to eliminate the reactance in problem 13. Show from 14 that this can be done without changing the resistance by insertion of a short-circuiting wire on the center line at c2 = 11.39 cm from the end, so that M° is real, provided its radius is 4.5 mm and assuming that our formulas hold for such a radius. 16. A rectangular wave guide, a = 1, b = 2, and closed at z = 0, is excited by the application of equal and opposite potentials to the ends of a rectangular loop in the x = 2 plane. The legs of the loop are at y = z and y = 4, and their length is not given. The loop is driven at the frequency of half unit length free waves. Show that only the following waves can be transmitted: T/1/12, TM32, TM's, (TE10), TE12, TE16, (TE N)), TE32; give reasons for excluding other waves. 17. A rectangular wave guide is filled with a medium 14161 when 0 < x < c and 142€2 when c < x < a. Starting with 13.00 (2) and choosing (We..)1 = A1 cos pis sin (IV t..)2 = A2 cos [p2(a
my
e-i$' p.'
—e- „„= x)] sin 1 rY
show that 13p„ may be found by solving the transcendental equation p2ei cot plc = —p1e2 cot p2(a — c) as in 13.05. Indicate how the cutoff frequency may be found. 18. If the divisions between the mediums in the wave guides in 13.05 and the last problem are at y = c, show that the equations to be solved for 0',,„ are --Aiq2 tan qic = g2q1 tan q2(b — c), —61q2 cot qic = 62q, cot q2(b — c),
for TEX for T MX
19. Show that the variational expression for the normalized susceptance of 13.06 (7) is
B° = 8b z
[focE(y) cos (nry/b) dy] 2 F
X0 n=i[n2 —
(2b/Xo) 911 E(y) dY] 2
PROBLEMS
549
Use the trial E in the quasi-static expression of 4.22, namely, cos (17ry/b) 21 cos (147) [5m 2 arc /b) - sine (try/b)]1 (cos — cos 0)1
E - . where
q5
= 7ry/b and 0 = 7rc/b.
With these coordinates show that
1 focE(y) cos (117Y) dy = -b[ P. cos b 2 b
P._, cos
1 rC bL„b 2
Using the fact that the trial Ey is exact at cutoff where X = co and B° is given by 13.06 (7), write the variational upper limit in the form 17-C 2b B° = -{41n (csc —) + b X,
,rn 2
[(70 - 4b2V,2 )-1 - n-1](/,„1 } b
n=1
Note that the static value provides a lower limit. 20. The rectangular wave guide in 13.02 is obstructed by a thin conducting iris in which a slit of width d with its center at y = c runs from x = 0 to x = a. Calculate the normalized shunt susceptance as in 13.06. Verify that the transformation W = sin-1
[- cos (7rzYb) +cos (77-c/b) cos (ird/b)] [sin (7rc/b) sin (iird/b)]
gives U = -ii- when z' = 0, 0 < y' < co and when y' = 0, 0 < x' < (c - id); V = 0 when y' = 0, (c - id) x' (c id); U = 17 when y = 0, (c id) x' b and when x' = b, 0 i y' < .0 and that when y' co, instead of 13.06 (6), the stream function V becomes Try irc 17rd) V = - - In sin - sin b b b The potential difference is
U2 — U1 =
B° =
7r. Thus as in 13.06
b irc 17rd) In sin - sin — b X,
This is the quasi-static approximation to the "Waveguide Handbook," la, page 218. 21. One-half of the iris in Fig. 13.06c is removed leaving a gap of width d. Put the right-angle bends at +1 and -b in Fig. 13.06e and take the origin at the base of the iris, so that the transformation and magnetic induction in the iris plane are i rz ( irc W = C cos — sin 2 — - sine a a a 13
a {cos (17rx/a)[sin 2 arc/a) - sin° (17rx/a)111
jBy = C1
ax
Proceed as with 13.06 (13) to show that the quasi-static normalized susceptance is B° =
X, ird -17 rd) cote — 1 ± csc° a a a
550
WAVE GUIDES AND CAVITY RESONATORS
22. Show that the upper and lower limits for the B° of the last problem found by insertion of its B. and B. into the variational formulas 13.08 (10) and 13.08 (16) are
[
B° = _ Xg a
1—
m =2
20) ± 13„,_i (COS 20)]2 2 cost 1,G
(2a)2 Ii[ P„,(cos2-y) — P„,_i(cos 2-y) 2 2 sin 2 -y X
a z [ m2 X, m= 2
1 Bo
2a 2 lir Pm (ma)
where yG is bird/a and -y is irc/a. Then assume, as proved for the symmetrical inductive iris by 13.06 (14), that when X = co, the above formulas give the quasi-static values of the last problem. Thus write the above limits in the more rapidly convergent form
B° =
X, cot 2 0[1 + csc2 a
1 + - sec' 4
2a 2 ]i — 1}[P„.(cos 20) {[ 1 — ( ) mX
m= 2
Pyn-i (cos 2012} 1 — =
°
B
a
cot 2
[
1 + sect -yr + csc4
Xp
m —
2a)2
ti
m-1}[Pm(COS 2y)
=2 — P„,_1(cos 2-y)]2}
23. The inductive iris in a rectangular wave guide is a centered strip of width c. Nate from the symmetry that identical areas on the top and bottom of the guide must charge via the strip so no current circulates around the windows, and the stream function has the same value, V = 0, on strip and wall. Show that in the plane of the strip
B. + jB. = C cost —
—K
sine —c — cos2 —
)-i
where E and K are complete elliptic integrals of modulus k = cos (iirc/a). show that the quasi-static normalized shunt susceptance is
B° =
Hence
X, E — 1(1 — k 2)K a E — (1 — 1k 2)K
24. Show that the upper and lower limits for the B° of the last problem, found by insertion of its B. and B. into the variational formulas 13.08 (10) and (16) are X, B° = -- [1
a
f(B) -
n=2
4a 2 (2n + 1)2X2
{P,,[cos(rd/a)]
.A 0),
.., —a 1 4a2 r — = — [ X2 (2n + 1)2 — — 1(0) B° Xg n =2
P„_i[cos (,rd /a)] 2[1 — 2(E/K)]Pn[cos Ord/al2 cos (rd/a) + 3 — 4(E / K)
Assume, as proved for the symmetric inductive iris by 13.06 (14), that when X = the above formulas give the quasi-static solutions of the last problem. Thus write
PROBLEMS
551
the above limits in the more rapidly convergent forms ..s — Ag{E—'1(1— k2)K ]i 4a2 B° = 1 (2n ± 1)2x2 — 1)f(0)} a E — (1 — ik 2 )K + Z ([ n= 2
1
—a— {E (1 — 1k2)K g E 1(1 k2)E +
4a21-1 x x2
1) ([ (2n + 1) 2
n =2
2n 1+ 1)f (e)}
25. The wave-guide section is a right isosceles triangle with legs of length a. Show that for the simplest TE wave the cutoff frequency v, is iv/a and that 2
1 a =
iz-yvaS[(
1
—
2
(1 + 21) + (3 ± 212)P11 — 72 P2
where v is the velocity of a free wave in the medium that fills the guide. 26. The section of a wave guide is a right triangle with two sides of length a. Show that the simplest TM mode cutoff frequency v, is 1 51v/a and that a=
(2 + 2i)7[1 — 02]-1 Ava5
where v is the velocity of a free wave in the medium that fills the guide. 27. A circular wave guide of radius a that extends from z = — 00 to z 00 is excited by a current element of strength 43I o podct,e'"` at po, 0, zo. Show that the amplitude of the TE modes in 13.09 (5) and (6) is = ..7(2 — 6,° )A / c90.,p0 J:„(0”„po) eis,. 02 273„„(3L„a2o M 2)[J„,.(13„„a)] 2
=0
—
where J'„,(13„,„a) is zero, and that for the TM modes in 13.03 =
Atio
dOOMIrrt(OmnP0) 13,
(8) and (9) it is =0
,[(3/3„,.a4,(0,..a)pe
where J.,(13m.a) is zero. 28. A circular wave guide of radius a that extends from z = — 00 to z = 00 is excited by a small current loop of magnetic moment kr dSe"' that lies in the plane z = zo at po, 0. Show that only TE modes appear whose amplitudes in 13.09 (5) and (6) are tomn},
j(2 — g„)AI dSOL„J„,(0„,„p 0)
=
71-0 (f32,„,„a 2 — 2,„„
=0
M 2)[Jm(13mna)1 2e
29. A circular wave guide of radius a that extends from z = — co to z = 00 is excited by a small current loop of magnetic moment 4)1 dSei'" that lies in the plane = 0 at z = zo, P = PO. Show that, if z > zo, the amplitudes of the TE modes in 13.09 (5) and (6) and of the TM modes in 13.09 (8) and (9) are, respectively, omni„ =
r Omnion =
— mAi dsJm ( unpo) 7,-poo2mna2 — m2 umom.012e
"* 2
jµ 1dS(2 — g„).1:0 (0,n,,p0 ) .0, e'' 27/3""'13mniaJ,n (0m,,a)p
[0-1 . = 0
[i5m,d,m
=0
552
WAVE GUIDES AND CAVITY RESONATORS
30. The plane cb = 0 in a circular cylindrical wave guide is an infinitely thin conducting sheet. Show that the lowest TE cutoff frequency is given by = 1.1656
or
tan (27rv,av-0 = 4irv,av-1
and that, if the sheet has finite conductivity, the attenuation is infinite. 31. Show that the lowest TM cutoff frequency in problem 30 is iv/a and that the attenuation is infinite if the sheet has finite conductivity. 32. A uniform current I cos wt flows in that portion of the perfectly conducting cylinder p = a which lies between two infinite parallel perfectly conducting planes normal to its surface. Show that the phasor vector potential of the outward moving electromagnetic radiation between the planes is
A%
110(Rp) — iY 0(3p)]
2rai3[Ji (Oa) — Yi(Ra)1 where R2 = w2Ale, and from 5.293 (5) Y i (Oct) approaches 2/(r0a) as Oa approaches zero. Note, from the Jahnke and Emde tables, that if a < 0.0181X then [11(0a)] < 0.01[Yi(13a)] 33. Show that in the last problem the wire's input resistance and reactance are Ri = ,r 2ga2m,l'
X. —
Aii[Jo(fici)Ji(Ra) Yo(fia)Y1(001 27raMEi
where M is [Ji(fla)12 [171(0a)]2, and 1 is the distance between the planes. 34. The planes in problem 32 are connected at p = b by a perfectly conducting cylinder. Show that the phasor vector potential between the planes is
=
arY 0(3b).100p) — .10(0b)Y 0(3p)] 277-a0[J 0(0)171(0a) — 110(0).1-1(0a)1
and calculate the input reactance. 35. Electromagnetic waves move radially between two parallel perfectly conducting planes normal to the z-axis. If the electric field has only a z-component and is independent of 95 and if the ratio of E, to Ho at p = a is 2, show with the aid of 12.16 (2), (4), and (5) that at p = b the ratio is
H.s
=
26 =
Jo(Ob)Y1(0a)1 j2.[J1(13a)Y0(0b) of.10(0a)Yo(ob) J0(00170(13a)] ev2.[Ji(fia)Y 1(0) — Ji(0)Y1(fla)1 .iVo(Oa)17 1(0b) — J1(13b)Y0(1301
36. A wave guide of radius b transmits the T/1/01 wave of Fig. 13.09 when v > POI. For this wave to pass a gap in the walls, a field must exist across the gap to transfer the induced charge. Thus show from the previous problem that the transmission of the frequency v, > volis blocked by covering the small gap with a coaxial cylindrical box whose plane ends terminate at its edges, if the box radius satisfies the relation JI[2-n-(w)i orb] Yo[271-(1.te)iv,a1 = Jo[270.1.04v,a] Y1[2/r (me)1 v,b] 37. It is desired to insulate two sections of the outer walls of radius b of a coaxial guide for d-c potentials. Show from problem 35 that this can be done without interfering with the transmission of the principal mode at frequency v, by leaving a small gap between sections, provided each terminates in a plane flange whose outer edge of
553
PROBLEMS radius a is coaxial with the line and satisfies the relation Jo[217-(.1, €)1vb] Yi[27r (µe)i vx] = Ji[27r(w)ipca)Y0[27r(w)iv,b]
Assume there is a magnetic or current node at the flange's edge. 38. If I is the coaxial line current in the last problem, t is the gap length, and v is (w)-i, show that the potential between the outer flange edges is
It
Vo — ,2 ,,,Eab[Ji(wa/v)Yi(cob/v) — JI(wb /v)111(codiv)i 39. A sectorial horn formed by the planes z = 0, z = b and q5 = 0, 4, = a' is excited by a uniform current kI cos wt in a thin wire at p = c, 4 = a. The wave is completely absorbed at p = oo. From 12.16 (2) and (4), find suitable forms for A: when p < c and A, when c < p. Set up the Fourier series by equating A: to A, at p = c, integrating Bo — B' over the surface p = c, and applying the magnetomotance law as in 7.31 (9).
M I and K1 and K2 are complete elliptic integrals of modulus — mi)1/1111 and (M; — m:)1/M2, respectively. 53. A cavity is bounded by the planes z = 0, z = d, and by two circular cylinders orthogonal to them of radii a and b whose axes are a distance c apart where a > b c. Show with the aid of 4.13 and 13.22 that for the pth principal mode
Q=
d(a — b)[(a b) 2 — c2] {1 + 2µ'S, 4a b[(a 2— b2)2 — 2(a2 b 2)c 2 c4I1cosh-' [ (a 2
b2 — c2)/(2ab)]}
2a is partly closed 54. A rectangular cavity of width a, height b, and length 2d at the center from y = c to y = b by a thin conducting iris and excited in the 7'E10 mode. Consider it as a transmission line 7' section short-circuited at the ends with the capacitance of 13.06 forming the leg of the T, and thus show that the lowest free space resonance wave length is given by 2aX1 (4a2
X1)1
d) 4b 1 = — In csc -II — tan tan — XI 2b X1
55. If a = 2, b = 1, c = I, and d = 2 in the last problem, show that X = 3.692. 56. Obtain an approximate solution for the fields in problem 54 as follows: Use fields TE to x from problem 3 but with a cos 13;„z cos wt dependence on z and tin 1471,„ Take z = 0 at the left end of the cavity and assume the quasi-static result of 4.22 (6) for E at z = d, so that Ey
sin (rx/a)
21V0cos (17y/b) b[cos (iry/b) — cos (71-c/b)]1
or
Ey sin (7rx/a) — 0
according as 0 < y < c or c < y < b, where Vo is the electrostatic potential across the iris. Thus get the coefficients in problem 3 by using the Mehler integral, see 13.08 (11), and so obtain for the electric field in the cavity, when 0 < z < d and
556
WAVE GUIDES AND CAVITY RESONATORS
u = cos (irc/b), sinh Ific„z1 niry cos cos wt [P.M P'-'("sinh
irx{ sin (3',0z
Ey = Vo sin b a 2 sin $,,d
ax Vo a Ey = — b sin —
n=1
nir cosh 'Oilnzl Wiry ,„dI sin — cos wt [P„(u) ♦ P„_1(u)] , nh 101 l sinh b 101 n=1
where = Ico2A€ — (nir/b) 2 — (ir/a)2I and co is given roughly by problem 54. Note that in the window Ey is small compared with Er but does not actually vanish, so that the electric field line which leaves the edge in the z = d plane strikes the y = 0 plane normally but slightly displaced from the z = d plane. The boundary condition on the iris is rigorously met, thus the result is best for small windows. This also gives one-quarter of the field in a cavity of height 2b with a centered window of width 2c and in a cavity with height 2b and a centered strip of width 2 b — 2c. 57. Use the method and notation of the last problem but assume E, — 0 sin (r x/a)
or
E, sin (irx/a)
21Vo cos /b) b[cos (re /b) + cos (iry/b)]1
depending on whether a < y < c or c < y < b. Thus get the coefficients in problem 3 by using the second Mehler integral in 13.08 and obtain sinh It3',„z1 airy 1/3,„,b1 [Pn(u) + P,-1(a)l cos—cos ca a 2 sin 1#„dI + n = 1 b cosh 113c.dI n7
V0 ax sin Viozl E, = — sin
b
Vo rx E, = — sin — a
n=1
airy , cosh I81„zI [P„(u) + P„_1(u)] sin cos cot cosh 1(4.dl
Here E, is small compared with E., on the iris but does not actually vanish as it should so the iris surface is not exactly plane. The boundary condition E, = 0 is rigorously met in the window. Thus these expressions are best for a small iris and a large window. 58. A rectangular cavity of width a, height b, and length 2d > 2a is partly closed at the center by a thin iris having a slit running from y = 0 to y = b with edges at x = 1(a — c) and x = 1(a c). Proceed as in problem 54 to show, with the aid of 13.06 (16), that the lowest resonance frequency corresponds to a free space wave length X given by X=
2aXi 2a 2rd rc , 1 — cot — tan' — (4a2 XD1 2a X1 Xi
59. If a = 2, b = 1, c = 1, and d = 2 in the last problem, show that X = 2.7488. 60. Obtain an approximate solution for the fields in problem 58 as follows: Use the fields of 13.06 (9) and (10) but with cos cot time dependence. Take z = 0 at the left end of the cavity as in Fig. 13.06c and assume the quasi-static By at z = d given by 13.06 (14). Thus obtain for the magnetic fields in the cavity, using relations in 13.08, =
By =
ax cos Wiz' , sin cos lOidl a +
m= 3
A„, cosh 1/3„%z1 . nirx , sin a cosh I13„,41
Al
71-sin Ii3;zI irx , 1,31 d. cos I0,al cos a — I
-k A „,rmr sinh max 0,mo cos — a AiI4a1 cosh
In= 3
—
PROBLEMS
557
where ,1 0'„,21 = Ice2A€ - (nor/a)21, is given by problem 58, m is odd, and [ 2n + 1 1)" n(n + 1)
A2"+1
irc [ 7iC sin —Pi cos — a n a
Note that on the iris B is small but does not actually vanish, so that the iris surface deviates slightly from a magnetic flux line. The boundary condition is rigorously met in the window. Thus this result is best for wide windows. 61. Obtain a second approximate solution of the last problem by assuming the quasistatic at z = d. Thus obtain for the magnetic fields in the cavity
By
=
-
cos r
sin Adl sin 113
sin
a
Am10„,al cosh
+ rt =3
cos
sin A di
rX
sin , Amur sinh 10„,d1
irx
A„, sinhl4z1
a
Al sinh 1(3„,d1
,
cos
max
a
mrx
ni= 3
where the notation is that of the last problem. Note that in the window the Bz is small but not actually zero, so that if the image field is drawn in the region d < z < 2d, there will be a discontinuity in the slope of the flux lines in the plane of the window. On the iris the boundary conditions are rigorously met. Thus this result is best for narrow windows. 62. Use the second form of the stationary variational formula 13.28 (3) to get the resonance frequency of the spheroid (p/a)2 (z/b) 2 = 1. Take for the components of the electric field the divergenceless form, which meets the symmetry conditions,
E, - A - Bp2 - Cz2,
Ep= 2Cpz
Determine A, B, and C so that E is normal to the spheroid, substitute into 13.28 (3), and show that [ 14(3a2 + 2b2)]OA ___). 2.788 a—+b a2(5a2 + 462) 63. If d cBoby transforming to a system of coordinates moving parallel to the axis of the cylinder with such a velocity that the magnetic field disappears. Show by this means that it is impossible for an electron, once leaving the filament, to return. 3. Treat the preceding problem rigorously when E5< cB,s by transforming to a system of axes moving parallel to the axes of the cylinder with such a velocity that the electric field disappears. Show that the relation such that electrons just fail to reach the anode is 271-moc(A1— /4) = A/q(c2— A )1(1 —
b a
In —
SPECIAL RELATIVITY
588
where A = v12 v': and v', and v', are related to the initial velocities vi and v2 in the x and p directions by the relations ,
Vi —. COI V1 =
f
. V10 1 1— — c
V2 =
v2(1
ODi
—
1—
V101
c
where vi = Ep/Bos and we neglect the initial vo. 4. A vacuum tube contains a cylindrical cathode of radius r0 which is surrounded by a coaxial anode of radius ri. This tube is placed with its axis on the axis of rotation of the confocal hyperbolic pole pieces of an electromagnet. Show that the potential which must be applied in order that electrons, in the plane of symmetry, reach the anode, if the magnetic field in this plane is given by B0b(r2 b2)-1, is rigorously
moc2({ B 2q 2b2 2 2 [(r? + moc2r q 1
b 2)1
+ b2)1 j 2
— 1)
5. A parallel beam of electrons which have been accelerated through a potential
V carries a current I. If the section of the beam is circular of radius a, show that the acceleration, normal to the beam, given to an electron at its surface due to electric and magnetic forces is
=
I
y (1+ Vq 11 (2qi[(1 + Vq ) 2mc 2 mV / mc2
6. A point charge q moves in the field of a stationary point charge Q. Using the conservation of angular momentum p and of energy and the expression 14.07 (2) for kinetic energy, show that the equation of its path is r-1 = A + B cos yq5, where (IQ )2 —1 (
4reocp 7. A line normal to two plane sheets at a distance a apart in a vacuum is concentric with a hole of diameter b in each sheet. A beam of charged particles, all having the same energy, but having a maximum divergence angle a from the normal, emerges from the first hole. Neglecting a compared with unity, find the set of values of the magnetic field which, when applied normal to the sheets, will make these particles pass through the second hole. Show that the maximum diameter of the beam between the holes is then [2a/(nir)] sin a b, where n is an integer. 8. If, in the last problem, all particles comprising the cone originate at the same point on the axis and the magnetic field includes this point, show that the minimum value of the field which will bring these particles to a focus at the image point beyond the second hole is 222-mva[q(aa b)]-1, where a is the angle subtended at the source by any radius of the first aperture. Show also that this field increases the number of ]-2 ions passing the first aperture by the factor brb/[2(aa b)]1 2(sin [47b/(aa 9. Two similar circular parallel conducting cylindrical shells carry charges +Q and —Q per unit length. Inside each, a wire is so situated that when a current I flows through them in opposite directions each shell coincides with a magnetic line of force. Show how, if cQ < I, it is possible to transform to a set of moving axes such that only a magnetic field appears. .
589
REFERENCES
10. In the last problem, show that, if a charged particle starts from rest on one cylinder, the component of its velocity parallel to the cylinder at r1, r2, is given by rir20 Q ln = / ln rir2n( r2rio r2r10
27rmoy1
here n o and r2o are the distances from the starting point to the two wires. 11. In a two-dimensional magnetostatic field where the vector potential has only a z-component, let the total velocity of an ion of charge q and rest mass mo be v and let its z-component be v,, at a point where the vector potential is A i. Show that, when it is at a point where the vector potential is A 2, the z-component of its velocity is given by v, — q(moc)-1(c 2— v2)1(A2 — A i)• 12. Show that, in any two-dimensional case in which the electric and magnetic fields are orthogonal, it is possible to choose a set of moving axes such as to eliminate either E or cB whichever is the smaller. 13. The position of a charge q is given by s = a cos wt. If wac-1 1 and a a then aais positive and ab negative and re(U I-U 2)=-107:(a/b)soq=re(U1-U2)/2/70/0 Thus ca
e(u1 -u2) (x 2+y 2) 4/2 a22/11(13/a)
crb-
e(UI-U 2) b 2P/n(b/a)
Ans.
Chapter IV.
Page 111. (107)
Page 33.
-46C-
n(' + :10)=On at a-t t=—co
t
x
4- t=-a
0= 0
t=-17a
0= 0
0= 1
0=1 a --In t an- 1 (k/h)+22-k gn(-14-(h 2+k 2 )h -2e-Uil =
4
r
r,
Length increments are:L/11= DC
-Uo
+Ian
+_
+-3-1-1-t an -1111=
1 t 2 +ktan-111 +Itan-11`) Ans. -27T--E (12/n *
-61-
Ys U°0 4
k h 2+ka 2h h h 2+k 2 2ktan -1 h fc , 640 Ti2m—Wr + r tan r 4k a U
2k
Erk h2+k 2 , h
" uo
2b,
2bn,
2b m,..
1 n o,
z -7-- 7_,.-,il, z1 , x = --ir,ri, y= -r.i-goi , zi = e-2-
4--
14-la i --.)
,n(i.. 0
1
-Tra /b
, a1 - e 2
1
4 2e---) Tr —rah. dz , a i b1= 1. — Then by 4.29 (5): dzi= 2be 2 1• , „z i dz1 U0 U0 .. 1 r ETT(z+a)/b -17a/b ) K sn te ,e 121 3- a 1--°J. j Ans. W - K(bi/ai)sn' fz', ai a i K 0.,/(b2i-zf)(af-z?) rrUo di.,L dW dz i= Uorr 12-17z/bte -na/b_ errz/b,-ir na/b rTz/b1. Ans j te -e ' 2bK(edz dzi dz 2bK e "a fb)losh(rz/b)-2cosh(rra/b) 1 -zrra/b , r / -, From 4.29 (9): C= 2EK{ (l-e Ans. )2 j/Kte-rr -a- b j
b1 = e
- -17a/ b
-
2
-62z1 dz By 4.18 ( 7 ) .71--"V2_1,2, z=liFq;at z i =0, z= Pi= ibi uzi z i LI. c -hf. so b1 =h. When zi =a i , z=d so d=,/aFq or a1= jItr
(2„...a., _7.4, 2 2 rr r/2 -U0 -bi V=0 b Y
Substitute z 1 , al and b1 for z, a and b in 4.29 (5) and 4.29 (9); z =h{sn2 (KW/U 01 b/c i -,./7---Fh 2 )}= jh cn(KW/U0,h/,/314-h 2 ) Ans. C=
h air 2+h' /K( d / cFtl 2 ) Ans.
a U0
V=0
-uo
ib f-d-4
Chapter IV.
-63rKW b „ ,-1 3,4.29(5). W=U° - sn-i (sb1 ,11 ) or z 2 = bi snt— u o al
dz dzi N/T-aF
Page 39. IT Y1
Page 115. U0 1 -
0 -01
Uo
By Dw 768.1 dzi /dW = (b i K/Uo) cn(KW/U0) dn(KW/Uo). dzidW = (dz/dz1 )/(dz1/dW)
V1 al
b1 Ay
13 2 KW KW zi- a2 —a2rdz c--i sn 2 (—)+ 11= -aidna(—)by Dw 755.2. — cn 05 - , 1213 Do dW ja iUo "U o 1 a? Uo C r1CW, KW r KW, By Dw 785.2, 71- cnt—id—= 7-cos - tdrr-„ 3=z. When W= O, z=0 since al Uo U0 uo do 0=1 and cos0=1. When W=U0 , i dn K=(1 - (b i /a1)23 2 by }IMF 16.5.3.
-d -c -U0 V=0
U0
d >x
z = c so (Cl]) cos-1{1-(bi /a1)2}1= (C/j)sin-1-1-311 =c. When W= Uo +jVo , then by 4.29 (4) d = cos -1(dn(K+jK')3=cos -10 by IIMF 16.5.9, so Cl] = 2d/ft and bi /a i = sin(jc/C)=sinticr/d3 2d KW . cr , ERA_ 4 cuo= 4EI ____C(sinIcr iLcil r Thus z = —cos "ltdn(—,sin--)3 Ans. Capacitance is • Ans. rr Uo 2d V I - VI K(cosicr/d) -64A Y1 U=Uo z i = Bsn 2 (KW/110 ,,5) 4.30 (1). dz/dz1 = C1 (z 1 --1-(1+B)1a(1-z1)(zi -B). T i lT TT/2 V=0 a U=0 By Dw 380.011: ftz,/,./(1-zo(z i -Wdzi =-/777 + 2(1+017771-1dzi B ,t2
(11-13)/2
so z = -00/77: +C. When z1=0, z=0 so C= jC0/13. When z1 =1 or B, z = b so b = C1 =-jb/J, C= b. When z3 = (1 +B)/2, z= b+ jh
V=0
V= VI
=(jb/)(1-B)/2 + b, so ,,k= (-h ±./h 2+b 2 )/b. Take + sign SO k=j3+ (.Jhz±b 2 _11)/13. Z= (jb/k)•11 - k 2sn 2 (KW/U 0 ,k),/k 2 sn 2 (KW/Uo ,k)-ka + b. Use Dw 755.1 and 755.2 so z= b(1 + dn(KW/U0 ,k) cn(KW/U0,k)) Ans. The capacitance is C= 2EV I /Uo = 2EK( 1:17)/K(k) Ans. -65Invert about z= D- 2jR with radius of inversion 2R. Then from problem 64: z-
4R 2 z 1-D+2jR
-
4R 2 Bsn 2 (KW/Uo ,k) - D + 2jR From Figure:
D= 2Rcoty, B- D= 2RcotP, A - D= 2Rcota, B= 2R(cot Y + cot f3 ) = 2RM
U= U
U=0
A= 2R(cot y + cot cx) = 2RN k = 0/7/i =
-2jR
2R z
Msn 2 (ICW/Uo ,k) - cotP + j
Ans.
Then. from 4.30 (2): C=
Uo V1
2e' K(JITi7)•
2E' =
E
Ev
Ans.
>xi
2R
›X
Chapter IV.
Page 115.
-66r -(z 2-c 2)dz (z 2-c 2)dz xlpositive if z1--L1(z 2-a 2)(z2-b 2)41(a2-z 2)(b 2-z2) 0<xk. Integrate B to C; -jb= jfidy= j u vic Let s in 2U = k 2+ (1-k2) COS2* so cosU= piT=TcTsin*, dU = -V57:17Icos*Obik2 - (1-k2) cos2*, sin 2U - k 2= (1-k2)cos 2V. When sinU=k, cos*=0, *=r/ 2. When sinU=1„ cos*=1, 111=0. -b = CI I / 21(1-1(2)COS 201k2+(l-k2)COS
cl4i= C1::/1(1-k 2'-(1-k2Sill240 )/NA (1-k 2)s in 2*] d*
= CI(E( 1=171)- k 2K(A/I:172)3 so a/b= (E(k)- (1-k 2)K(k))/(k 2K( 1:177) -E(N5=(7))
Ans. (1)
E(0,k) - (1-k 2)F(0,k) +jb Ans (2) Charge densities are, since V=0 on bar, E(k)- (1-k 2)K(k) 1 Nrii.7. E(k)-(1-k2)K(k) = E dW — = dW dz, On a sinU < k aA2 ..sin2 u U < U/ a dz dzi dz 5.,/z21 -1 Co /;T:17
z- a
E dW
dz
=C
1
2 U-k 2
E(k) - (1-k 2)K(k) ar/l-k2-cos 2U
On b sinU > k.
Page 116
Chapter IV.
Page 41.
-68-
1
:1
dz _ cl 241 - )(10 z=ai sn-1(111)=Csn- 3k} W 2rrc z1 - x10 +jyio • dzi Ai(b21 -zi )(a12 -zf) -a1 -b1 aidxi -b+ja pb rb 0 < x1< b1 < Dw 781.01 b= J o KL-1 al 3, b=CK(k) 1A./(bf-xi )(at-4) al aidx i 0 < b1 < x1 < jb Dw 781.02 ja=i jaiK{,,/1-(bi/a1)2} ../(x?-bi)(ai-x?)
, 7,12 112 tDi ' al rb+j a
-
'q
so a = CK(k'). Let u= xo/C, v = yo /C, then (xio+Yio )/h = c+jd = sn(u+jv,k) c+jd snu cnjvdnjv+snjvcnudnu 1 - k 2 sn 2usn 2 jv
sn(u,k) dn(v,10-1- jcn(u,k)dn (u,k)sn(v,W)cn(v,k') cn 2(v,V) - k2 sn2(v,k)sn 2 (v,kl)
Since: Dw 759.1, sn(jv,k) = jsn(v,k')/cn(v,k'); Dw 759.2, cn(jv,k)= 1/cn(v,10); Dw 759.3, dn(jv,k) = dn(vI k I )/cn(v,k'). zi = bi sn(z/c,k) = bi sn(u+jv,k). -q 242 sn(z/c,k) - c - jd W Ans. 2i sn(z/c,k) - c+ jd -69(1 )w .
La /o . TT
:727-al,77 1 ,2.. z 2 -JvC1 -
. 2Vo slrin .1.0
-2V 0
vb2,2 ..., 1
"
11;r 1 7 zi ui
-a 2
9 Vo ___,. 1 —,..
2V When 0 ra incomplete Beta Function, z =B ty„--2-3. When z1 =1 it is the CN.f z1 ra complete Beta Function Btr,Ti, see Dw 855.1. Thus V=0 0 ra 1 b = BBtr,73, a = B117,71, and by HdMF 6.6.2 and 26.5.1 the ratio is the function —
(Et'121 which is tabulated. Tables of the incomplete Beta Function. K.Pearson, a = B r Biometrika Office, University College, Camb. Univ. Press, Cambridge, England. 1948. Thus the modulus AJ = k of sn(KW/U0) is known. Since k < 1, HdMf 26.5.4 gives
B
1713(2 /747Ti frr —2 3 = aB(a/ 7,1/2)
13(1+(a/r),n+l t n+11
1+ E r
I BL;+(a/r),n+1
The potential difference is the same as 63 but the charge is twice as large so that C =
4Ev1= 4EK U0 K( )
Ans.
Chapter V.
Page 223. (199)
Page 48
-10_ a b ( b4+ 2a 2 b2+a4+b4_282b2+134 0 = -agI t/3 1 + 1 li 1617E132' q2 (134 -a4 )2 4TTEb - 0 2 /112Lb+(a 2 /b -Fq 1 2abb b 1 2ab(b4 +a 4 ) 1 —r . a=g• Ans. 2i --b2 9.1 _a )2 . If a ;a
dI
21- in+ 2a+z]-1 4.13 It
'2a+2b n z -1 1 [-F[r' + ;7213] 2b-z7 -1 1.1 c° V-Cri+ 26.21) J + n1
cr'
-
)
cif +[n+ 2a+2b - zil_ 13 1
Sum by HMF 6.3.16 and by HMF 6.3.5
2a+2b
eg' • 1 1 1 cc -z 1 -z + E -1 ---+z n }-- z+ iE-7----= nnz) z 11(1+z)+ Cl= -T(z)+C'. Thus V becomes:
V
8TTE (a+b)
DV2 (2a+b)/(a+b)]+Y[iz/ (a+b)] +'1[4(2b-z)/ (a+b)3+11[12-(2a+2b-z) / (a+b)]) Ans.
i
Chapter V. Page 227. (203) -32E-E then: p = —iv
00
d2x - trreit-i'
dxA dx dx _ 16rev x2 = -Tilde' dt
laq 2
Page 54. or`
a 4rrev (2x) 2 .1., v 1 1 d Integrated ( _13q2 ) -1, = d'jv /T:la to x= O. dt = (--L-1dx and integrate from t= 0 to /t1/./ a -3150 -4 —: dt Prrevma a-x If
c+
c) ix )0a = -Ira, a 3 = pq 2d 2 /.(2r3 evmv 2 ) Ans. -,113q 2/(877 Evma)div = {-x(a-x) - 2a • 2 tan-14.(a73— - 33CBy Gauss Theorem charge on tube of radius AB is fq+q(c-cv )/(c+cv)31(1-cosa) =12-(1-cosf3) . Thus 1-cosp= 2e(1-cos ct )/(e+Ev ) or sin2 -ip = 2e sin 2lia/ (e+ev). Ans. - 34C-
el
1
el
4r- a---,. 4.-kql/ OP 4= 2a •c4/(%/Tx%/a 2+4c 2 )• -017; Add Tre(WS-5) from problem 43. Potential is - 4/a. Total charge is
1-4c 2 ). The capacitance is then -3c 2)+(2c/477F.c 2 )-(4cAra-Firre([(4/45)-53-(2c/a)+(4c/fa-27— 2 2 -46CFrom problem 42, o=2cri+eV/a; Integrate, Q=2Qi+(4Tra-S/a)V0E. As S approaches „-------.% I zero, Q approaches 4'n'eaV so Qi approaches TeVo S/a.
Ans. I
/1
e
i
-47C-
, - ..... % ...
f
Invert, by 5.102, segments of earthed conducting plane lamina under the
-470 '. • . \
n
1,
influence of charge -417c at inversion center. By problem 12 and 5.04 (1),
0,4 .I
before inversion, o1 -o2= cro= 4ear-3. By 5.102 (1), alo =+(r/a)3 o0 =4€/a 2 . For potential Vo , c',3= 2aVo • 2e/a2 = eVo /a.
Ans.
-48CNote that e= 1-(b/c) 2 and A= TTbc= 'Rea] ..1.7;1. In 5.02 (4) let 0= b20, a2=0, factor out c and let c=(1-e2) 7 . Then C-
811/Ad0 ,,gT (1-e2) 40(140)(1+(1-e 2)0)
Use Integral Transforms Vol.I, No. 23 page 310.
Jima' r (21) 2 2F1 (4. ) 411 ; e 2 )"" rt( 1 + rife2+ e4 +2-2-5% e6 +• • ) by Handbook Math. Functions 15.1.1. By Dw 5.1, (1-e2)1 /4 = (14 2 - Ae4 - -ffice° -••) so (1-e2) 01:=TT(.1-1-[-14]e 2+[11-4-?-6- 2 ]e4 Je8 + • • 1= (1-144e4-1114- e 8- • • . c=8\FIii/E II( 3 = •17%Fr8 [ 1+ .1k4 + 1 +[ -9-8...14 .4 ee 256 )
Ans
-
Chapter V.
Page 228. (204)
Page 58.
-49CSince ellipsoid and sphere of radius r have the same volume: r 3=abc=(1+a)(1+0)(10)r 3 so a + p +11 IV-(743 + a Y + 13y ) 1 neglecting fourth order. (a+ 13 4. y) 221Clt 2 _Fp 2+y 2+2ap+2py+2ay ::::,(ap + py +ay ) 2 P-.. 0. Thus 2(Ci+ p + yyz( a 2+ ps+y2), gap+ py + ay ) p,-.,, _ (as + fit:4. y 2) . From 5.02 (4), if x= r-20 +1, r 2 dx= de, 0 < 0 < 03,1 < x < co. Let Z = C/ (8Tre) then C = {j0 (4,/[(1+ a) a r s+Eci[0.+0 ars+ p][(1 +y)2r2+63) dE31 1.1 = i,7(4,/[1+( 2a +a2 )/x ][1+(213 +13 2)/x][1+ (2Y +y2)/xj) (r4/30) dx 7(,I1+[2(a +p+y) +a a+ p 2-1-Y 2]/x + 4 (a131-py+aY) /x 2 )-1(4,0) -14 -ir Pe, ;(c°1 (4/1+(a 2+13 2+Y 2 )(2x -1 - 2x-2 ) ri(Nrx7)-1d3]-1 21471,174(a 2+13 2 ±y 2 ) (4./7b--4/27 )] d ) = r [ 2- (d 2-113 2+y 2 ) (4.- 4)] 1 =ir [1- (a2+13 2+y 2 )].
-1
Thus C = 4TTEr[ 1 + 115 (a2+13 21-Y 2) ] Ans.
-50See 5.27, 5.271, 5.214.
V= C0Q0(j C) = Ccot-1 C -f-Irc-e > C/Cc i /r = q/(4rer).
so V= qcot -1C / (4Trec 1 ). By 5.27 (2) : a 2 =cfCc21 b2 = ci (1+0), cal= 1,2 -a2 , C20 = a2 /(b 2 -a 2). -q -q e to.,r7-c ri q -1 a=
161 _
h2 bcCo- c 1A 2+c i 4reci i+cg 4rrco/40.+0)2- p2 -477c 0/134 (b2 -a2 )"1- p2 2 b4 2Ib rbja. q2k(b/a) 2.11p dp rb q2 rCY2 dS 2 F=J 2E 32TT 2 E (b2 -a2 ).10 b4 (b 2-a2)-1 - p2 ) JO r b2 -a 2 P 0 • F-16rE(b2-e2)
Ans.
-51CSee 5.28, 5.281. V= CQ0(71) = an C(11+1)/ (71-1) 3 m.: 2Cc 2 /r = q/(41Ter) so V -014--(471rre! i (r1-1) I r a2 = cpi e ) b 2 = c22(r)2-1), Ti g . a 2/(a 2_132) , ci = as -1)2 . a = (E/112)[N/all] at n .rio ! . F .rcradsz q q q -2 1 J 2c a= 8T7c a 'ni -1. h2 -41% 2,1C 2(118 - §1) (111 -1; - 4rrC 4/p + b4 / ( a 2- b 2 ) q2 b rb 217p d p q2k(a/b) F- 32r s e (a s _bs )..J0 p 2+ b4/02 _b2) .1 j'o = Pnitp 2-F a 22 b : b l b° . F . Ans. 16TTE(a 2-b 2) -52CIn 5.03(4) : r=b, y=0, z=0, b=a.r tan-145aab2-a2-4,/(b2-a2) 2-4b2a2 3 = (2 1.70 / TT) tan-1 (a/b). This is potential of •spheroid of minor axis b.
Q - 8eaVo 47ea 4rEa 4ree External spheroid Ci =--r V1 -V0 V1 -V0 Tr-tan-1(a/b) cotNa/b) angleBsc capacitance is C 2 8eaVo _ 41TE a 4rea w_ _g2_ + +Q,) 2 Q2 •angle.ags,+ CcH-Q,) • anglesBc C2 = tan-1(a/b) anglesBc' 2C1 2C2 8rrea
Ans.
-53C2 As in 5.19: z2+p 2 (1-e2 ) =c2 , ri =c1 (1-elsiniK--2:=c1 (1+gei)- ic aPa (p), r a c 2 (1 +aged- Te a, 204,,,• V=A+B/r+(Cr 2+D/rJ)P2 (p). C,D, order of e. VI=A+B(1-1e1)/c1, 04Bc1 +Cc 12+D/c? , V2=A+B (1- fei)/ca , rbV 0=3Bc2+Cct-I-D02 . B= (V 2- Vaic (1--2p 2)-c11(1- §ei ))-1 = c1 ca (Va c3 -a( ea c i -e ic a )). 0=-cj-57dS fl li P 2 (p)dp=0 so Q=4TTEB. Thus B=Q/(417E) . C=411eB/ (V a-V1 4Ttec ic 2 /(e1- c2-4(e 2 ci - e a cd) c f _L a eac - eic , _ 4rre i c 2 ci - c2 i(e2ci - cies)3 Ans. L 3 C1- C2
C1-C2
(C 1 - C 2 ) 2
Page 59.
Page 229. (205)
Chapter V.
- 54CLaPlace's Equation in cylindrical coordinates is, referring to 3.05 (2),
_L
by b 1 b 13 bp k"IP bp' ' bz
bv\ = 0, bz,
K'Kiev =--p- . .
. b 2v_
Thus bp: bz2 - u.
This is two dimensional La Place Equation.
A solution giving ellipses is 4.22 (4), z2/(c2cosh2 V) + p2/(c2 sinh2V) = 1. Let a1 = c coshV1, r2_ 4c2e2V, V-4 Onr+62(2/c)
b1 = c sinhVi, a2 = c coshVa , b2 = c sinhV2 . For charge let
Q = SKI (aV/br)dS= forr(EvKL2TTr 2 sine) (pr ) ide= Ev 21TK/Zde = TOKiev . VI = coshia/c) or. .
V i = Pin( (a l/c)-1,1(a 2 /c) 2 - 1) = Pgr((ai-Fb i ) Lc), 112 =071((a 2+b2)/C ) .
Q
2TTIKtE„
111 -‘12-2nUa 0 14/ (a a+b 2) }
Ans.
-
-55C-I r/2 -1 5.02(4): C=817e/r 0 (0(a 2+0)(b 2+0)3 7cle= 8rre/ro (cos40(a 2+a 2 tan 20)(134-Fa 2 tan20));95 a rr/2 CV = 877€/j o 02sin20+1)2cos 20)-2idszs. From55.03(2) . 03(2): c= 4rrab [14 a+ 2b • Invert elliptic disk on radius K=c. x 2 a'2 +y 213-2 =(x2+y2 )r-2= 7P-12/EV X2 v 2. .21A -i2 QVc =cQ/6-T so that 14— But cP7c= cP/OP and --61b2 = 1 0242 a 2a+-1 4. 007152_typ27z2 CO2_157,2)tip2_(Ep2_M2)75(12 c 2a5p2 -'6(12) - cQ 20P 2-cP at)t) 2 cQ 2-0 2 OQ 20P 2 OP 20Q 2 cQ 20P 2 .
qi -]
From 5.102(1) a'=
C 3 a CV C 3 cQ OP 4rab OP3 c,Arp 247:/(T) 2
-
c -cQ 8rEv a tirrabI OPaNt(572...7Q2
Q -2-2 From 5.102, if V= 4r cl inducing charge i s Q so a l =-1 4rfaoI Or -,vopcg 2 -57Q 2 e -56CO E(cnr -n-l+Anrn,, ,,,‘ v _ q co From 5.153 (1) Vi = 4r 4fre ci e 0 , J.L--n k../, 0 — E Bnr -1.1-1Pn (II) n -n-1 nv -n-1 n 2n+1 . v + Ana = Bna since potential is continous. , Bn= c +Ana At r = a, c a n -n-2 n-1 -n-1 n , / // //////// e v C-(n+l)c a 1.-e(n+1)Bna .-e(n+l)c a-n-1 -c(n+1)Ana n-1 +nAn a n-1 (n+1)(cv -E)cn n -n-2 (evn+(n+l)e)=(n+1)c a An a (ev- €) ,An— (e n+E(n+1))a 211+1 • /,, Field inside at r=c 0=0 due to dielectric displacement is cn E .----2- ci A_P—rn_ci (n+1)(Ev - E)ncn-1 a2n+1 . Force on q is qED . o cvn+e(n+1) Dbr 4rev D 4rev o _ (E-eu)q 2 ; n(n+1) r c ian+1 Ans. F411Eve 2 d Evn-FE (n+1) ' a' -57Add to 5.16 (1) the induced charge potential Vi, so total potential is -1 -n a2n+lc-n then V = (Q/(4rrec)q(rn c +Anr-n )Pn (coscl)Pri (cos 0). If An=Egb at r=c, 0=ci due to Vi is -cos ccovi /Or +r-sina bvi /be. Ez= (Q/(411Ec))o-a 2n+ic-n((n+1)4Pn (p.)-(1-1.12 )Pn'(13.)1c -n-2Pn (cosa), pg•cose.
V=0 at r=a.
...(Q/(4rrec 2)) f(a/c) 2n+1 (n+1)Pn+i (cosa) P (coga ) From 5.154 (5) Ans. Closed form in ep (n+1) (a /c) 2n+IP (cosa)P (cos a) F = QEz= -(Q/ (41TEc2)) i n n+1 Problem 18.
Chapter V.
Page 229. (205)
Page 60.
-58-
co n V o =(27/ ev)4( sin a (r/a)nPn (cos0)-- r-"-lAn )Pn(u), Vi= (.1T/ev)pnr Pn(11). n n -n n-1 An= Bb .(1) . evbV0 /OR= eOVi /Or so At r=b, Vo = Vi• so b a sin a Pn (Pcp+b n+2 n+1 n-1 -n-2 n-ln = (2) .b sin a Pn(Ace )- E y(n+1)Anb (2). (1) .neb a , = neBnb n evb
r (cos CX)=4(1 +1) gv + nelA,, so that n(e-el )bm-fi a-n sina ci, rn n 11(E-e)13 211+1 Vo = Tivan ( (n+i)EvillE)anrn+1 since Pn (cos ix ) Pn (cos())
Ans.
-59By 5.16(1) charge element is dQ = a0 211c 2 s in Cx, and its potential is
cc
dV = 277c 2cros in ada (4rrec)"i c (c/r)ri +1Pn(cos a ) Pn (cose) . Then by 5.154 (5),
Sol aPn (cosa)d(cosa) =2(Pn+1(cos 00- Pn _ 1 (cosa))/ (2n+1) so total potential is cc 1 cn+lr-n -1 pn (cos 49) (cos011)) (2n+1)(P (cosa) - P V = c e -' ao (cr-1 cosa +E n even n+1 n-1
2n+1 -2n-1 r P (cose ) Ans. P2n-1 (cosa)) (4n+1)-4c = ce-icro(cr-lcosa+ t c 21P 2n+1 (cosa)2n -602n+1 2n+111 ;.,°' , 1‘ In 1.3. •(2n-1) [- ( 2n +14117TI -11 - 2n(t) 5.17 (1) gives V so a= E 5.17 = it P 2n (P) Or r=b 4na V --''' 2.4- . . . 2n E = 2 . -. QE,-pri 1-3..(.2(1-11 4n+1(Al 2n hi P (u) Ans. Closed form from generalized solution: e 4rE o f -l) 2'4••••2n b2 - 2n ' 4ad r(s2+b 2 'W Qk Ans. E(k) - F (k)] where k2 E= 4r2 E,Aab L4ad(1-k 2) a 2+b 2+2ad Where d is distance from field point to axis General Solution. Internal ring. b, c,d, z coplanar R = b2-1-o 2 -2bo oosa+a 2 -2ac sioacos0. Ring charge Q Charge on dO is q = Qd0/ (2r).
v=
Q(8r20-1: 02 1-2dO
.
External ring. b',e,d',z' are coplanar. -t-D 4 c'2-2c1 (1kosa R' 2. (bt7z ) 2±f'2. (bLe) 27 a l 2 • +4_ 2a'd'cos0 = 27 + a's in•a cos 0 ) V'= (81.720-11 R-Ido . di= c'sina, z'= c' cosCx.
d=c sin a z= ccosa - b
Let 0=11-2*, do= -20, cos 0= 2sin 241-1. 0
Need field along c„
V = Q (4112 erirr/ t a 2+b 2+c 2+2c(asina-b cosa)- 4acsinaSin 20-12 20. Q r.a2-1-b2.+c2.482 +b2+c2 A-2c(asina -b c osa ) -4ac s in as in24/3 -Oc-4n20 61a 2-fb 2+c2+2c (as ina -bcosa)-4ac since s in 2 *
Ec
7/2
= Q(072E )-2(yo (a 2.1.1) 2+c)p
-
3
204014-1 P -01J where P= r 2+a 2+2ad-4ad sin2*
Since r 2= b2+c 2 -2bc cos a , d = c sin a. Also true for r' , b' and d ' . For internal ring,
k2
4ad r 2+a 2+2ad
Qk
[(a2+b2 +0)k2
E c- 8r2 ea 4ad (1-k2)
E(k) - F(k)
Z
'
0 Image relation
is N/(a 2+b 2 ) (a ' 2+b' 2 ) = c 2 . Then from similar triangles, a' = ac 2 /(a2+b2 ), b' =bc 2 /(a 2+b 2 ) 1A7.4...a 12.2(r 24.a2)/( a 24112) . Substitution shows that the external k equals the internal k. (continued)
Page 230. (206)
Chapter V.
Page 61.
-60- (continued) c( , - , a wa 24-b 2 ) ; (91a a k (a2+b2+0)ka E(k)-CF=(Ek'C).3 Total E or cr/E on sphere surface is E= 4r2 ad c 4ad(1-k2 ) Q'=-cQ/(a 2-1-b2 ). For the external ring,
where a is ring radius, d is distance of field point P from axis, r is distance of P from ring center and c is the sphere radius. The modulus k is k2d
4ad 4 ad r2+a2+2ad
where p is distance from
P P to farthest loop point. From diagram at left p 2 = a+c] 2+r2 r2+a2+2ad. -61
For q only: r b, Vq = zirrqeb E(b/r)n+1Pn (11)- 11=cos0 From induced charge: Vi= q(49Eb)'1E (Anrn+Bnr -11-1 )Pn (p) (1) An cn+Bn c -11-1+bn+le -n -1.,-. ID n -n-1 at .r = c and Ana +Bn a +an b -n = 0 at r= a . Equate the An values to get 2n+1 2n+1 2n+1 r. n , 2n+1 2n+1)), An .. . (a 2n+l -b2n+1)/ ( bn (c2n+l -a 2n+1)) Bn = a (b -c )/L b ( c -a a' -n 2n+1 2n+1 2n+1 2n+1 -1 n 2n+1 -n-1 r b: 411EbV = q b (b -a )(a Ans. -c ) (r -c r )13n (cos 13) -621 ce 2n+1 -2n-1 For q and image in plane: V= q(217Eb)_ ; r (0) = 0. (cos 0), since P b P 2n+1 -Fol s 0. p. Inc ' 2n+l -2n -1 -2n2fTevbV 0= 4 (r b +r An)P2n+1(P)' 217ElibVi=qEBnr n+1p2n+1(p). a2n+1b-2n-1+ -2n-2 2n+1 At r=a: V0=Vi, a An=Bn a and EvbVc,/br = ebVi ibr. Let K=E/Ev, then -2n-3 2n -2n-1 -2n-1 -4n-3 2n (2n+l)a b An and - (2n+2)a Ari = K(2n+l)a B +a Bn = b rr -2n-1 -1 -4nn-l -2n-1 r i a -6V KBn= b-(2n+2)(2n+1)a /t2n+2-1-(2n+1)Ki. - — at 0=-. An . Thus An = (K-1) (2n+1)b ev= r60 2 By 5.154(4.1),5.157: bP 2n+1(cos0)/b0= -(2n+1)P2p (0)= -(-1)n (2n+1)::/(2n)..: at 0=17/2. Thus
_ 1 -
b
i_
Et
.1-) n+1 (K-1)(2n+1)(2n+2) (2n+1):: a 4n+ 3 i (2n+1)K + (2n+2) (2n+2): ! b2n+2r2n+3
a- 2r ,i(b2 +r 2 )3 -F o -
Ans.
-63b 4EV 4EV 217 m' I., P 2 (0) r c 2n+id c By 5.03 (2), (3): 2a= _ /--1 lki coib 2 c 2 P2Th (cos0) See 5.16 (1) (1=17/2. nvb2 -p2 • v =17 470; Tir+ Let c= bsin0, dc = case bd0=.1b 2 -c2 d0. y0u2 s ion+10 de . (2-4.6-2n)/(1.3.5- • •2n+1) Dw 858.44 -2n-1 2n Thus V= (2V 1 M)E (-1)n (2n+l)b2n+l r P (cos0) Ans. -645.16(1). Get -51 S 5pat Pat 0=2 by formulas below Fig.5.157. and write 2n for m-1. Thus n(2n+1): 2n+2 be - QQ1(417Ea2)-1(-1) F=Q't,Vp/(.960)=-Q/(477Ea2)-Isinin+lapm(cosa)OPm(12) aP2n+1(a°sa) Ans sin 0=17 =2a2 /(217Ea). 20, 2 c 0)=2a 2 0) d0=-2d0, (1-cos2 2 (1-sin Closed Solution: cr 1=C 1 a2 (cot2 a+ 2-2sin2 0) = 2aa k-2 (1-k sin 2 0). 477EV=2aifori:d295 z 2 =a 2 cot 2a, z2 +c2 = + 2 617 _Q i ger zdo Q1(22 k3/ 2 acota7/2 Q i Q 2k 3/ 2 cota E(k) de 16774a 2 2r 2 cJo (z2+c 2 )3/ 2-2r 2 E 8a 4 o(1-k 2 sin 20)2 / 2 1-k 2
Use V
The integral is from Problem 18. k 2=2/(2+cot 2cx)= 2sin2 a/(1+sin 2a)
Ans.
Page 230. (206)
Chapter V.
Page 62.
-652n+11 +1 a: +An rnj P (p) V = " n 1 4rEvc r 717r1Pn(P). p=cose. V° -417e c o kr/ n+l -n-1 n -.7111- ) . + Anb = Bn (bn-a2n% V1 =0 when r=a. At r=b, Vo=V i so c b ,, ,n +1 bV, n by0 a E = E--' at r=b, so Ev F(n+1) (t3;) + nAnbri = 'Ern [nb +(n+1)-d vbr Or c n+1+nAnb2n+11622n+1 _ a2n+13 : ecnb2n+1 2n+1 ,. n+1 .. • 2n+11J—_evt - (n+i)C +An D )1. c Equate B values +(n+1)a n+1 2n+l 2n/.1 n+ltr 2n+1 a2n+13 EC nb (b + E v (n+l)c -a ) +(n+1) Solve for A - 1 • Then 2n+lcnb2n+1 n Evnb2n+1 2n+1 2n+1 +(n+1)a2n+1 (b -a ) - cb .. n 2
nA cn-1 Ans. F= 0sur ?;)— [4reqv i A nP c o n r n(0 r=c-417sv 3 n -66ce = Vo=q(4rrevcrl Ea(rnc-n+Anr -n-1)Pn (p), Vi=c1(411EvC)-1 EB (rn-a 2n+lr -n-1 )Pn (u) • _ ap os_101:1) 2: +cib on Vi is zero at r=a . At r=b, V0 =Vi and evbV0 /Or= tbVi/br so bnc -n+Anb-n-I=Bn (Y.' -n-1 n -n 2n+l -n -1 =e (nbn+(n+1) a ex,)B (nb c -(n+1)An b b n . Equate A values to obtain i . 2n+lc _n _ enb2n+1 t . ev( Bn (b2n+1?-a 2n+1 ) ..b2n+lc -n3 . evn (n+1)- b +(n+1) a 2n+1 1 (n+1)-1B 1 2n+nl+(n+1) (E-ev)a 2n+1, -I Solve for Bn=(2n+1)Evb2n+1 c-n ([(n+1)E+nEJb v 3 Put into Vifor Ans. -67Cv=E(r-a 3/-2)cos0. o=e— at r=a,a =3tEcos0=3tE,./a 3 -co 2 /a . dF=2E-1/42dS 2 so that br F = 2 EE 2 a- 21a (a 2-p 2)2TTpdp = 477eE 2 a 3 Ans. -68CTake origin at center of a. c is small. By 5.153(1) 11-' = 1-1(1 + 1)13.1 +. • ) so R P.,' b ( 1 iP1+ • • ). V= T^ ' 2 1'1(1.0+ C + DrPi(11), Q= - Ci g-dS = 4Trr 2 tA/r 2 . i +1 so A = ---247E . At r=a, V= 41,rt so C=0, B= -Da 3 . At r = R, V = Vo so that _1r g + 3 Q ca 3 Dso_bL-„1 a p). Thus Dv0_4rr 4Tfeb2 411E(b3 -a3 )' D- 4r (b3 -a 3 ) bV 2ca 3 3c cos a = [e J + --r-C Jcose) = —q-C+ b 3 -a03 3 Ans. br r=a= 417--(1 a2 + C a3 (b3 -a3 )-a 4r a° 3 (
+Q cos0 sign depends on choice of 0 = 0 line.
-69c-n-2 Let V= ErPi + Br-2P1 +Cr-nPii _ i+ Dr Pn+.1 . When r= a (1+8Pn), V=0 so, if C,D are order of 8, n-n+1 2p. By 5.154 (1), 0 = Ea (1+8Pn )Pi+ B(a-2-2a-28Pn)Pi+ Ca n Pn.1+ Da +-11Pn+1. P1Pn-2:+1Pn-1+2:11-4n-2D . 0 (2). (Ea 8i12.1:_jaB8)n +Can= 0 (3) Thus Ea + Ba-2= 0 (1) . (Ea 8- 2a-2 B6)(n+1) (2n+1)-4aFrom (1) B=-Ea3 , From (2) D=-3(n+1) (2n+1)-1 an+3E8, From (3) c= -3n (2n+1) _ian+1E 8. 81ra an+lr-np + (n+l)an+3r -n-2P so V =E{ (r - a3 r-2)P1 - 3 (2n+1)" n-1 n+1J3 Surface charge density is ce[71.7] =€,,/(bV/br) 3 +(bVirb0)2 , neglect (OV/rO0) 2 . i[r a= EEC (1+2a3 r -3 )1711 +26(2n+1)- man+lr -n - 1pn _1+(n+1) (n+2)an+3r -n-3Pn+0). Write r=(1+8Pn)a. +(n+1) (n+2)P 3). Substitute for Pi kl by 5.154 (1) . n+1 rI-1 33. Ans. = EEC3P1 + 3n8 (2n+1)-1[(n-2)Pn-1+(n+1)P n+1
= tE{ (1+2-2. 3 813n )Pi+ 3 8(2n+1)-1[n2 P
▪ Chapter V.
Page 231. (207)
Page 63.
-70CV1. A = r-1+Br-2n - 1Pn+ CrnPn+D. At r=a(1-1-8Pn ) Vi =V. Vo=r-3E + r -r1-1FPn . At r= b(l+nPn ) li. = Vo and EbVi/br = EvbV0 /6r . BI C and F are small of the order of ri, 6. (Ba-n-1-fr_a un)Pn+D., Aa-1+ D= V (1) -Aa-I 8 + Ba n-1+ Can= 0 (2) V= Ae1(1-6Pn)+ (A-E)b-1(1-TiPn)+((B- F)b n-l+Cbn)Pn+D = 0, A-E+bDJ (3) . -T1(A-E) + (B-F)b n+ Cbri+1= 0. (4) (EvE-EA)(1-271Pn )b-2+ (EvF -EB) (n+l)b-n-2Pn+necbn-lPn = O. (EvE - EA)b-2 = 0 (5) n-2 (n+l)b (evF-EB)+ necb11-1= 0 (6). From (5), A= ev EC, from (1) D= V-EvErla-ly from (3) E= EvEe-i+Vb - EvEbria-1or Vb= ERE -Ev )a + Evb} (ea)-1so E= eabV/C (E-ev) a + evb) • A = evabV/t(E-Ev)a+ evb). From (6), C= -(n+1) (evF-EB)/ (neb2n+1) . Substitute in (2) and : n+l 211+1 b solve for B na Bnb 2n+1+(n+1) a 211+1 (EZV 1(11;El e l bl :71F a: EvIT: n+1 evbV,8 EuF ev -nb (n+1) a (n+1) E•ty bV8 CEan+1 nb 2n+1+(n+l)a 2"1 [(E-Ev )a+Evb + ngb 2n+1 (n+1) an 1 ' [.] [(E-Ev)a+evb
1 -
Substitute for (A-E), B and C in (4) . n+l n+1 n n+1 n+1 n+1 n+11 b 1 eF na (n+l)a b +(n+1) a b ievbV6 F 0. -nabV(Ev -E) + (na Zn-r-1 }RE-9 + 2n+1 a ( C- Ey) ai-evb Crib2n+T+(n+1) "1717 1 E a +(n+l)a 2n+1 nb a+e b) 17.1 nb p[neb 2n+1+ [nb2n +1+ (n+i)a2n +I-1 2n+1 2n +1 evb b all C(E- 0 Ti (n+1) e a 2n+1-(n +1) Eva J+(2n+1)Ev anbn18) 0= Ebn cvb (E-E v )a (c_ev)r1tnb2n+1 anbn+1.5] EabV +(n+1) a 2n+1)+(2n+1) Fso and ev) a+Evb tne+(n+1) ev jb 2n+1 (E-Ev ) (n+1) a 2;1+1 n n+1 2n+1 +( e-nb vo _(E EabV 1 (2n+1)6Eva b + (n+1) a 21.1+1 }bn Ans. •—ITT r -cv ) a+Ev b [r In e+(n+1) Ev 3b 2n+ 1 + (E-Ev) (n+1) a an+i - 71C-
CO Cl. q so at r = co . Let V= Take conductor r = A(1+1)Sn) at potential V0 finer 4- rP+1 SP co V = q(4Tter) -1.Thus Vo =qt (1-1)Sn) (4rea)-1 + ii Anr- P -1 Sp ). Equate coefficients, 4rrEaVo = q, ")
0n , so that Vp= q(4rrer)-1[1+11anr
nSn (0,0) . Apply Green' s iirea•Vr as V Reciprocation Theorem, 2.12(1). Induced charge is q' =-Tri:l qi'3=I- - r t 1+Tlenn (0,0)) Ans . q -72Cn-1 co MT T r c 2rraT `13 r rn c2n+1 ., n—n--+ Et = ft At r=c, a= (la). (0) _ EL--+-IP 4rre Let cc = -1-17 in problem 57. V = or 2 0 a a o an anrn 1 n , 03 m (2m-1):'. 2m -2m c2n+ a (4m+l)c a (-1) ,, P - -e1' E "-) P n(0)P n(p) . Pn (0)=0 if n is odd, so a = (n+1)Th-r+-(2m).. 2m(11) • o ac
An=ann/ (4170, Ap=0 if p
„, a
3 r 14
1
(see problem 60) cr= -Lt1 - 1L.I3J 132+ 9•i•TLiJ
C 1e 73-[-V 13 6-1-• • •) P 4-13• i• T3• 5
Ans.
- 73C2m -4m-1 A2m+ a sina P 2m(0) • In problem 58 set a = yrr, Pn (0)=0 if n is odd. Let n=2m, B2m=b (- Om 1. 3. 5. - • (2m-1) (4m+1) ev - 2m (E-Ev ) .,_ ;I P2m (0) T =-g— la a2m ' 2.4.6- (2m) + 2me. 2rTa' -2m-(2m+1)Ev + 2m e ' J a 2m - (2m+1)Ev 1
q
F., + E,(-1)111(4111-F1) Ev •
V•= 14rrea t.1.-
1 (2m+1)Ev+2mE
1.3.5- • • (2m-1) . srlam_. r 2m (cos 0)] 2. 4 . 6 • • • (2m) La'
Ans.
Chapter V.
Page 232. (208)
Page 64.
-74C-
v. Vi=EB(r-air-2 )cos B. At r=2a, V0=Vi, 8a3+A=B(8a3-a3)=7Ba3, EvbV0=4_2. • br ' ev (8a3 - 2A) = eB(8a 3- 2a 3 ) = 10BEa 3 . B= 12ev /(7ev+5 E). At r=a CI= 3CEB=P017:COSe, br dS=2TTa 2 sin0d0. Q= 2E2IviV 7/2ka sin Ocosed0= 36ra 2 cevE/(7cv+5E) Ans. 7 ev+5 c 0
V 0 = E(r+Ar -2)cose,
-75CVi=A+Br -1 + (Cr-a+Dr)cose, V0=Er-1+Frkose. C, D, F are very small. bV0 By Gauss Theorem: € j bVi. --tI 1 =ev is2s- dS2 so eB = ev E . At r=a, Vi=V0 Si br S so A +13a-1 =V (1), aD +a-2C = 0 (2). Thus Vi=V+B (r-i-a-1)-1-C (r-2-ra-3)P 1 , V0= eev -1Br-1+ Fr-2P 1 . At r= (b+yPi ) : Vi =V0 . V +B(13-1 -a-1)=eev -1Bb-1 so abevV _r b3 -, B= a(e-ev) +bev (3). -13-2Y B+C (b-a - ba-3)= - Ec'elyB + b-2F or y(e-ev )B = -evc..1--3 j+Fev (4) a OVi= E ---OVA • -ec [2- ±1 }=-2e F or b3 2a 3y(e-ev)B from e---v, Fe = cC{1+ — } (5) • C= v or ' or b3 a 3 13 4 v 2a 3 e(2a 3+b 3)-2ev (a 3-b3 ) (4), (5) so C=
2a 4 bYev(e-e,)V bV B 3C a + --s a cos0. Thus fa(e-ev)+bev)(2a3(e-ev)+b3(e+2ev)1' a = - e&-lr =a so cr=--j -
beveV 6 a 2y(e-c„) cos° 3] at a(e-c.v ) + bey] .+ 2(e-ev)a3+(c+2Ev)b
Ans.
-76 CSince evls i nTi/ br dS1 =e1cOVd/Or dS2 = evls3bV0 /br dS3 ; 4revVi/q = r-1+ A +BrPi ,
4reVd /q =r -1+Cr-2P 1 + DrP i+E, 4ttevV0 /q = r-I +Fr-2 P1 . B,C,D,F are of order ofc . r = a -cPi , r' = el+ c a -2 Pi , Vi=Vd, ev011i / Or = EbVd/br. cca-2+Bac=evc a -2 + Da ev , olio ELAd 1 1 cP1 7 -c C(Ev+2E) + (1). r= b+cPi , 7. = 1-) + Tr, 170=17c1' cv& = br B = -2Ca-3+D, so Da 3 a3 (e-ev ) E c E C 1 Ev+2 E EC EF -2C so IF+ It +evbD= Ti+b -i (2), Tr +D=-2b3 F (3). (1) in (2) evC E-s+-7 --7] lDi a (E-Ev) - (E-E
by
L. + Ty EF
_C_. 3
a
r
(13 3 (tv+2E) -2a3(eE +2e 21 c -2F L a3(c _ ev) b3 (4). (1) in (3) C I—v— —- a-3 - b3 (5) or C a3 b3 (E-eV)
chi
=ca-3- 213-3F (6). b 3 ( (4) - (5)• ev ) gives 3ev C=(e-ev )c + (e+2ev)F. Substitute for C in (6) C _ 2F c[133 (ev+2E)-2a 3 (e - Eve + (e+2ev ) [133 (e+2e) - 2a3 (e-ev))F S o lve for F 3eva 3b 3 a3 b 3' 3 eva3 b3 (e-Ev) 2c (b 3-a 3 ) (e-ev)2 cose 2c (b 3 -a3 ) (e..e,)2 1 Ans. FV° ci [1 - 4rrev r + (2a3 (e-ev) 2 -b3 (c+2cv) (cv+2c)1r2 2a 3 (E-Ev ) 2 -ba (e+2ev) (ev+2e) ' -
-77an tis,n q co 4TrevV 0= e-E o t— en +1-+--1-3Pn (cos0), 4TTeV.3.=(cos0). At r=a, V0 =Vi so e-yi + n+1 -6vBnan • c EBnrnPn 0 c a n 2n+1 .23 eLi V0= ELS m a it1+115,n -n(c-ev)a 61.71 1 -(q/c)2 s., n(e- Eli)(n+1) ral2n+1 B A = - nan Ans vor Or ' el anal n' gln f n e+ (n+1 ) cv1 • F- (1.67 = 4trev V ne+ (n+l)ev InJ w rrn____Ja___ A
-78s
A (r -a
2s+l-s-1
r
1
)P s (cose) satisfies LaPlace' s Equation, and gives a
potential positive if 0 < 9 C o,
V =-12--E(2n+1)Qn (jC)PnOCOPn(g0). If C Co V=(jq/(4rea))E( 2n+1 )Qn(iCo)Pn(to)Pn(jC)Pn(C)
)
(g)
n
C< Co
- 87Ring potential from 86 is Vine(jq/(4rEa))pn(4n+1)Q 2n (jC0)P 2n (g0)Q 2n (jC)P 2n (D Vr ing=-Vind at 6=0. Only even Vring terms.remain so that B anctan(i• °) "P 2n (i'0) E(4 +1 )(22n(iCo)P2n(§o)Q2n(jC)P2n (g) Ans. B 211=-2P2n(i•0)/[-iTTP2n(i•0)). V ind=7,-4-zr- ea 0 -88Only odd n values remain since V and g change signs together. On spheroid V is function
co 1 C o V= ..AnPn (g). Multiply by Pn (Ddg and integrate -1 to +1. 2VO4Pn (§)d§ An _211 -1-11 .n-1 21. : ‘31. = Ani:(Pn (g)) 2ds so by 5.154 (5), T 21..ifpPn ( :1:21 ) Vo,n odd. 1 -Pn-1]0 --a 2:+1. . . tim Q,n4., (jC) n 4n+3 1.3.5...(2n+1) r Thus V = Vo f,(-1) Ans. of g only. At
:
u/ Q2n+I(..iC0) 2n+1 2.4-6 ••• (2n+2) '2n+I ■ - 89-
.15 s.2i74 1;'1 °II/1i- -;TVIIs I -laii-%;),,,
IT, (1)
( -1 rn+.(nim:) 1
V° -
CI-MT, 01D)cg 01) smn
Of (n-m) ! 2_ s(n+m) . 01. 1)
(
6,
)
igl,
yi - (":.(+.: 1;M)' (rg- 1)4(710)11,.(n)siT (4) on (5) "I 'Io Sr _ zPrshi( g) cos s(0-90, and multiply thru by and cr=rE TI2 -E
where 4=CmnP1,;(g)cosm,(070m). (3) Let Amn=
(2)
PW(g)cosm(0-0m)h i h 2dgdo, 6 . = (2n+1)(n-m): r r r j! j 2aP .m( 0 cos m (0- 0m)h i h 3d drzi and integrate over 'n 0 . mn 2rc 2(n8-1) (n+m)! 1 D n Vi=itMmn P1.11 (-
0m)) (9) Vo=itNmnQT(71)4ni(g)cosm(0-0m) (10) -0m i r,v, (-1)m( 2-5 m) (2n+1) Fli.,2„9 2 (g)cosm(0-0m)h ih 2dgd0 (11) M (n )1: j 0oPm , Qm n n 0 4rec 2 L(n+m). mn
(8)
Ninh- Q P1Ini:(0110 1°)) Nmn (12)
• Chapter V.
Page 234. (210)
Page 67.
-90Take q at fa = 0 0 , t = g o . In last Problem: PT(C0),Efahih 2dtd0 = py::(g o)rsadS = q1 (C0) .6j20 [ (-1)m(2-8%) (2n+l)qrki, 'mn n (gc), Nmn4TTEC 2 (n4m) Qrn n (71°) Pui L(11410 PIR (rid PR( 4rec 2 CO CC
03 CO
Vi=E Ellinn P(TOP(DCOS M(95-00), ri"no • 00 - 91. [..11 (11r a)I. CD ... .N Axial charge from 5.298: Trea 2Vp=fsinhEpr (L-c)3 sinhprzJo(prP)(pr sinhprLrl P is point on ring. qq+0•Vi)+Q•0 = 0•Vc+q•Vp+Q' •0. Replace p and z in above formula by b and c to get potential at z=c on axis due to ring at p=b,z=c. Potentials reducing to this on axis are: If z c, write (L-z) for z and c for (L-c). Ans. prcji oran •. TTea2sinhprL
Second solution: Set q= q'd0/(270 in 5.299 (1). Integrate from 0= 00 to 0 = 2rr. All terms but m=0 vanish, giving above result. Force downwards is the limit as z approaches c of 21qE (oV/bz)
+(bV/bz) J. Numerator of first factor is prsinh[pr (L-c)Jcoshprc minus z>c z‘ c Or sinhprc cosh[pr (L-c)J which gives prsinh pr (L-c-c) so the force is F=Eq 2 /(2rrEa 2 )4 sinhpr (L-2c)cschprL [Jo (prb)/..T 1 (pr a)J 2 Ans. -92Solution finite on axis and giving V0 on cylinder when Jo (pr a)=0 is V =Vo +E(Akcosh pkz+ Hi'c sinhpk z)Jo (pkP). To make potential V0 at z=c take V = Vo + EAk(sinhpkc cosh pkz -cosh pkc sinhpkz)J o (pkP)=Vo + EAksinh[pk(c-z)]..To (pkp). When z=0, V=0 so -Vo= TAksinh pkc Jo (pkp). Multiply thru by pJ0 (pkp) as in 5.296
4-- 2a—* and integrate. 5.296 (4) gives As= -2V0 (a 2 sinhpsc[J i (psa)J 2 3-11:pJo (pkp)dp. From 5.302 E the integral is ps-2fo'Is a xJ 0 (x)dx= p-s'aJi (ps a) so V= Vo [1 2 sinhIpk(c -z)).10(PkP)] Ans. uksinhpkc J i(pkp) A potential giving proper value at z= ±c and z= 0, and finite on axis is V= Vo C(z/c)+ EAn Io (nrp/c)sin(nrrz/c)J . Multiply by sin(nrrz/c) and integrate from -c to +c. pc
c
c
V oJ o sin (nrrz/c)dz = VoJ o (z /c) sin (nrrz /c)dz +An Io (nrra/c) j 0 sin a (nrz/c)dz An=2Vo/CnTTIo(nra/c)) . Thus V= Vo [(z/c)+ (2/1)E 10 (nrp/c)sin(nrrz/c)/(nIo (nTra/c))3
Ans.
Extension: 172 = EBkSinhpkZ,To(Pkp) (1) 14=170-1- EAksinhilk(C -040.1k0 =E4+Aksinillik(C - 4J0 (10) (2) where Ck=2Vo(pkaJi(pka)) by HTF, 7.10.4 (54). At z=b, VI=V2 and bvi /bz =6112/6z so Ck+Aksinhpk (c-b)=Bksinhpkb and -Akcoshpk (c-b)=Bkcoshpkb, (3) and (4). -2Vocoshpteb 2Vocoshpk(c-b) Eliminate Bk : AkBk-iikaJi(pka)sinhpkc iikaJi(pka)sinhpkc' VI=V0
coshukb sinhpk(c -z).4 ( 10)) t1-2E pkaJi(pka)s inhpkc
V 2=21.70 Z
coshpic(c-b) PkaJi (pka)sinhpkc sinhpkz Jo (pkp)
Ans.
t t
Ans.
2a --*
Chapter V.
Page 234. (210)
Page 68.
-93-
Solution giving V= 0 on walls and the z = 0 plane is V =TAksinhpkz.ro (PkP) where Jo (pka) = 0. When z=c and b> p> 0 we have V=Vo. Multiply thru by pJ o (pkp)dp and integrate as in 5.296. rib Ak= 2V0( a 2 LJI (pka)] 2 s inhpkc ',I 0p Jo(pkp) dp-
2bV„TiOeb) ra s inhpkcLJ I (pka Pka 2 ,
Ans.
Boundary conditions are also met by V2= Vo[(z/c)+3FBn I o (nrrp/c)sin(nrrz/c) 4-2134
if p b. At p=b,V 1=V 2 . Multiply V I (b) = V 2 (b) by sin(nrrz/b) and integrate z=0 to z=c. 4An[Ko(nria/c)I0(nrrb/c)-I0(nra/c)K0(nrrb/c)]= - (-1)n (c/(nrr)}+ iBncIo(nrrb/c)
(1)
At p=b, OV I /Op=bV 2 /6p. Multiply both sides by sin (nriz/c) and integrate z=0 to z=c. An ko (nrra/c)I;(nrrb/c) - Io(nrra/c)K8 (nrrb/c) i= Bn I8(nrrb/c)
(2)
SUbttact (2) • I o (nrrb/c) from (1) • I“nrrb/c) and use 5.32 (7) to obtain
n 2I,;n(n rrnb/c) ApI o (nra/c)No(nrrb/c)I;, (nrrb/d) - Ko (nrrb/c)I0 (ntrb/c) 1 = Ap(I o(nrra/c))/ (nrb/c) =(_1) So VI = 2(b/c)V0 E(-1)n { Ko (nrra/c) I o (nrrp/c) - Io (nrra/c)K0 (brp/c)ll I 1 (nTTb/C)/I0 (nrra/C)) Sin(1117z/C) V 2= (VoiC +2bE(-1)9 Ko (nTl'a/c)Ii(nrra/c)+ Io(nrra/c)Ki (nrrb/c) (To (nrrp/c) /Io(nra/c)isin (nrrz/c)] An: 94From 5.298(4) for lower charge : q2 Force is qE- — 2re a
e
2rea2 le -2prb { ji(pra ) _ 2. Ans. }
PrJikPrai
E
e "Pr z q oV Oz-2rrea 2 7(J 2 (p
q•
-95-
(e-ev) Use above potentials and plane face image laws. Dielectric image at -b is (e+ev) lz-b I Ans. VI = q(2rreva 2 )-1E[e (E -Ev)/ (E+Ev)3 e-Pr( z+b) ]cibi(Pra)}-2 To(PrP) i e-pr(b+ (pr o)1 -2 J0 (PrP) Ans. V= q(2rrev a 2 )-4E2tv (e+Ev )-962 ( Pr a) By 5.102 invert charge q=4Tre at 0 on cylinder axis. V= — Ee p aa J141r 2e1,0 e -Prz bV -E-513 at p=a gives a- a2Tt ji (pop 5.102 (1) gives a' = OP 3 /a 3=a/cos 3(x.
z = tancx, dS = 2r(acoske) (a/2)dB, but e= 2a so dS= 2rra 2cos 2a da. Q.2.1 2a i d_S a Q= 8rEE[Ji(i}s)}-1,17/2e-pra tan secada. Let tana =sinh0, sec aada= cosh0 do u-,. a sinh 95 d0j/Ji (pr a), where Jo (pr a) = 0. c o°e-• seta da = cosaill+tan 2ad0 = d0 . C---q v=aQ= area ELr so C = 8rrea E S o, o (Pra)/JKura) . Here S0,0(pr a) is a Lommel Fctn. HTF II (50) p 84. -97a2 p e sin 0, Dipole M= 4TTEa 3E inverts into uniform field E. Take M in cylinder p=a. rr'=a
z = rcosel= (aa/r 1 )cos13. From 1.07 and 5.297 (4), Vd= 2E4 ellkz Jo(likP)/{J i(11ka)}2 . From 5.10(2), V'=
f,V= -(.2E8 2 /r 1 )Ee-
pka 2 cos Oir
r Jo t (pka a sine)/r 1 }01 (pka)}-2
And.
Chapter V.
Page 235. (211)
Page 69.
-98From 5.298 (4), potential of +q at 0=0, p= b and -q at
0=
r, p=b is, since cos sO
(2-44)e-Pkizink- 1 Ds+1 021( 03-2 Js (pkb)J s (110)cos s0 • - cos(17-s0) =2 cos sO, V= --T 2, rgea i 1.111 where s is odd. As b --->0, Js (pkb)-",-(pkb) s (2ss!)-1 by 5.293 (3). Neglect s>1 and let 2qb=M=4rea 2E as in 97, then V= 2aE ie -Pk IzIJI(PkP)L-T2 (11ka)r acos 0. Invert as in 97, 2Ea 2 l e -pka2cos 0 /r. J 1 (pka 2r -lsin 0) Ans. cos 0 r 1 (J2(Pka)) 3 -999 By 5.35 (3) axis potential is: Va=4TrER 21.1 2 SccskzKo (ka)dk. A solution of V' -
LaPlace's equation with this axis value is.-* .1. 12efo Ko (ka)I0(kp) cos kz dk, (ka)K0(kp) cos kz dk if p> a. Ans. 2T if p r write rPr -n-I for rnr0 -11- land differentiate, eliminate Prinl i (p0 ) to obtain
il!( 2 4 -) (4=24 1Pm ( p ) Pm( c 0 s 0 ) c 0 s m0 Ans. n+m):( m+n ) q-cTIF n-1 ° n
V =7 1-, D 4rer6 oo
-1162n+1 TITI P (p)cosm0 Use 115 Ans. (Ev fore) for Vm and []. Vi=[]Amn PT(p)cosm0, V0=Vm+[ ]Bmn{E?) Amn. Bmil i01 2n+1+1, rr.. nIlBmnq) 2n+1+1, so Bmn=-n(K-1)a 2n+1hTgri-n+1] ro 211-" • LatKAmn= At r=a:
rn-1 I BmnSina Pri'll(cosa). Torque is At dipole: Er br ------AB +1)04(cos00, Ee= ri b li = [iril r ir ar ern mn MErsina+MEe cosa= M[]r31 8min since[(n+1)41+110PV . Use 5.233(8) . Torque is T =Msina[ ]rot Bum( (n+1)Pri+( 1-par 1[ (n+l)pplin+(m-n--1)1.1041+1 ]) =14 [ ham /rsinal ( (n+1)Pri (cosa) -
(
+ (m-n - 1) cosaTiT+1 (co;a) 1 Ans. Radial force is Fr=(ti.V)Er from 1.07(3). E =r1B ltifFsin+24(cose). M0=-Msinia, Mr=Mcosial Fr={-since (bEr /rb0)+cosa(bEr /br)) r=r 0 ' i'T " mn 1-0"-' ni = - (n-m+1)1141 by Fr= ME jBmn (n+l)r-02 (s in2aPinr°(Co sa) - (n+2) c °sot II(c os a). (1-11 2)Pm n '- (n+1)On 5.233(8) so
Fr=-Mc 3B=(n+l)ra 2(COSa Pn(CO sa) + (n-m+1)13;141 (cosa)1
Ans.
Page 239. ( 214)
Chapter V.
-117sinhui W dz z - a cot _ ., From From 4.13 (2); v - - coshui - cos u2' zj du
Page 73.
2_ -a a2 kdu u)i (cosh urcos u2) 2. 2 j sin 2(-ju/2) ' (d
sinhui au .t_l_ b r sinhui m 2U NJ } In 5.11 (6): V 2 U- 6 I' 6.11- coshul -cosupuis ' bu2'-coshui -cosua Ou2 sinhui (coshui - cos u 2 ) sinhu sin uaALJ a m U - 0. Oul R sinhul R4
sinh LI, i6 2U + b aU l+r coshui sinhau ,OU R 2 ' bu ? bui -3 " R2 R l ?Jul
.a
If R---VCOSh UrCOS112
let U = RW, bU/bui4R-1(sinhui)W + RbW/bul , bu/bua = -1/R-1(sin u2)W + RbW/6u2 ,
6 au r cosu2 sin2u21,..,_,_ sin u2 :?1_47 + Rb 2W o 2u r coshui _sinh 2u w+ sinhui bW +Rb 2w so that R Oui dui y.,1= I 2R 4R 3 i " -'- R Oua 64 FCITI 2R 4R3 o 2 1.1 + b au _ cosh ut+ cos u2w÷ sinhuiOW + sinuat231 +Rr o 2W + b 2W.I. Substitute in third line abov∎ ouf bu 2 R bu2 1 bu 1 6112 I 4R R Oui 172U=
.I. sinhui I. b 214 + b 2w1+ coshuiOw +r 3coshu,sinhu,4-cosu2 sinh u t sinhaui +sinhui sinaui _ m2 211 2 Rsinhu1 I R bui 1 R 1- but twil 4112
In the above all 12-3in the OW/Ouland bW/bu2 coefficients cancel out eliminating bW/bu a Simplify the W coefficient and multiply thru by R/sinh ui . Equate V 2U t 0 zero to get. 62w b 2w 2 1 bw ., 2 + 1 ---ff, + COthtliF + ( 4 . . 2 3=0 Let W= W IT,42 and v 2 (W I W2 )=0 separates. Thus ou i C.W. 2 Ui sinn ui 2 1 b 2W1 ,_ cothulOW I +1 1 baw m sii -T11u-+ w ---2 6u2 '= 0 yields two total differential equations ci:Ful ' WI tali 4 1 2 2 d21071
1
M2
dW
___i + Coth uidui du ? i
t n 2--4 1 +sinh 2 u1 3W1 -.7. T
°
(1)
2W and d du 2 --= -n 2W 2
(2)
2
-118-
0 AnPn _i (coshui ) cos nu t, 0 .1.12< 2 rr . V =^/cosh ui- cosu2 E At ui=u0 , V= Vo . Multiply by cos nu2du 2 , integrate 0 to 2/1 or let u 2= Tr -93, due= -dO, cos nut= ( -1)ncos 110.
N
US) .......-
-A/ N1
.
1....."
1
, I ► \ i ... „1—L.....t. \ I ....„ ! , N. / „ I NI, , --i -- r/ 1 ... I , Ii -... I I t -7- --1--. I ! I /` , , // I %. I ....e .• ---- 1.- -I -
Write cosh uo= zag.7. so sinhuo=1//j71. f---- Tr r 2r cos nu,du2 _ (-1)n2Voksinhuo V0 rcosn0d0 A n rip 1 (coshuo )Jo ,IcoiEuo-cosuo TTP 1 (coshuo )40./Z+(z2-1) n-n-T 2 This is Laplace's integral 5.23 (15) so that n n n a (-2)2Vo Pn_ q2(coth tici) ( 2n+1) Au _ el ) 2Vo P- 3 /, (c oth u0)/... .1 111.10 P 1 (cosh uo) (2n+1) :1. P 1 (cosh uo )4.1.4.• • (n-1) n- 2 n-T -119I I By 5 .23(15): P 1 (cosh ui )TrMp-i.1-17(coshui)11-id0 -,(coshul )n T -el. n--2-
112=3r/4
1
7
-
.
=17/2
Ans.
At u 2-4' 0, y= a sinhul / (cosh u1 -1)1.7i1,e au i / (1-iut • • • -1)---w- 2a/ui
A/cosh url -0- u1 1,11= A./1 al y so V
4/2 -i- E An= 0/ (4rtr) since y=r in ring axis plan( o I 2) n 24/2(coth uo) (2n+1) From 4.13(3): a=bsinhu,3 . Thus C= 2.--- 8rte.12sinhuo b E `Ans. Vo 0 (2n+1)V.P (cosh uo) --** 1.1 1-1. 1.1 2-v 0
n-1/2
Page 239. ( 215)
Chapter V.
Page 74.
- 120By 4.13 (2) x= a sin u2 /(coshui - cos u2 ) . V is zero i3 ui =u0 or u2 = 0 or n. (coshuo innu2 For uniform field V= -Ex= -Eoa sin uo(coshui -cosu2 )i= EAn P o n-2 Multiply by sin nu2 du 2 and integrate 0 to 217. rr 3 -Eoa sin nu,sin u2du2 Let u = sin nu„, du= n cos nu2 du 2 so that A n -rirp (cosh uo ) (cosh uo-cos u0 3/2 v = -2/4/cosh uo-c osu2,dv = sin ua du2 (cosh uo - cos u2 ) 2 r1.• 1/ 2 r2rcos nu2du2 (-2r 4nEoaP12 v_2(coth uo) (2n+1) 2nEoa Ans. (from problem 118) 11_1/2 (cosh uo)•0,/coshuo-cos u P _ (coshuo )(2n+1). 2 n This potential cancels uniform field potential at torus surface cosh uo = R/b.. -121-
co V is to be zero when ui =uo . V= Epcos0 +4/cosh ui -cos u2 oAn P1 (cosh ui )cos nu2c0s0
3
By 4.13 (2), p = a sinh u1 / (cosh urcos u2) . At ui= uo , V=0 so -Ea sinh u1 (cosh urcos u2)-7 =131 1 (coshuo ) cos u2 . Since the left side is an even function of u2 , it is clear_ T n r,2rr that sin nut terms cannot enter for jo is zero. Multiply by cos nu 2 du2 , integrate 0 to TT. rr - 2Ea sinh uo cos nu2du2 cos nu t due - Ea sinh uo r 21" An - 9rPn11 / 2(cosh u0 )1 0 (cosh u1 - cosu2) 3/2 - 17P/1.1 _ 1/ 2(cosh u o ) (sinh u0) 3/ 2brtz-,fir=I co su a )3/ 2 pri 3/2. p nin so The z-substitution is from 118. In 5.23 (15) n is -I, but from 5.22(1) n n n+1 Ea P/,(coth u]) -2 (-1) EaPv,(cothuo) /(. . 3 ... [n _ 3 3) _ (2n-1) (2n+1) (-2) A = Ans. 2 2 2 2 n ,,./sinh uo (2n+1):: P/1-1/2(COShUo) v2(cosh U0) a = b sinhuo , cosh uo =R/b. -122From 117 1 dz/du 1 2 = a2 /(coshu1-cosu 2 )2 , x= a sin u2 /(cosh ui-cosu2 ). Substitute in 5.11 (6). mzu sin u2 1 bU + O sin u2 1bU - 0 sin u2(cosh urcos u 2) bui coshui cos ua jbul Ou cosh th-cos u2 bu2 sin 21tU m 2U sin u 2U t 21 sin u sinhu bU .4. [cos u 2----1— --,2 + - 0 bUt Otq R4 bu t Re sin U 2 R2 R4 R' oul -
Let U= 4/cosh u1-cosu2 W = RW. Derivatives were evaluated in 117.
m2 , sinuar O 2W + b 21,41+cosu 2OW +(sin u2 (cosh ui+3cos u2) sin u2sinh2u1 1W = 0 R 1oul bu P R bua 4R 3 2R5 Rsinu2 big ILW In the above all R-3 terms in the Pl. and coefficients cancel out eliminating -6— 1 . I Write 1-cos 2u 2 for sin2u 2 , cosh 2u 1 -1 for sinh 2u1 and simplify so that m2 6 aw b W bw j_ x 2 +-i, —2-+ C 0 t U2'571 23W - O. Let W = U l (u i )U 2 (u 2 ) and V 2W=0 separates. Cill i UU 2 C 4 ' sin2u2 72u _
-
2
m2 1 d U 2 +cotuadU_ 2 _ 1 2u1 + 1 -- -= 0 yields two total differential equations U2 du22 U2 du 2 T sin2 U2 UL out 2 U 2+ 12 d 2U m2 CO t u (12.2 + (n-1-1 ) (1) and —13= (n 2+n++) (n + U dui 2dU2 11 ' 7.7-711172= '
(2)
Chapter V.
Page 240. (215)
Page 7-5-.
-123From 4.13 (2) z = a sinhui / (cosh ui - cosu2). To cancel field potential Ez on sphere ti1=u0 and give it value Vo write U1 0
V0=Ez + REAn sinh(n+4)uo Pn (cosuo ) where R= ,icosh ur cos u2 so that co
:
11"4 Vo - R-3aE sinhuo= EAn sinh (n+4)uo Pn (cos u2 ) (1). If we let cosh u°-l= (a-b)2 = 2 sinhaliuo , ° ±u /2 r7-14 , / r- fu /2 cosh uo+1 = (a+b) 2 = 2 coshaiu o so that a= e ° /v2 b= e - 2 /N/2. Then a/b = e- ° and i 1 i -(n+ .0 0 expansion by 5.153 (4) ields Ri= (cosh o-cosu2 ) "E= (a 2 +b2 -2ab cos u2)72-=4/Yi pn(cosu2) lc° - isinh uo(coshuo -cosu2 ) 7 = b(r -')/0u0 =-29 (n+i)e-t1÷7)u° Pn (cos u2) (3) Substitute (2) i ‘ and (3) in (1). Collect coefficients of Pn (cos u2) : E(./27V0 - (2n+1)EaJe - 'n u ' t u°-Ansinh(n+i), 3 1 r (2n+l)u 1 •Pn (cos u2 ) = O. So An = 22(v0 -(2n+1)Eathe ° -13 Ans. From 119: a = bsinhuo , coshuo=1 02
SO
V=
Ez +./cosh ui - cos u2 ; An sinh(n4)u1 Pn (cos u2)
Ans .
-124From 4.13 (2) and problem 123 p= alr2sinu2, z = ar2sinh Li" R 2 = cosh u1- cos lir In 123 V = 0 when z = 0, u1= 0 so sphere charges are equal and opposite and all flux between them crosses the u1 =0 plane. Thus for no charge 21710 (61/s /bz)pdp = 0. Vs is last term i cos u2[bvs1 bvs by tu _ O. cos u2 ) 3/2 c°(n+ 7)AnPn i in 123 answer. — -s •---1 b (cos u2 ) Oz - Oui 6 z u1 -0. 0 NO ui -3.0 a a -
-
But (pdp)ui =0 = (1-cos u 2) -2 a2sinu2du2 =-a 2 (1-pr 2 dp, where p= cos u2 . Thus -folews/Oz)pdp pi
= -L i (1-p) 2 E (n+ 4 )An Pn (p)dp = 2-faE j_ (p)f(n+-1-)AnPri (p)dp, use 5.153 (1) (a=b=1), o 1 s i 1r =2"fa fo (n+-2-)Anf i tPn (p)32dp= 2 2 a pin . by 5.155(3). Substitute for An from problem 123: E(2n+1)Ea(e(2n+1)uo _1)1. Q = 0 = EAn or E Vo (e0n+1)uo 1)-1 = By problem 119 a = bsinhuo , o ..13E[ e (2n+1)uo.1311-1 V o = Ea (2n+1) He (2n+l)u Ans. o -125-
g
Integrate stress across u1=0, z =0 plane. From 1.14(2), F = ler( (aV/t)z) 2 -E 2 )2Trpdp. By 124, -pde = (1-0-2 a adp where V is total potential of 123 and by/bz is from 124. i co rl 2Ea co 1 3. (n+-0AnPn F= TrE Li (:7 (p) + (1-p) [F3 (n-F-i) A n Pn (II) 3 941 = ItEt1 -27EaAn+(n+1 a )2Atlf_i[Pn (p)]adp pi co ,, i i -E(n+-i ) (s+ 1)AnAsj_ipPn (p)Ps (p)(p))• Since from 5.155 (1), (2) j_I Pn (p)Pm (p)dp = 0 if min. o 1 -1 but j, (P_ (p)) 2 dp = (n+ i) . Expand pPs (p) by 1.154 (1) and last term becomes pi co , 'a:. ,. „ E(n+ii+AnAs t s j iPs _ 1( p.)Pn (p)dp+ (s+1)11Ps+101)Pn (p)dp) only s=n+1 in first integral o co
r
0
and s=n-1 in second so = -2-.(n+1)AnAn+i+r3lAnAn _ i j= (n+1)AnAn+i , since 7A 0A_ I =0. Thus F = ITEE An -27Ea + (n+-10An - (n+1)An _ i ) o (
Ans.
Chapter V.
Page 240. -126-
Page 76.
(126,p216)
Potential inside sphere has form V(r,0) = iaAnrnPn (cos 0). If potential at r=a is f(p'), r.1 1(n+-ii)rn a-nPn (p')Pn (u)dp.' (1) then, by 5.156 (2) this is V(r, 0)=.1.2 f Note that co summation may be written 41'reaViQ-16(17Vp ) /Or (2), 4TTEVp=Qpn a -nPn (cos a) Pn (case), r qn116/b = eAnK i(nrrp /b)2Trpnr Mn r na K tang \ 1-1 b 1 .1
1
An =2eb roKi(nro /b) 2E11 21- b
Mn 2eb 2
by 2.32 (5).
Invert with R= b = 2a, choosing M= 4rre (2a) 3E so by 5.102 (1) (2a)3., 1 r OP 3 = oos 3aL cozi.
4Tre(2a)3E 7 o / . _tana. a cos3a2e(2a) 3. 2a 2172 eE n so a = 3 E(-1.) n 2K 0 (nr tan a) cos a
2
nn (-1) K °(nrt a lia Ans.
-128Spheroid potential from its charge in spheres presence is, by 5.40(7), U c° 2s+1 P 2s (cose) . Vin= iAn[Q2n0C0)/P2naGo.)3P2n(iC)P2n(t) i rslAnCsn (c/r) g) = L, i V ex=AnQ2n(jC)P2 s+1 n-° (r/a) P 2s (cos()) by 5.40 (25). Csn=(-1) s (2s):/t(2n+2s+1)::(2s-2n)r.) , = j-0 s=0An Cns(n/a) Cns= (-1)n(2n+2s-1 ):'1:? 2n (jc o)(a/04s+1/1 (2s): (2n-2s )!!P2n (go)), Crin=(Csn+Cns3s=n' Sphere's Nn 2s+1 r 'AC charge potential must be V' =P 2s (cos0) to cancel Vin at r=a, so ex N n.-9 s4:1 n ns (c/r) 2s+lr , ,a , 2s _ a ,r ,2s+11 total V t between them is V = E E ( / ) !P (cos0). But ( ) ) tkrt ) t n=0sF0An Cns c a 2s r)2s+1 spheroid's potential is maintained at P 2s (cos()) BnC sn (c / V=3302n (.5°)P2n() 40 st 2s+1 The sum of (c/r) P (cos0) due to V and V' must equal 2s ex ex the original V coefficients so N 0 A 0 0 00 + A i Co 1 + A 2 CO2 +• • •+ ANCoN = jo BnC'on 1
A o C io
+
A /CI1
+ A 2C 12
+...+ AN% =nti BnC in
N
A oCN0
+
A 1 CNI
+ A 2CN2
+. • •+ ANCNN =
BNC
NN
• Chapter V.
Page 241.
Page 77.
-129Potential of sphere due to its charge in presence of spheroid is N = 2n+1 from 5.40 (10) V x=j0An(a/r) P2n(cos0) =nosEhAnCsnQ2s(jC)P2s") 2n Vin=ntoAn(r/a) P2n(cos0) =
from 5.40 (27) 0AnCnsP2s(jC)P2s(4) . 1-Al2n+1 (-1)si(2s+1)(2s+2n-V.1 c „ _/-12n (-1j1(4s+1)(2n):: CneCsn+Cns 'sn kci (2n)' (2s-2n):: ns "' (2s+2n+1)2(2m-2.5)::' at s=n To cancelVi . at C= Co , spheroid's charge potential must be N n n So Vtbetween them is qx -jOAAn Cns(1)2s (jC°)/Q2sa C°)1Q2s(-1C)P2s ( C)• N n (g). But sphere's potential is kept at Vt=nE0 s-E-0AnCnsEIP2s(iC0)/Q2s(i")3Q2s(iC) -P2s (j6)3P2s N = 5.4) (10) The sum of the Q (jC )P (C) V = B P (cos0)= ° 2s 2s j0:F411Bri CsnQ2s(j‘ 0)P2s(g) I n 2n 'xmust equal the original BnCsn coefficients so that Vex +Ve coefficients due to A0000 + A1C01 + A2CO2 +*"+ ANCON = 1 COOBO AoC i o + AICII + A2C12 +
.
•
•
+ ANC1 N =rigo C ln Bn
-
•
•
•
•
•
•
•
•
•
•
•
•
•
•
n A0CN 0 + A 1 CN1 + A2CN 2 + + ANCNN =ngoCNnBI
-130From 129 and 5.214 (3) V= When COD, a= -ehbVoc • nliOsikAnCsj(2j/11)Q2s(j4)-4)2s(jC))P2s“). a= ---4 16, C ((2j/11)Q 1(i*0)-P' (i 0))IP (g) h -oc.g by by 5 27(7). Then by 5.2142(3),(5: c i , n=0s=u n sn 2s 2s .".IT 2s . 2 0)) so that P2S(j.0)Qs(j.0)-P2S (i .°)Q2S a.°) = 1 =TSP2S(i .°)C (2j 1")qS(j.°) -PiS (.j • 2e a= je.21 __1211L A n P p2s((j) ) A C (§) Ans.5.214 (3) Trc i t n=0 s=O 1-1‘s cl r n=0 sWIn'sn 2s -() k /n sn (2s-1)” P2s -131N N n 2s For spheroid: Vin=nEcAnP 2n(jC)132nM nE0sFoAnCns(r/c) P2s(cos0),5.40(7) = co N A1-.. islAncsn(?)2s+1 _2s 5.50 P (cose) )3 Q2n(j C)P2n(D=n) Vex=nErn"E2n(j"/Q2n(j" (7) s n 2s+1 _(-1) (2n+2s-1):: at _ (-1) (2s):(c./0 = + where C ns (2s): (2n - 2s):: ' Csn j(2n+2s+1)::(2s -2n):: ' Cnn Csn Cns s= n N = To cancel Vex at r=a need V! = E En-An C sn(r/a)2s P2s (cos0). So Vit between in n=0s= N Z AnC 2s+1 2s , sphere and spheroid is V E ((air) -(r/a) jP (cos0). The internal spheroid = sn N 0s= = n= 2s t N potential remains nE0BnP2n(jC)P2n M (r/c)2sP2s(cos0) so this must be the =nE0sEnBn interior value of the potential needed to neutralize the induced charge on the sphere, 2s namely -Vin. Thus the coefficients of (r/c) P2s(cose) must be same in Vin and V'in so • • • • • • AoCoo + A1C01 + A2CO2 +...+ ANCoN = nEen Cno • -
A0C10 + A1C11 + A2C 12 +'"1• • • • ANC1N = nkBn Cni ' A OCNO + A1 CN1 + . • • • • A2CN2 + 0 + ANCNN = • •
•
•
BN CNN
Chapter V.
Page 242.
Page 78.
-132This is a special case of problem 129 and the result can be obtained by the substitution of n+1for n in that problem and rearrangement. -133cle Pin{l + (f) )2 -Icos0} Potential of line charge q at 0=0, F=b is VI =4; q °:., pn cos n0 q ''' , nn sinne cosn0. If rnf 0,0) satisfies V 2V=0, so n nbn -27rE nt'l 1pi 2rre 1114 1 r-n-lf does (0,0) . Thus is solution giving V=0 on the sphere r=a is V.
E rcy__4r ( bra 21 sinnn0 cosno = V1 + Vs. By Dw418 new term is
2rE n=11. -13
a 2a2sin0 a 4 sin2 9 cos0) . The addition of V1 gives =-1 Pm(1+ bara Vs TTE br a a2n+1 n • b2 (b2r 2 +sa sin 2e- 2a2br sin Ocos0) r V= q sin0i pin ri a 2 sin 0 s = -:— qa 41%272 (boaa/r(ba+ra si n 2 0 - 2brsin Ocos0) • At 0=0, V 217er rl4.1 2re nbnrn+1 2n+1 n 1 -q a p 1 pndz rc° F = 9 j_: By Dw 601.1: Vs=— p sdz= 3c- r 2re Z 1 nbn (p2+z 2 )n+ a • At p=b, 600 (p2+z2)n 2 re, 1 nb q2 OD ral 2n+lr rno)+ ( 1'1(112 2 co 2n+1 bp -n r.00 xn '' ldx 1 /)2) Use Dw856.11 Let z-2_ - , F-i-ra 77 E 1-13-1 o 2 1 J (1+x) 2 277E 1 nbn -57 q 2 a2 2 sa 1,a fa .12n , = 22_.2 ., a Ans. -a2 sin-1b ' reb 4'1' (2n-1):. I Ei - Tre b2-a2 tan- 4/b2 -a2 We •sr:77-l
L
i- _1
E---a Tr_ 6
-134Potentials inside Viand outside Vc, which are equal at r=a are now
1
CO r r i nsinne Sinne (4 C° rin ,(Ar 4)a 2n n c os no, . - — V • = ---1- E A,LT, J brirni-i n cosn0 1 2i 6 170— 1 C oviv bv = Evo--° For c l. at r=a give An the value An = (2n+l)ev /(ne + (n+1)ev ) so that tn. c7)ln r sinn0 °D 2n+1 V.= -qcos ng3 n r< a 1 2r 1 n{rie + (n+1)E v }L q 2n+1 sinne (e -ev)a rn cos nr6 V r> a 0 2rev n (ne +(n+i)ev irnii- —97 , r pri . q E 2n+1 n a n(E-ev) cos n0 = V +V s 2rev [pbn (ne 1 (n+l)ev ) nbn (pa-Fz 2)11+ 2
1--1-r My 1 [
'—
--
So Vs = n(E-ev)Vs /{ne+(n+1)Ev } where Vs is from problem 133. Thus the force on the wire
is from 133,
F
E
(2n-2): :n (e-Ev) rev 1 (2n-1):: { ne+(n+l)ev
ral2n+1 Ans.
Chapter V.
Page 243.
Page 79.
-15 352L
1 L
1R
From diagram Right side images are: rti+ =(4nb-z)2+p2,
2R
1 .12 1...=(4nb+2c-z)aiP
2, 0.oo Io(ka) 10(4n ra/b) co Thus V"= q(2rtb)-T0 (1'nrro/b) sin (iniT) sin(4nrz/b). Even n is out. Shift origin to z1 =b. .-q(211 tb)-1 K.0(0+ -i) Tr (3/ bj(-i)SS in{(s+12-)17z/b}
---->
t.
V —q(2rrsb) -1 EK0 (s+frP/b}cost(s+i)rrz/b),
1
{
r rs+112n, rs±orro] (2s+1)rrz .....„ an [ q ] q q 12E01P2dcose, (1) , cos 6 z 2n 2nEb s=0 2b 2b " 2b r-41:-) -67174rrzr - 4rre Ozan r - 4rre r In+ 1 a2n+1 - 2C c (2s+l)rrrnir (2s+1)1 (2s+1)Trz 2n+1 so Cn(?) cos P 2n (cos0)=- 11---2b (2n)! b s0[ 2b ''°[ 2b pa-12)21 ilE±117E 2n n ral2n+1 2C = n 7 13 cos Ans. E (s+1) K o (2n. [ j 2b 2b s=0
o 2nv _ q
-01n11
c°
r2
1
This field has the same multipole sources as problem 136 and is zero at z = ±b. Proof of (1) Or bz
op 1 - p 2 6 [Pn (p) = 31' Oz = r , Oz rn 1
un(P) + (1-P)Pn(P) rn+2 . Apply 5.154 (5) to give
(n+1) rn+2
4(21 (n+1)Pn41(p) = ..(n,1)EN.11 +(n+1 )Epn_ so bz L[L3-1= r. 2 141+2 rni" Thus O 2n (l/r)
Let 5.39 (4) be written:
/6z 2n
(20! (1/r 2114-1 )P
2n
n+1 P ri+1111) rn
(cos0) Q.E.D.
-138
co -2s-2p-1 (cos0). To cancel this E M (a,b) r P 2s+2p s0 p=0 sp
co V= E
at r=c, the interior potential of the sphere charge must be cc co 2n+2p -4n-4p-1 V = - E E M (ab)r c P s=0 p=0 sp ' 2s+2p(cos9) Let re ,he the radius vector to a point on the cylinder. Then the V' on the cylinder must n+2pr-4n-4p-1 (cos()) and the internal form have the external form r P2s+2p 7+2p -4n-4p-1 r r P (cos0). These may be expressed in cylindrical coordinates by 2s+2p c 5.40 (35) and 5.40 (36). This gives a correction factor to as and ce of 5.39 (1) since the tangential components must vanish on the cylinder. Thus new values of C 1 and C 2 will be needed.
Page 274. (256)
Chapter VI.
-1For a lower limit put conducting disks on column ends. Capacitance T
T
of one side to earth is, from 5.03 (2), 4ea. By 6.06(7), R1=Fra-2+1-7; Ans. For upper limit insert coaxial circular cylindrical thin shells ending in confocal paraboloids at both ends. resistance of upper half of parabolic shell whose base has radius x and width dx is 477;1 R.2 Te_ 2TeV 4rala 2-x2 2Te_Tfa---2--T 1 =%ra {..r[ 2 • + 33 2x dx 2Trxdx 2rxdx.8ea 1 0 C Q 2 xdx Ru 21Tra R.
2a 1+ Q,71 1 x dx Td o +Th/a 2 -71 T ra 1+1m .
Tr Ru - 2(ra - tan (1 + reit)) -2C-
Consider outer cylinder to lie on equipotential of freely charged cylinder of section A. By thin insulating layers isolate two tubes of flow of identical crossection, b and c. Transfer a layer of resisting material from c to b. Resistance of tubes t' t' becomes (1 ----)R and (1+-- )R respectively. Resistance of tubes t ° t 0' t ir in parallel is now T1. 1-(-) JR0 which is less than original resistance. TR° so net resistance is decreased. Removal of insulating lamina further decreases resistance. Therefore outer layer should lie on natural equipotential surrounding cyoinder of section A for maximum resistance. From geometry we have minimum area for conductor if the boundary of A is circular. Therefore concentric cylinders give maximum insulation. - 3C-
qr
This is just the field of Fig.4.13(b). Half the flux between
2a
2
cylinders passes inside orthogonal circle and half outside. Thus by Te T 2T d 2-a 2 2 6.08(4) and 4.13(5), -zR= c= -237 cosh-1 or from 4.13(2) R=-rr cosh2a 2
■
- 4z-a bil z-a 1- 'Tr +gm a..i.2.i 17 + • . . +onz - a - • rr -+ ... -• • + Image solution: w=2//2 — + z+a z+a+jr z+a+2jr z+a-j1' --- -2(z-a)2) • _ gn j(z -a)(1-17-2fj(z-a)) 2)0.- (21.r2(i (z - a)) 2) • t-a -*-- a 4 (z-a)(1+17-2 (z _a) a) (1+(LB) -072 (z+a)(1+17-2(z+a)2)(1+(2r)-2(z+a) 2 )• j(z+a)(1-11-2ij(z+03 2)(1- (21-2 1.i(z+a)1 2) • . + =Pnitsinj(z -a)/sini(z+a)3Ans. Conf. Trans. Sol. giving U=0 if r1 =1 is " W=k ((z1-a1) /(z1-ail))-Fina l 44(z1-as)/(a izi -l)1, z= z1 , x=ignri , r= 101 2 • upper half plane in strip of width 11/2. e z_e 2 aa sinh(z -a) + _pm W _art y esz+2a -1 + sinh(z+a) Y1 sinh8 , , f If x=a-±8,y=0;V=0 or rr.u1 U=0 i Pin sinh(2aR6) - 4 r/2 If x=-a±45,y=0;V=0, r. u 2=-1.71 so as 6-00 -. .+ - '+ 1t4- Xi -0 ■ % +1 ,\ taii ii ... I 1.11 -U 2 -* 22m(6/sinh2a), Lv]= 2r. So by 6.06(5) A' 1 U -U - - - -.-,--y; x al /"'4-a --3(--a-4\ wn x1 aiR=T —21- - Tn_sinh2a Ans. ..."
N]
ft
8
-
t r
i r Tr
i(
7
- + i'
Chapter VI.
Page 274.
Page 83.
-5- (46,p256)
Yi
• __1_ ,
z =Pinzi , x =Plltri , y= 01 ; zi=eia, z;=e-ja, z;=e j(r-a)- -e-ja
+ •/
. +ii,
13
zy4 = ei(r-a)=-e ja . Images will also work. 1 ,,4\ a 1 „; 1 i ) a% z ja z : tanhT(z+ja) +e ) _o (e (e -e i -z1) t ..,.. _ .442 (z 1 -z1) (z ja 1 t W= 2n (z 1-zi)(z1 -z4) . tanh1(z-ja) ft (ez-e-l a )(e 2-Cia) ; :).• a 2:A 1 When z= j(i--1-6) )1.11=124 tani(2a+ 6)cot76), if z=-j(a76)) 132--g-P1' '. - ■ _ - ''' '14712p Ans. T(Uvi-U 1 ) 11 la .[V] = 2r, By 6.06 (5) R = —r2--- SO 1.11 -1T2 2 072L4L 0 )
-
W
z i+ja i
_ on
- 6- (47,p256) sin z + sin ja_em sin(z+ja) cosl(z-ja)_ ontanT(z+ja) sin z - sin ja sin-i(z-ja) cosT(z+ja) tanT(z-ja)
zi= sin z
1 T
taonh a . When z=j (a7-6) U1=pin tanh(a36) 4 0 If z—j(a7 6 tanh.1.6 2 r T 2 tanh a Ans. 112= -U1 . [VI= 2r By 6.06 (5) R=1'114 -1.11 1 / - Tr zffl 6
a
f— Tr --0
-7C- (48,p256) By 6.04(3),(4) and 4.04
-2C0C I I -1 072 " T.T W = 1-"IiWz-0+;1)-3 )+C ° 2r o+C i cz- a2 " 1-217(Co +anz(z-c -8C- (49C ,p257)
p =c cosh = f(z) (6.15), u= ji0xN/sinh 2 (x/c)+1/(c costs) dx=. From problem 4: F(z)=(-12-TI/r)einsinhYL, (11 ups strip width to 2r) d ...2rI n_ J1 • 1 -LA 4_.3cosh----si-L-. W= -72-11:e2 sinh [1c-21:0+if7°] real Lsinh?cos° 2 2c _..TI fQ u 2,72EinhaSocos29LC2 " roo+sxhx23-2co _ ccs — 21.0+ cosh2 — x -x°sin2- 41 = 2TTT yy 2 2r 2c 2c 2 j Ieil 02c TI n _ [ ,x xo_ Ans. 7r KM cosn cos (0-950)1+C = -712/72 co s 01-6) c .,/ye+zj N/701 4r, ) - yy0+zzo Since cos (0-00 PPo -9- (50,p257) .... 2. 1
Use Fig. 4.22a. To find V value on outer boundary set x=0, y=a U=0 in 4.22 (3). ja=c sinjVi= jc sinhVi, so VA = sinh-1 -a Set y=0, x=a,U=fr/ 2 so a=c cosh Vu,Vu=cosh-1 §. From 6.06 (5), -t3=-V=0 U2 U1 since upper half has twice the total resistance R=Tand U2 -Ul=r. -
L
Thus
Tr Tr C
V= ra2 k=1
sinhpkL
Ans.
1k:0:()Za(1)1 1 Jol(1
Ans.
Pk(Jo (Pka) )
-28Rke c TI c° In 27 let 1.,=0D, c= land a=b so, if 0 < z< c, V=r -r-a-a jie 2sinh pkz
(1-1k0)
ukJ o(uka)
c If e -Pkd -e "Pk(cm d)}. Then, as p approaches a, Let z=-2 Id , d
u
317/4
xr _
+.i.c°dz= jCS: dei so k= Tij err . f i-,h.-i. i'dz = jCS17.,rada1 so h = 74jC AnIT. ritic
Thus a = h2 k-2, C=T2jk17. Put in a and C and integrate. Verify signs by direct 11({-jtanj.,,M+ tan-kg --A-tan-V differentiation. z= 2: -stp
- tan-kg/; }+ C 1 Va
= 2Frick tanh-V.i+k tan-VP -h tanh-'kJi/h- h tan -1kiP/h3+ C1 . (continued)
2h--)`
Page 89.
Page 278.
Chapter VI.
-32- (continued)
u=4,0D, z=k-h+j(k-h)+Cl :
z1 = 0, u=a= (h/k)2 , z=-jco+CI : zi= a-e, u=0, z=j.0+CI :
u=1, z=jm+k-h+Ci . Thus C1=h. If z1= e40, u2->a 2(1- (cia)2)(1+0-> a( 1-2ai-E2):
,Vt( 14(1-a 2)(e/a) 23 . yh -4 (MO fic tan 11-ksra-+k tan-l ra-- 2hk2+12-111242 (1 (1-a2)(E/a) ) - 2thr) = (2/r) (k tanh-1(h/k)+k tand(h/k)+(h/ 2)2/Ai (k4-0)/h4 3-Fhkc- -1Thr)
NIE"
When z1= R-)co, u2= (1 - a21372)/ (1-R-2), u
1+3(1- al)/R2,
y k-> ( 2 prf) t-ilain2-1kintr÷ (1-a 2)/ R -41cr-htanh-Vi-htofkli W=0nz vU=Onxv[V] =fr. At wide end Uk=e/nR=-Z-1 :+ rt U -On e-Tjh - k an1-1-11' + tan-11 .. 12n2112--14 + 4 4 8h 2 k k 2h h h-
1+4(1-a 2)/R 2 . Thus {107471141- +kkR-1-1-4-, `"-h tanhl -h t k8 4_hh4 . At narrow end
,f . Total resistance is (Uk-uhqr
6.06 (6), ( 7:
Wide part resistance: ifk-I{yk-(k-h)), Narrow part resistance: iykh-lir. Thus effective resistance of tapered part is k
1
k4 h4. h tan-1 --+27i k 4h4
d5R = R_rf.Kkih) +
2 +.0 r [1: tanh 1 h + k2
r
hk
k
hk
0•=
[hZ2 tanhl
tan-1 h +241k4 411 k 4h4
tan-1 t
Ans.
-33Abstract of paper "Ideal Flow along a Row of Spheres" by Paul Michael in "Physics of Fluids, Volume 8, July 1965, pp. 1263-1266. Current flows between parallel planes around hollow sphere(half shown by heavy line). The flow is the same as around an infinite row of spheres. Solve by images. Use the method of 6.18. Assume the source distribution on (1) which gives inside sphere j= each sphere is of the form k= E (4n+3)CnP12114.1 (cos0) n= 0 2n+1 (2) The subscript zero indicates +iLri-2n-2PI 2n+1(cos() )3 Cn Cr A li n-O the portion due to its own sources and the subscript j that due to image sources. The 2n-2 (3) Equations total external vector potential is A0e= ilo CnicLr3 P12,14.1(cosej) (2) and (3) correspond to 6.18 (1). From 7.02 (8)the vector potential at ID= co must (4) when r o >a: A =a [2-+C-1 sine , A 03 -jFr2 a- sin 4- D172 + Es sine. =1-14 0 1 4 r2 a b Or ) 2_ (p pD)a3= 2(p2 C -pE)b3 (2) At r=b, A1=A 2 so a3 (1-D)=b3 (E-C) (1): -(--- 1_ rAr ir PXc 3 -133)3 3 At r=c, a 3(F-D)= c3E (3), (-p2F+pD)a3=2pEc3 (4). C- (112 -11X112+2 (1 2+ 2P)(2112+11)c3- 2(113-103 TrIa 4 p(K„-1)(Km +2)(c 3 -b3 ) Additional Flux is 217a 4 4 .C.et LetKm-112 ii. • N - 3 2b ((+2)(2Kr3+1)c3-2(Kin-1) 2 b3 3
Ans.
-15CIn 7.27 (3) designate the last term, which alone contains Km and so arises from the ItiA l 7.26(11). magnetic materials, as A. Then the force on the wire at p =b, 0 = 0 is I*=--z bp n 1 . 1.2..v. L icl b Ectp] p . So at : F . Ey I2 Km-1 E b by Dw9.04, write b for p to get Kin+ln= lb p La2 2r Km+1 a 2 p= a2 • 0=0 ' n 21 F . _EvI 2 Km-1 1 This is the force between I and an image I(Kin-1)/(Km+1) at a 2r Km+1 alb-1 • distance d = a 2 b-1-B from it. -16C-
ri22,
From 7.02 (7) : A0 = 4Br sine outside, A0 2 = 4B[C1r+rgi9Sitle, Aoi = ZBC3r sine. From 7.21 (3),(7) p2 b(rA0)/br =pb(rA03)/br and A0=A0 2 . Thus b3 = C1b3+C2 (1), 2.1221, 3= 2pb3C1 - pC2 (2) Cia 3+ C 2= C 3a 3 (3), 2p1a 3C 1 - p1 C 2= 2p2a 3C 3
(4).
From p(1)+(2) and 111 (3)+(4) get C3= 111(1+ 21.12)/f p.(11 +242) 3 From 2p(1) -(2) and 2p1 (3)-(4) get C3=p113 3 (p-p2)/(Pa 3(111 -112)). Equate C3 values, 21.12.±E . b3 Since a, b, p, If p<pa then p1 p.2 then pi>p2 . (6) Ans. Since b 3> I: 2p1-2p2 p i +pp2 -ppi>2p1- 211211+41112-ppi . So p>pi . From (5), if p2 >p then p2>p.>pi .
Ans.
Chapter VII.
Page 321. (301)
Page 93.
-17C-
A B(r+Cr-2 ) sine, A0i= 1B Dr sine, a3+C=Da3, p2 (2a3-C)= 2a3D . (1124 110-31 sin28 2(p1 -p2 )a3+(2p1+p2)C =O. So 21rr sin0A00 = TrBrz +-11. is flux through circle. But r sine= Nix2 +y 2 so equation of tubes is Ans. [1 + 2112=211(411 (K 2 + y 2) = constant 112+ 21-11
A00=
-18CFrom 7.17 and 5.12 C2= Arn Sn . Magnetization Potential zero at infinity so that 2n+11.s.ca 2n+1 (1) -S).:0=pii/jCT.Ii s C.20= (Arn+Br -n-1)Sn , ni=CrnSn . At r=a, nb=ni so Aa }roa r 2n+1 I 2n+l +1 ° 2n+1 po (2n+1)Aa -po(n+1)B=pinCa 2n (2). From (1) and (2) po(2n1-1)Aa =(npi+(n+l)po)Ca (2n+l)po 0 Thus CkAns. npi+(n+1) po 19C
nj = (Arn+Br -n -l)sn , n2 = (Crn+Dr -r-1)Sn , C23 = Er Sn . Let K = pi /p. Then
2n+1 2n+1 2n+1 2n+1 +B=Cb +D (1), nAb -K(n+1)D (2), -(n+1)B=KnCb 2n+1 2n+1 2n+1 2n+1 (4) At r=a: Ca +D=Ea -K(n+1)D=nEa (3) , KnCa 2n+1 (1)and (2) give {n+K(n+1)}Ab (5) +(K-1) (n+1)B=K(2n+l)b2"4.1C 2n+1 2n+1 2n+1 2n+1 (3) and (4) give K(2n+1)a C=CK(n+1)+nla (6). (1)and (3) give Ab +B-Ea 2n+1 2n+1 2n+1 2n+1 2n+1 2n+1 )(K(n+1) +n) +(b -a -Ab =C(b -a ) (7) . (6) and (7) give B=Ea (2n+1)K Substitute in (5) for B and C: 0= (n+K(n+1)-K(n+1)+n+11b 2n+1A +(n+1) (K-1) a2n+1E (02n+1-a (n+1) (K-1) 2n+1 )fK(n+1)+nl-fK(n+1)+n3K(2n+l)b2n+1 E. This simplifies to (2n+1)K a 2n+1 r 0 = b 2n-/-1(2n+1)A+C- (n+1+nK)(K(n+1)+nlb2n+1+n(n+1) (K-1)2 JEt (2n+1)K3-' At r=b: Ab
Thus E/A = (2n+1) 2 K/[ (n+1+nK)(K(n+1)+n)-n(n+1) (K-1)2
(a /b)2n+1]
Ans.
-20CFrom 7.02 (6) : Uniform Field Az = -Bx= -Br sine. A3=-BC4 r sine, Let K=pi/p. A2=-B(C2r-i-e .)sin9, A1 =-B(r4)!ine. At r=b: b2+C1 =C2 b4C3 (1), Kb2 -C i =b2C2-C (2). At r=a: a 2 C2 +C3=a2 C4 (3), a2C2 - C3=a2 C4 K (4). (3) and (4) give (K-1)a2 Ci=-(K+1)C3 . Substitute in (1) to get 132 +01 = (b2 - Iiia2}. Substitute this in (2). Kb2 -KCI=f132+ TMa2)C2 . Write p for (K-1)/(K+1) then (K-1)(K+1)(b2 -a2 b 2(b2 +8a2 -K(b2 -Ba2))+Ci (b2+Ba2+K(b2 -Ba2 ))=0 so CI- ((44)2132_((_1)2a2 and flux ratio is 5_1 b2 (K14) 2 b2 K-1) 2 a2 RR 2AZ R Ans. e=2 , r=b. = 2A1 at 21([ (K+1)b2-(K-1)a23 )
-
(
Chapter VII.
Page 322. (301)
Page 94.
-21-
aPt., (g) _ PAM Current density due to nth term is in= COPn / (hi bg) . From 5.23(6) liCja r.1f.2cos0 OPn(t) From 7.04 at 0=0, An= 4rr j 11-1 3d t 46. From 5.27(6), (7) hrs/T-TiT = ha.TFTI. o h 1-p-J o R Also hih2dtdos = dS, PAM cosse =S11 so An = 4
JA- dS. From 5.274(6), (7) at 0=0, C j -11Cn Ans. A9in=.1-74n(n+1) 1 (1+3 Qt (i°)P11(i °Pr" = --jj484j-1--n(r;!2— F1) CA(i")PniCi C); ) -22Let current density be I/ (hi d t) if g o
0 Bp= -76 -kn z Bp = E oknCn Iti(knp) e (6) Multiply through by pRi(knP) d(kn p) and d"..ii— lkr1.1.11n.P.L 4:1 4 integrate a l to a2 . On left side have knia aB pRi(k odp -2 4a i gn(a 2 /a i) 2 4aiefft(b 2 /131 ) ai P n 0 by k2n,16a 2 pRi (knp) do = Yo (knai)Jo(kna) - Jo(kn a l) Y 0(knad- Ya(knai)Jo(knai) +Jo (knas) Yo ( (na,i) Ro(kna2) = 0 (4) a ukn I pRi(kr,p)dp So Sa2 Bn pRi(knp)dp = - Cn r 2 0( RI Ow))) 2d (kn P) 2 101gi2(b2 / b 2 / 131) 2&z(b al 4. 1(nCnta22ERI(cna2)] 2-a? CR1(Irna1)] 2) by 5.296 (5), (7) since Ro(krra2)= RP (kra a l ) = 0. 5.293 (9) can be written Jn (v)Yn _ i (v) - Yn (v)Jn _ i (v) = 2070 -1 so (5) is Rocna2) yo(kna2)
_f aQcri a211112.na2)1 Yo(kna 2) Substitute for every ratio from (4) so that 2
Lrrkn a2 Y0(k na2 )
Jon, ao I 2 "sn Trkn a 2 Jo(kna
(nikano2()akon a 2si c) !-2)]
4,10(k„a i ) 4_ Jo((rr32)YArrao)JpOrnaL Yo(k n_a_i)Ji(kn a2)Y0(kna) _
( a ) c r2 r_Yo(kn:2) a TT 2k 1(1::{2:dik .1c na22)12 2J0(irnai) (kn n aai2)) 21:(1,T (kri j Or RiOCria2)n 2d(kn o)= C11—Fr 7• n ptR (k p),I rrkn a2Ji (knad. But Ri(knai) - m nai ai 1( x) 1 2 2 Cn= 7T 21111.31(kna232 fRa (knq Ro(kabi)3/ [42//2 ( 32/b i)(knai)] - [J1(k n a 2)] 3..] Ans.
111(k na2)
rrkna2 Jo(kna
Jo(kn a2) 2
-32Add 22-pIz(p2in(a 2 /4)-1to (1) above. Alt=luIz{Pk(a 2/a 1)}-i + SoCn Ri (kn p)a -kr z -33-
Ans.
2
4 = t1Z Cn te -k-n (z+c)+ e -kn (z-c)}- iiiII[z+c+z-cl(POn (a2/a1)1 1 (e'kn(z+c)+ e-kn(z-c)1_ A j ( a 2 /ai ) J =nE1Cn So 45=2 E Ans. e -kn zcos h knc Ri(k np) -}az(Pk(a2 /ai )) n= 1Cn /,; = 2 IA Cne -knccoshkn 2 Ri(knP) -PIC[P62 (a2 /al )) Ans.
c1
Chapter VII. ci 42 4,r/ 2 TT/2 fr/21DI _ 171
- 34-
)71
7‘
Page 324.
dz/dz1 = (k.17F-14/77ca )-1=6/7-Z14,r7F cazt
i=se'lz i
zi= sn z, sn(K+jK' )=k-1, ma=K, mb=K I , a/b = K/KI.
W=2--Pat[(z -c + id )(z -c -id )]= 21.2fizt( -0 2 +d 2)=-1-11071 ((x 1 - c i)2 -yt+d ?+2j (xi -c y i ). If ci+di =zi 1 2r 1 - 1 1 1 -1 2r z1 2r +jd,= sn (mc + jmd) c+jd=z so from Handbook Mathematical Functions 16.21 p575: sn (mc) dn (md)j sn(mc)dn(mc)sn(md)dn(md) + . U=Pert((x -q2-yf+df]a+4(xi-ci) a Yi) 4r cn2(md)+ k sna (mc) sn2(md) cn2(md) +k sn2(mc) sna(md) )
=111-247C(x1-02+y213+2[(xi-c1 ) 2-yi]di+df) Ans. Subtract a similar expression for -I1at fiand g 4Tr This -I cancels the singularity at y=b produced by +I. Then from the above reference: sn(mx) dn (my) sn (mx) dn (mx)) sn (my) dn (my) Ans. cna (my) +k sna (mc) sn2(md) Y1 - cn 2(my) +k sn 2(mx) sn2(my) -35-
-bi dz p_„_thL
w=-41.iirrign((zrjb)(zi+ Jkl)(zi-1)
+WI ) = - 21T wil z?-1 ' dzi z sinh(2x/C) +jsin(2y/C1) i is rea1 if z=jc z = C' tank' z1 , z = tanh_, -r --lif--f-I u cosh(2x/C1)+ cos (2y/C') ' z _1_ 11_( 12 ilTtz/c)+t an2(-10113/c) . rb w.pionta 2c sin--0 = 0, CI= -F Vi cr-. . If zi= jbi , z= jb so jbi= tanh -- =j tan-27. 217 tanh 2(i-Trz /c)-1
112 a must be Ii(Np) cos Nz and K l (Np)cosN z . The potentials must match at p=a and, from 7.02 (5), A must be proportional to
0
when r is small. Hence the form given. -42-
B = (417)-111r2 ds x
ds = i ds sine +j ds cos°
ds x RI = i dscose sin* Ids sine sin* ,/
/1
+k ds cosS(cose sinfzi + sine cos0) cos0 = (p-s cose) /ro , sin0 = s sine/ro ,
-
sin4f= z/R, cos* = ro /R
dsi x R1 =
(i ds • z cos° - 1 ds • z sine + lcds • p sine) /R. laz cos() de pIz sine de _pIp sine de d Bx = --5-- ds, d Bz ---T-s---4ff Tr rids 4m1 *Tr ds' dBy- - 417R3 • _pIz rw r r-a cos() de ds 3 , 31122Slef ma sine dsde 3 B x-417 2.1 0,1-a (z2 +p 2 +s 2-2p s COO) 2) .°Y7417 Co a (za-i-p 2 +0 -2p s cos()) a in ededs ds 1 3 2. where By Dw r 3 B 0 a (z a+ps 2+s2_ 2pscos0) 2 z Y .857.02 ..10 0-1-p 2+s 2 -2pscose)l - r(r-pcose) ra=p2 +z2 I
B z 4rr
(r+p)cot(a/2) z rr-a cosede p.I Ez - 4172p r ( r-fa-a) + 2( tan-1 4r2r4-a r-pcose 1
.r..(t
4r 2p
r
an
___I/c_ ztar c/2)
tan
r+p
(r+p)tan(a/2)
tan-
(r+p)Etan (Ce/2)]].]
,0a. = nEiCnIi(—E-)Ki( c )cos Ao=jiCnII,( c )K1(--F-)cos-E- if pc must choose same sign for ATA 3•7 1 and N 2 -1 to eliminate radical so that 1 s in pvotan -11 . Ans. Chose - for ± since M=0 if a=17/2. M= pv seca(c±c sin a )=Pvc sa -
If c >a take opposite signs for 4/E171 and B2 -1 and - for t to give M = pv secafc-A/c 2 -a2 cos 2a) = pv{c sec a Thic Isec2 a - a2 }
Ans.
-8For uniform field Az0=B(r + CI r-1 ) sine, Azi=BDr sine. At r=b: Azo=Azi so bz+Ci = Db2 , - PI6Azo/ar= 1126Azi /Or so p.i (b2 -CI )=u2 Db2. Thus I'.21/1 / (111+112 ) , = ii 1 jj ( 217r ) •8F14(4) d Fn= i [Li1; C I=C(pi-122 )/(p1+112 )lb 2 . For 11 Be .11 11 2 r p--111 - .1 2 Pa r.217 p r Bs ine]2+ 411? B 2 c os 20B 1 (Pi -11 411 f [ I (pi +p 2) 2 . Fx= No F ncose de =0, F ' = bj F sine de +p2 Y 4nr Jai o n 1.11 —ET 2 r "b21Bsine Ans. F' is on surface of insulation. =2(p2 - P1)j0 Orr (pi +112)sine de= IB [P2Pil Y 11 2+11Li - F”Y = 2 1-1plIB Force on inner wire is IB x 11 14-p Ans. Total force on system is 2 1+1-1 2 = Ft +F" = IB . Ans. FYY Y -9 See VII-17. A00= iB(r +Cr-2 ) sine, A10= -1-BDr sine, C= 2a 3(112 -PO/01 2+211 0 . Flux through sphere is rra 2B1= 2rraA0 at r=a, El=r/ 2.Bi=312213/(12 2+ 2p1). From 8 . 15 AW=ALi/22ppv u-uv 417,33 2 (p-pvi 3pB 2 (p-pv 4ra3 )4rra 3B 2 =____ B B. Ans. DL= 3 2ppv (11+2 ilv) 2111.1v 3 liv (11+211v)
B • Bid v -
- 10SOlve 7.15 (6), (7), (8) and (9) for A2n+i . Substitute in 7.15 (3) for shell Al. (0_01(a4n+2_0n+2) PvI ' 1 rcl4n+2 A•• TT n=02n+11-71 01 _1102 a c61- 1_ ( 31+1102 b 4n-rt • Integrate AI along two wires of unit lengi 4n+2 4n+2 -b 2pv 1. 2_-_07 ) I 1 rs_ 4n+2 Ans. OL = 2A =-- rr L J 4n_ (1111v) 2 b 4n4.2 (g-pv )2 a a+2 n=02n+1 a
Chapter VIII.
Page 345. (323)
Page 104.
- 11CA 10 = +pva 2 (Ci r+r-2)sinO, A20= ÷pva2 (C2 r +
A30 = 4-gva2C4 r-2sinO.
At boundary A10 = A20 112 NrAios )/Or = p1 b(rA 2 0)/br so at r=b: C1b3+ 1 = C2b3+ C3 (1)
11 ( 2C1b3-1)=1-1V ( 2C2b 3 C) (2) . At r=c: C2c3+C3= C4 (3), pv (2C2c3-C3 )= -pC4 (4) (3), (4) C2(p+2pv )c 3+(p-pv)C3 =0 (5) (1), (5) (p-pv )(C2b3+1) = (u-pv)b3- (p-2pv)c9C2 (6) (1), (2) (p.v+2p) C2b3- (p-pv) = 3p.vC 2 (7) . Cia3([3Pv(11-Pv);(11-P-X11 +211nb3+ +21 ) (11v + 2p) c33 11-11v)i.-(2Py+P)vb3+ (2P +P)a3/ =-3pvb3(p -pv ) - (11-Pv )2 c3+ (1-1- P)(11+ 2320C3. C1b3LS: 2 (11-'1 12b3 + ("+2)1v) (171.1v+2P)C3.1 • c3-133) :"2:v' Ans. AL= 2TTaAA1 at r=a so AL=Pv7Tatil-Pj(2A7-" 2b3[(p+2pvXpv+2P)c3- (P-Pv)- b (
2
-12Set t o=g=t, and 11 0=71=71 1 in VII-25. Multiply by 2Tra, setill..-011-1 =a /c 2 , Then
AL=
2n+1 (11-14Ptigo) Ori Cno)1 2PAM c)Pn o) Ans. ITAva 2 I c 1;1n2(n +1) PvCM (71 0 )Pii o -PQn(rl o)Pili (710) -13-
Set t o=
= a i /c. Multiply by 2ra. 11 0=71 i , t= 2 1 1l to in VII-25./117-T1 2n+1 (1 1-1-1v)P1; (ti)Pri(t2) acrIOQA(T10131-11(71o)Pr,(71o) n=1n2 (n+1) Pv4n(Tio)Pn( rlo) -11Qn (rlo)P/11(710)
Ans.
-14Write p=c and z=s in VII-28, multiply by 2rc, and set I=1.
These give
AM = 2pvacro fb(k)K 1 (ka)K 1 (kc)cosks dk
Ans.
-15Write p=a and z=0 in VII-28, multiply by 2ffa and set I=1.
These give
PM
DL = 2pva 2j0cl, (k)(K 1 (ka))2 dk
Ans.
-16dk 2ra 2 --1411(kal)K1(ka2)coskzcoskci rx By VII-29 d z 2 EMF,C, with 11=1 is —1. .11va dz current is Cdz/(2TTa2c) so I 2 =rf4edz/(217a2c7). Total Resistance is 2c R= 2Tra 2c/(2c2). Effective EMI' from 11 =1 alone is IR so that z 24i ia2k M=lva a2 ecrIi(ka)Ki(ka)cos kzsinkcid 411(ka0K(kaa sinkc i sinkc2T , 2 Sk c2 -17-
116---- 2 a
In VII-30 set z=0, p=a so additional self inductance is, taking A at z=0, p=a, - 2k t -2kb -2kt -2kb , F(p2- (1) 1 702e _Itt 2 2iTaA = TTPv a-Jo 01+1.\)2_ 61_42 e - Zkt (ka)) 2dk= Trpva2 131:` {J1(ka)} dk. Expand t-tel -pkt -2kt -4kt 4 -6kt by Dw 9.04. 1-e =1- (1-132 )(e -1g e +p e +•••) .Trpva 2 krocte-2kb_ (1_ p2)n boo2 (n-1) e-2k (b+nt ) 3 ji 12 k . By VII-27, M for coaxial coils -kzr_ 8.06 4a2 of radius a at distance z apart is Tr7a.va2 jo e (ka)3 2dk --=-17[(1-2-)K-E)k2= 1 ka 2 2 (1),(2 4a 2+z2 a _ 1-1 V . So LLD= p[mo (1_0) E R 2(11-1) 1,4n3, Ans P---'n=1" a2+ (b+nt) 2, p+pv 2
_
Page 105.
Page 346-(324)
Chapter VIII.
-18r.co r 2 - 2kt,By VII-30 M 1 2 =2TrbA 3=rrpvab(1-81)j Ji(kb) Ji(ka)e -keLl -8 e idk. -k(c+2nt) =rruvab(1-132) nio2ni e VD) Ji(lca )dk . By VII - 27 , rrpva brJi(ka) Ji(kb)e-lczdk ?ri m k2 4ab k2 (a+b)2±c2 by 8.06(1), (2) . Thus M12= 2us/Z(1-82) -k-;-{(1 -T)K - El ,
where kn = 4ab/( (a+b) 2+(c+2n0 2 ),
An s
13 = ( Pv ) / ( P+P ) •
-19(2 sink) -1) an Current in element du By 7.10 (3), for loop, A0= Py— ITTd 017/2 `r z2+d 2 cos2 0
d
is I '=Nc-idu, in dz it is I=Nc--;dz. Flux linking dz is (Prrd/c)rAfzs (u)du. For unit current dM4Nd 2c If;zfolle2 sin 20 -1)( u2+d 2c os 2ElrOdedu iaNd 2
r11/2
r
0 ( 2sin2e -1)1sinh-C(c -z)/(dcos0)3+sinh-l [z/(dcose) ]) de. M= (N/c)SocdM
so M=pd 2N 2 c -2f 77/2(2sin2 0-3).rsinh-1(z/(dcose))dzde, since arc sine integrals
dz
0 72(2 sin20-1)(csinh-i[c/(d cose)]-..✓ c 2 +d 2cos 20 +d cose}de are same. m= pv2 d2c -2: For arcsinh integral let u=csinh[c/(dcose)] , du= (c2 tanekica+d 2 cos 2 0)de. dv= (2 sin2 0-1)de, v=-sinecose. ludv= luvi 1712 -Svdu=0 c247112 (sin2 eac 2+d 2 cos 2 e)de osin20 =4/7:C c 2+d2 cos29= la7 i.(12 Let k2= d2/(c 2+d2) , ,/ -d2 P so that M= }1(Nd/c)2 ( c 2 +d 2)-1/2(.172[c2 p-isin 2 0 -2(c2 +d 2 )Psin2 e (c2+d 2 )133de+ de} -1)3 - k2 sin2 ecos2 e/P = (2-3sin2 e)P - (1-sin2 0)P 4 so that rrr/ 2p-lsin2 ode. d 2 /c 2 k2/ 2 04A-k2 s in2 Ode = tri2Pd0 - -kr/ 2P- de +to (1-k2 ). Thus
But t (Psinecose) /be = (1-2sin2
(1-k 2 )sin2 0 4P1 2sin2 0+p]do di_ PN 2d f(1- k2)E+ ( 2k2- 1)K- k3 ) 3k(1-k2 ) 3 3P 3P 3 uN 2d k2 If k=sina, (2k2 -1)/(1-k2 )=tan2 a - 1, 77 1 (2 - tan2a M= 3 [csca[(tan2 a-1)E÷K]-tan 2a) Ans.
M=p(Nd/c)2 U rri 2{
-20riD+2 , A0dz, where A is vector potential M = 271,9146, F = -II ' bz= -2TranIV-6-p-j of ring dz at z with unit current I. so F = -2TranII (1141 04 ] - 11-( 2 [ (1 - --i2)K2 - E2 ]) Ans. [ (1 - 2 F = 2grar where k =
4ab (a+b)2 (p+L)2
4ab da
4ab 11=(a+b)2 + p2
4ab
c2
Page106.
Page 346. (324)
Chapter VIII.
-212 sin20-1)de]dudz by 8.03 (2) and 7.10 (3) where p=a=d/2. +d cos a 1 c-a+z c-a+z .. . /1d2nmr/2(2 sin2 0-14r0 t sinh" sinh-1-2— 1 +z-ldz] de, Let uetc. 2 o dcose dcose dcose'. f,a_ rr /dcose . ..1 r(a+b)/dcose -dcose u du sinh-lvdvj= fc sinh-id—C— Lj(cc-a)/dcosesinh Jo cose 4/c 2 +d 2 cos 1 0 jb/dcose
t41/20 pc-a+z 1 lid 2mn j 0 j M = -1 0 .1 b+z
- (c -a) s inh-11=41-- +A/(c -a) 2 +d2COS 203 - (a+b)s inh- a+b +W(a+b)24-d2cos20+bsinh-1-1-dcose dcose dcose r rr / -Wb2 +d2 cos2 O. All integrals have form j (2sin2e-1){Psinh-l— P -A/P2+d 2 cos 2 e}de,„..._ d dcose
The value of this integral as found in VIII-19, if tanft= r=cotal k=sin8=cosC4 is P2 (tan2 13-1)E-K tan 2agcot2 a-1)E-K) d
3d
sin
3cosCt
- 3tseca[(1-tan2a)E-tan2a1G . Thus
M= (Pnmd 3A)nti( -1)nsecag(1- tan2 ari )E(kn ) -tatl aCtnICTI (kr, )
Ans. Where kecos an ,
t ana l = cid, tana2 = (c-a)/d,
a4=
tan a 3 = (a+b)/d,
tan
b/d. Ans.
-22From figure in last problem observe that if one coil is displaced axially 6,13=k,c so that OM/bb = OM /ac. Differentiation of the first expression in last problem rT1/2ra 2sin2 0 - 1 gives, since b and c appear in limits, F = II ' OM/bb = 1 j_ j u 0 c-a+z) +d 2 cos E sinh- a+b snh_ 2sin29-1 dzde= liud2 nmIff:72(2sinae-1)(sinh-idcose c -sinks d cose 'd dcose d cose rrT/2 sin 2 ede _ co_tp [K(k)-E(k)) k=sinp From VIII-19: ri2 (2sin2 0-1)sinh-1---C--de tio c+ • 77= - sing d cose (-1)rcos8 tan13=d/c. Thus force is F =21pd2nmIr sinpn 'nfE(kn)-1((kn)) . If tan cc'= c/d etc. as in VIII-21 above: F =
(-1)n sec2 asinan(E(kn )-K(kn )), kn=cosan -23-
From figure: At= AIcosa = A (c cos0-b)(c2+b2 -2bc cos0 )
_ =-Ai bR/bb.
By VII-26 I= 1, A i=(1.1a/11),10 I1 (ka)Ki(kR) cos kdodk, M= brAtdczs. K i(kR)oR/ob = olCo(kR) /(kbb) so M- V0°I 1(ka)(OS0 Ko (kR)d0/abi cos lad, By 5.35(4): K0(1(11)= K0(kc)I 0 (kb) +2mE1Km (kc)Im(kb)cos A 0-integration, 0 to 2r, removes the summation so that 217j:1 i (ka ) (kb)K0 (kc)cos kddk . Ans. From Watson b2m-2s+l a 2s+lt2m+2 Bes. Fctn. p148: I I(t 11 ( tb)= =m-0 s- 04m+I (m+l-s):(m-s): (s +2): s: • By 5.40 (3): $0t2p cos ztKo (ct)dt = i(-1)1311(2p): r -2P-1P (cos()) 2p m+l 2m-2s 2s OD m (-1) a (2m+2)'. P (cos()) b 2rrt+ 2 M 4.4a2 b Ans 2 TTMEOSE0 4m (m+l-s )1. (s+2)'. r2M+3
Ans
Chapter VIII.
Page 347. (325)
Page 108.
-29M12= =
ds •ds By8 .03 (2) . ds i , ds 2 components 0 12 I " are: parallel to c, ds sino and ds' sine (1) .
normal to c in a-plane; ds cos, ds' cos0bosa. (2) So ds•de = ab(cos0costecosa - sin0 sine )do =-ab(cos0cos0' cos + sin0 sinfind0 de. R component parallel to c: Re =c-acos0-bcos0'. R component perpendicular to c and vertical: Bcos a +A-b sine sins = - B cos 8+A-B singe sink. R component perpendicular to c and horizontal: B sina -a sin0+b sinsecos a= B sin (3 -a sings - b sinecos 8. Square R components and add :
R11,= c - 2ca cos0 -2cbcos0' + a 2 cos2 0+b2 cos201 + 2ab cos0 cose R2
B 2 sin 2 B-2Ba sin8 sin0 -2Bbsintkos (3 sine +a sin 20+ba sin2 0cos 2 +2ab sin0 sine cos
RCILL 2 =B2cos 2-2ABcos 13+A2-2Ab sinfitsin0l +2Bb sinficos 13 sine +112 sin20' Si.28 R2 = D 2 -2( a (cos0+B sin8 sin0) + b (cose +A sinfl sine)) +a2 +b 2+2ab(cos0 cos0' + s in0 sinecos 8) Ans.
where R_ =r i 2
-30z =e ja
a .!_rf 2‘ a /a -Trzia - 2sin /a) __pi a zf-2z i cosa+1_ a pelee.rrzia. (z 1 -ej a) ($ 1 -e -ja) pI W= Pe2 i 2r 2rr z f+2z i cosa+1 (rr-")) (11-a ) (z1 2rr (z +e a+2sin(irc /a) z
= _a_e/Izz Ty
11
e
1
/
r =enx/a . If _ a-c z- 2
)
U is Az , the vector potential. Let x=0, y=i(a-c)+E so that -cos(yrtc/a)sin(TrE/a) U_ _g_gn cos (ry/a) - sin(ric/a) _4 -pIgm 2rr • 2rr cos(ry/a) + sin(rc/c) 2sin (-11c /a) For currents without walls U1 = -1142/n . The AL is the Az difference at the wire surface with and without walls Tic pi (17E a)cos (re /a) c I Tic • e = - on(--cot-3= , , I = - Pili 1U2 2r 2r 2a 2r 2a 2sin(ireis)
so that 2a Trc (—tan--) Tic 2a Ans.
,a
Preceding page blank
Page 348.
Chapter VIII.
Page 109.
-31m (-1) a 2m+1F(m, -m-42;b2 a-2 ) 2m+2 k . m=0 4171+ 1m (m+1)
From Watson Bes. Fctns. p 128; Ji (ak)Ji (bk)= b E From Hd. M. F. 15.4.14 and 8.2.5.: b 2 m+1 -1 a2 +b 2 b2 a F(-m, -m-1;2;72) =131. 1--l a 32 Prn.+1 {1717.7
}
rso+13 2 ue (a 2 -b2 )m+1 (m+2): Pm+1Laa-b 23 ba 2131+1
Substitute in VIII-25 m m-Fl co (-1) (a2-132 ) aa +b re° 2m+2 -kd M= 27Tuab E k e Jo (kc) dk P1 (-3j m=0(2m+2)::(2m-1-4):: m+1 a 2 -ba o From Int. Trans. VI (9) p182 in the integral let 2 = 2m+2, V = 0, a =c, E =d, r2=d 2+c 2 -2m-3 and cos a = d/r. Then the integral becomes (2m+2): r P 2m+2 (cos a) so i m(2m+1)::(sa -b2 )m+1 1 r aa+b2 M= 21Vab 3P2m+2 ( c o sa) Ans. (2m+4):: r 2m+3 i m+1‘ a 2 b2 -32(2m+2): By Hd.M.F. 15.1.20: F(-m, -m-1;2;1)- (m+i),. 01+2): J1(ak) of VIII-26 to get {J1(ak)}2= from VIII-31:
M=
2 2m+3(2m+1):: . Insert this in (2111+4)2
co 2m+2 (2m+1)::]ka /fm: (m+1): (2m+4)::). Then m= E0 [2a
m: (m+1): (2m+4)'.:
2m+2
-33Like VIII-31 but with a factor (4AB)-1 and a new k-integral. In VIII-28 write -kr2 L e 2+e -kC2 3-e -1( 4 ) where 01=-A-B+d, 02=-A-B+d, no=A+B+d sink kA sinh kB = 1. (e and Q4=A-B+d. This gives four integrals of the form of VIII-31. Thus we get rco 2m -k t o k Jo (kc)e k s i
4 -2m-1 P 2m(cos as ) sdk = s;1(2m)!rs
4 -2m-1 2 2 where r s =Os +c2, cos as =C2s /r. So VIII-31 integral is 4-(2m): sEirs P 2m(cos as ) • m+1 (_,)sp (cosa) a 2 4 ,2 (-1)m(2m-1)::(aa-b2) q rg - rrmabmn Ans. P' 3 rim2 T 1 m+1 a 2 -b 2 s= 8AB m=0 (2m+2)(2m+4)::
t
-34Use of (J1(ak)3 2 as in VIII-31 gives (2m+2):a2m+3 /{m!(m+1):} for b(a2 -1)2 ) (-1)m (2m-1): (2m+2). a 2m+3 (-1) P(cos a) in VIII-33 so M- namn 8AB m=0 rn: (m+1): (2m+2)(2m+4):: s=1 r 2111+1
1P110.÷i{ a12. Ans.
21
Page 114
Page 363. (435)
Chapter IX. (XII)
1/4
'1
1,
1/4
Lx = lOsin30°sin30°= 2.5 ft. x=212.5 ft. h= l0sin30°cos30°=5sin60°=4.33fi: -2Starts from rest at 0 0 . d2 0/dt 2=-C sine. -1 (dO/dt)2 =C sine e = C(cos0 - coseo) r0 j i,(cos0-coseo ril2 d0 =.../FCT/4. Let 0=20, A/1-cos00 -1+cos0 =Arksin2 0 0 - sin20. Let sin00 do _ singocosu du _ 4/sir1 20, -in20clu If 0=0, 0=0, u=0. If. E*00,0=00, u=17/2 so that Ni1- sin O0 sin u 0.-sina0
sins
,TET/2="517/2 (1-k1sin2u)-Iodu= K(k), k= sin0 0=sini00 . If CI and C3 are two torque constant values, when started at same 00 ,and Tiand T2 are unmagnetized and magnetized periods:
T 2
Ans.
= N/3mg1 + pv4ra 3 /VA/TaiT.
/F271= 3-
dx fz 2 . Get slope dz' ,ci:7— Since u= 1000pv, VII-17C becomes,' setting y=0, (1.+2(a/r) 31x 2= C, r =d2:_3a3rzi:12c _ .512c- .0015=a= 5' Ans. a3 +2d i c = O. (f)3=. 001, x=z=Afi, so -62r° + z lx 2 +(l+ 2 (f.)3 1 2xd-2' dz dz dz dz r r4 -4Replace, as in 9.07 and 9.08, by two spherical current shells. From 9.08 get outside ima 3.43 _ 3 -mq(ba -a3 )._. From 9.08(1), stream function is dipole field Md = 11 = Ma(Pi (cos0)-P0 (cos0)). Inside put n=0 and n=1 in 7.12(4) (P;(3)=0) so that by 7.02(7), Ai=- 3pv Ma(r/a)sin0 represents a uniform field independent of a. Therefore
e
in cavity fields cancel. There is no field. Ans.
I
-5CForce on a from b is Fab .
M al
0 1 2= M 2 COS0 2 /(4TrIlvr 2 ), C2 21 = MI COS0I R4T1pyr2)
I( -cosOrTe - sin01m 6 '1 r- )n ai ) (F,dy =m igg- br on012.1= 4rrilvr4 1112 at e2=2 . (F 21 "Y =-2rbEy )
3M sin201 1 3M M ( 1 =M 6 2rb A i4rf • 2pyr 3 'Or 0 4Trpyr4
T
_m r.
12 = "11. r6032 )411p.vr
-
T 21
-1;171,:r2 3 =-142 (-k21 3= 4
So down force on 1 at r/3 and up force on 2 at 2r/3 are of magnitude 31.1111 2/(41fur3 ) . Ans. -6CWith polar coordinates r, 0, 0; r being the radius vector, 0 the colatitude and 0 the longitude; the megnetomotence of the fixed magnet is clearly independent of 0, i.e. sin0d0 =0. Thus its field lies in the plane PQ and C, so that 2Mc os an M sine' H 3 .Thus tans = tans=-He /Hr' --e . Ans. 411`pyr r rue 4Truvr3
H0 =
on_
-7CFor equilibrium at magnet B there must be no field normal to it. Thus H
rA
=H
rC
sin30° -H
A __ . _
ec
cos30-. H
rA
LcmAcosem _2mAcossoo DiAsln322. br t. -W I. .az —.1 41Tur 3
4ITUr3
,
cos() 2M -b f ljgcos2a l.mr sin30° So 2M =M (1-2)4/372=-■ H =- bm fcci_ c cos30° H ,5M /2. ' A C rC br L Limit. a ' 4rrpr 3 ec rb0L 4-Tur a i 4rpr 3 C ' /4. But M =M so MA :MB :MC J:4:4 or MA /MC Ans B C
Page 364. (436)
Chapter IX.(x11)
Page 111.
-8C-60 M sin Hen=T6os in (0-0) = 3 sin (0-13), r cos(0-0)= d, 4rtpx obi sinesin (0-0)1 . -61-3u 3cos2(0-9) sin(0-0)(1
tin 2Mcos0 Hrn=-F; cos(0-0) - 41.Tpr 3 COS (0-0) N-
mcos 41Tpd 3
)
-Ø " (2 COSOCOS (0-13)
(0-A)
o-e
0=cos2(0-0)(3cos(0-0)sin(0-0)cose
-4-Esin 2(0-03)-cos 2 (93-0)]sine)• cos2(0-0=0, 0-9=1:TT/2 is minimum so -I sin2(0-0)ccite-cos2(0-0)sin19=1 is a maximum.
Thus 3tan2(0-0) = 2tanO Ans. -9C-
Resolve each magnet into components as at right .
Ml sinei
Forces normal to r come from components shown. When
Mi cose i-o
-
-314 COSO
2
1M2 sine2
-3MiM p since i+0
magnets are parallel forces are along r. From IX-5C:
(F12)n-
4rrp.r 4
since
3MiM sine cose 2+ cose sine 314iM p sin (A 1 +9 2 ) (F21)ii= — 4ngr 4npr 4 41lir 4 From 9.12 (2) : OW MiM 2 _ Mit4 (sine2 cose i +2sin0i cos02)=(F2i)ri,T2 =-Oleos; +2 sine2 cos0i) = - (F12 )r 2 1.1 = -601 Lrtiar 3 602-4mir3 (sin sine2 cosei +2 sine, cos02r sine,cos() +2 sine cos() So r i = r2 = r. Add, ri +r2=r 3 sin (01 +02) 3 s in (91 +02) r i sinel cosei +2sine, cos02. tane,+2tane1 Ans. r2 sineicos02 +2sine2 cose1 tane1 +2tane2 -10CLet M= rni M, Z1 =
r = rl r. For force along M need mi-VW. Use vector forms in 1.071.
4rtliF=m1 .Vr-3M.M I - 3r -5(141-rmi.7(M-z)+M-r mr7V.,01-3M- r M'-rmilr-5. But riLA•tr-s= -312.1.1 -Er-4=-3r-4 ose 5 • Vr-5 = -5rni•rjr-e=
-5r"cos0, tn -17 (M-) = (m i-yr) = M•miMj • (mi-vr) =
M•E= Mrcose, 1..r = Mercose',
cosei , M-M'= MMI cos E.
(M er) = tim)
Thus we get
4rrp.F = -3MM'r" cosecos -3r-5(MM'rcos0' +MMIrcos 0 cos E)+3 • 5r-6MI M 2I r 2 cos 20cosel F = 3/-241 (5cos 2 0cose' - cosec - 2cosOcose) /(41Tpr 4 )
Ans.
Differentiate 9.12 (2) . T = -614/4= MN' sine sine' sin*. -mia+r +m'ib+d=0 (1) But m i . di = 0 and 71,1•cli=0 so d i - (1) gives r•cl1 +d=0 or r i •di= -d/r. x mi=di sin€ . Therefore r1 -(mix4)=r 1 - d4sine = - (d /r) sine r))=5.1
(El
r isinesine"sin*= (ri xmi)x (r2xnei)=K2(r
=-EA (d/r) sine. Substitute in T. T= -MM'd sine/ (4Trur4 )
m ixrd Ans.
-11Cns2iM a n2ai sinea sinei =sin(a - 0 2 ) In equilibrium moments on two magnets balance +or(mMi/imsai: 2: 9 = =sing. cose2 -cosasin92= 4/1- (M1 sin91 /ma )2 sina - (M1/M2) sin% cosa, ssin2ecti(1+ (M1/M2)cos a = 4/1- (Mi /M2)2 s in 201 s in a, sin 201(1 + 2 (Mi/M2) cos a sine MI +1.1? + 2M1M2COS a)
Mz
s in 2a,
sine ' M2
sine2 _ M1
N/Mi+Iv11+2M iM icosa
Ans.
Chapter IX. (XII)
Page 364. (436)
Page 112.
-12CPut I' = 0 in 9.12 (3): W = Mrs? /(4rrpr 3) (sinesinee- 2cosecos9'). Couple on second is zero. 4r5C, -bW /be' = 0 = -MCsinecoset+ 2cosOsine')/(41Tur 3 ). Thus 2tan9' = -tan°, cos9'= 2cose/14 OW 1 (cosesin042sine cose') sinel= -sinea1+3cos 2 0. Couple acting on first magnet is 4rrpr be - -MM 3MM' sin20 -3MM'cosesine Substitute for -MM' -cosesine + 4cosOsine Ans. C= 417pr 34/1+3c os 2 0 - Einpr kil+3cos 20 417pr 3 cosel and sine' A/1+3cos 2 9 -13CTotal potential energy of system due to mutual interaction of field
"'re TB
and magnets is, from 9.12 (3), if C = M 2 / (4npr 3 ),
R 6W 0 OW W=C(sinesine1 +2cosecose')+MH(1-sine) + MH(1-sine') be = be' when 0=9'= 2 since: tow bw = ' C(sinecosel +2cosesin01 ) -MHcose' . 60 be 6 2WC(cosesine'+2cos0' sine)-MHcosO and — 6 2W oe2 = C(-sinesinEr+2cosecosep+MHsine and w,-2 = C(-sinesine'+2cosecose')+MHsine' •
b2 w ye-67 - C(cosecose' ow [ - 2 sine sine')= -2C if 0=9'=n/2. 2w ]2 02w 02w. oo m From Wood p117 or Osgood p178 for a minimum be_ and el be2 be' 24( 0 .' rr 02w 02w Li 10 2WI When 0= Er= - tie2 = ag 2 • Thus 2 rb0b2eil taLi/l 1-1tEb3riJl0;4 < Or la 1602 1 < 2) 2w/68 2=6 2w/be'
When 0=et-17/2
-C+MH.
. 5,
,
3Mr3 Ans. So 2C-MH+C < 0 or H > 3C - 4rrp -14C-6 licos9 Mcose z ..j-Fv2 The.yz component is r 'cos0= y., sine ar 417pr 2 - 2' 771-3 ir ' y Myz z Mz2 4Mcose ,Vc77zT _ Mcose NA-77 7-71 Aff2.___7 5. 5,E =pE so Ery- rtpx3 • -, r . rz rz rpr 4npra ' r +z rpr 2) ricosEll Msine -2Msine M(3 247 2 ) E Msin0 2,22se Myz 'sine= µ rbe 417pr 2 - 4rpr3 ' E0z-417 r 3 2nUr5 ' ey 2nur3 4572.77-2npr 5 M(2z -a -y ) M(2yz+yz) = 3 Angle with z axis Thus E z2rur 5 'E 2rrpr 5 2 5• 3 y2z Thus 0= tan-- 2z2 Ans. Position is stable with magnet in direction - a -y2
r
-15CFor 1 and 3, 9=
2.
For 2 and 4, e=0,17•
1 r2
2 =
3
-b rcosel rbe 4frpr f
MsineM H 417µr -4npr? -
[Mcosel Mcose M roe 4rrprazi 2rrpx s - 2n 7.732
r2 = 1.26 r1
H
31
Ans.
44,
-Page 365. (437)
Chapter IX.(XII)
Page 113.
-16 CTake components parallel and perpendicular to AD =4.b secO. F' = 411pb 3F = 2Mcos30 ° • 8cos3 30°=16Mcos 4 30°=9M F' =Msin30°• 8cos30°=-VSM AL F ' = 2Mcos30° • 8cos3 30° sin30°=-9-2 M F1 =714cos30°- 8cos30° cos30°=V3-M Br Br_L =-Msin30°• 8cos3 30° sin30°4/3 M F130 = -Msin30° • 8cos 330°cos30 °=-41 FB0.1 =-2Mcos30°•8cos 3 30°cos30°= Fc~~ = 2Mcos30°• 8cos 3 30°sin30° 4 F'
FA
FCO =Msin30° • 8cos s 30°cos30° 41 Fc' al.=Ms in30°• 8cos 3 30°sin30°413-M F =
it 41703
(9
e ___,_,_ 9 e el
4.7-M-71
18M, 4Trilb -Y F-1- = 514 4‘ --1"t..3 '— : ;r--:1
-
-
41b3
141324 + 27 #551M 62 0 , -14011 F sine sine. , For pendujum 4rrgb 3 41"rp.b 3 : IFtr - -M 4770 3453i 00 b 20 4 rri.ab 3 d mg/ sine or -oT.sine. = E . Same period for same w so Ans.
F= jr7FF- F1 2 II ---d 20 ML 2 dt 2
-17C-
m2 S
_/42
n(150 ° 0)cos (8 -30°)+2cos(150°-0)sin(0-
TAC 41.4. 1r 1r Isin(6-3°)33s(15°°- 150o, C 30°)). TAB-OTT-4+2cos(0- 30°)sin(150°-13)). Add , equate to 0; 0= - c os(0 -3/3)sin(150°-0)+sin(0-33°)nos(150°-at,-$
So sin (20-30°-150°) =0, sin(20-1)=0, 0 =0, 1/2, TT, 317/2. Ans. Positions e=77/2, 3r/ 2 are stable for each magnet lies along lines of force of others. Positions 0 =
are unstable for each points against are
field of others.
-18C-
Write potential energy W from 9.12 (2) .
a
41tua 3W = M 2(2_ 2[-cowl cos02+2sine1 sin03 -cos02 cos04 -1-2sin02 sin04 ]
da A 7-02)cos(7 - +03) -• sin i n (f0i)sin( Z4-02)- 2c os (7-0)cos (Z+0) -sin (E-03)sin (;c03)-2cos (7rr rr rt r Tf rr rr - s in (.- 03 )sin (4+04 )-2cos (4-03)cos (4+00-sin (T-04)sin (4+0)- 2cos (4-04)cos (4+00 3 '
Pick normal modes.
. Let 0 = 01 = -02= 03=
g‘i
r -04 .4rrlia 3W= M2(2 7(3sin2 0-1) -69.n 0440)cos (E-0)-6sin(;;O)cos (44))
= 3M2 (24-(sin2 0-4) -2s ir; • 4mk 26 20/ot2 = -OW/a0=-3M 2 2 Tsin20/ (41%1a Ans.
so T = 2111M-ksligrumk 2 a 3 2172/3 II. =
a
3M 2132-ri (4fiaa.3 )
3
Let 02=04 =0 , 0 = 01= -03 . 41nla =11.2( -2i(2+ sin 20) 2sin
(
-0)+sin (111-0)+2col;-0)+2ccs
3
3
/4,12{..2 7 (2+s in 2 0) -6 c os0}. 2mk2b20/0t2=-6W/b-M2(-2 7sin20-1-6sine)/(41Va3)PJ1428(2 7-3)/(21Va 3)
so T = 211M-1N/4111imk 2a 3 / (3-2-3/2)
Ans.
III. Let 0= 01=02=03=04 . 4Trpa 3W= M 2( 2 7(3sin20-1)-4[ s (E-0)sin (;-0)+2cos ( -0)cos(+0)] = m2{ 4(3 3 in 2e_r) 6cos20}. 4mk 2b2 0/Ot2= -6W/oE*-3M2 (2 3sin20+4sin20)/(4Trila 3 ) -3M2 0 (2 7+2)/(Inia 3 )
T = 2114-1,•/41Tumk 2 a 3 / [ 3 (2+2' 3/2 )]
Ans.
Chapter IX.(XII)
Page365.(437)
Page 114.
-19CIn passing around the path there must be at least four points a,b,c,d at which the needle lies tangent to the path and has irrevocably completed a quarter turn. Since lines of force can cross only at a charge or neutral point, those lines adjacent to a must enter on one side and leave on the other. Similarly for b,c,d. In passing from a to b we come to the boundary: line between those looping a and those looping b. We meet the same line again between b and c, between c and f and between d and a. Thus this same line of force has one or more branch points, each of which is a neutral point. -20CFrom 9.07 a uniformly magnetized body's potential is reproducable by a shell coincident with its surface around which current flows in planes normal to x, its density being uniform in the x-direction.Each current filament pi is replacable by a uniform magnetic shell fitting the surface and bounded by this filament. Thus the whole potential is reproducable by a superposition of such shells giving a variable strength shell Mx. -21CFrom 9.08 (2) and 9.12 (5), steel sphere gives same field as a needle of moment 4
-srpvMa°, where M is sphere magnetization. If needle moment is M, then from 9.12 (2) 4fipvMa3 m W= (sine, sines - 2cosei cos02 ) since \P is zero if they are parallel. The couple is 4Trpyr Mma 3 = - 3 (COSei S itle2 2sinel cosed T= Ans. ui r d(m/R) {toss 6 r > a 4rpd0o= dx a br
-22Csine 6 a: raini+i pn(cos 0)= m -e _n i+2 -7 j (-(ni-F)pPn(p)+(1 ar a
(p)
from 5.154 (5). Refer to axis 01 ,01 by 5.24 (5). Write
= ma-3n iZ0-
m co A.,n+1 0 n for n1+1. 4rrpvdn0=---i a nrinCri (2 6°)-(1 .1111 1 Pm(cose1)411 (cose)cosm(0-01) (m inn ): = -85nEint-gin+ 5n (0140. But m = Si (000 a2 sinei del d01 = -a aSi (el,951)dpidgii . Therefore m(
cose) E0 (2-8°)-(inral! EI rcosmo1 cosm(01 -0)41 5.231 (3) m 2 1+ 1 (im): b o 4rri a 1+1 2i 0 2rr P (cose)mk Amcosszi = 1 21.+1 \r/ 2i+lO si (" ) ' n°-(2i+1) ;Iv r S i (e4) Ar " M 1 m SP r 1 r < a 47rpvd0o= 2.; cnicosOlh(p)+sina OC(p))=-1rz P (p). a ni 0 a a n0 a nrl Write n+1 for n1 and refer to axis el , 01 by 5.24 (5), n m cct rn 4TTpvdc)0 a gn= =— E0 (n+1)(rn ) E (2- 6 °)-(1 P (cosed Pn (cose)cosm(0-03). This is same as m= 0 m (n-Fm): n ry ,\ r for -i.C.ai i+1 before with (i+1)(--) so 0 _ Ans. a °- (2i+l)pv a Si‘"P"
4rivao=itfi
P
Chapter IX.(XII)
Page 366.(438)
Page 115.
-23CFrom 9.08 a uniformly magnetized sphere has a dipole field. Hr ( r -2cos0) /Or -2cos0 - 2 cotO = 2tano. tan8=-r Ho rO(r-2cosO)/be - sing
Ans.
-24CLet 0 = E(An rn + Bnr-n-l`c•
(1). Horizontal component along meridian is Hm=a-1 60/o9
so Hm = E(Anari-l-Bn a-n-2)OSn /O0
(2). Since An and Bnoccur in same combination, they
cannot be found independently. However, if magnetic field is due to internal causes, An=0 and Bncan be determined from the simultaneous equations (2). -25Cds = 0. If no magnetomotance is present, this equation may be satisfied • on a closed curve through the unknown point and a known point by assuming a uniform field covering both. Values at the unknown point estimated in this way for several known points, weighted inversely as their distance,are: (110p=(r1+r2+r3)(r 1
and (Hy)p =(rl+r241.3)(r11
. Ans. If, on a
circle through the unknown points, the line integral i ti:weig hted x according to the length bx 1...y ‘ AC bx by , BC + m )- -r‘ty 4K-- +Hy v-)-of arc it controls: 0=0-1-1R-+H., rfor the line 3 os3 2rr .eutsi - 71 bs1)21Tr 4-(Hxib;2 Y2 os2 2 rr XVS3 integral around the circle. This gives one relation between NI, Hyl, Hx2, Hy 2, Hx3, Fly3 . Ans. -
-26CMsin (0/2) ,2Mcost(17-F93)/23 sin(V2)( 1 .14 H Hb= 4rua 2 4rub 3 41711 a3 b 3 ' Mcos(0/2) ( 2 1 ). 0 tan 3(95/2)-2 Hb b 33 -2a tana- Ha.E777stan2 2tan 41111 a 3-175 3 (S/ 2)-1 6 =i1T+22-0-a, 0= 11 x,8=111 ik a. tan (6+4Xtan( -a)=(tana+1)/(tanCX-1). let T= tanCI\ 2 X -1+8T+6T 43T4+1-8T-6T 2-3T 4 -16T X 7 Ans. Scot (8+-2)= cotr6tarr2-3tan3X tan( -6+7X/ =- 1+8T+6 T 2+3T 4- 1+8T+6T 2+3T 4 --2+12T 2+6T 4 -1 2-
-
-
-
-27CThere are 2n+1 rational harmonics Sn(0,95) of order n, so if 0= is1+: 11_ S2+1. - 2S F13 ( al Ii+f -arfS2 there are 16 terms. H, V and 6 values give a Wbr, , b0/60 and an/b0 at one station, so 4 stations give 12 relations and 8 at 4 more give 4 more relations. 28Cr
,..,
2
ri= 1f1
„6 Si+t= r3 S2+1.-a r iS3+tfi S4 . From 27C this requires 3 observations from each of 8 station r
„,
4
e
-
29CI'Mr-2Pi(co se) + Nr -4P3(cos e). (Mr-2- -}Nr-4)c ose + a 1 Nr-t os . 60/rbe= -1Mr4+20-4--Ncos 203r-5sine =- (Mr-3- 1Nr-5+ 3Nr-5) sine +3Nr-5sin 30. But sine = cos), so A=-(Mr.' 11-6Nr-5 , B= 2.Nr-5; N = 2a 5B/ 15, -A=Ma -3+ 4B/5, M= - a 3 (4B+A). 60/br = - 2Mr-3131 -4Nr-5P3= (-2Mr-3+6F5IT )cos0-10Nr-Aos31 4 or V= 2(-Ma-3 +3Nalsinx - 10Na-5sin 3\ = 2(A + 4B)sinx - -B sin 3 X Ans.
3
Page 366 .(438)
Chapter IX. (XII)
Page116.
- 30CFrom 9.09:
-3/42 2 rrp2x) (
11+11v
a M2(11-111) w= -F dx - 32,rtpv a3 (p.f pv) - 31-
Force and torque same as for image real magnet Mi= M(p-uv)/(pi-pv) sin% = -sine" cosOecos0 . From 9.12 (2), for fixed magnets
Ans. k--------al I( a , A( _ / e1
W = M 2 (p.-pv)(sine1 sine2 -2cose1 cos%) /(41iir 2(.1+12.7)) From 9.12 (3): F= 32(-singe, - 2cos 201) (11-AO 4rtuva 1 (u+p.v) T = - bW
a Al
(co sei sine, +2sinO, cos02) (p-pv) 4Truva I (il+iiv)
641p.a 4 (u+ uv) -M2 (u-pv) sin20 3217pva3(p+pv)
Ans.
Chapter X. (XI)
Page 408.
Page 117.
-1C- (p415) 1.1 6B From 10.00 (9) 7r- FE2 =7 2Bz where B is a function of x only. As in 10.01 jak jcut 6e -'1JPW'rx +DeN iPaVrx . When x=-±h, Bz=B0 e a solution is Bz =77/ , C=C0 e , Th+Boetr iu / ii= Do ejCpt soBo=A0 e-nv -1" - (I) 7 h and. A0 =B0 . Solving for A0 gives 3/ 1 ec o sh ((dLl ./.;1 ,3 h)) eja it 1/7h) so that B - B0 Ar71A0 =B0 (e Nij'll tr h+eN/WrhI 1= B sech(ju
X
(1)
3ITTrr = (1+j) -if. Multiply (1) by conjugate complex and get amplitude by Dw 655.2.
But N/
cosh2mx+cos2mx cos(pt+8) where p=co and ma=4pip/T. Ans. cosh2mh+cos2mh Find p by multiplying numerator and denominator of (1) by coshC(1-j)3c A.1177rhj so that B=13
°
latter is real. By Dw 651.08 numerator is ifcosh[4.Wr(h+x-jh+jx)]+cosI[i7iTr(h-x-jh-jx)]) Ta Expansion by cosh(a-jb) = cosha cosb -j sinha sinb gives sinhrm(h+x)Jsinrm(h-x)]- sinh[m(h-x)] sin[m(h+x)] imaginary tan real - cosh[m(h+x)]cos[m(h-x)]+cosh[m(h-x)]cos[m(h+x)] A.
mh and mx small:
B. mh and mx large:
B = B0 coscut since
= 0.
B—P Bo e —InhA/COSh2MXCOS2MX
Ans.
Ans.
cos(wt+p) and when x=0 this gives
the field at strip's center falling off exponentially with thickness. Near the surface -m(h-x) h and B-0-B0 e cos(cot-8) so that field falls off exponentially with distance from surface. Also tan8=m(h-x) so phase changes linearly with distance from surface.
-2- (p416) Outside the sphere the potentialof loop and eddy currents is, by 7.13 (2), n -n-1 1pni(oe jtut . Ae6=ECAn(r/c) +Brr Inside by 10.04 (10) the potential is = E(CTI NT3In _4(iliir)Pnl /on + B) n l= so An(a (Cn -
-
eia)t . At r=a : A = A In
and lib (rArjs) /or = pvb(rgis) /Or .
(4/jp a) (1) • and p( (n+1) (a /c) liAn - na-II B n}
= pv(21 1/5/-5- +Nrjr: 7•11 1 11=Pv (Cni:/-5)(i-I -(n+i)I + 2 n+-F n+7 n-I-2 n-1-2 An( -nlIvI...
+17151.1v ain
uF 2
p (n+1) I 11 (a /C)n= Bn (1111vI
i) n- 2
-Ar5pv aI
so that -
la -n-1 .
n-7 2n+Ii+T
4137311vI_-(nliv+n(n+1)1I+ a
Write I for I i and I for I so B + n-T n+T n (jam-p)rd+ r- (p.„-g)nI+-Nrip ThvaI -b\r— aI_-fnuv+Ii(n+1))I+ Cn =4/a I+C(1.31-pv )nI+ jppvaI.)
cr
(2n+l)p.11(a/crAn (p-pv )rd+ +
Putting An value in from 7.13 (2) gives
2n+1 I+(4/75r)P 1 cos a n=1 n(n+1) (p-pv)nI+( jpa)+
A;z1= ppvIVa7 27sina
Am
n ejczt
Ans.
Chapter X.(XI)
Page 408.(416)
Page 118.
-3At t=0 surface eddy currents keep zero field inside. Field at infinity unperturbed by sphere. Solutions 10.08 (4) and (5) give uniform field inside plus dipole field outside maintained by surface eddy currents. At t=o the inside field
-F
A B just equals the static field inside due to a uni- form field of induction B. Superimpose
A and B for C. A is 10.08 (4) and (5) at t=0. B is 10.06 (10) and (11), a static field C is the suddenly established field at t=0. Thus the fields are: at 3pBr rA Br (P-Pv)a 3 B -qst Jo(ksr)e ‘a sine + sine, A 2 E ie A0i-2(p+2pv) t 4 00
Ans.
From 10.08 (4) and (1): A00=0 at r=a when t=0, and from 10.08 (5) and (1) A0i=0 for all r values when t=0. At t== get static values of 10.06 (10) and (11).
Ans.
-4From 10.00 (9), (p/T)N5jot= V2Z or (p/T)dBz /dt= 6 2Bz / op 2 +bBz /00. This is identical with 10.02 (1) so 10.04 (4) and (5), with B for i is a solution. From 10.00 (5) Pt= V41, and, since Bz is a function of p only, 1 bBz. Nr _12 — It _WITp)..ja) berWTp)+beii(N/FP) B0 cos(wt4a) -p Io N3Ta)."- wr. beri(V-p-p)+beig(A5p) i0 = -g op tang=-
ber0 (45a)(beri (N/Fp)+bei l (ViSp)1+bei0(V15a)(ber,(mo0)-beil (VTp)) bero (Vpa)tberi (Vrp)-bei l (Vrp)j+bei o (Vpa)tber i Wg0+bei l (450).1
Ans.
Ans.
5 ( 1-e -R(T+t)/Li c_tiv Isinatin Use 10.16 (6). f(t)= P'(cosa), Ans=Pil,(cose . ' n 2n(n+l)an n 6 -C(T+OrTe (C-N)T 1 Let N= (2n+l)o- R dT Ilvb 1 L = C LARis_it=0 =CnAnsTe 'Jo CCnAns r -CT -NT, ric° .E-CCnAns e-CT{ e (C-N)T..1 } . r Le -e J. Momentum is M=JFdt, so C-N C-N o TA:le -CT (Prnsina+BA,cosa)(N-E2C)(N-C) M= 2rasinaICear(1-e -C e-NT) dT=2raICdha i C- 0 _1 2CN(C+N)(C-N) n+1 -n-2 ._b n+ sinaLr_ 2,P1/11 a Cnsinatcsc^6(sinePC21 Brnaina= b u--5 ---- p=a ---7 171 Opt .' 4 )41 1 n+1 -n-2 x r r 1, =b a Cn sina n(n+1)Pn (cosa.) by 5.15 (1). Bencosa=-Cnbn+1 cosaPj(cosa)1.11.7 1rag h 2n+1 OPn(P) =11-0Ticom/sira bp • B (3 PA(cosa)sin2a{n(n+1)Pn+npPO O=la an np 2an(n+1) a 1 1 2n+1 = i pva_ I(n+lr'sina(b/a) P,1(cosa)P' n+1(cosa) -N-2C f(2n+1 t1.+2pvbR3Pvb gv b{(2n+1)TL+2pvb0 2N(N+C) 2(2n+1)q(2n+1)fL+pvbR3 2(2n+1)q(2n+1)IL+pvbR}
r
M=
sin2 a 0312n+1(2n+l)a+2pvbR E 2S n=1 (n+1) (2n+1) (2n+l)fL+ pvbR PA(cosa)P4i(cosa)
Ans.
Page 119.
Page 409. (416)
Chapter X.(XI)
-6From 7.10 (1): A= pvMsinecosca/(4rfr 2) and, in 10.10 (4)
ff(t-T),P,(2+2f11-1710) sincgt-T)dr or V pv>>.cur / =-etr-1 pvMp{p 2 +(c+z+2filv-I T)21 1cosw(t-T) • A0 - 417 °14 : 1D ( p 2+ (C+Z+2r ).1. 71 T) 2 ) 2 • F or I 4 f2 u,T2r+f pyl(c+z) 3 u„copM sin= I dT 1.103 ,pMsincut sinco(t-T)Zsinwt so Adz 4rr 4rr roc{ p 2+ (c+z+2sliV .02} 2 4 4r 2 11/47 2 p2A/p 2+Kc+z +2sPVT- 32 c+z 1 41.114sinta I - cose 2 s (c+01 p kiwpM sinwt [ sin=[4f2 _A‘DOM l sine 8171r 2f p2 . 417 477•4f 2 p 2 2f ,442+(c+02 2+(c+z) aj 8rrBr = qcoM(fr)isinciftsine(r sine)-1 =Nuti(fr2)- inuft, Be=0. Total fields are Br+14. = p,Msine ).6Mcosfio EvMcose sin= Ans 4 rtr Be rd Be = rs- sinciatf sin= Ans. --T8r fr 21 -7 'typos incot 2 P..& C2 1 -IhicoM sinat r From 10.09 (3) i0=pv - Ibz z=o 1 4rIoNfp2-1-c 2 4-1' 4- 1.17 c2J -4r,co a+c2)3/2. p4IP.4 2(13 2 2r14 20,2 [-1 1.„ .. Eir 2c (0 2.2rpdp Ans. 32r rr 2 J 0 (32+c2)3 -- 32rTr2 12c a +217C7 641Tic2 -
,
01
to
r
-8pvm(T-t+T)p
dT • b ( PA0) From 10.10 (5) A0 P . 4rfrcp 2+ (C+Z+2f p;1T) 2 / 2) pap t 2p.„.m t F T)d'r IlvMdT r PSM T) BP = S n ot 4iff(c+z+2Vp&'11.)3Ip=0 Jo 2T1T(c++2fT)3 EriTfc+z+2f u • 3 }
-
- rsI 2 P= J-7 dA.
- -
o. a a t p'
)+ 3 02rih(cM+tzt)li{11(vc(+c+ zz )+ft2f t) 2
t
When t>T, field dies out as if z increased at constant rate z =z0+2Sp.71 (t-T) from value at t=T, sb 130=22.rr -1 )4M{pv-(c+z 0)+f(2t-,r)ilpv(c+2,:)+2f(t-T))-2(pv (c+z0)+2T3T2 Ans. -9M Pv114 9 , Fx-4TTre 3131/MIM x . By 1.071 (6) for electric dipoles: F=-1--f. For magnetic, by 7.00 F-341.rr By 4rrer If x=0 when t=0, x--vt so by 10.10(5): Fx-
311,142 Orr
V (t T )d T prd 2 )3/21 . tJ 0 Cv 2 ( t - T) a+ (2c+2
for integrand: 6/6T = -6/tit +(f/p.v )6/6c, SO
Note that
_r_ p
pvLT& 4(0+4 f 21.1).1T 2+8 c f piT/T+4c 235/2] By Dw 380.019 xdx -3116M2v p4 so F and 380.013 Jo (a2x2+aix+a)6/2- 24c3(gyNra-2-+21)x 211(2c)4 2 +2f32 Ans.
r
10 Split dipole into components M sin0 in direction of motion and Mcos0
/c
normal to it. Msing5 exerts no force on Mcos0. Calculate separate forces and add. From -9- above: 1F4< - 3/1"M2v(t-1") sin 20= sin 20dF0 4r1R 5 x..Livt-12 c22g. c 62 COSO pvM2COS2 0 x 311,1,42c 0 s 20 co se (3-5cos20) 411,R4 Orr Ox2 R3.1 -
6x2 R2
- 4rr
J__
MsingS Mcos0 1.1
_J`c
:C
‘---12rT --
i2sT
_ v(t-T)
where cose= v(t-r)/R, sine= 2(c+r11v -1T)/R, R 2 = v2 (t- T) 2 +4(c+rlivir)2.
5cos3OdT 5v 3 (t-r)3 ciT 26 r (t-T)diNote that: — to R4 - bti:Iva(t-T)2+4(c-l-fp /-71T)137/ 2 v 6v OtJofv2 (t -T)44(c +W-09 5/2] so that Fx= KLX+FIIX= CF 0sin ays+C3F0 +1.7 26 (F0 v-1)/Ov]cos 201 +
2cos 20 [1
n 2v 2
0 = [1 (v4v9-41‘+2f)1/44v2-1-45 D.F
LIS COS 20
a 4. 4s.2] F0
Ans.
Chapter X. (XI)
Page410. (417)
Page 120.
-11C(cose). If )144,pliso From From 7.13 (2) A'= 3E1 n=ln(n+l)g111 (coset)Pril 22 By 10.1 _ pIansing a _ uuMIan r e°(2rH-1) an MI Pn(cosu)cose cos (cot -En ) c 4f" fn & 10.14' n 4rf-A-2. w=411af2 n1.20 -12In steady state, fields and eddy currents rotate uniformly. Replace rotating field by two alternating fields at right angles and 90 °out of phase. Torque is constant so find it when By is zero by using interaction of Bx with eddy currents produced by By. From 7.02 (7), in 10.16 (4): n=1 and C0=+B. Set cot=41T. In 10.16 (15), i0=k1 B sine mg sine = -gico arB sine/ (9r 2+ptra 2o3 2) . Torque on ring element is BITR2iad9 so that g ,,17/2 B.Tragsin-u.7koaSBsine 9 T?B2 CUa4 s in:0 de 67B2Ccoa4 . T dT= ade = 9f2+4a2m2 a 2 co ) 2 ( 9 Br2
= 2f dT -
Ans.
-13- 21TS B2 2 aa2 W2 317fB 2(02a 4 By By 10.16 (16) : n=1, C1=-2-, cos ei-49.0+ P4 v af-D sza)7 . PL---r 6 • 4 • a 952+4,92032 912+34s2a)2 Pv -14sine 3B r B 11.16 (13): Aio= 2 By sin sin (cot- e) . B= B sine sin(cot- e)=49/4 —T-TTff iva co sin (cot- e)
Ans.
Ans.
-15In 10.20 AI=Bp sine,coscot sinE=-2r/Af 2+(o2 p4a2 .
-S2 i sine sine sin(Lut -e), tane= n=1. Inside field is A'+A pB uvom Ans. so Bi= -2 $13 sin (wt-e) /A/452+(o2utoa
162A 2oB 2Ssint3 2A • /iyalacos(ent-e). Set cot=3 as in -12- above. From 10.20 (2), at p=a i==---'--• uva uva ,v4S•+w-pi•ja. 2r 2 T 8tura 3B 2
452 4tow a2rosin20
41703ra 3 B 2 dB- 43.2+ coap4s2 Ans.
-17-
2n+l [12] cos (2n+1) 8 jat From 7. 09 (1), Alz= Lvegz--1-'2 . From 10.20, =-Ve z . From 4.02, Al IT r 1 2n+1 n=0 P 2v/ E (b/P)2n+2 2 (2n+1)5 si __ -2(2n+l)r sin encos(2n+1)0 sin (cot - en). tanen Ao+A'=- 17 n=0 2n+1 (2n+l)hp+utrcoa' n_ •4/4 ' " /iva)a Ao+A,. SI E 1b1 2n+1 cos (2n+1)0 cos (cot-en ) Ans. 11 n=0,,,(4 (2n+1)4 V +Ti fito‘
Chapter X. (XI)
Page 410. (417)
Page 121.
-18By 10.16 for ring, p=ho, e=-1= -±§c i{Trrh 3(A401-4
2rdh Ns
h )
-1-2---2 bPi(g) .g 4.c2 -oc ,= g=h 2 g . From VII-21, if 10 =C1,./-(
d g so -±d (2+A°)—
srir8Ci 1Pi(jC0)
(1)
C)Pi( g) if Co< C
LP .J...(1) LITER 241._ipt C-4C0) _ 2 c1Ao ( a t ‘ =C 0) crii2+0 ilic1Ai1+t§V-40PRiCo)Q1 (iGo) hsPi (go) (iCo) c1Vg 2+C 2 At surface of shell, :When t> 0, d(A)/dt = 0 from (1). 1•2 151 h, 2c, go dt =A0e2gotifjp(l+C8)Pla Co)Q1aCo) 2g1, dt I A0 I C=o jP(1+C 2)11 (jC0)Q1(jC) rID 46*
Imposed field is A;25 = 2ric1 BP1(jC)P1(g) so that at t=0 AO, A0 where
Ao 4jc i BP1(3)PICA*10C0)}-1Q1(jC) • =
BPI(g)PI(i
By 10.00 (4), i0 = -(dAoidt)/Kti lg o ) at C = Co.
2g0 ti( jp(l+C DPI (iCo) Qi(iCo))
10 h2g o 2.thi-( 1+CPV.i D go)4(.40) e
1.Co) e pct Arrcm2+c8Q10
where
3
-2cati(a-Fcg)Taco)) Bisql+GN/57Ie -Ngo t Ans. N-F Iig 2+C 2
2
3
C2)3/2QP CO
-19From 7.02 (6) potential of x-directed field is A=Bp sings eicpt so that outside cylinder,
AZ +Az= B (p+ air') singie jult (1) . From 10.03 (3) inside Az=ZI1 (..4115p) sin0 di:14 (2) At p=a, AcrAi so
CI 1 (Nrjr3a)= g(a+r)a-1)=2e.r.5a-6(10 -12) (3). Also pvoAi /6p=p6A0 /Op so that
1.1011:FII(Nri-P-a)=P1.3- ( 1-3a -2 11v4.1iPa(I0+1 2) (4). (4)+(p/A) • (3) eliminates D so that e .4pB[VIT(01+4v)10 - (11-11v)10]-1 Ans. D=a2((p-liv)Io-(P+Pv)I23/[(Pillv)I0-(1-1-11v)I2) Ans. -20By 10.00 (4), i=T o t-
cu2 CC 1 ci i(VIT)p) sin0e jept. By10.04,Pav- 2T
Lk 12-4Thera ber 2 ()=01+1102here+"1(p+pv)2bei11-2 (p2- p4)beiobei2+(p-pv)2beil =((n+pv)ber0+(p-pv) be r2) 24(p+pobeio+03-1.0bei
aE-k=16p 2B 2p -IR1.1±}1v) 2101*o -(P 2-i4) (101*2+I*0 I 2) + (p-pv) 2 ber
sin20 pdpdta=7,F CC j:I iI iPdP•
' 1.
81pacuB2(ber ,(/a)beilifip-a)-berikr5a)bei(.,,rp-a) From 5.323 Or 10.04 (11) Pav Nig[(p+pv)te ro-F(p-pv)ber 02-i-t(p+tly)bel +(p-p.v) be i 2]23 4rpa 2cuB 2 (bern (Vga)bei2 (A/Ta)-beic(ViDa)ber 2 (Vip-a)) ° If p is small, by Dw 828.3,828.41 820.3 [(p+pv)bero+(p-pv)ber2] 2 + ap+Ildbe io+41- p.v) be i2 I and 820.4, ber 2 O , beyr.(x/2) 4)=-Pa 2 /8, ber0=1, bei0=Pa 2 /4 so that Trisza 4B 2 noco2a 4 B 2 2B 2.2.21 4r _ Ans. 11' 13a P av 20141) 2T 8 (p+Pv) 2 2 01+Pv) 2 T -21A rotating field is replacible by two alternating fields at right angles 90°out of phase 15P)e iC25 eitpt (real part). Ai=idIi(4.75p)cosoe ja1t-f-ni(.15p) sinfzfe j (c11t+4")I (real part) = ..d.I2(4/3IT= 2 * , =-7-dai /dt=- T I i (V3ISP )e j°e jcpt at any instant. Pr A° v t=f1ZerS pdpd0=Thr . —C J pdp
o
ro This is twice the average power in the last problem so that, since T = PAvt /c131 81 ►pa 2B 2 (berkitia) bei2(Aia) -be io ( T5a) ber, (447 a)) TAns. ((p +gv)bero+(p-pv )ber 2 1 2+((p+pv)bei0 +(p-pv )bei 2} 2
Page 122.
Page 411. (418)
Chapter X. (XI)
-22-
m nrz T° 3s in.--c-, A 20= InfoAnii (q)K i rEll sinn7z planes : A1 0= In E0 Anl i fIV)K i t72cL give Bz=0=P-1 6(PA0)/OP at z=0,c. At p= a, nrz _ 1% _An a sin__c_.inio _ A nr A Lim. 23K0Erlic... 71 n0 I. .L.n`0"11c This is zero except at loop. Multiply by sin (prz/c)dz and integrate from 0 to c.
E
B 1z_ B3z
"
. nTtz
C.-
i(Biz-132z) z.ds in (pTTz /d)d z = s in (prd /c)pI = -(I/a)Apc/2 so Ap = -(2pa/c)sin(prd/c) otiris so A10 =-2pac-IIriioI l(nra /c); (nr p/c)sin (Wird /c) sin (nrz/c) Planes alone. To make Bp =0-7i at p=b, add A40 =2Pac-inE0 Ii(nra/c)Ki(nribic)Ii(nr p/c){I i(nrb/c)}-lsin(nrd /c) sin (nrz/c) Only Ir(nr0/c) are permitted since KinVP /c)=co if p=0. For self inductance
d
decrease compute 2fra (Allimages+40) at p=a, z=d. Positive images at distances -±2nc, -m" V= 0 4-2c-4
Zk = rle/c= 1TIKCoostrrc/(2d)3) /Ksin[rc/(2d)]} 0,1
C from IV Prob.63.
- 28C from Prob.64-IV. k=(X27Fb2 -h)/b.
E—L=1- -. VT:7 0 142-,I2h
Zk= ne/c= rac(k)/K(45:17 .1 ) - 29-
C from Prob.65-IV.
M=coty+cotp, N=coty+cota, V is potential.
Zk =r1E/C= 411C( 7:(7/N))/K(g5) A/i4,
U= 0
- 30C from 4.29 (9), b a from IV Prob.66. Zk= TIE/C=r11((b/a)/KPr-73/a).) - 31-
d U=
C from IV Prob.79, k= cos(rc 2 (1+X)(213 2)-1 ), X from 4.28 (6) with b2 ,c2 for b,c. ..4/7-72 ) Z k =71E/C=4;r1K(k)/K(] - 32k= cos(-1.2-17(1+X)), C from IV Prob.80. X from 4.28 (6) with
for c and -liN/F for b.
d
Zk = Tle/C= iTIK(k)/K(1 .17:172)
U =0
U0 CO
/71, 2,1
Page 446.
Chapter XI.(XIII)
Page 133.
-332 )= bK(k) From IV Prob. 82, (5), (6) and (7) (a-c)K(NT-7 determines k and ki=45:-72 . Let P = 2akI K(k')/(1713). Then
-4P 21
B _ 1[, -
14(1+Pa ) 2 kI 2 le P 2
ne • `k
rliC(45TA C 4K(411-(B/A) Ans. -34- (17,p466)
To make an E loop (open circuit) at far absorber boundary need reflector (2n+1)X/4 meters beyond it. From 11.15 (13) to match at front face of absorber, Zk = qcothfcl. If d is very small this becomes Zk = Z4/(lid). From 11.15 (15)
Zk =
4= jad/(a1 + jut),
= at+ jpi
and (1) becomes
= icull'ACe+13')2(1 3= {(yt + yze)(1}-1. This is satisfied 111. Ans. only if led >> Jule'. In the latter case y'd
7
V=0
r Uoc+.
U= 0
Chapter XII. (XIV)
Page 134.
Page 482. -1-(1,p501)
wt -13r) . Then by III Prob. 2, jpr bcobes203 811EVx:i zz= 9(2) zze jca te [(13- jr-1-j 13rJ sin0cos0-(pr -2 - jr-5e-Nlj (a3t-DY) =0 e = VX(V)d) -3jr-3 +3pr-2-Fj B2r-13 sine/cos°. From 11.01 (9)
From 12.01 (13) 8rrei z=q 22(13r -i-jr-2)(r cos2 0 - sinOcos0)e j
(s in 2 0 cos0) /be
ry
Er =q(2)2(3cos 2 0 - 1)( 3pr cos (cut-pr) + (3-132 r2) sin (cut-pr))/ (817er4 )
Ans.
8TrZEr = 42)z ej (13t -Pr) -3jr-3 +313r-2 +jp2 r-1(r sine)
81EE - -q(2) eiwt-l-in20bC(-3jr -2+ 3pr -1+ j ps)ii 13r 3/ (rbr) zz 2 Ee= q z sin2e( (6 pr -p3 r3) cos (cut -pr) + (6-302 r 2) sin (cut -pr) )/ (1617Er4 ) Ans. e i (cot -Pr) 3jr-3+ 3pr-2+ j par -1)-isin20 8TreB0-- j Flee) zz B0= citz21;1psin20[(3- p 2 r 2 )cos (cut-pr) - 3prs in (cot-13/)3 / (161Tr 3) Ans. -2-(3,p501) From 12.01 (16) 16rEr 2 Ziy = cgy (pr-j)(i sin0 + j cos0)sin0e j (wt-Pr) . From 12.01 (11), (12) 1617432cy = qTy (f3r -i-jr-2)(r I sin20sin20 + sinOcos0 sin20 +0sinecos295)e j(wt-Pr) . From III Prob . 2, 161re (vxzicy)r .cgti(p) (sin2 A) isin860]cos20 - cos06(sin20)/o0)e3 (cot-Pr) 0. ( r component) = 42) xy'(13r"2 jr -3 )(2cosOcos20 - 2cosecos295)e3(cut-Pr) = 13r -sinecos20bE (13-jr-)e jPrDrorie jwt . 16re (VxZ32(y) e=qTcy ((pr -2 -jr-3 sine (sin20) /O0 -
=4y sinOcos20(3Pr-2 - 3jr.-3 + jp2r-1)e j ((a-Pr) 16rre(VxZ,2cy)0=q1x2)y[sinOcos0 sin20 b[(13- jr -1) e ifir]/ror - (13r-1- jr-2 )b(sin2 0)/roesin20e-jPr le jult j (cot -pr) -cos0 cos20 • 16rre (VXV ) ) e . =q;c2)y sinecosesin20 (-3(3r-2 +3jr-3- j132r1)e From 11.01 (11) B =
(v)a) /6t = jculie(WZ)=jprIe(VxZ) from 12.01 (9).
02) pri j pr -2+ 31-3- 13 2r -1)sinecos20 ej (ca "13°' e -- xy
1617B
130 = -Be cosO tan20
Be = cicdyprif (3-13 2 r 2 os (cut-pr) -313r sin (cut-pr) )sin0cos20/ (1617r 3 )
Ans.
Ans. Br = 0. Ans.
From 11.01 (9) 'E. =Vx(Vx'i ) and from III Prob. 2, Er =[o(sin0A0)/o0-bAe/b0] /(r sine) . 1617CEr= cicg (3pr-3 - 3jr-4 + jp ar-2)(- cscO[a(sin 2 Ocos0) /be] s in20 - ( cos 20 ) /60 e (wt-¢r) =q 2 (313r -3_ 3jr-4+i fizr -2 )sin20 (1-3cos2 0 + 2)e j (a3t-13r) = 413,(3f3r-3-3jr-4 + j (32r-2) 3sin20sin29sei (c13t " xy Ans. Er= q()2c)y 3sin20 sin20[3prcos (cut-pr) + (3: P 2r 2 )s in (c13-13r ) ) / (16rer 4 ) jcut From III Prob. 2 and 11.01 (9), E0=b(r (VXZxy )0)/rbr=4)ysinOcos20?)((3(3r-l- 3jr-al-j132)e jr3r)/rbr ierti (wt -Pr)/(16Tre) E0- cf2) x ysinecos20( -613r-3 +133 r-2 + j(6r-4- 3f32 r-2)}e j Ans. = qxy 2 sinOcos20( (P3 r3-613r)cos (cut-13r) + (3(32 r2 -6)sin(cDt-Pr))/(1617Er4) Ee= 46 cos() tan20 Ans.
Chapter XII.(XIV)
Page 483.
Page 135.
-3From XII Prob. 1 need sin 20 in Ee. bCP2 (cos0)3/60 = Pi(cos0) = z sin20. By symmetry need even function TM waves independent of 0. From 12.11 (5) for expanding wave: Wtm =ECa P2n (cose)k2n (j8r). Then from 12.14 (1): Ee = jcuAe = -ja)b 2 (rW tm )/rbrbe = -j(DE-Cn(b[rk ar (jfir)]/r6r}Pin (cos0) by 5.23 (6). If jiia>a, Hio(R)z No(p)• Ez = -jakz so .
Ez = IfoBo cos0' eiwtHP(Bp)la.sig:"AdY0 y 130:2 '‘/IThO y so Ez+P2a2E0 cosfe HP)(13p)eiwt . Ans. 1 If le =N/R2 +yg -2Ryo cos (ir -0)#p, then HP)goes under integral. By figure, le cos0!= Rcos0. By HTF I (15) p177, Cacos(417-0))=secgisinfn(Ift-0)), so from HTF II (28) p101 cosfeHlakiliN) = 2nsintn(i1,-0)3y;I JEI(By0)Hga(R). But x-lJn (x)=xn" sEApx 2P so integration -a to a removes even n-values. Write 2m+1 for no then cos (2ur+1)0=sin[(2m+1)(41T-0)}. Thus integral is Posfellf2)(pRIWT-Tridy0 =2E (2m+1) cos (2m+1)0 Iran 2ig. 2 (R) where Ita= 2,1ao y0-1.17;TJ 2ak+2 (fly0)dy0 -a ,1 •xJ 2÷i(paAN)dx, where yo=a4fcco dyo = adx/(2vg). From 5.293 (3) x -1 ,1 2m +2(BaN/27)dx =ajo x 07---(-1)r(13a)2m+ar+1 p I
- E r:rci m+r +2) 2 2111+2r+1,1.\7 2 -7 2 xm+r-'dx. By HTF I (11) p 10, t (1-t)v-idt =B(u,v) 2M+ 2r+ I B (M+r±1 e•Dr ( 13a / 2) 21- a ,= aEtho so that =B(m+r+D-12.). Thus I= rE0 r: (2ur+r+1)'. CID Cl
Ez=4.BawB o eiCptin
k (2rn+l)cos(2nr3-1)0 H2)1(R) Em. Ans. If we take r=m=0, B(-12.1 4)=77.,
then Ez=4-13 2 a 2curB oeia3tHi2)(R) which, with B0 =E0/a) gives original approximate result -44From 4.23 (2), if V is the potential function, W=4E(z+4zr), c2 . In problem 42 set f i(y0)=4y0E0 /.177q. Ex=-[aW/bz]i.p= ixE0 /,./-7 re dYo • HP)(1310 HP7(13 (R- Y0 sin0 )) aYo HIV (PR')/N/P-T Aci=.+NIICE0eici5t Hi2j(BR)-Byo singi[H12) (13R)}'+ • • • Neglecting yi compared with R 2 , only second term survi d integration from -a to a so A4;64BsinizINtraEoeiwtHP(BR)r a _y8=43\rire Sill0E0e ja3tHfal(13R)Trasid E
=-2-j1713 2a2E o singS R b3R)elwt Ans. 00 -45b (R'dA,A) -— 4 =Ce-' CHP)(13R1)/N/77Thdy° . dBZ- RIbRI ,-From 4.22 (3) W= sin-1 (z/a), bW/bz=(a 2 - z 2) 2 A l 0 a R-1 o[pRi }qv( pR)] /b (13R)=13110(2)(fiRD . At slit center Rt =yo. Let yo =asine, dyo = a cos° de 2 pa HU) 77: d 7-4.72- = dO so Bo=BC j a a2-y2 dy0 =0C.I174-42)(I3a sine) d9=BCSJo(Basine)de- jfiCfY0(Pa sine) de
=14
r/A_ //n(- ceBa sinEr)dEr where fla is very small so 132a2 terms =13c jr fr/2de- 2.1TT2c j wi (a Pa sinO) dO - -1•1IT1-2117/ 0
are neglected, El' =-0, em(-1)=j1,, ka= -0.11593. Use Dw 868.1 so Bo = 13C(17+21170 / d0-V,Vr(af3a)d0
jliS TT orksine de) = pc( 21,- 2 jk(apa)+ 2 .j0n2)= 213E {1,-jk(aPa/2)) so C=B0 /(213[TT-j242(-1a0a)]) and d
jcot jraHiM (WR2"02-2Ryo sine) dy0 A --P• TrCe jCi3tHi (13p) since 1 7.7 =Tr . 0 -- Ce o✓a -yo2 R=P -SP°2 a a -y , -a B = LLELL-qa) _ _____", Rc1 Te j(cot f3R 41r) Ans. If neglect (Ba)2 get B z Re) z P.Ce r1BR 1 1 ___110 {212j(cut-BR-rt) If Pa is neglected: B --0. z psco Um (pa) BR e (2)
(Opp)} )1 (pp)
-21
-
-
Page 147.
Page 489.(507)
Chapter
-46-
_44_
-45_
By 44: Bz=-1j17132a 2 B0sin as ing$ Hid(PR) e ja3t Bcara
Be Els ina-M-)
jpt By 45:B -1occsa lcf2)(8R) e zzr j2n(YPa/2))
11T
I L11- - - -
Add amplitudes. Take PR >> 1. Let 13a be so small that 2(17 - jOn(y P a/2)).cz -2 jk (pa). Then ,
Bz= 447130C-13i a2 71's in a s ingIeir" + jcOsaf2A(13.3))127(rTPR)ei (a't -PR-TIT) since, as R becomes infinite rPR) e j0R+jr/21 Hiv oR)..0je3r/2HotajcoR) so that Hp) ( f3R) e TT/ 2,/-7— 2/(1 (wtBz= ijBo(-12-132a 2 sinasin0 + cosCak(Pa)}✓217/(PR)ej PR-Trr) . TT varies with B2 so sina IT= 1-7-( [cos a /k (Pa)] - 4132a 2 sin a s in0)2 no 2I3R -47By 43, Ez= T17 1 P2a 2E 0cos 0 11/2)(13p)eialt . B30 sina is zero on screen. As p-'›ce ,
Ri8N 4_ 3 .,,,,
IT
by 5.295 (8), Hi(2)(f3P) = NriT(T--V3Pe j PP+ T irr• Ez-4132 a2E0 posps 2rei4-"" . T3-, E0=Eicosa. 17is proportional to E2 so ft= 4-17P 3 a4 R-1 cos2 0cos2a fro Ans -48,Ec E = -1 by' =-Ec1 =_LACC ._LE .-1 _ _ -EP r, )+-9-. By By 5.272(6): V=Ez(1 ---m 17C t hiogic=o 'Rh 1 G=0 TT/c 2 -p 2 At great distance (nXEt)xr i = Et(Oi cos0 +ft)Js in0cos 0) . By symmetry A0=0. j (wt-pr) rr E pe p cos0ei"cos0 sine pdp cif/3 . Neglect p compared with R rzR-0 oos0 sine 2rvi) JJ (R-p cos0 sine) 4/C2 Subs. in 12.18(1) Ao= 1T in amplitude factors (denominator) and observe that real part of exponential for integral range -Tr to 11 is zero so Aia= [1 j cos0 sin ( pp cos0 sine) Of] dp ei (43"13R) . By 5.302(:. u corn( ca p2 o r11COSt PSinO)dp ej (Lot -(3R) ICP 2 _.1 J ( 1.........___ J i (z)= j —sin(z cost) dt so AA. =--. Let p=csint, dPa,;277FT=dt so u o 4c .•17 2 TT , PEc 2 rri 2 . 2 J3Ep a A0 =- ayftR j sin tJi(pc sine sint) dt J312 ( Pc sin() ) , by IT II (19) p361. 03R.I2Trfip sine o But by 5.31 (3)T.77(S)J 3 /2 (x) =3C-1 (x -4 sin x - cos x) = x -2 ( sin x - x cosx) . Thus sine) - Pc sine cosaic s ine)) e j (wt-I3R) ____? 13 2c 3E singej (wt-pR) A - pEc 2 ( S in (I3crwRf3 aurR 2c 2 s in 2 0 3c -00
Ans.
-49See 5.271. Static field form is
n= Bciopi(jc) + csQl(jC)) Pi(g) poso.
_-----
.-E.C2 , Qi(jC)=TrViTT 5.23(11). -------....„,..--• At C=00, Pl(jC)=jN/7-C i- 2 , Q 1 (jC)=0. At C=-40, Pl(j6)= j./i nc.. -7r-Bp cos0 = Bx. nc= Bx(-1+17C) = 0. so C=r-1. n= Bei[jpia 0+ 40 a 0)pl(g)coso IT Bc 1x. 3.i.-r(-7 + tan-1 C +1—142) C-- AFT-C2,ff-Ticos0= which becomes by 5.23 (11): n = -- (TT -cot-1C 1 C+11 +1+412 2Bx 2Bx 1.42 2Bx 2c1 =_ Bx r 2 V1. FC 2 an B In hole at C=0 Bt=Bx=-, _4..2 ..., C +. 2 circ imic 2-p2 irrfr ca-y2 . 2 Bn=co/V+C 26C C-sOcirL1+0 1 r 4/E -
Where B, the standing wave value is twice the incident Bi. From 12.18 (3) bEy /bx=yoBz . Ey=-2ja3BT7-17-72 =- 2 j pE orriNc777. (2yE)xr1 = (Ai sin0 +0 cos0 cose)E. r ^4: R-p cos0o sin°. (continued)
Chapter XII.(XIV)
Page 490.(508)
Page 148.
-49- (continued) By 12.18 (1)A 0 if r >> a
21132 E0 sin0e j(cot-13r) Scr 217Alr— laeiPPcc69"in°0dod 21r2co R - p cospisin0 oo
By 5.302 (3) jrcros (zcost )dt = ITJ0 (z), frs in (zcost)dt = O. Thus Ao=(2 j 0 2 E0s in0 /(TUR))ei ((zit- r3R) rcAlj:7 0/ 4(13P sinO)pd0. Let p=csint, do=c cost dt. The integral is then Hd.M.F. 11.4.10, ,
0.
j=ric3.3.0 ( 3csin0 sint)cosat sint dt = TT/2{13c s in0) 3 / 2 J312(Pc sin 0)G3. Use 5 o A0 _ 2 j0 2 E 0 c3 s in0 j(cot e - f3R) sin (13csin0)1c sine cos (f3c sine) 0 3 es ins@ 11(.0R (rraA))eifmt-13r). E0sin0 / 3 2c P AO = cot0cos0 Ao !See cnX9s,,r,i sine) - pc sinOcos(fic sine) ___-0 j132c 3Ex os0 cose j (cot-0R) 2 jE, cos0 cosO j tot-13R) sin e . A0= e f3c +0 MLA singe TTE3u3R -50If the incident wave electric intensity is ED, then from 49, writing 2E0 for E0 , E0=jcoA0 =213 2 a 3 cosa cos0coseE0/ (11TR)„ E0=213 2a3cosasin0E 0/(17R)
E6+? _ 404a6cos 2a E0
r R2 2 2
(sin,fs+cos'20 cos291To = !i142421172R2 22 (1-cos 20 s
V2
rFrds=sro S o (1- sin 29cos 2 0Rsn ?
.
Bcos a
0-521-
2r 2r 1604 a e cos 2aCR4 (1-§cos0d0=CRrr) 2so P3r T6 52-
Take entire B tangential and normal to incident plane, let 0 be measured from this plane. It is then Ift-0' where 0' is the 0 of problem 49. Write 2p2 a 3 E0 cos0 j(cot -13R) 2f32a3E0cos0 sin0 j (cot- DR) 2E0 for E0 . From 49: Eo= e , E0= e TTR rR There is also Esina normal to sheet so by 48 E0
TT= ((q +E4)/Eg)lfo=C paa3/(rR)) a s in20cos 2 0
f32a3E0 s in0 sinsj (cut - f3R)
rit
e
2COS0 + sin a sin0)9fT0
Ans.
-53
2r,. = CR2.10 t-sin20 .sitds cs2r2(4s +4cos20- Tfc os0s in izs cos2 0 + (-2cos0 + sine sina) )R2sin0d0 3 o + Is in 2 a) do = cR2 (jr-4- 4r + irsin 2a)= i1TcR2 (4+sin2a) . Ans. 15 = +04a6r-1(4+ s in 2 a)-Tro -54Assume slot fields to be: E= t-al'sinp(2-z1) cost z1>0, E=I?sinf3(L+z1)coscot z1tclo,
/ 2d
The wire and its close image is a short section of the bifilar line of 2X 0 2 ../ c 46 +.1c 2 -14-1,_,••• 221127: ,„:"2 11.18 (4) so its reactance X0 =coL is given by TI Onfir+Y0 (2pc) 7=ff utr wpb-T1 r 1,-.11 c which agrees with 9. By HMF 9.1.79 V= ol a =3: o ( 4 32n aa2 +432c2)=11-4 t ( 23n a)J2m( 2 pc ) . youpna)J3 y2 221 ZY0 (20na)-Y0 (2pna)-4-P4-" --- 2y2(2pnoW..—1aYI -2 (2pna)=Y0(23na)-Yi ( 2 Pn5a-(-1)mY (2pna)by 5.294(6) neglecting (13c)4 terms. Combine first and third summations in last problem, omitting .n=0 term already included in X0 . In same way:Y0(44320e-d)2+4132c2)---yof2Kna-03 p2c2Y1(213(na-d)) na-d Yif2p(na-d))] Thus X= 11211EC 0 11.-1 TTf3c 2 (213na)+11c2 Ans. n=1 'na 2r n=-02 2 (na - d)
EY
If d= 4a:
X= -IPa 12 cosh-1 e 2 r 213 ra 211 1 (-1)InY m" (nigia)
Ans.
Chapter XIII. (XV)
Page 547. (554)
Page 154.
-11- (38) a x, 13.02 (3) t= 13.02 (1) Ute=ECrancosuirr
--T-ce i Ninz 4 c1 11"Cain sinm
x3e-ii3,Inz. min. m2
13.02 (4): B= E c.mfii"mknsinm14.c+kg,„cosargr— x2 a te m 1 n Set z=c, multiply through by isin(pyTx/a)dx, integrate x=0 to x=a, --=-1.1. 73/2111 C40 . sine Bxdx =-4sinErgTd 3-731 V:sina 11 -21 rxdxC.I;in=::•11--
-+1. i",„E 31Tsinq-Td .
3 ~mc = 2 sinh jj3thc. C =LlP1 e3l3mc sin ad . For closed end add image -I at z=-c. m m10 (3 j2a4mElsin(m7d/a) :L arx-e-J56z If c >z, interchange c arAariggng ign When z >c gte= sinhjgle sinni When z> c ilte= -(2pI/A)miisin(m1Td/a) sinh(jf3Lc)(-i sin (mrx/a)+k am17/131a)cos(mrx/a)) 0
involve x and there is no incident Ex so higher order iris waves also have no Ex . By 11.01 (13) an x-directed magnetic Hertz vector with the incident wave x-dependence is rx -1171; z rx n (2) z e Anz (1) From 11.01(13) g =- Erm0 sin--cos b e n -m ir on sinlreos 3.7 =a zn a -2 !Tx .n172 Trx . Eta jpin z b i-j-Ecoslrsin b e Jen 40322n+ b2 (3). The variational formula
n
13.07 (4) is the same forevery thin slice normal to the x-axis and looks like a slice for a parallel plane guide for which (2) and (3) become bEy=-E(2-68)Vncos(nry/b) and bHx=E(2-6DIn cos(nry/b) (4).
By 13.00 (11), (12) and 13.02 (2), with .a infinite,
I/V= 1/Zk=N/T e gzjb/xg)/NA12-(zbAg)2 (5). Ey=0 on the axis and Hx=0 e /A/1- (3 .g/Xin)2 =4/7— in the window so multiplication of (4) by cos(n11y/b) and integration from 0 to TT gives b pc Vn=j0E y (y)cos(nry/b)dy, In =jc Hx (y)cos(nry/b)dy (6) From 7.01 (2) and 1.06 (5) the current In charges the iris to potential Vn . In=-YnVn and Vn=-ZnIn (7) where Yn and Zn are the iris admittance and impedance for the nth mode. If
0.4:y4ze,
Hx in (4) is
zero so shift the n=0 term to the left side in (4), and substitute for In from (7) and
rc cos b cos b aa' „ (8) E (y )cly = lYn.-1 Compare 13.07 (2). Multiply by E(y) and integrate over the aperture so that for Vn from (6) to obtain
-12-I0V0 =-nkYnaE(y)
s°= -n=1YnVncos 27Y b
cosni -3-7cosn1 -2 -7E(yt )dydyi (9) Divide by Ve from (6), use (7) and (5)
rfic
12
0 B 8b cc 1,10E(y)cos(nrep)dYi for 14-l and 11.15 (15) forte. B°= = Yk %g n-1 4/112-(2b/Xg)2[4E(y)dy)2
An E/y value exact at long wave lengths appears in 4.22 now carry primes aperture is at V' = 0,
(10)
(I).
The
y 1 =0 in the right half of Fig. 4.•22b. Insert V! and y' in 4.22 (6)
and take oU l /bx' with a'=c. Write Try/b=0, TTc/b= 0 and dy = bdOPIT. Then 4.72Cos(40) E ,- cos(4ry/b) (11) 4/sin2 (41%/b)-sin 2 (41/y/b) 4/C00-cos° The (10) numerator integral, expanded by Dw 401.06, gives two Mehler Legendre function integrals, HMF 22.5.36 and 22.10.11, so that ZE(y)cos(nry/b) dy is
-DO
4fib ro cos (n+4)0 dO 4. 4,5b recos (n 2r.44/C-00-cose ' 211'io4/cos0-cog0
Tb[Pn[cos ( rrc/b)]+Pn _i[cos(Trc/b) ]3= ibIn (rc/b)
The (10) denominator integral is bP0[cos(trc/b)] so that B°= Note that, when
Xg = 1:0,
s..;i1 k1L1(.7 ))2
2-
(12)
(2b/Xg) 2 (13)
(13) must equal 13.06 (7), which gives
407qcsc(.i1Fc/b)3=ni1{Ln (ffc/b)} 2 /n
(14)
Add the left side of (14) to (13) and subtract the right side to get 2b rc (15) B°=-(4,0ncsc]F+ E [(n 2 - 2,2„13511.- n-131 L'n X D n=1 2 g
In all cases tested four series terms or less gave four correct digits in B° . The static
B°of 13.06 (7) is a lower limit.
Chapter XIII.
Page 549.
Page 158. ya
Y3 Ui0
-201 U=-17/2 U=11/2 3§ Shift right, z2 =z3+7(d 2 -c2 ). a2 =a3=4(d +c 2 1 V=0 z3=a3 sinW 4.22(3). z2 = a 2 sinW+-f(d 3 -c2). ftz i Bend x2 axis up at ±b2 by 4.22 p90. za =b2 sinff;. Let b2=1. = a2-17(da-c 2) sin (12-rzi)= a 2 sinW++ (d2-c2 ). When zi= 14= FIT so sindrdi
E—b1
When z1 =-"c1 , W=-717 so -sin(i'ffci/b1)=-a2+ i(d1c2). Add and subtract to get s in( .-12-17di/b1)-sin(iTtcl/b1)=d 2 - c 2 and sinerdi/q+ sin(rc i /b1)= 2a2. Thus sinW=
U -Tr/2
U 11/2
2sin07zdb,)-(d2- c)= 2 s in(;Trzi/b,)- s in(01d ,/b)+ sin(+17c,/b,) 2a1 tin(ird1/b1)+ sin(71c1 b)
V=0
zi= z-b1 shifts origin to iris bottom. When zi=bi ,z=2bi=b. z=0 at z1=-b1 . c so 2b1=b. c+id=1)1+di or di=c+i(d-b); c-id=.b1 -c1 or c1 =4(cH-b)-c.
db -1=-cosf4. sin(-1-11di /b0=sinS(c+ 2 )]
sin(12-17zi/b1)= sinElTr(zi-bi)/bj=sin
VO 4-d
.sin[11 ,(c+ 1d)-irr].-cos[ - 77(2c+d)/b]. sin(ircA)= sin[I-T(V2-c)]=cos[211(d-2c)/b] - s in(12-Tidi/ + s in(ici/b1)= cosCi11(2c+d)/b1+cos[4rr(d-2c)/b]=2cos (17c /b) cos (-in:lib) by Dw 401.06 s in (-irdi/b3) + s in(iricA)= rcos[417(2c+d)/13]+ co s[1211 (d-2c)/b] = 2s in Mc b) sin (ird/b) by Dw 401.07 Thus : sinW= (-cos (Trz/b)+ cos (re /b)cos (% 1-11d/13)//f sin (ffc /b) sin(Ind /b)
Ans.
Insert W=jV, z=jy. Let V-► co and y-go, then only first term in numerator survives a nd_ sinW -> jeV=-erY/b csc(17c/b)csc(i1Td/b). Take logarithm, neglect Efftj and k(-1), then Trc rd, B0.,4b Bel a V= b -ktsinTsitr-53- 1 Ans. Potential difference is U2 -1.11 =17 so _x:72(sinihin35 :3) Ans. 13.06( -213- cos (11c7a) dz' _ C cos(rre/a)- I- 2 c os (rre/a) where 1) 1 1 +cos(rY ca) da l Nizt-1 Nii7.51 ' z1l+cos(Tre/a) Check: If z1 =0, z1=-b; if z'=c', z1 =-1; if zi =a, z1 =+1.
NiZ771 =
3g
!cos i/14'= 2Cbos;zi n/siniC a: - s in2 rx1 . 2 2 1-rc_ Trx b Omit primes. -c< x0 so in 13.11 (3): -sin m(0-480)+sinm(0+80)=2cosm0 sin180 Pe•rn80cosr0 so coefficient of cosm0, -Dmn in13.09 (9)0 comes from Cmn of 13.11(3) with eqg omitted. The result is -Dmin which cancels the arc terms above. Thus NO TM wave. 8c orprin cJ4,(filmne 6c erc rta)] -8mn (c---2-)4[Pinn(c-TE WAVES. For arc elements finin (c+T)4[11mn (c+2 2-)= d({3miic) Q (Pronc) =C8mncJm "(fininc)+Jm t n (pinnc))8mn dc. Substitute in -27-, letdS = c dc do Cm'n= juIdS (2-41)(134nc24(8mnc)+pninc4(13Tanc)) ei 131:11nz°/ (217c2 gin (81na2-m2)4(13mna) . For dpo sides when 0›.0, current is negative so in 13.11 (2) and 13.09 (9) for B 0-component, -Dmn(-sinm(0-450)+sinm(0+-100=-2Dmncosm0 sinm = -mDmn 80cosm0. So coefficient of cosm0 in 13.09 (9) comes from Dmn of 13.11 (2) multiplied by m80. Write 2-8% for 2. C;l'in=-jm2 p.I(2-54)dS Jm (filmnc)eif34Inz421fc 2 13m1n(Brgn =-8rfinc2 Jm (8mnc) by 5.291(3) so Cran=Crnn+Cm "n-
-m 2) [-Tm (8mn a)]2) Add 13m 'm2Jm 2 nc2 4-F8Enn c.Jm
- j pIdS (2-8A) flignJra(f3„,,c)
217131n(f3An aa -m 2) pm (puma)] 2 e
j Om'n zo
Ans.
-29-(34,p553)
, a, TE WAVES come from dpo only. By 13.09(6)0 13.10(3): ei/34m`zo-1:2)-illtnn(204) ,.e iPmn zo -muIdSJni8„,nc) i prrin 6e j (3Mnzo . Put into 13.10(3) j8/ zo n Ans. Dmn" rc(81na2-m2 ja ( mne)32 e Let dS= adpo TM WAVES. As above, for radial 8 write j8thn8e 3/31nnz° for e jl3r6z0 in 13.10 (4). so ;it n = ndS (2- 8/) (I2mnc) e nz°/(2178mn[8a4(13mna)] 2 ). Outer radial 8 is negative so -Jm[f3mn(c+1)] Jm[Poin(c4).]= - Pmn6 J42 (13mnc). Substitute into 13.10 (7). Let dS=6dzo . qin= -juI8mndS(2-8/)Jmi (8mnc)e il3ron z° /(21,BL[13a4(13inna)] 2 ).
Prim +
Pmn 0dal
_Orini+0An _ 2 Pmn0riln 0mnPln
so r
-ipids(2-6,00,141(13mnc) u.r1, -Cln' ''rnn 2r0mn1341n[aJM(Pmna)32 -30-(46,p555)
Ans.
From 13.09 (1),(2) Ute= CJI/2(8no)cos12-0, 42 031 0)=0. E0=jus, Ep= 16 a((.0)• From 5.31 (2) J1/2 (v.,r277(-; -s1-1:71"-T +cosvlivV= 00 so tanv= 2v. J;i2 (v)-).. s/2inv
n-
CJI/2 (13 1P)3 2 -4. 811/(217p). Thus the integral for ate is logarithmically infinite at the lower limit so the attenuation is infinite. When tanitu a=2Pna, 80211.1656 Ans. -31- (46,p555)
sin 1 pZ From 13.09 (1),(2) Utm= CJ//2 (80)sin10, Ju2 (pn a)=0. From 5.31 (2) Joi (13np)=JI 01713 2 p so 811a=1'7=21TV11 a/v by 13.00 (8). Cuts off at VII = iv/a. By 13.01(7),Wirdp at 0=0 2 C 2 sin2 811 0 occurs in atrn . But -0so aimlogarithmically. 21703 0 P 50 0=0
3 p-0 -P 2rp
Chapter XIII . (XV)
Page 552. (549)
Page 163.
- 32-(8) By 12.16 (2), 5.295 (8) an expanding wave depending only on p is Wtm= Ce 3cut{Jo (0)- jY0 (13p)3= Ce jCptH6,2)(130), so B0=133C e imt(4 (PO- jY,')(13p))=-Rs cePt H(12t130) by 12.16 (5) . At p=a, 271aBo=pieica from 7.01 (2) so -217a8 311M8a)=0I, C=
27703HIV(8a)). From 12.16 (4) Az=-132 Wtm=1I4,23(8p)/(217a8Hitat8a)}
J1(0.036217)=J10.1137 0.0569, Y1(0.1137)=5.69, so 1J1(pa)J nrrib. Az Thus cp > nr/(b) or v> n/(2b) Ans. -
43 (19) -
z, B2= b(pA0 )/pbp and from 12.16: p>c Wt e = e j6t3tECnIC0 (Pn p)I i (pnc)sinn: jwt nrrz rCnKi(pnc)I0(11np) p>c = -6W/OP pc, A0 = (2cp.I/a)ni/Ki (pnP)I2(Pnc)sinn4Td-sinn
fq= (nr/a)2 -wape = (nr/a) 2- (277/X)2 . pn= (2rr/x)A/4-(nrria)2 -1
Ans.
-44-(20) From 12.16
= f32 _ crt/ a )2>o,
= (nr/a)2 -02 > 0 if n>1. If X.=a, P?=(2r/a)2 -(r/a)2 =3(Tria) 2 .
If X=2a, 14=0. if X=a, pi=0, so piand 132 are just on the verge of propagation. H, (P io)Ji (plc) sinr; tic2cnKoor,
inonosin -n-;E)e jwt
A00= -6W/bp so that 1-.+z)ejwt _ A 0 --(cill p lit 2) 03 PV itc)sinill n12 PnCriK1 (PnP) Ii(3nc) sin z-, e jwt . p < c, These give A0=A i at P=c ci Hp.)03,0J,(13,0sin7-+ 2 f3nc„K ,03 ,G3no P>t ' W te = ( Ci
Cn is evaluated in problem -43-. Use first term: IBzil - 1Bzo l p f-c- (CA Do(1310 ..710:310- jlio(fif)4 1310 - J0(131C) J1010+141310 Yi(1310Sit114e jcpt which by 5.293 (9) is e jwt 2jf3iCisin7-/ (re) . Multiply by sini1;-, integrate, sinr4§Bds = pi° eiwt sinV -Vsinn--1:?) "-1 K (13np)Ii ((3nc)sinn = -11 C ejwt . So X =-j-rliI0 (c/a)(42)(P i P) J1(P ic)sin-ac n2 1 TTc I 0 rd TT 'rrcp.I0 pi=7,:14-(X/a)2 , Pn =-7.4/(nX/a) 2 -4 Ans. Here CI=
Ans.
-45 -( 2 1)
0, If a ■ .0 but if n=0: C1 = (1+cos7) = 4cos2 (i-rc/a) (2)
Let Bx=B X /C I, B Z=B Z/C1 then .8, sin81 z cosrx+ ci pmCmsinh f:qz cosmrrx cospiz_si rx+ E Cmcosh Orlizsinmrx Bx= a cos= Ans, a coscut Bz-- filicosfq z cosfq d a m=3Cicosh 8;c1 a m=3 f3; Cicosh 86d -61Write Cm for Ampm in 13.06 (9), (10) where m= 2n+1. Thus
E
Bx= ((pi /131)C1cos 13'iz c sc 81d sinria-x + (13/h/l3m)Cmcoshp,hz csch 13'ind sing!) coscut ... m3 mrx, rrx Bz = (Coin f) z csc8;d cos - mE 3cmsinh 13/In z csch8m'd cos7-3cos cut . Assume Bz=0 on the iris and has the static value in the window, where, in the notation of the last problem, Bz =Nrisingsacos0-cos0 if a/2< x cB0) then in the xy plane Ey >vBzand we can choose 13/c =Bz/Eyso 131 vanishes. Thus there is always an accelerating field for the charge in the O'system so it continues to leave the x-axis and never returns. -3Similar to last problem except that cB0>Eco. Consider xy-plane then from 14.16 (4) and (5 Ey ' =x(Ey-vBz), Elz =x(Ezi-vBy), 14= 24(By+ -Ez), Bzr =X(Bz
u
.E)7). Ey -
Jai
Ygn(bia) 1 jjz= 2ITY • To eliminate E' NrTBz =1 ...r.:7 321.3.1/(21715)• Y take v=Ey/Bz=21TV/[ pIrm(b/a)] so B1 -K(1-132)Bz=1cravY.1„.._ q,v7zr_P_L ] dx i . These are the equations of motion for S% d(lcm‘qc) Jr.:niEL 1.).'L 32___ dt -(1' " 2Try dt ' dt " 2ny dt Integrate, putting 1/x= N17-1307=4/c2-vp-v 12 2/c, where vi and NTI2are the initial v; and v;„ velocities for 5. Then 2Trmo(v-v) =c/1-n24(T.:17.712IPni. If vyis zero when y =b, then
vX = nc is entire velocity at p =b, so that 2rm o(n-v1)=AIT/1-
which gives
2 a , where, from 14.02 (3) and 14.02 (4): 217m0(.j1 2+v12 2 -v1)= laq4c 2-v1 2-v 12 2.117132/n-12 1.7= (v 1-cP)/(1-1.71 13/c) and v'2 =v2,17137/(1-v1 13/c). Here vl and v3are the initial velocitie in the x and p directions and 13= cEy/Bz=cEp/F0 • -4-
d(mr2 e) -Bobri vc1;17 (1) From 14.04 m=m0/1-(v/c)2(2) Integrate, dt assuming that
o=o
at r=ro . r26=-Bqb(.4F117-49i71,7)/(mox) (3)
From energy conservation, 14.07 (2), (x1 -1)moc2= qV
r /
(4)
Velocity is constant so at outer cylinder, where r=riand v is all in A direction. v= r1 0= Dil-(1/k)2from (2). Substitute in (3): ric,A17-7 1=(Bqb/m0)(17i-b2-4,/ /17 0) so Xi = 1 "IL 1-b2 4 1 /T1-b2Put )2 (5) (5) into (4) to obtain: mor ic] (N/717
/m ac2 B2 b2 V = M 0 C 2 (X1 - 1 ) = C 2 r2 q
m Dc2 b 212 --
Ans.
Chapter XIV.(XVI)
Page 176.
Page 588.(581) -5-
I
2a Beam moves in x-direction with velocity v. Vq =(x-1)ma c2 by (1 T Va 1+1 1,7 / (2m 0 c2) 14.07 (2) so x =1 +--'-2-, p2 = (x2 -1)/x2 =2Vq . Let S' move with the beam. 'floc K 'floc
) 3 -
He then sees only an electric field cr'/(2rrevp) where cr' is the linear density in his systei His observed acceleration is 1:0=
2 (t-- . The density seen by stationary S is xcr l = a 2r evmaa 1 q I = I/v = I/(c0) so that, from 14.02 (6), since ux' =0, ior x2.2fremaa • xc0 • __AI_ [1+ vci-2 ina.slif;+_is vi_i +-1711 2}..1 Ans. moc J 2Vq 2Tryn a ac 2m 0c2 L mc2.1 ano c 2j - 4TfEva m a V
J
-6-
dr dds e (2) r if S . Central force so d (mr 2 e)ldt = O. mr 2e = po = p. (1) --7 :rib= pr= r2de — 2 Wk 1 (3) where Wk is kinetic energy. Thus if Wt -7 3i Then from 14.07 (2): x 2 = --I +1 -1 [m ac 1 is total energy: W k= Wt - qQ/(4Trevr)=Wt - cps/ (4rrev) (4). But 132 -c20e2+i. 2) m 21.9 2/ r 2 ± m 2i. 2 3 /r )2 +O. 1 S 2 p 2 i-pr2 (13/0 2 + Pr* 02 (6) - ( m 2 cZ / 1.,p2 1 (5) m2c2 m8c2 1-02 - mic2 .
a
Note that m=mj/(1-02 ). Substitute in left side of (6) from (3) and (4), in right side
qQ/ (4rre) 1 2
P del l • Only variables are s,0. - [s2 + — = 1 + 11 de nio l---moc 2 12 rds1 2_ ir qQ 1 -1 ) 52 - 211e mclQp[1 11t-1 12-115-2-[ 1 +---2] . Differentiate with respect to e a +IsieJ _orrEvpc i ma c rly 21 S 4- p 2
from (2). Then: [1+ Wt -
and dixiue out
ds d2s
dO" d0 2 =
inacIQ [1 +-iLt 1 ] -v2 (s-A). Thus s=A+Bcos(s0+8)
1'Tg E2 vpc) 1 s 1- 4revp 2
ma c 2 - '
If s is maximum at 0=0, 8=0, and s =A + BcosyO Ans.
where y 2= 1- [qQ/(4rrevpc)]2 Ans.
-7Each particle travels on spiral on surface of cylinder. In worst i
i
-M2p ir--
case, to get through second hole, particle must make a complete number of loops. Time needed is t = 2nrrp/(vsina)=2nrrn/(Bq) sincot 4:/a my sina/P=Bq. a = t •vcosa= 2ffmnvcosa/(Bq). But cosakrl, sinaAla, Ak.
Ans.
Independent of a if a#0: Maximum beam 1E--- a 4mva aq2aa ; diameter is Ib+4p DAlb +4mva/ (Bq) = -q--2rminv+b- nff +B so B 2.12rmnv/(aq)
.0"
= b +2a sina/(nrr) if a3 terms are neglected. -8-
A
a
Minimum B gives one complete loop between A and B. Write a+ 2d= (aa+b)/a for a in last problem. B 211mva/[q(aa+b)] Ans. Particles going on a helix of radius p will get through if the arc chord for arc length pe is
2 less than b/2. 0 _ 217b e arc 2 0 2Tra + 2d' - a a +b' F- a2 chord 2 4sin2÷0 Ans.
4
Chapter XIV. (XVI)
Page 588. (582)
Page 177-.
-9Let observer move along x-axis. From 14.13 (4) and (5) Eqc =Bx = 0, BYt =x(By+-PEz), Bz=x (Bz-fEy), E)Ic=Ex=0, 4=x(Ey-vBz), El=x(Ez+vBy). But E = A6 [---q--6t1 -11 B =-F6 [Ellej-l so Ey=- 11Z. E =4-- [---2-k-r-1-1 Y -..,y 2rre r 2 ' z -.37 2if r 2 2 I pe Z Z 21TE r 2.,' 02 / _I -6 [ E4n I ll so E -----B Let v=-pei so Ex I =Ey 1 =0 and IC = 1BYThz 2Tr- r2 ' ° z -peI Y' [ utI21 ]
Since v
