Permanent Magnet & Electromechanical Devices: Materials, Analysis, and Applications (Electromagnetism)

Permanent Magnet and Electromechanical Devices Academic Press Series in Electromagnetism Edited by ISAAK MAYERGOYZ, U...

Author: Edward P. Furlani

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Permanent Magnet and Electromechanical Devices

Academic Press Series in Electromagnetism Edited by ISAAK MAYERGOYZ, UNIVERSITY OF MARYLAND, COLLEGE PARK, MARYLAND Electromagnetism is a classical area of physics and engineering that still plays a very important role in the development of new technology. Electromagnetism often serves as a link between electrical engineers, material scientists, and applied physicists. This series presents volumes on those aspects of applied and theoretical electromagnetism that are becoming increasingly important in modern and rapidly developing technology. Its objective is to meet the needs of researchers, students, and practicing engineers.

Books Published in the Series Giorgio Bertotti, Hysteresis in Magnetism: For Physicists, Material Scientists, and Engineers Scipione Bobbio, Electrodynamics of Materials: Forces, Stresses, and Energies in Solids and Fluids Alain Bossavit, Computational Electromagnetism: Variational Formulations, Complementarity, Edge Elements M. V. K. Chari and S. J. Salon, Numerical Methods in Electromagnetism Göran Engdahl, Handbook of Giant Magnetostrictive Materials Vadim Kuperman, Magnetic Resonance Imaging: Physical Principles and Applications John C. Mallinson, Magneto-Resistive Heads: Fundamentals and Applications Isaak Mayergoyz, Nonlinear Diffusion of Electromagnetic Fields Giovanni Miano and Antonio Maffucci, Transmission Lines and Lumped Circuits Shan X. Wang and Alexander M. Taratorin, Magnetic Information Storage Technology

Related Books John C. Mallinson, The Foundations of Magnetic Recording, Second Edition Reinaldo Perez, Handbook of Electromagnetic Compatibility

Permanent Magnet and Electromechanical Devices Materials, Analysis, and Applications

Edward P. Furlani Research Laboratories Eastman Kodak Company Rochester, New York

San Diego San Francisco New York Boston London Sydney Tokyo

This book is printed on acid-free paper. Copyright 2001 by Academic Press All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt, Inc., 6277 Sea Harbor Drive, Orlando, Florida 32887-6777. Explicit permission from Academic Press is not required to reproduce a maximum of two figures or tables from an Academic Press chapter in another scientific or research publication provided that the material has not been credited to another source and that full credit to the Academic Press chapter is given. Academic Press A Harcourt Science and Technology Company 525 B Street, Suite 1900, San Diego, CA 92101-4495, USA http://www.academicpress.com Academic Press Harcourt Place, 32 Jamestown Road, London NW1 7BY, UK http://www.academicpress.com Library of Congress Catalog Card Number: 2001089410 International Standard Book Number: 0-12-269951-3 Printed in the United States of America 01 02 03 04 05 06 EB 9 8 7 6 5 4 3 2 1

This book is dedicated to my parents Edward and Helen for their love and sacrifice, my wife Karen for her patience and encouragement, and my two children Edward and Amanda for the joy they give me.

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Contents

Foreword Preface 1. Materials 1.1 1.2 1.3 1.4

1.5 1.6 1.7 1.8 1.9

1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20

Introduction Units Classification of Materials Atomic Magnetic Moments 1.4.1 Single Electron Atoms 1.4.2 Multielectron Atoms Paramagnetism Ferromagnetism Magnetostatic Energy Demagnetization Field Anisotropy 1.9.1 Magnetocrystalline Anisotropy 1.9.2 Shape Anisotropy Domains Hysteresis Soft Magnetic Materials Hard Magnetic Materials Ferrites Alnico Samarium-Cobalt Neodymium-Iron-Boron Bonded Magnets Magnetization Stability

2. Review of Maxwell’s Equations 2.1 Introduction

xi xiii 1 1 2 6 7 9 14 17 19 23 23 27 27 31 33 36 39 44 46 48 50 51 53 55 66

73 73 vii

viii 2.2 Maxwell’s Equations 2.2.1 Constitutive Relations 2.2.2 Integral Equations 2.2.3 Boundary Conditions 2.2.4 Force and Torque 2.3 Potentials 2.4 Quasi-static Theory 2.5 Static Theory 2.5.1 Magnetostatic Theory 2.5.2 Electrostatic Theory 2.6 Summary

3. Field Analysis

Contents 74 75 77 79 82 82 85 87 87 89 91

97

3.1 Introduction 3.2 Magnetostatic Analysis 3.2.1 Vector Potential 3.2.2 Force and Torque 3.2.3 Maxwell Stress Tensor 3.2.4 Energy 3.2.5 Inductance 3.3 The Current Model 3.4 The Charge Model 3.4.1 Force 3.4.2 Torque 3.5 Magnetic Circuit Analysis 3.5.1 Current Sources 3.5.2 Magnet Sources 3.6 Boundary-Value Problems 3.6.1 Cartesian Coordinates 3.6.2 Cylindrical Coordinates 3.6.3 Spherical Coordinates 3.7 Method of Images 3.8 Finite Element Analysis 3.9 Finite Difference Method

97 97 102 110 112 116 118 126 131 135 141 144 144 153 161 162 175 180 185 190 200

4. Permanent Magnet Applications

207

4.1 Introduction 4.2 Magnet Structures 4.2.1 Rectangular Structures 4.2.2 Cylindrical Structures 4.3 High Field Structures 4.4 Magnetic Latching 4.5 Magnetic Suspension

207 208 208 218 265 268 273

Contents

ix

4.6 4.7 4.8 4.9 4.10 4.11

282 286 306 312 318 325

Magnetic Gears Magnetic Couplings Magnetic Resonance Imaging Electrophotography Magneto-Optical Recording Free-Electron Lasers

5. Electromechanical Devices 5.1 Introduction 5.2 Device Basics 5.3 Quasi-static Field Theory 5.3.1 Stationary Reference Frames 5.3.2 Moving Reference Frames 5.4 Electrical Equations 5.4.1 Stationary Circuits 5.4.2 Moving Coils 5.5 Mechanical Equations 5.6 Electromechanical Equations 5.6.1 Stationary Circuits 5.6.2 Moving Coils 5.7 Energy Analysis 5.8 Magnetic Circuit Actuators 5.9 Axial-Field Actuators 5.10 Resonant Actuators 5.11 Magneto-Optical Bias Field Actuator 5.12 Linear Actuators 5.13 Axial-Field Motors 5.14 Stepper Motors 5.15 Hybrid Analytical-FEM Analysis 5.16 Magnetic MEMS

A. Vector Analysis A.1 A.2 A.3 A.4 A.5 A.6

Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates Integrals of Vector Functions Theorems and Identities Coordinate Transformations

335 335 335 337 338 341 354 354 358 361 363 363 366 376 383 393 402 406 413 421 437 446 455

469 469 473 476 479 485 491

B. Green’s Functions

495

C. Systems of Equations

497

C.1 Euler’s Method

498

x

Contents C.2 Improved Euler Method C.3 Runge-Kutta Methods

502 503

D. Units

509

Index

513

Foreword

This volume in the Academic Press Series in Electromagnetism presents an in-depth, self-contained, and up-to-date treatment of permanent magnets and related electromechanical devices. The emergence of modern rare-earth permanent magnets has led to the tremendous proliferation of devices that utilize these magnets. Currently, there does not exist any book that presents comprehensive treatment of permanent magnet technology and covers material aspects, field analysis, and wide-ranging device applications of permanent magnets. This book, written by Dr. Edward P. Furlani, represents the first and successful attempt to give the exposition of all these issues within one volume. The book reflects the author’s extensive experience as well as his intimate and firsthand knowledge of the area of permanent magnet technology. The unique feature of this book is its broad scope and the proper and delicate balance between theoretical and applied aspects of permanent magnet devices. The book is intended for readers without extensive experience or knowledge in the area of permanent magnets. It reviews the physics of magnetic materials and the basics of electromagnetic field theory. Then, it presents in detail the analysis and design of the wide range of permanent magnet devices such as magnetic gears, couplings, bearings, magnetic resonance imaging devices, free-electron laser magnets, magnetooptical recording systems, permanent magnet motors and stepper motors, and, finally, magnetic microactuators (MEMS). The book contains a large number of practical design examples that are completely worked out. This feature will help readers to follow all the computational and design steps involved in the development of new devices and will be very beneficial for graduate students and practicing engineers. I believe that this book will be a valuable reference for both experts and beginners in the field. Electrical and mechanical enginers, applied physicists, material scientists, and graduate students will find this book very informative. Isaak Mayergoyz Series Editor xi

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Preface

Permanent magnet and electromechanical devices can be found in a wide variety of products that have an impact on our daily lives. Examples of these include audio and video players, telecommunications equipment, personal computers, printers, copiers, automobiles, appliances, power tools, data storage devices, and biomedical apparatus. Furthermore, the use of modern rare-earth magnets is proliferating at a remarkable rate, which is giving rise to new products and markets and stimulating research into a variety of new devices. However, despite the substantial theoretical and practical interest in this technology, there is no single comprehensive text that adequately treats the subject matter required for the research and development of such devices. Instead, one needs to extract crucial information from a variety of texts, each with its own notation and style and emphasis of presentation. It is hoped that this book will eliminate the need for this cumbersome approach. This book is a comprehensive and self-contained exposition of the theory and methods used in the analysis and design of permanent magnet and electromechanical devices. It is intended to be used as both a text and a reference for researchers, professors, students, and engineers who are engaged in the research and development of new and conventional permanent magnet and electromechanical devices. It is written to provide the reader with a first-principles understanding of the underlying theory of such devices, and to enable him or her to analyze, design, and simulate practical devices. The salient features of the book are as follows:

· Extensive discussion of magnetic materials including the physical

·

principles of magnetism, domains, hysteresis, soft and hard magnetic materials and their properties, and the magnetization and stability of permanent magnets. Comprehensive presentation of analytical and numerical methods for the analysis of steady currents, permanent magnets, and xiii

xiv

Preface

· · · · · · · ·

magnetic circuits. Topics include magnetostatic analysis, the Maxwell stress tensor, the current model, the charge model, magnetic circuit analysis, boundary-value theory, finite element analysis, and finite difference analysis. Over 60 examples of practical permanent magnet and electromechanical applications with detailed solutions. Analytical analysis and design formulas for the field distributions of the most common rare-earth permanent magnet structures. Analytical analysis and design formulas for the performance of rare-earth magnetic couplings, gears, and bearings. Comprehensive presentation of the theory of electromechanical devices with numerous solved examples. Analytical analysis and design formulas for linear and rotational actuators. Analytical analysis and design formulas for permanent magnet brushless dc motors and stepper motors. Presentation of a hybrid analytical FEM approach for the analysis of electromechanical devices. Introduction to magnetic microactuators.

This book consists of five chapters and four appendices. The organization of the material is illustrated in the following diagram. Materials Review of Magnetic Field Analysis Theory $ Maxwell’s $ Analytical and Properties Equations Numerical Methods Appendices Permanent Vector Analysis Electromechanical $ Magnet $ $ Green’s Functions Devices Devices ODE Systems Units Book Organization

The first three chapters provide a theoretical background, while the final two chapters emphasize the analysis of practical devices. The first chapter covers magnetic materials. It begins with a brief review of the different units of magnetism, followed by a summary of the major classifications of materials. This is followed by a discussion of the basic mechanisms of magnetism starting at the atomic level with an analysis of an isolated single electron atom. The atomic magnetic mo-

Preface

xv

ment is principally due to the orbital and spin motion of the electron. The discussion is extended to multielectron atoms where the spin and orbital moments of the constituent electrons couple to give a net atomic moment. The theory is then generalized to the case of an ensemble of atoms. If the atomic moments are noninteracting, the ensemble exhibits a bulk paramagnetic behavior. If the atomic moments are interacting, there is a cooperative alignment of the moments (ferromagnetism) below the Curie temperature. This ferromagnetic behavior is combined with the concept of magnetic domains to explain the B-H (hysteresis) curves of bulk magnetization. A survey is given of commercially available soft and hard magnet materials and their properties. The chapter concludes with a discussion of the magnetization and stability of permanent magnets. The second chapter contains a brief but thorough review of Maxwell’s equations. The field equations are presented in both differential and integral form along with the constitutive relations and boundary conditions. The scalar and vector potentials are introduced and shown to provide an alternate and often more tractable formulation of field theory. Quasi-static field theory is discussed. Here, there is a partial uncoupling of the field equations that simplifies their solution. Static field theory is also discussed, and it is shown that the magnetic and electric fields uncouple into separate magnetostatic and electrostatic field equations. The third chapter builds upon the second. It consists of a comprehensive and systematic treatment of the various analytical and numerical methods used for the analysis of steady currents, permanent magnets, and magnetic circuits. Magnetostatic field theory is presented in some detail, including a discussion of force, torque, energy, and inductance. The current and charge models for magnetic materials are introduced. These are used to reduce a permanent magnet to an equivalent source term that can be analyzed using the field equations. Magnetic circuits are discussed, and the concept of reluctance is introduced and used to transform a physical circuit into an equivalent lumped-parameter circuit. Various analysis methods are also discussed, including boundary-value theory, the method of images, finite element analysis (FEA), and the finite difference method. These methods are demonstrated via solved examples. The final two chapters cover permanent magnet applications and electromechanical devices, respectively. These chapters follow a problem solving approach in which numerous practical examples are worked out in sufficient detail to be of immediate use to researchers and practitioners in the field. Many of the examples contain design formulas that cannot be found in any other text.

xvi

Preface

The fourth chapter deals with permanent magnet applications. These include bias magnet structures, high field structures, latching magnets, magnetic suspensions, and magnetic gears and couplings. Analytical methods are used to derive design formulas for many of these applications. Specifically, analytical formulas are derived for the most common rare-earth permanent magnet structures, and the performance of rareearth gears, couplings, and bearings In addition, various miscellaneous applications that utilize magnets are discussed, including magnetic resonance imaging (MRI), electrophotography, magneto-optical recording, and free-electron lasers. The fifth chapter covers electromechanical devices, with an emphasis on magnetically linear, singly excited devices with a single degree of mechanical freedom. Quasi-static field theory is used to derive Kirchhoff’s circuit laws for actuators with stationary circuits, and circuits with coils that move through an external magnetic field. The circuit laws are combined with the laws of rigid body dynamics to obtain the coupled electromechanical equations that govern linear and rotational actuators. The theory is applied to numerous practical devices, including magnetic circuit actuators, linear and rotary actuators, permanent magnet direct current (dc) motors, and stepper motors. The chapter concludes with a brief introduction to magnetic microactuators. In addition to the five chapters, there are four appendices. Appendix A contains a review of vector analysis. The topics covered include vector calculus in the Cartesian, cylindrical, and spherical coordinate systems, line and surface integrals, Stokes’ theorem, the Divergence theorem, various vector identities, and coordinate transformations. In Appendix B the free-space Green’s functions for Poisson’s equation are introduced. These are used to obtain the field distributions of free-standing permanent magnet structures. Appendix C contains a discussion of various numerical methods that are used for the solution of first-order initialvalue problems. Specifically, the Euler, modified Euler, and Runge-Kutta methods are discussed. These are used for the analysis and design of electromechanical devices. Last, Appendix D contains a summary of the different systems of units and conversion factors that are used in the analysis of permanent magnet and electromechanical devices.

Acknowledgments Finally, I would like to acknowledge and thank the individuals who assisted me in the preparation of this book. First, I would like to thank

Preface

xvii

Svetlana Reznik for many insightful discussions on soft and hard magnetic materials. I am also grateful for the patient and diligent work of Christopher Devries in the preparation of numerous drawings and figures. Special thanks go to Dr. Isaak Mayergoyz for his encouragement and support during this project. I would also like to thank Marsha Filion, Julie Bolduc, and the rest of the editorial and production staff at Academic Press for their help and guidance with the preparation of the manuscript. Last, I would like to thank the research managers at the Eastman Kodak Company for their support. Edward P. Furlani

a


CHAPTER

1

Materials

1.1 INTRODUCTION In this chapter we study the principles and properties of magnetic materials. Our study begins with a brief review of the basic units of magnetism. This is followed by a summary of the major classifications of magnetic materials. Next, we consider the basic mechanisms of magnetism. We start at the atomic level with an analysis of an isolated single electron atom. Our analysis shows that the electron’s magnetic moment is due to both its orbital motion and spin. The same analysis is then applied to an isolated multielectron atom where the spin and orbital moments of the constituent electrons couple to give a net atomic moment. After discussing single atoms we consider a collection of atoms. Specifically, we apply classical statistical mechanics to determine the bulk magnetization of an ensemble on noninteracting atomic moments. The ensemble exhibits a bulk paramagnetic behavior. A similar analysis is applied to an ensemble of interacting atomic moments. In this case, there is a cooperative alignment of the moments (ferromagnetism) below the Curie temperature. Following this, we introduce the concepts of magnetostatic energy, magnetic anisotropy and domains, and then use these to explain the B-H (hysteresis) curves of bulk magnetization. After describing bulk magnetization, we discuss soft and hard magnetic materials. For soft materials, we give a brief description of the most common materials and list their key physical properties. This is followed by a survey of commercially available hard materials (permanent magnets). The chapter concludes with a discussion of the magnetization and stability of permanent magnets. The topics covered in this chapter are summarized in Fig. 1.1.

1

2

CHAPTER 1 Materials

FIGURE 1.1

Organization of chapter topics.

1.2 UNITS There are three systems of units that are commonly used in the magnetism. These are the CGS or Gaussian system, and two MKS or SI systems that are referred to as the Kennelly and Sommerfeld conventions, respectively. Throughout this book we use SI units in the Sommerfeld convention [1, 2]. In the Sommerfeld convention the magnetic flux is in webers (Wb), the flux density B is in teslas (T) or Wb/m, and both the field strength H and magnetization M are in A/m. The units for the SI (Sommerfeld) and CGS systems are as follows: Symbol

Description

SI

CGS

H

Magnetic Field Strength

A/m

Oe

B

Flux Density

Tesla

Gauss

M

Magnetization

A/m

emu/cm

Flux

Webers

Maxwells

(1.1)

3

1.2 UNITS

The conversion factors for these systems are 1 Oe : 1000/4 A/m 1 Gauss : 10\ T 1 emu/cm : 1000 A/m

(1.2)

1 Maxwell : 10\ Webers Additional conversion factors are discussed in Appendix D. We can obtain an intuitive feeling for the magnitude of the fields (units) described in the foregoing text by considering the following physical examples: 1. Field strength H: Consider a straight, infinitely long wire carrying a current i : 2A. The conductor generates a tangential field strength H : 1 A/m at a radial distance r : 1 m from its center. As another example, consider a long solenoid that has n turns per meter and carries a current of 1/n A. It generates a field strength H : 1 A/m along its axis. 2. Flux density B: Consider an infinitely long conductor that carries a current i : 1 A perpendicular to an external B-field. A force of 1 newton will be imparted to each meter of the conductor when B : 1 T. 3. Flux : Consider a single turn coil with 1 Wb of magnetic flux passing through it. One volt will be induced in the coil when the flux is uniformly reduced to zero in one second. The fundamental element in magnetism is the magnetic dipole. This can be thought of as a pair of closely spaced magnetic poles (Section 3.4.1), or equivalently as a small current loop (Section 1.4) (Fig. 1.2). A magnetic dipole has a magnetic dipole moment m. In the Sommerfeld convention m is measured in A · m. In the Kennelly and CGS systems it is measured in Wb · m and emu, respectively (1 emu : 4 ; 10\ Wb · m). Magnetization M is a measure of the net magnetic dipole moment per unit volume. Specifically, it is given by m M : lim G G , V 4 where m is a vector sum of the dipole moments contained in the G G elemental volume V.

4

CHAPTER 1 Materials

FIGURE 1.2 Magnetic dipole: (a) magnetic charge model and H-field; and (b) current loop and B-field.

If a magnetic dipole is subjected to an external B-field, it acquires an energy E : 9m · B (1.3) and experiences a torque T : m ; B.

(1.4)

In both the Kennelly and CGS systems these are given by E : 9m · H and T : m ; H. In the Sommerfeld convention the fields are related by the constitutive relation B : (H ; M)

(Sommerfeld convention),

(1.5)

5

1.2 UNITS

where : 4 ; 10\ T m/A is the permeability of free space. In the Kennelly convention, the constitutive relation is B: H;J (Kennelly convention), where J is called the magnetic polarization and is measured in tesla. Notice that J : M. The constitutive relation (1.5) simplifies for linear, homogeneous and isotropic media. For such materials, both B and M are proportional to H. Specifically, B : H,

(1.6)

and M : H, (1.7) K where and are the permeability and susceptibility of the material, K respectively. These coefficients are related to one another. From Eqs. (1.5), (1.6), and (1.7) we find that or

: ( ; 1), K

(1.8)

: 9 1. (1.9) K The constitutive relations (1.6) and (1.7) need to be modified for nonlinear, inhomogeneous or anisotropic materials. A material is magnetically nonlinear if depends on H, otherwise it is linear. For nonlinear materials, Eqs. (1.6) and (1.7) become B : (H)H, and M : (H)H, K respectively. A material is inhomogeneous if is a function of position, otherwise it is homogeneous. In inhomogeneous materials the permeability and susceptibility are functions of the coordinate variables, : (x, y, z), : (x, y, z). Last, a material is said to be anisotropic if K K depends on direction, otherwise it is isotropic. In anisotropic materials Eq. (1.6) generalizes to B : H ; H ; H , V V W X B : H ; H ; H , W V W X B : H ; H ; H . X V W X A similar set of equations hold for Eq. (1.7).

6

CHAPTER 1 Materials

1.3 CLASSIFICATION OF MATERIALS Magnetic materials fall into one of the following categories: diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, and ferrimagnetic (Fig. 1.3). Diamagnetic materials have no net atomic or molecular magnetic moment. When these materials are subjected to an applied field, atomic currents are generated that give rise to a bulk magnetization that opposes the field. Bismuth (Bi) is an example of a diamagnetic material. Paramagnetic materials have a net magnetic moment at the atomic level, but the coupling between neighboring moments is weak. These moments tend to align with an applied field, but the degree of alignment decreases at higher temperatures due to the randomizing effects of thermal agitation. Ferromagnetic materials have a net magnetic moment at the atomic level, but unlike paramagnetic materials there is a strong coupling between neighboring moments. This coupling gives rise to a spontaneous alignment of the moments over macroscopic regions called domains. The domains undergo further alignment when the material is subjected to an applied field. Finally, antiferromagnetic and ferrimagnetic materials have oriented atomic moments with neighboring moments antiparallel to one another.

FIGURE 1.3

Classifications of magnetic materials [16].

1.4 ATOMIC MAGNETIC MOMENTS

7

In antiferromagnetic materials the neighboring moments are equal, and there is no net magnetic moment. In ferrimagnetic materials the neighboring moments are unequal, and there is a net magnetic moment. Rigorous treatments of these materials require a full quantum mechanical analysis that is beyond the scope of this text. We are primarily interested in ferromagnetic materials, and for our purposes it is sufficient to review some of the key principles and results. The interested reader can find more detailed presentations in numerous texts [1—7].

1.4 ATOMIC MAGNETIC MOMENTS In this section we study the magnetic moment of an isolated single electron atom. We start with a classical analysis. Consider an isolated atom with an electron of mass m and charge e moving in a circular orbit C of radius r with a linear velocity u (angular velocity : u/r) as shown in Fig. 1.4. The electron completes an orbit in : 2/ s and has an orbital angular momentum L given by L:m r;u C : m r. (1.10) C The orientation of L is given by a right-hand rule in which the fingers of the right hand curl in the direction of the particle motion, and the thumb points in the direction of L (Fig. 1.4). The orbiting electron defines a closed loop of current i where e i: .

FIGURE 1.4

Atomic system.

8

CHAPTER 1 Materials

The current i defines the flow of positive charge and therefore circulates in a direction opposite to the motion of the electron (Fig. 1.4). The current loop gives rise to a magnetic dipole moment (Fig. 1.2b). In general, if a current i circulates around an enclosed area ds, it gives rise to a magnetic dipole moment m, m : ids

(magnetic moment).

(1.11)

The vector ds defines the orientation of m relative to the circulation of i. This is given by the right-hand rule, which states that if the fingers of the right hand follow the direction of current, then the thumb points in the direction of ds (Fig. A7). The magnitude of the moment is m : ids,

(1.12)

and is measured in units of A · m. Let m denote the magnetic moment of the electron due to its orbital * motion. The magnitude of this moment is given by Eq. (1.12) m : ir * e : r e : r. 2 This can be expressed in terms of L. Specifically, from Eq. (1.10), and the rules for determining the orientations of m and L, we find that * e m :9 L. (1.13) * 2m C Notice that m is antiparallel to L as shown in Fig. 1.4. Amazingly, Eq. * (1.13) holds even at the atomic level when L is replaced by its quantized expression. If the orbiting electron is subjected to an external B-field, its moment m acquires an energy * E : 9m · B * e : L · B, 2m C and experiences a torque T:m ;B * e :9 L ; B. 2m C

9

1.4 ATOMIC MAGNETIC MOMENTS

FIGURE 1.5

Precession of a magnetic moment about an applied field.

As shown in Fig. 1.5, the torque causes L to precess around B. In classical mechanics, L can take on a continuum of orientations relative to B. However, when quantum effects are taken into account, the orientation of L is restricted to a discrete set of values as we shall see.

1.4.1 Single electron atoms Atomic systems are governed by quantum theory with electronic states specified in terms of wavefunctions [8, 9]. The wavefunctions are obtained by solving Schrödinger’s equation

9

; V : E , 2m

where m is the mass of the orbiting particle, V is the potential energy, and E is the total energy. The wavefunctions are indexed by a discrete set of quantum numbers (n, l, m , m ) that distinguish the allowable J Q energy levels, orbital configurations, etc. The principle quantum number is n and this specifies the energy of a given orbit or shell. Observables such as energy and angular momentum are represented by operators that act on . The observed value L of an observable L for a given state

10

CHAPTER 1 Materials

(n, l, m , m ) is obtained by computing the expectation value of its J Q operator L , L

:

*(n, l, m , m )L (n, l, m , m ) dv, J Q J Q

(1.14) where * is the complex conjugate of . It follows from Eq. (1.14) that observables are also quantized and indexed. We now discuss the angular momentum of an atomic electron. Consider a hydrogenlike single electron atom. There are two contributions to the angular momentum: one due to orbital motion and the other due to spin. From quantum mechanics we know that for a given value of n the electron’s orbital angular momentum L has a magnitude

L : [l(l ; 1)] .

(1.15)

In Eq. (1.15), l is the orbital angular momentum quantum number with allowable values l : 0, 1, 2, . . . , (n 9 1), and : h/2 where h is Planck’s constant (h : 6.6260755 ; 10\ J s) [3]. Electrons with l : 0, 1, 2, 3, . . . are referred to as s, p, d, f, . . . electrons, respectively. For example, the n : 1 shell has an s electron, the n : 2 shell has s and p electrons, etc. Notice that the magnitude of L is restricted to a discrete set of values. This is in contrast to classical theory, which allows for a continuum of values. The orientation of L is also restricted. Specifically, its projection onto a given axis (say, the z-axis) is given by L : m , (1.16) X J where m is called the magnetic quantum number, and is restricted to the J values m : l, (l 9 1), . . . , 0, . . . , 9 (l 9 1), 9l. Thus, for a given value of J l there are 2l ; 1 possible orientations of L relative to the axis (Fig. 1.6a). An electron also has a spin angular momentum. The concept of spin was originally proposed to explain the multiplet structure of atomic spectra. Initially, it was thought that spin could be explained in terms of the electron spinning about an internal axis. However, predictions based on this model do not yield the correct magnetic moment. Therefore, spin is viewed as a purely quantum mechanical phenomenon. The spin angular momentum S is quantized and has a magnitude S : [s(s ; 1)] , where s : 1/2. The projection of S along a given axis is also quantized with S : m , (1.17) X Q where m : L KL K\L ; KL> KL KL\ : . (3.319) KL x y This can be further simplified by choosing x : y : h, which yields

K>L

; ; ; 9 4 : h . K\L KL> KL\ KL KL

(3.320)

Equation (3.320) represents a system of linear equations. To solve this system, we need to represent the unknown nodal values in the form of a column matrix (vector). To this end, we introduce a new index p Y m(N ; 1) ; n for m : 0, 1, . . . , M, and n : 0, 1, . . . , N. Using this index, Eq. (3.320) becomes ; ; ; 9 4 : h . N>,> N\,> N> N\ N N

(3.321)

This expression is only valid at the interior points m : 1, 2, . . . , M 9 1, and n : 1, 2, . . . , N 9 1, with ; and ; . It is K,>>L KL K,>>L KL instructive to examine the vector : ' , ,> ,> [] : ' ,>, ,> ' ' +,>,

(3.322)

204

CHAPTER 3 Field Analysis

In Eq. (3.322), the boundary values are identified by m : 0 $ ( , , . . . , ) (x : 0), , m : M $ ( , , . . . , ) (x : L), +,>> +,>> +,>>, n : 0 $ ( , , , . . . , ) (y : 0), +,> ,> ,> and n : N $ ( , , , . . . , ) (y : H). +,>>, , ,>>, ,>>, As a final step, assemble Eq. (3.321) into matrix form: (3.323) [A] · [ ] : [(], where [A] is a sparse matrix (many zero elements), [ ] is the vector of unknown nodal values at the interior points, and [(] is a vector containing the nodal source values and the known boundary condiN tions. There are many approaches to solving Eq. (3.323), and the most popular ones can be broadly classified as iterative methods and direct matrix methods. The iterative methods involve making an initial guess [ ], and then solving iteratively for successively more accurate solutions [ ]I. The direct matrix methods entail the direct solution of Eq. (3.323). For example, if [A] can be inverted, the solution is given by the following: [ ] : [A]\[(]. A detailed description of the various solution methods is beyond the scope of this book. However, they are described in detail in several texts [17, 18]. References 1. Moon, F. C. (1984). Magneto-solid Mechanics, New York: Wiley. 2. Lowther, D. A. and Silvester, P. P. (1986). Computer-aided Design in Magnetics, New York: Springer-Verlag. 3. Woodson, H. H. and Melcher, J. R. (1985). Electromechanical Dynamics Part II: Fields, Forces and Motion, Malabar, FL: John E. Krieger Publishing Co. 4. Johnk, C. T. A. (1988). Engineering Electromagnetic Fields and Waves, 2nd ed., New York: John Wiley and Sons. 5. Morrish, A. H. (1983). The Physical Principles of Magnetism, Malabar, FL: R. E. Krieger Publishing Co. 6. Craik, D. (1995). Magnetism: Principles and Applications, New York: John Wiley and Sons. 7. Reitz, J. R., Milford, F. J., and Christy, R. W. (1980). Foundations of Electromagnetic Theory, 3rd ed., Reading, MA: Addison-Wesley.

3.9 FINITE DIFFERENCE METHOD

205

8. McCaig, M. and Clegg, A. G. (1987). Permanent Magnets in Theory and Practice, 2nd ed., New York: John Wiley and Sons. 9. Zahn, M. (1987). Electromagnetic Field Theory: A Problem Solving Approach, Malabar, FL: R. E. Krieger Publishing Co. 10. Silvester, P. P. and Ferrari, R. L. (1990). Finite Elements for Electrical Engineers, 2nd ed., New York: Cambridge Univ. Press. 11. Sabonnadiere, J. C. and Coulomb, J. L. (1987). Finite Element Methods in CAD: Electrical and Magnetic Fields, New York: Springer-Verlag. 12. Hoole, S. R. H. (1989). Computer-aided Analysis and Design of Electromagnetic Devices, New York: Elsevier Science Publishing Co. 13. Salon, S. J. (1995). Finite Element Analysis of Electrical Machines, Boston: Kluwer Academic Publishers. 14. Sykulski, J. K. (ed.) (1995). Computational Magnetics, New York: Chapman & Hall. 15. Sadiku, M. N. O. (1992). Numerical Techniques in Electromagnetics, Boca Raton, FL: CRC Press. 16. Furlani, E. P. (1994). Analysis of a sintered NdFeB magnetic lens, J. Magn. Magn. Mat. 134: 117. 17. Smith, G. D. (1985). Numerical Solution of Partial Differential Equations: Finite Difference Methods, 3rd ed., Oxford Applied Mathematics and Computing Science Series, New York: Oxford University Press. 18. Ames, W. F. (1992). Numerical Methods for Partial Differential Equations, 3rd ed. New York: Academic Press.

a


CHAPTER

4

Permanent Magnet Applications

4.1 INTRODUCTION Permanent magnets in one form or another can be found in a wide variety of equipment ranging from consumer products, to industrial machinery, to research apparatus. In particular, they are used in audio/ video equipment, personal computers, printers, copiers, automobiles, household appliances, power tools, industrial motors and generators, and various biomedical apparatus. Permanent magnets have several advantages over conventional (current-driven) electromagnets. The fundamental advantage is that they can provide a relatively strong magnetic field over an extended spatial region for an indefinite period of time with no expenditure of energy. Of course, the field they provide is fixed whereas the field of an electromagnet can be changed by adjusting the current. Another advantage of permanent magnets is that they can be fabricated with a wide range of structural properties, geometric shapes, and magnetization patterns. They are also relatively inexpensive on a per unit basis depending on the material used. Permanent magnets have an additional advantage over electromagnets in that their performance scales well with size. Specifically, if we change all the linear dimensions L of an electromagnet, while keeping the field strength at all the rescaled observation points fixed, the current density must be adjusted by a factor of 1/L (Section 5.16). Therefore, as the size of an electromagnet is reduced, a point is reached where overheating precludes viability. Moreover, if the dimensions of the conductors are increased to compensate for this, the field strength at the

207

208

CHAPTER 4 Permanent Magnet Applications

rescaled observation points will decrease. By comparison, if we change all the linear dimensions of a permanent magnet, the field strength at all the rescaled observation points remains constant (assuming the magnetization is constant). Thus, there will be a dimension below which a permanent magnet will be the only viable field source. In this chapter we discuss various permanent magnet applications. We consider applications in which a magnet is used alone as a passive field source or in conjunction with other magnets as part of a functional device. Specifically, we discuss bias magnet structures, high field structures, latching magnets, magnetic suspensions, and magnetic gears and couplings. We also discuss various miscellaneous applications that utilize magnets including MRI, magnetooptical recording, electrophotography and free-electron lasers. Throughout the chapter, we present solved examples of various applications. Many of these contain formulas that are useful for design and optimization.

4.2 MAGNET STRUCTURES Bias magnets are used to provide a prescribed field distribution over a given region. Such magnets are usually either rectangular or cylindrical in shape and have bipolar or multipole magnetization patterns. In this section we study the most common bias magnet structures. We derive field solutions for these structures and demonstrate the solutions with practical calculations.

4.2.1 Rectangular structures The most common permanent magnet structure is the rectangular bar magnet. We developed a limited field solution for this structure in Section 3.4, where we used the charge model to obtain the B-field above the magnet along a line of symmetry (Example 3.4.1). In this section, we derive general field solutions for two different rectangular structures, an infinitely long magnet and a finite bar magnet. The field solutions are derived in the following examples. EXAMPLE 4.2.1 Determine the B-field outside a long rectangular magnet (Fig. 4.1). Assume that the magnet has a magnetization M : M y . Q

209

4.2 MAGNET STRUCTURES

Rectangular magnet.

FIGURE 4.1

SOLUTION 4.2.1 Assume that the magnet is infinitely long. In this case the analysis reduces to a 2D field problem. Choose a reference frame as shown in Fig. 4.2a. We use the current model to determine the field (Section 3.3). First, we reduce the magnet to an equivalent current distribution. We apply Eq. (3.95) and find that the volume current density is zero, J : ;M : 0. To determine K the surface current density j we need to identify the unit surface normals. K These are n : and n :

y (y : h) 9y (y : 9h)

x (x : w) 9x (x : 9w).

As j : M ; n and M : M y , we find that K Q 9M z (x : w) Q j : K M z (x : 9w). Q Thus, the magnet reduces to two parallel infinitely long current sheets of height 2h as shown in Fig. 4.2b. The field solution for a current sheet was derived in Example 3.2.5. We repeat it here for convenience,

K x ; (y 9 h) 2hx B (x, y) : ln x ; 2 tan\ y . 4 x ; (y ; h) x ; y 9 h (4.1) Here, K is the surface current density (A/m). The solution (4.1) applies to a current sheet at x : 0. However, for the magnet there are two current sheets,

210


FIGURE 4.2 Infinite rectangular magnet: (a) cross-sectional view with reference frame; and (b) equivalent surface currents.

one at x : 9w, and another at x : w. We obtain the field of the magnet by taking a superposition, B(x, y) : B (x ; w, y) 9 B (x 9 w, y).

(4.2)

Notice that the minus sign in the second term reflects the fact that the current at x : w is flowing in the negative z-direction (into the page as shown in

211


Fig. 4.2b). We evaluate Eq. (4.2) and obtain the field components,

M (x ; w) ; (y 9 h) (x 9 w) ; (y 9 h) B (x, y) : Q ln 9 ln V 4 (x ; w) ; (y ; h) (x 9 w) ; (y ; h)

(4.3)

and

M 2h(x;w) 2h(x9w) B (x, y) : Q tan\ 9 tan\ W 2 (x;w);y9h (x9w);y9h

(4.4) where we have substituted K : M . Equations (4.3) and (4.4) give the B-field Q at any point outside the magnet. Calculations: We apply Eqs. (4.3) and (4.4) to a magnet with w : 10 mm, h : 5 mm, and M : 8.0; 10 A/m. This value of magnetization corresponds Q to sintered NdFeB with a remanence of B : 10,054 G. We evaluate B and B P V W along two different horizontal lines that span 92w x 2w. These are located 0.5 and 2.0 mm above the magnet, respectively (Fig. 4.3). Notice that closer to the magnet (0.5 mm above the surface), the vertical component B is higher near W the ends of the magnet than at its center (x : 0). However, farther away the opposite is true. The reason for this is as follows: Near the magnet, the field is predominantly vertical across the entire surface and has a greater magnitude closer to the current sheets, which are at the ends of the magnet. However, as you move away from the magnet, the vertical projection of the field decreases dramatically near the ends of the magnet and this results in a peak value in the center. Finally, notice that the horizontal component B peaks above the current V sheets (x : <w) as expected. ) EXAMPLE 4.2.2 Determine the B-field outside a rectangular bar magnet (Fig. 4.4a). Assume that the magnetization is M : M z . (4.5) Q SOLUTION 4.2.2 Choose a coordinate system centered with respect to the magnet (Figs. 4.4b and 4.5). Let (x , x ), (y , y ) and (z , z ) denote the positions of the edges of the magnet with respect to the x-, y-, and z-axes. We use the charge model to determine the field (Section 3.4). First, reduce the magnet to an equivalent charge distribution. From Eqs. (3.105) and (4.5) we find that the volume charge density is zero, : 9 · M : 0. However, there K is a surface charge density

M (z : z ) Q : K 9M (z : z ). Q

(4.6)

212


The components B and B vs x: (a) y : h ; 0.5 mm; and (b) V W y : h ; 2 mm.

FIGURE 4.3

The B-field follows from Eq. (3.107), M B(x, y, z) : Q (91)I 4 I W V [(x 9 x)x ; (y 9 y)y ; (z 9 z )z ] dx dy I ; . [(x 9 x) ; (y 9 y) ; (z 9 z )] W V I

(4.7)

213


FIGURE 4.4

Rectangular magnet: (a) geometry and polarization; and (b) refer-

ence frame.

The x-component: B follows from Eq. (4.7), V M W V (x 9 x) dx dy B (x, y, z) : Q (91)I . V 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I Integration with respect to x gives

M W dy B (x, y, z) : Q (91)I>K . V 4 [(x9x );(y9y);(z9z )] I K W K I

214


FIGURE 4.5 Cross-sectional views of a bar magnet with reference frame: (a) x-y plane; and (b) x-z plane.

The remaining y integration can be evaluated by making a change of variable to : y 9 y. The resulting field expression is M B (x, y, z) : Q (91)I>K ln[F(x, y, z, x , y , y , z )], V K I 4 I K

(4.8)

where (y 9 y ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I F(x, y, z, x , y , y , z ) : . K I (y 9 y ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I

215


The y-component: B also follows from Eq. (4.7), W M W V (y 9 y) dx dy B (x, y, z) : Q (91)I . W 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I Integration with respect to y yields

M V dx B (x, y, z) : Q (91)I>K W 4 [(x 9 x) ; (y 9 y ) ; (z 9 z )] K I I K V . The remaining x integration is evaluated using a change of variable : x 9 x. The resulting field expression is M B (x, y, z) : Q (91)I>K ln[H(x, y, z, x , x , y , z )]. K I W 4 I K

(4.9)

where (x 9 x ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I . H(x, y, z, x , x , y , z ) : K I (x 9 x ) ; [(x 9 x ) ; (y 9 y ) ; (z 9 z )] K I The z-component: B is given by X M W V (z 9 z ) dx dy I B (x, y, z) : Q (91)I . X 4 [(x 9 x) ; (y 9 y) ; (z 9 z )] I W V I The x integration is performed using a change of variable : x 9 x,

M W V\V z9z I B (x, y, z) : 9 Q d dy X [ ; (y 9 y) ; (z 9 z )] 4 W V\V I M : Q (91)I>L>(z 9 z )(x 9 x ) I L 4 I L W dy ; , (4.10) W [(y 9 y) ; (z 9 zI)]((y 9 y) ; b

where b : (x 9 x ) ; (z 9 z ). The remaining y integration is performed L I using a change of variable : y 9 y. This gives M B (x, y, z) : Q (91)I>L>K X 4 I L K (x 9 x )(y 9 y ) L K g(x, y, z; x , y , z ) , (4.11) ;tan\ L K I (z 9 z ) I

216


FIGURE 4.6

Bar magnet.

where 1 g(x, y, z; x , y , z ) : . L K I [(x 9 x ) ; (y 9 y ) ; (z 9 z )] L K I The formulas (4.8), (4.9) and (4.11) give the field at any point outside the magnet. Calculations: We apply the theory to a magnet with a : 10 mm, b : 20 mm, and c : 10 mm. We set x : 0, x : a, y : 0, y : b, z : 9c, z : 0 mm, and M : 8.0;10 A/m (Fig. 4.6). We evaluate B across two different surfaces Q X above the magnet, one at z : 0.5 and the other at z : 2.0 mm. The computed data are shown in Fig. 4.7. Notice that closer to the magnet, B dips near the X center of the magnet, but farther away it peaks there. This can be understood from the current model. Specifically, the magnet can be reduced to equivalent current sheets along its edges. Near the magnet the field is predominantly vertical and has a greater magnitude closer to the current sheets, which are at the edges. However, as you move away from the magnet, the vertical projection of the field decreases dramatically near the ends of the magnet and this results in a peak value at the center. ) The field solution derived in Example 4.2.2 can be used to determine the field of multipole rectangular structures such as the quadrupole structure shown in Fig. 4.8. This structure consists of four block magnets and the field is given by B(x, y, z) : B (x, y, z), G G where B (x, y, z) is the field due to the ith block. G


FIGURE 4.7

at z : 2 mm.

217

Magnetic field above a bar magnet: (a) B at z : 0.5 mm; and (b) B X X

218


FIGURE 4.8

Quadrupole magnet structure.

4.2.2 Cylindrical structures Cylindrical magnets are the second most common magnet structure. They are used in devices such as encoders, actuators, motors, electrophotographic copiers, and magnetic couplings and gears. Cylindrical magnets can have various magnetization patterns including bipolar with magnetization either through the cross section or along the axis, and multipole with radial or axial magnetization (Fig. 4.9). In Sections 3.3 and 3.6.2. we developed partial field solutions for the bipolar cylindrical magnets (Examples 3.3.1 and 3.6.3). In this section we consider the full range of magnetization patterns and derive general field solutions that

Cylindrical magnets: (a) bipolar magnetization; (b) multipole radial magnetization; and (c) multipole axial magnetization. FIGURE 4.9

219


FIGURE 4.10

Bipolar cylinder: (a) geometry and magnetization; and (b) refer-

ence frame.

are valid at any point outside the magnet. The field solutions are derived in the following examples. EXAMPLE 4.2.3 Derive an expression for the B-field outside a bipolar cylindrical magnet of finite length (Fig. 4.10) [1]. Assume that the axis of the magnet is along the z-axis. Let R denote it radius, and let z and z denote the positions of its bottom and top edge, respectively. Furthermore, assume that the magnet has a magnetization M : M x . Q

220


SOLUTION 4.2.3 We use the charge model of Section 3.4. First, we reduce the magnet to an equivalent charge distribution and then determine the field by treating the charge distribution as a field source. The equivalent charge distribution follows from Eq. (3.105). We find that the volume charge is density zero, : 9 · M : 9 · M x : 0, and that the surface charge density is K Q : M · n K : M x · r Q : M cos(). (4.12) Q The B-field is computed using

X L M cos() Q B(x) : 9 R d dz, 4

x 9 x X where acts on the unprimed variables. Radial component: The radial component is given by

M R X L B (r, , z) : Q cos()[r 9 R cos( 9 )] P 4 X ; g(r, , z; R, , z) d dz,

(4.13)

where g(r, , z; r, , z) :

1 (r ; r 9 2rr cos( 9 ) ; (z 9 z)

.

We evaluate the integration numerically using Simpson’s method and the z integration in closed form. We obtain M R , B (r, , z) : Q (91)I>S (n) cos((n)) P 2N L I ; [r 9 R cos( 9 (n))]I(r, , z; R, (n), z ), I where N is the Simpson’s method mesh number (even), (n) :

n 2 N

(n : 0, 1, 2, . . . , N),

(4.14)

221


and

S (n) :

(n : 0) (n : 1, 3, 5, . . .) (n : 2, 4, 6 . . .) (n : N).

Function I in Eq. (4.14) is defined by

X

g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z ). I X I Evaluation of Eq. (4.15) gives

IY

where

(4.15)

F(r, , z; r, , z ) I if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z ) I if

(4.16)

r : r, cos( 9 ) : 1, z " z I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 ) Notice that the functional form of I depends on the relationship between the coordinates of the field point (r, , z) and the source point (r, , z). Specifically, I is given by F as long as r ; r 9 2rr cos( 9 ) " 0. If this condition is violated, then the lower ratio is used. Azimuthal component: The azimuthal field component is given by

M R X L B (r, , z) : Q cos() 4 X ; sin( 9 )g(r, , z; R, , z) d dz.

(4.17)

As in the radial case, the integration is evaluated using Simpson’s method, and the integration in z is expressed in terms of Eq. (4.16). The resulting

222


expression for B is M R , B (r, , z) : Q (91)I>S (n) cos((n)) 2N L I ; sin( 9 (n))I(r, , z; R, (n), z ). (4.18) I Axial component: The derivation for the axial field component is similar to that for the previous two cases. Specifically,

M R X L cos()(z 9 z) B (r, , z) : Q X 4 X ; g(r, , z; R, , z) d dz.

(4.19)

Once again, the integration can be expressed as a discrete sum, M R , B (r, , z) : Q S (n) cos((n)) X 2N L X ; (z 9 z)g(r, , z; R, (n), z) dz. X The remaining integral can be evaluated analytically. This gives

(4.20)

M R , B (r, , z) : Q S (n) X 2N L ; (91)I cos((n))g(r, , z; R, (n), z ). (4.21) I I Calculations: We demonstrate the field solutions (4.14), (4.18) and (4.21) with some sample calculations. Consider a solid cylindrical magnet with R : 2.54 mm, L:50 mm (z :925 mm, z :25 mm), and M :4.3;10 A/m Q (B : 5400 G). We evaluate B and B at r : 3.8 mm for a series of angular P P values : 0, 10, 20, . . . , 180°. These data are shown in Fig. 4.11. Next, we analyze a cylindrical shell geometry with inner and outer radii R and R , respectively (Fig. 4.12). The hollow interior of the shell is taken into account using the principle of superposition. Specifically, we take the field due to a solid cylinder B (x), and subtract the field due to an oppositely magnetized inner 0 core B (x), 0 B(x) : B (x) 9 B (x). (4.22) 0 0 The cylindrical shell is symmetrically positioned with respect to the x-y plane

223


FIGURE 4.11

The components B and B vs for a solid cylinder. P

(center at z : 0) with its top at z : 3 cm. The parameters used in the analysis are as follows: M : 4.3 ; 10 A/m Q R : 2.0 cm

(inner radius)

R : 4.0 cm

(outer radius)

L : 6.0 cm

(height).

The components B and B are computed at z:0 and 2.7 cm at a radius r:5 cm P for a series of angular values : 0, 10, 20, . . . , 180°. These data are shown in Fig. 4.13. Notice that the field decreases in magnitude as the observation point moves away from the center (z : 0 cm) towards the end of the cylinder. Finally, the field components B and B are evaluated at r : 5 cm and : 0° for a series P X of points z : 0, 0.25, 0.5, . . . , 5 cm. These data are shown in Fig. 4.14. Notice the behavior of the fields as the observation point passes over the end of the cylinder (z : 3 cm). )

224


FIGURE 4.12

Cylindrical shell geometry and reference frame.

EXAMPLE 4.2.4 Derive an expression for the field outside an infinitely long cylindrical magnet with a linear second quadrant demagnetization relation of the form B : H ; M (r, ). (4.23) Q The magnetization M (r, ) is well behaved but otherwise arbitrary [2]. Let R Q and R denote the inner and outer radii of the cylinder, respectively (Fig. 4.15). SOLUTION 4.2.4 Since the magnet is infinitely long the problem reduces to a 2D boundary value problem (BVP) in cylindrical coordinates. Boundary value problem: The magnetic fields H inside the magnet, and H and H in the regions V and V outside the magnet can be represented


FIGURE 4.13

The components B and B vs for a cylindrical shell. P

FIGURE 4.14

The components B and B vs z for a cylindrical shell. P X

225

226


FIGURE 4.15

Cross section of an infinite cylinder with arbitrary magnetization.

in terms of scalar potentials , and where K K K H : 9 HI : 9I (k : 1, 2). (4.24) K K The BVP for and I follows from: (a) the constitutive relation B : K K H ; M (r, ); (b) the field equations Q · B : 0 · BI : 0 (k : 1, 2), (4.25) (c) the boundary conditions for B and H at the media interfaces; and (d) the constitutive relations BI : HI (k : 1, 2), (4.26) which apply outside the magnet. Specifically, inside the magnet we solve : 9 , K K where is an equivalent volume charge density, K : 9 · M (r, ). K Q

(4.27)

(4.28)

Outside the magnet we solve I : 0 K The boundary conditions are : I K K

at

(k : 1, 2). r:R I

(4.29)

(k : 1, 2),

(4.30)

and where

I K 9 K : I n n

at

r:R I

(k : 1, 2),

(4.31)

227


I : nI · M (r, ). (4.32) K Q Here, nI : (91)Ir are the outward surface normals at r : R . We also I require that (0) -, K

(4.33)

and (-) -, (4.34) K which restrict the form of the solutions in regions V and V , respectively. Separation of variables: We employ the method of separation of variables to solve the BVP (4.27)—(4.34). Accordingly, we represent the solutions to Eqs. (4.27) and (4.29) as follows: (r, ) : [V (r) sin(i) ; U (r) cos(i)], K G G G (r, ) : rG[V sin(i) ; U cos(i)], K G G G

(4.35) (4.36)

and (r, ) : r\G[V sin(i) ; U cos(i)]. (4.37) K G G G The volume charge density and surface term can also be expressed as Fourier series, (r, ) : [C (r) sin(i) ; D (r) cos(i)], K G G G

(4.38)

and I() : [AI sin(i) ; BI cos(i)], (4.39) K G G G where C (r), D (r), AI and BI are determined once the magnetization M (r, ) G G G G Q is specified. Substituting Eqs. (4.35) and (4.38) into Eq. (4.27) yields the following ordinary differential equations for the unknown coefficients V (r) and G U (r), G dV 1 dV i G ; G 9 V : 9C (r), (4.40) G dr r dr r G

228


and dU 1 dU i G ; G 9 U : 9D (r). G dr r dr r G

(4.41)

These equations are of the form df(r) 1 df(r) i ; 9 f(r) : g(r), r dr r dr with homogeneous solutions f (r) : rG and f (r) : r\G. Solution: We use the method of variation of parameters and obtain general solutions for Eqs. (4.40) and (4.41) of the form

f ())g()) P f ())g()) d) ; f (r) d), W(f , f )()) W(f , f )()) 0 0 where c and c are constants and W(f , f ) : f f 9 f f is the Wronskian. Application of this method to Eqs. (4.40) and (4.41) yields f(r) : c f (r) ; c f (r) 9 f (r)

P

1 P ) V (r) : E rG ; F r\G 9 G G G 2i 0

r G ) G 9 C ()) d) G ) r

(4.42)

and

1 P r G ) G U (r) : P rG ; Q r\G 9 ) 9 D ()) d). (4.43) G G G G 2i ) r 0 The unknown coefficients VI, UI (k : 1, 2), E , F , P , and Q are determined G G G G G G by imposing the boundary conditions (4.30) and (4.31). For example, the following group of simultaneous equations is obtained for the coefficients UI, G P , and Q : G G i UiR\G> ; P iRG\ 9 Q iR\G> 9 S : B, (4.44) G G G G R G UR\G : P RG ; Q R\G 9 S , (4.45) G G G G and URG : P RG ; Q R\G, G G G

(4.46)

UiRG\ 9 P iRG\ ; Q iR\G> : B, G G G G

(4.47)

and

229


where

1 0 S : ) G 2i 0

R G ) G ; D ()) d), G ) R

and

1 0 R G ) G 9 S: ) D ()) d). G 2i G ) R 0 Equations (4.44)—(4.47) can be written in matrix form: 0 1

RG

9

0 1

9RG

91

1 0

91

9R\G R\G

1 0

9

U G U G : P G Q G

S B G; G RG> i R 9RG S G 0

.

(4.48)

B R\G> G i

A similar set of equations is obtained for the coefficients VI, E , and F . The G G G method for solving the BVP can now be summarized: Given M (r, ) inside the Q magnet, evaluate Eqs. (4.28) and (4.32) and determine the coefficients C (r), G D (r), and AI, BI (k : 1, 2) in Eqs. (4.38) and (4.39), respectively. Next, G G G substitute D (r) and BI into Eqs. (4.44)—(4.47), and C (r) and AI into the G G G G corresponding equations for VI, E , and F and obtain two sets of decoupled G G G (four by four) simultaneous equations for the unknown coefficients UI, P , Q , G G G and VI, E , and F . Finally, solve the equations for VI and UI and substitute G G G G G these into Eqs. (4.36) and (4.37) to obtain the desired field solution. Calculations: The analysis outlined above is applied to a cylinder with the magnetization as shown in Fig. 4.16. Such magnets are used in the electrophotographic process (Section 4.9) [2]. The magnetization is

9M r Q M r Q M : Q 9M r Q 0

(9 9 ) (9 ) ( ) ( 2 9 ). This can be expressed analytically as follows: M : M()r , Q

(4.49)

(4.50)

230


FIGURE 4.16

Magnetized cylinder and reference frame.

where

and

M() : M cos(i), G G

(4.51)

2M Q (2 sin(i ) 9 sin(i )). M: G i

(4.52)

A solution is sought for the region V outside the cylinder. Therefore, the coefficients of interest are V and U. From the symmetry of the problem, the G G coefficients V : 0 for all i. The coefficients U are determined from Eq. G G (4.48), which gives H(M , R , R , ) G U(M , R , R , ) : G G G G(R , R , ) G

(i : 1, 2, 3, . . .),

(4.53)

where

1 ; 1 K(M , R , R ) H(M , R , R , ) : RG> G G G R M R G ; G 19 9 1; i R M R G> G ;2 ; h* (M , R , R , ) , G G i R

(4.54)

231


with

M R G G K(M , R , R ) : R 9 R , G i(i ; 1) R

(4.55)

and with

M 1 R G G 91 h* (M , R , R , ) : G G i R R 1 R G R 9 R 19i R ; R R ln R

and

R G G(R , R , ) : G R

(i 1) (4.56) (i : 1)

92 ;1 9

;2 ;1 . (4.57)

The field components follow from B : 9 . Specifically, K B(r, ) : ir\G>U(M , R , R , ) cos(i), P G G G and B (r, ) : ir\G>U(M , R , R , ) sin(i). G G G The parameters used in the analysis are as follows:

(4.58)

(4.59)

M : 2.0 ; 10 A/m Q R : 8.0 mm R : 12.0 mm : 30° : 60°. (4.60) The value of M corresponds to injection-molded barium ferrite material. We Q compute the field components B(r, ) and B (r, ) at r : 16 mm and : 0, P 5, 10, 15, . . . , 180° with : . Figure 4.17 shows the predicted field data along with the corresponding data from FEA. We also compute B (r, ) at the P same observation points as in the preceding and with : 1.0 , 1.5 , and

232


FIGURE 4.17

The components B and B vs (; : FEA). P

2.0 . These data are shown in Fig. 4.18. Notice that lower values of give rise to stronger fields outside the cylinder. ) EXAMPLE 4.2.5 Determine the B-field outside an infinitely long cylindrical magnet that has an alternating radial polarization and a linear second quadrant demagnetization curve of the form B : H ; M (). Q

(4.61)

Let R and R denote the inner and outer radii of the cylinder, respectively (Fig. 4.19). SOLUTION 4.2.5 As the magnet is infinitely long, the problem reduces to a 2D boundary value problem (BVP) similar to that of Example 4.2.4. We use the solution method presented there with the magnetization inside the cylinder given by M () : M()r , Q

(4.62)

233


FIGURE 4.18

FIGURE 4.19

Component B vs . P

Infinite cylinder with radial magnetization.

234


where M() : M cos(i), G G

(4.63)

and

M , (2k 9 1)i (2k ; 1)i M : Q (91)I sin 9 sin . (4.64) G i N N I In Eq. (4.64), N is the number of poles. The expressions (4.63) and (4.64) give the Fourier decomposition of M(). First, we determine the representations for () and I () by evaluating the coefficients C (r), D (r), AI and BI in K G G K G G Eqs. (4.38) and (4.39). From Eq. (4.28) we obtain () : 9 · M () K Q M() :9 . r From the orthogonality of sin(i) and cos(i) we find that C (r) : 0, and that G M D (r) : 9 G . G r Similarly, from Eq. (4.32) we have I() : (91)Ir · M () K Q : (91)IM().

Therefore AI : 0 and G (k : 1, 2). BI : (91)IM G G We seek a solution for the region V outside the cylinder. The coefficients of interest are V and U. From the orthogonality of sin(i) and cos(i) we find G G that V : 0 for all i. The coefficients U are given by Eq. (4.53): G G H(M , R , R , ) G U(M , R , R , ) : G (i : 1, 2, 3, . . .), (4.65) G G G(R , R , ) G where H(M , R , R , ) and G(R , R , ) are defined in Eqs. (4.54)—(4.57) G G G with M given by Eq. (4.64). Once the U are known, we compute the field G G components B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.66)


FIGURE 4.20

235

The components B and B vs (r : 12 mm). P

and B (4.67) (r, ) : ir\G>U (M , R , R , ) sin(i). G G G Calculations: We demonstrate the field solution (4.66)—(4.67) via the analysis of an eight-pole cylindrical magnet with R : 1 mm, R : 10 mm, M :4.3 Q ; 10 A/m, and : . We evaluate B and B at r : 12 mm for 0 P 180°. These data are shown in Fig. 4.20. Here, : 0° corresponds to the center of a north pole. Notice that B peaks at the center of the pole, whereas B P peaks at the transition between neighboring poles at : 22.5°. ) EXAMPLE 4.2.6 Determine the radial and azimuthal B-field components outside a finite, radially polarized multipole cylinder (Fig. 4.21) [3]. Assume that the magnet has a magnetization M : <M r , (4.68) Q where the < term takes into account the alternating polarity of adjacent poles.

236


FIGURE 4.21

Radially polarized multipole cylinder.

SOLUTION 4.2.6 The geometry is three-dimensional and, therefore, we need to solve a 3D field problem. We use the current model of Section 3.3. From Eqs. (3.95) and (4.68) we find that the volume current density is zero J : ; M : 0. K

(4.69)

Therefore, Eq. (3.93) reduces to A(x) : 4

1

j (x) K ds,

x 9 x

(4.70)

where j : M ; n. K

(4.71)

The B-field is computed from Eq. (4.70) using B : ; A.

(4.72)

Instead of determining B for the entire magnet, it is simpler to evaluate Eq. (4.70) first for a single sector shown in Fig. 4.22, determine B for the sector, and then compute the total field as a superposition of the contributions from all the sectors. As a first step, we determine j for the various surfaces of the sector. K From Eqs. (4.68) and (4.71) it follows that j : 0 on the back (r : R ) and K front (r : R ) of the sector because the magnetization and surface normals are

237


FIGURE 4.22

Sector geometry for a radially polarized cylinder.

either parallel or antiparallel for these surfaces. There are four remaining surfaces to consider:

r (1) r r (2) Q Q top surface : (1) (2) Q Q z : z (2), Q

(4.73)

r (1) r r (2) Q Q bottom surface : (1) (2) Q Q z : z (1), Q

(4.74)

r (1) r r (2) Q Q left side : : (1) Q z (1) z z (2), Q Q

(4.75)

238


and

r (1) r r (2) Q Q right side : : (2) Q z (1) z z (2). Q Q

(4.76)

The unit normals for these surfaces are

n :

z

(top surface)

9z

(bottom surface)

9

(left side)

(4.77)

(right side).

Therefore, assuming the radial magnetization given in Eq. (4.68), the corresponding surface current densities are

9M Q M Q j : K 9M z Q M z Q

(top surface) (bottom surface)

(4.78)

(left side) (right side).

Taking into account Eqs. (4.73)—(4.78), we write Eq. (4.70) as M A (x) : Q (91)H Q 4 H ;

XQ PQ

XQ PQ

z dr dz

x 9 x : ( j) 9

Q

PQ

PQ

Q

Q

x 9 x

r d dr . XYXQH

(4.79)

In this expression, and for the remainder of this example, the subscript s denotes the contributions due to a single sector. Notice that the unit vector in the integrand of Eq. (4.79) is itself a function of , : 9sin()x ; cos()y .

(4.80)

239


Therefore, Eq. (4.79) can be written in terms of Cartesian coordinates, M A (x) : Q (91)H Q 4 H XQ PQ z ; dr dz

x 9 x : ( j) Q XQ PQ P Q sin( ) x ; Q r d dr

x 9 x XYXQH PQ Q P Q cos() y 9 Q r d dr . (4.81)

x 9 x XYXQH PQ Q This can be rewritten in terms of cylindrical coordinates by computing the following projections:

A (x) : A (x) · r , PQ Q A (x) : A (x) · , Q Q

(4.82) (4.83)

and A (x) : A (x) · z . (4.84) XQ Q We apply these projections to Eq. (4.81) and obtain the following explicit forms: M A (x) : Q (91)H> PQ 4 H PQ Q sin( 9 ) r d dr, ;

x 9 x PQ Q XYXQH M A (x) : Q (91)H> Q 4 H PQ Q cos( 9 ) ; r d dr,

x 9 x PQ Q XYXQH

and

(4.85)

(4.86)

M A (x) : Q (91)H XQ 4 H XQ P 1 dr dz. (4.87) ;

x 9 x XQ PQ Y QH Equations (4.85), (4.86), and (4.87) give A (x) for a single sector. For the sector, Q B (x) is obtained using Eq. (4.72). As the total field is a superposition of the Q

240


fields due to all the sectors, it can be expressed as , B(x) : (91)Q>B (x), (4.88) Q Q where the (91)Q> term takes into account the alternating polarity of adjacent poles. For the derivation of B and B we use the function P 1 . (4.89) g(r, , z; r, , z) : [r ; r 9 2rr cos( 9 ) ; (z 9 z)] Radial component: The radial field component due to a single sector follows from Eq. (4.72): 1 B (x) : (4.90) A (x) 9 A (x). PQ r XQ z Q Substituting Eqs. (4.86) and (4.87) into Eq. (4.90) yields M B (x) : Q (91)H> PQ 4 H XQ PQ r sin( 9 ) dr dz ;

x 9 x XQ PQ Y QH P Q (z 9 z) cos( 9 ) r d dr . (4.91) ; Q

x 9 x XYXQH PQ Q The integrations in r in the first term and in the second term can be written in terms of discrete sums using various numerical methods. We use Simpson’s method. Specifically, let N and N denote the number of mesh points in the r P and variables, respectively, and let r(n) and (m) denote the values at which Q the integrands are evaluated. As r (1)rr (2), and (1) (2), we Q Q Q Q have r (2) 9 r (1) Q r : Q (4.92) Q N P and (2) 9 (1) Q . : Q (4.93) Q N

The integration points are as follows:

and

n r (n) : r (1) ; (r (2) 9 r (1)) Q Q Q N Q P m (m) : (1) ; ( (2) 9 (1)) Q Q Q N Q

(n : 0, 1, 2, . . . , N ), P

(4.94)

(m : 0, 1, 2, . . . , N). (4.95)

241


Finally, define the Simpson integration coefficients S (n) and S (m), P (n : 0) (n : 1, 3, 5 . . .) S (n) : P (n : 2, 4, 6 . . .) (n : N ) P and

(m : 0) (m : 1, 3, 5 . . .) S (m) : (m : 2, 4, 6 . . .) (m : N ). Using this scheme, Eq. (4.91) reduces to

(4.96)

(4.97)

M B (x) : Q (91)H> PQ 4 H , ; r P S (n)r (n) sin( 9 (j)) Q Q P Q L X ; Q g(r, , z; r (n), (j), z) dz Q Q XQ , ; S (m)(z 9 z (j)) cos( 9 (m)) Q Q Q K P ; Q g(r, , z; r, (m), z (j))r dr . Q Q PQ The remaining integrals in z and r can be expressed in closed form. In addition, the contributions from all the sectors need to be taken into account in accordance with Eq. (4.88). The resulting field expression is

M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I ,P ; r S (n)r (n) sin( 9 (j))I (r, , z; r (n), (j), z (k)) Q Q Q P Q Q Q L , ; S (m)(z 9 z (j)) cos( 9 (m))I (r, , z; r (k), (m), z (j)) . Q Q Q Q Q Q K (4.98)

The function I in Eq. (4.98) is defined by XQ g(r, , z; r, , z) dz : (91)I>I (r, , z; r, , z (k)). (4.99) Q XQ I

242


Evaluation of Eq. (4.99) gives

I Y

where

F(r, , z; r, , z (k)) Q if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z (k)) Q if

(4.100)

r : r, cos( 9 ) : 1, z " z (k) Q

(z 9 z (k))g(r, , z; r, , z (k)) Q Q . F(r, , z; r, , z (k)) : Q r ; r 9 2rr cos( 9 ) The function I in Eq. (4.98) is defined by PQ g(r, , z; r, , z)r dr : (91)I>I (r, , z; r (k), , z). (4.101) Q PQ I Evaluation of Eq. (4.101) gives

K(r, , z; r (k), , z) Q if r sin( 9 ) ; (z 9 z) " 0 or 1 r r (1) (r 9 2r (k)) Q Q p: (91)N 2(r 9 r (k)) 2 r r (2) Q Q if r " r (k), cos( 9 ) : 1, z : z Q or I Y r ; 2r (k) Q 2(r ; r (k)) Q if r " r (k), cos( 9 ) : 91, z : z Q or

1 r (k) Q if r : 0, z : z

(4.102)

243


where K(r, , z; r (k), , z) : Q

(r ; (z 9 z)) 9 rr (k) cos( 9 ) Q r sin( 9 ) ; (z 9 z)

;g(r, , z; r (k), , z). Q According to Eq. (4.100), the functional form of I depends on the relationship between the coordinates of the field point (unprimed) and the source point (primed). In particular, I is evaluated using F as long as r ; r 9 2rr cos( 9 ) " 0. However, if this condition is violated, then the lower ratio is used. The function I also depends on the relation between the source and field points. Specifically, if r sin( 9 ) ; (z 9 z) " 0 then the first term in Eq. (4.102) is used. When this condition is violated, the remaining three cases apply as indicated. Notice that the fourth term in Eq. (4.102) is well defined because r (k) " 0 for Q this case (the field and source points cannot coincide). Notice also that this analysis does not include the case in which the field and source point are one and the same. The parameters that appear in Eq. (4.98) are summarized here for convenience, r (2) 9 r (1) Q , r : Q Q N P (2) 9 (1) Q , : Q Q N and

(2s 9 3) (1) : Q N (2) : (2s 9 1) Q N s : 1, 2, . . . , N r (1) : R (inner radius) Q r (2) : R (outer radius) Q z (2) : top of sector Q z (1) : bottom of sector Q N : angular mesh number (even) N : radial mesh number (even). P

(4.103)

244


Azimuthal component: The azimuthal field component also follows from Eq. (4.72), B (x) : A (x) 9 A (x). Q PQ z r XQ

(4.104)

Substitute Eqs. (4.85) and (4.87) into Eq. (4.104) and obtain M B (x) : Q (91)H Q 4 H PQ Q (z 9 z) sin( 9 ) ; r d dr

x 9 x PQ Q XYXQH XQ PQ r 9 r cos( 9 ) ; dr dz . (4.105)

x 9 x Y QH XQ PQ Similar to the radial case, the integrals can be evaluated numerically using Simpson’s method as outlined in Eqs. (4.92)—(4.97). This gives

M B (r, , z) : Q (91)H Q 4 H , ; S (m)(z 9 z (j)) sin( 9 (m)) Q Q Q K P ; Q g(r, , z; r, (m), z (j))r dr Q Q PQ , ; r P S (n)(r 9 r (n) cos( 9 (j))) Q Q P Q L X ; Q g(r, , z; r (n), (j), z) dz . Q Q XQ The remaining integrals in z and r are evaluated using Eqs. (4.100) and (4.102). The total azimuthal field component is obtained as a superposition of contributions from all the sectors,

M , B (r, , z) : Q (91)Q>H>I 4 Q H I , ; S (m)(z 9 z (j)) Q Q K ;sin( 9 (m))I (r, , z; r (k), (m), z (j)) Q Q Q Q

245


, ; r P S (n)(r 9 r (n) cos( 9 (j))) Q Q P Q L

;I (r, , z; r (n), (j), z (k)) , Q Q Q

(4.106)

where I and I are defined in Eqs. (4.100) and (4.102), respectively. Calculations: We demonstrate the field solution with some sample calculations. First, we test Eqs. (4.98) and (4.106) using the 2D field solutions (4.66) and (4.67) from Example 4.2.5. Consider an infinitely long, radially polarized cylindrical shell with 10 alternating poles. Assume that the shell has inner and outer radii R : 2 cm and R : 4 cm, and a uniform magnetization M : 4.3 Q ; 10 A/m. The value of M corresponds to a bonded NdFeB material. For 3D Q analysis we need to set the length L of the shell. To obtain 2D results, the length must be much greater than the diameter of the shell (L 2R ). We set L : 80 cm with the bottom and top of the shell at z(1) : 940 cm and z(2) : 40 cm, respectively. The field components B and B are computed at a radius r : 5 cm P for a series of angular values : 0, 3, 6, . . . , 36° (i.e., from the center of one pole to the center of the neighboring pole). The data from the two models are compared in Fig. 4.23. Next, we apply the model to a cylindrical shell geometry with an axisymmetric magnetization (Fig. 4.24). We use the values R : 2 cm, R : 4 cm, L : 4 cm, and M : 4.3 ; 10 A/m. The cylinder is symmetrically Q positioned with respect to the x-y plane (z : 0 corresponds to the middle of cylinder). Field values are computed along several different lines. First, B is P computed at z : 0, 5, 10, . . . , 60 mm, with r : 50 mm (Fig. (4.25). Next, B P is computed along the line z : 40 mm at the points r : 0, 5, 10, . . . , 60 mm (Fig. 4.26). ) EXAMPLE 4.2.7 Determine the radial and axial field components at any point outside an axially polarized cylindrical shell (Fig. 4.27) [4]. Let R and R denote the inner and outer radii of the shell and z and z denote the axial position of the bottom and top of the shell. Assume that magnet has a second quadrant demagnetization curve B : (H ; M z ). (4.107) Q SOLUTION 4.2.7 We use the current model (Section 3.3). From Eqs. (3.95) and (4.107) we find that the volume current density is zero, J : ; M : K ; M z : 0. Therefore, Eq. (3.93) reduces to Q j (x) K A(x) : ds. (4.108) 4 x 9 x 1

246


FIGURE 4.23

FIGURE 4.24

B and B vs (r : 50 mm). P

Cylinder with axisymmetric magnetization.

247


FIGURE 4.25

FIGURE 4.26

Component B vs z (r : 50 mm). P

The component B vs r (z : 40 mm). P

248


FIGURE 4.27

Axially polarized cylindrical shell with reference frame.

As for the surface current density, we find that j : 0 on the top and bottom K surfaces of the shell because the magnetization and surface normals are either parallel or antiparallel there. There are two remaining surfaces to consider,

(4.109)

(4.110)

r : R inner surface : 0 2 z zz

and r : R outer surface : 0 2 z zz .

The unit normals for these surfaces are n :

9r r

(inner surface) (outer surface).

(4.111)

249


It follows that the surface current densities are

9M Q M Q Thus, Eq. (4.108) can be rewritten as j : K

(inner surface) (outer surface).

(4.112)

X L M R (j) d dz, (4.113) A(x) : Q (91)H A

x 9 x 4 PY0AH H X where R (1) : R , and R (2) : R . Recall that is a function of position, A A : 9sin()x ; cos()y . (4.114) Substitute Eq. (4.114) into Eq. (4.113) and obtain

M X L sin()x A(x) : Q (91)H 9 R (j) d dz A 4

x 9 x PY0AH H X X L cos()y ; R (j) d dz . A

x 9 x PY0AH X Note that A(x) has no z-component. It can be written in terms of cylindrical coordinates by computing the projections

A (x) : A(x) · r P

(4.115)

A (x) : A(x) · .

(4.116)

and

This gives

M X L sin( 9 ) A (x) : Q (91)H R (j) d dz P A 4

x 9 x PY0AH H X

(4.117)

and M A (x) : Q (91)H 4 H Radial field: The radial field

X L cos( 9 ) R (j) d dz. (4.118) A

x 9 x PY0AH X component is given by

B (x) : 9 A (x). P z

(4.119)

250


Substitute Eq. (4.118) into Eq. (4.119) and take into account both the differentiation in z and the integration in z. This gives M B (r, , z) : Q (91)H>I P 4 H I ;

L

cos( 9 )g(r, , z; R (j), , z )R (j) d. A I A

(4.120)


1 . (4.121) [r ; r 9 2rr cos( 9 ) ; (z 9 z)]

The remaining integral in can be written in terms of a discrete sum using a simple numerical scheme such as Simpson’s method. Specifically, let N denote the number of mesh points in the variable and let (m) denote the values at which the integrand is evaluated. As 0 2, we have (m) :

m 2 N

(m : 0, 1, 2, . . . , N).

(4.122)

The integration coefficients S(m) are specified in Eq. (4.97). Applying this scheme to Eq. (4.120) gives an equation for the total radial field component: M B (r, , z) : Q (91)H>IR (j) P A 2N H I , ; S (m) cos( 9 (m))g(r, , z; R (j), (m), z ). (4.123) A I K Axial field: The axial field component is obtained from

1 B (x) : (rA (x)) 9 A (x) . X P r r

(4.124)

Substituting Eqs. (4.117) and (4.118) into Eq. (4.124) yields

M X L r cos( 9 )9R (j) A B (x) : Q (91)H> R (j) d dz. X A 4

x 9 x PY0AH H X (4.125)

251


Equation (4.125) can be simplified using the same numerical scheme as in the previous two sections. Specifically, we obtain M B (r, , z) : Q (91)H> X 4 H 2 , ; S(m)(r cos( 9 (m)) 9 R (j))R (j) A A N K X ; g(r, , z; R (j), (m), z) dz . A X The remaining integration in z can be evaluated analytically and will result in

M B (r, , z) : Q (91)H>I X 4 H I 2 , S (m)(r cos( 9 (m)) 9 R (j)) ; A N K

; R (j)I(r, , z; R (j), (m), z ) A A I

.

(4.126)

In Eq. (4.126), the function I(r, , z; r, , z ) is defined by I XI g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z ). (4.127) I XI I Evaluation of Eq. (4.127) gives

IY

where

F(r, , z; r, , z ) I if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z9z ) I if r : r, cos( 9 ) : 1, z " z I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 )

(4.128)

252


Calculations: We demonstrate Eqs. (4.123) and (4.126) with some sample calculations. We use the following parameters: M : 4.3 ; 10 A/m Q R : 2.0 cm R : 4.0 cm h : 4.0 cm

(4.129)

where h is the height of the magnet (z : 92.0 cm, z : 2.0 cm). First, B and P B are computed at the points z : 0, 5, 10, . . . , 80 mm, for r : 0.25R , 0.5R , X and 0.75R . The computed data are compared with corresponding FEA data in Figs. 4.28 and 4.29. Next, B is computed along the axis of the magnet where X the radial component is zero (Fig. 4.30). Finally, B and B are computed along P X the line defined by z : 40 mm at the points r : 0, 5, 10, . . . , 80 mm. These data are shown in Fig. 4.31. )

FIGURE 4.28

The component B vs z (; : FEA). P

253


FIGURE 4.29

The component B vs z (; : FEA). X

EXAMPLE 4.2.8 Derive an expression for the B-field outside an axially polarized multipole cylindrical disk magnet (Fig. 4.32) [5]. Assume that the magnet has a second quadrant demagnetization curve B : (H < M z ). (4.130) Q The < term takes into account the alternating polarity of adjacent poles. SOLUTION 4.2.8 We use the current model of Section 3.3. It follows from Eqs. (3.95) and (4.130) that the volume current density is zero, J : ; M : 0. K Therefore, Eq. (3.93) reduces to A(x) : 4

j (x) K ds.

x 9 x

(4.131)

(4.132)

1 The B-field is computed from Eq. (4.132) using B : ; A.

(4.133)

254

FIGURE 4.30


The component B vs z along the axis of the cylinder (; : FEA). X

Instead of determining B for the entire magnet, we first evaluate Eq. (4.132) for a single sector shown in Fig. 4.33, determine B for the sector, and then obtain the total B-field as a superposition of the contributions from all the sectors. Equivalent current density: As a first step, we determine the surface current density j for the various surfaces of the sector shown in Fig. 4.33. The K magnetization for this sector is M : M z . Q

(4.134)

From Eqs. (3.95) and (4.134) it follows that j : 0 on the top and bottom of K the sector because the magnetization and surface normals are either parallel or antiparallel on these surfaces. There are four remaining surfaces to consider:

r : R inner surface : (1) (2) Q Q z (1) z z (2), Q Q

(4.135)

255


FIGURE 4.31

The components B and B vs r (; : FEA). P X

r : R outer surface : (1) (2) Q Q z (1) z z (2), Q Q R r R left side : : (1) Q z (1) z z (2), Q Q

FIGURE 4.32

Axially polarized multipole disk magnet.

(4.136)

(4.137)

256


FIGURE 4.33

Sector of an axially polarized disk magnet.

and

R r R right side : : (2) Q z (1) z z (2). Q Q The unit normals for these surfaces are

n :

9r

(inner surface)

r

(outer surface)

9

(left side)

(4.138)

(4.139)

(right side).

The corresponding surface current densities are

9M Q M Q j : K M r Q 9M r Q

(inner surface) (outer surface) (left side) (right side).

(4.140)

257


Vector potential: We use Eqs. (4.135)—(4.140) and rewrite Eq. (4.132), M A (x) : Q (91)H Q 4 H

XQ

Q

x 9 x

r (j) d dz Q

PYPQH Q X PQ r 9 Q dr dz .

x 9 x Y QH XQ PQ

XQ

(4.141)

In this expression, the subscript s denotes the contribution due to a single sector. The unit vectors r and in Eq. (4.141) are themselves functions of , and therefore need to be considered when performing the integration (Section A.2). From Table A.2 we have r : cos()x ; sin()y

(4.142)

: 9sin()x ; cos()y .

(4.143)

and

Therefore, Eq. (4.141) can be rewritten in terms of Cartesian components, M A (x) : Q (91)H Q 4 H

; 9

XQ

XQ

;

XQ

XQ

9

XQ

XQ

9

XQ

XQ

Q sin()x r (j) d dz Q

x 9 x PYP H Q Q Q cos()y

x 9 x r (j) d dz PYPQH Q Q PQ cos()x dr dz

x 9 x PQ Y QH PQ sin()y dr dz .

x 9 x PQ Y QH

Notice that A (x) has no z-component. It can be decomposed into its cylindrical Q components by computing the projections, A (x) : A (x) · r PQ Q

(4.144)

A (x) : A (x) · . Q Q

(4.145)

and

258


The cylindrical components are M A (x) : Q (91)H PQ 4 H XQ Q sin( 9 ) r (j) d dz ; Q

x 9 x XQ Q PYPQH X PQ cos( 9 ) 9 Q dr dz ,

x 9 x XQ PQ Y QH

and

M A (x) : Q (91)H Q 4 H

XQ

(4.146)

Q cos( 9 ) r (j) d dz Q

x 9 x PYPQH XQ Q X PQ sin( 9 ) ; Q dr dz , (4.147)

x 9 x XQ PQ Y QH Equations (4.146) and (4.147) specify the components of the potential for a single sector. The field B (x) due to the sector can be obtained using Q B : ; A . Since the total field is a superposition of the fields due to all the Q Q sectors, it can be written as

, B(x) : (91)Q>B (x), (4.148) Q Q where (91)Q> takes into account the alternating polarity of adjacent poles. Radial component: The radial field component due to a single sector follows from Eq. (4.133), B (x) : 9 A (x). PQ z Q

(4.149)

Substituting Eq. (4.147) into Eq. (4.149) and taking into account both the differentiation in z and the integration in z yields M B (x) : Q (91)H>I PQ 4 H I Q cos( 9 )g(r, , z; r (j), , z (k))r (j) d ; Q Q Q Q P ; Q sin( 9 )g(r, , z; r, (j), z (k)) dr , Q Q PQ

(4.150)

259



1 . [r ; r 9 2rr cos( 9 ) ; (z 9 z)]

The remaining integrals in r and can be written in terms of discrete sums using Simpson’s method. Let N and N denote the (even) number of mesh P points in the r and variables, respectively. Let r(n) and (m) denote the Q values at which the integrands are evaluated. As r (1) r r (2) and Q Q (1) (2), we have Q Q r (2) 9 r (1) Q (4.151) r : Q Q N P and (2) 9 (1) Q . : Q Q N

(4.152)

The integration points are n r (n) : r (1) ; (r (2) 9 r (1)) Q Q Q N Q P

(n : 0, 1, 2, . . . , N ) P

(4.153)

and m ( (2) 9 (1)) (m) : (1) ; Q Q Q N Q The coefficients S (n) and S (m) are P S (n) : P and

(m : 0, 1, 2, . . . , N). (4.154)

(n : 0) (n : 1, 3, 5, . . .) (n : 2, 4, 6, . . .)

(4.155)

(n : N ), P

(m : 0) (m : 1, 3, 5, . . .) S (m) : (4.156) (m : 2, 4, 6, . . .) (m : N). Applying this scheme to Eq. (4.150) and summing the contribution from all the sectors in accordance with Eq. (4.148) gives an equation for the total radial field

260


component, M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I

, ; S (m) cos(9 (m))g(r, , z; r (j), (m), z (k))r (j) Q Q Q Q Q Q K

, ;r P S (n) sin( 9 (j))g(r, , z; r (n), (j), z (k)) , Q Q Q P Q Q L (4.157) where N : number of poles (1) : (2s 9 3) Q N (2) : (2s 9 1) Q N s : 1, 2, . . . , N r (1) : R (inner radius) Q r (2) : R (outer radius) Q z (2) : top of sector Q z (1) : bottom of sector Q N : angular mesh number (even) N : radial mesh number (even). P

(4.158)

Azimuthal component: The azimuthal component also follows from Eq. (4.133), B (x) : A (x). Q z PQ

(4.159)

Applying Eq. (4.159) to Eq. (4.146) and taking into account the differentiation

261


in z and the integration in z yields M B (r, , z) : Q (91)H>I> Q 4 H I

Q sin( 9 )g(r, , z; r (j), , z (k))r (j) d Q Q Q Q P 9 Q cos( 9 (j))g(r, , z; r, (j), z (k)) dr . Q Q Q PQ Again, apply Simpson’s method and sum the contributions from all the sectors. This gives ;

M , B (r, , z) : Q (91)Q>H>I 4 Q H I

, ; S (m) sin(9 (m))g(r, , z; r (j), (m), z (k))r (j) Q Q Q Q Q Q K

, 9r P S (n) cos( 9 (j))g(r, , z; r (n), (j), z (k)) , Q Q Q P Q Q L (4.160) where the variables are defined in Eq. (4.158). Axial components: The axial component of a sector is given by

1 B (x) : (rA (x)) 9 A (x) . XQ Q r r PQ

(4.161)

Substituting Eqs. (4.146) and (4.147) into Eq. (4.161) gives M B (x) : Q (91)H> XQ 4 H

;

XQ

Q r cos( 9 ) 9 rQ (j) r (j) d dz Q

x 9 x XQ Q PYPQH X PQ r sin( 9 ) ; Q dr dz , (4.162)

x 9 x XQ PQ Y QH

262


Equation (4.162) can be simplified using Simpson’s method as above, M B (r, , z) : Q (91)H> XQ 4 H

, ; S (m)[r cos( 9 (m)) 9 r (j)]r (j) Q Q Q Q K ;

XQ g(r, , z; r (j), (m), z) dz Q Q XQ

, ;r P S (n)r sin( 9 (j))) Q P Q L ;

XQ

XQ

g(r, , z; r (n), (j), z) dz Q Q

.

The remaining integration in z can be evaluated analytically. We sum the contributions from all the sectors and obtain B for the entire magnet, X M , B (r, , z) : Q (91)Q>>H>I X 4 Q H I

, ; S (m)[r cos( 9 (m)) 9 r (j)] Q Q Q K ;r (j)I(r, , z; r (j), (m), z (k)) Q Q Q Q

, ;r P S (n)r sin(9 (j)) I(r, , z; r (n), (j), z (k)) . Q Q P Q Q Q L (4.163) In Eq. (4.163), the function I is defined by

XQ g(r, , z; r, , z) dz : (91)I>I(r, , z; r, , z (k)). (4.164) Q XQ I

263


Evaluation of Eq. (4.164) gives

IY

where

F(r, , z; r, , z (k)) Q if r ; r 9 2rr cos( 9 ) " 0 or 91 2(z9z (k)) Q if

(4.165)

r : r, cos( 9 ) : 1, z " z (k) Q

(z 9 z (k))g(r, , z; r, , z (k)) Q Q F(r, , z; r, , z (k)) : . Q r ; r 9 2rr cos( 9 ) The functional form of I depends on the relationship between the coordinates of the field point (unprimed) and the source point (primed). In particular, I is evaluated in terms of the function F as long as r ; r 9 2rr cos( 9 ) " 0. If this condition is violated, then the lower ratio is used. Calculations: We demonstrate the field solution via the analysis of a four-pole disk magnet with the following parameters: M : 4.3 ; 10 A/m R : 2.0 cm R : 4.0 cm h : 1.0 cm

(inner radius) (outer radius) (height).

(4.166)

The magnet is oriented with its upper surface in the x-y plane with one of its sectors symmetrically positioned in an angular sense with respect to the x-axis. The upper surface of this sector is taken to be a north pole. The field components B and B are evaluated along a radial line that is 1 cm above the surface of the P X magnet and centered with respect to this sector ( : 0). Field values are computed at r : 0, 5, 10, . . . , 45 mm. These data are compared with corresponding data from a 3D FEA as shown in Fig. 4.34. Similarly, the field components B and B are evaluated along an arc at the mean radius of the magnet, 1 cm X above its surface. The arc extends from the center of one pole to the center of the neighboring pole. Field values are computed at : 0, 5, 10, . . . , 90°. These data are shown in Fig. 4.35. )

264


FIGURE 4.34

The components B and B vs r (; : FEA). P X

FIGURE 4.35

The components B and B vs (; : FEA). X

4.3 HIGH FIELD STRUCTURES

265

4.3 HIGH FIELD STRUCTURES Conventional permanent magnet structures such as those studied in the preceding typically produce field strengths of 1.0 T in proximity to their surface. However, special structures can be fabricated that provide uniform multitesla fields throughout their interior. For example, fields of up to 4.0 T have been achieved in disk-shaped cavities 2.5 cm in diameter and 0.5 cm high with structures 15 cm in diameter [6—8]. Such structures can be used for numerous applications such as Hall effect studies. They can also be used to focus charged particle beams if through-ports are provided. The most popular high field structures are the ‘‘magic’’ cylinder and sphere [6—9]. These are made from high-strength rare-earth materials such as NdFeB. They are fabricated piecewise by cutting uniformly ` a magnetized block, and then orienting polarized slices of material from and assembling the constituent pieces so as to provide a predefined magnetization pattern throughout the assembled structure. The magnetization pattern is chosen to provide a desired field strength and uniformity in the interior of the structure. The ‘‘magic’’ cylinder is shown in Fig. 4.36. It has the form of a cylindrical shell with a magnetization that varies with the polar angle as shown. This structure can be built from a stack of cylindrical sections as shown in Fig. 4.37. These sections can be constructed as follows. First, cut a circular ring from a uniformly magnetized block of material. Next, cut the ring into the desired number of pie-shaped elements. Finally, reorder the pie-shaped elements around the circumference to obtain the desired angular magnetization distribution. The fabrication sequence is illustrated in Fig. 4.37. The flux density in the interior of a magic cylinder is given by B : B ln(r /r ), (4.167) P M G where B is the remanence of the material and r and r are the inner and P G M outer radii of the cylinder, respectively [6]. Notice that Eq. (4.167) implies that there is no theoretical upper limit to the field that can be obtained in this structure. However, the slow logarithmic dependency on r places M a practical upper limit of 2.5 T for cavities of more than a few centimeters in diameter. The mass of material required to obtain higher fields would be prohibitively large. The ‘‘magic’’ sphere is shown in Fig. 4.38. It has the form of a spherical shell in which the direction of magnetization varies with the polar angle as :2. This magnetization pattern is obtained by assembling a collection of melonlike segments. These segments can be constructed by

266


FIGURE 4.36

Magic cylinder: (a) idealized structure; and (b) physical structure.

beveling the ring structures used in the construction of the magic cylinder (Fig. 4.39). The flux density in the cavity of a magic sphere is given by B : B ln(r /r ), P M G

FIGURE 4.37

Fabrication sequence for a magic cylinder [9].

4.3 HIGH FIELD STRUCTURES

FIGURE 4.38

267

Magic sphere: (a) idealized structure; and (b) physical structure.

FIGURE 4.39

Fabrication sequence for a magic sphere [9].

268


where r and r are the inner and outer radii of the shell, respectively [6]. G M Such structures provide uniform fields of 2.0 to 3.0 T in their interior when the outer radius of the shell is only three to four times the radius of the cavity. For example, a magic sphere with B : 1.2 T and an outer P radius of 3.0 cm produces a uniform field of 2.5 T in a cavity with a 1.0 cm radius [7].

4.4 MAGNETIC LATCHING Magnetic latching is one of the oldest and most common applications of permanent magnets. In latching applications, magnets are used to provide a force of attraction between two members. Latching magnets can be found in a variety of products including household appliances, tool equipment, electrical devices, and toys. Such magnets are usually made from ferrite materials because of their low cost. Latching applications fall into one of two categories: contact and noncontact. The distinction between these two is important because the design approach is different for each. In contact latching the magnet is attracted to and in contact with another member (either another magnet or a soft magnetic material such as iron). The latching magnet is often designed to have numerous closely spaced poles to enhance the latching force. In noncontact latching the magnet exerts a force of attraction over a distance. Here, the poles of the latching magnet need to be separated so that the field will span the gap between the magnet and the adhering surface in order to provide the latching force. Therefore, the magnetic structures for contact and noncontact latching are quite different. We consider contact latching first. Consider a magnet with a remanence B that is in contact with a plate of high-permeability material P such as iron that is unsaturated (Fig. 4.40). The force of attraction between the magnet and the plate can be estimated using the Maxwell stress tensor expression (3.59), which gives BA (4.168) F: P , 2 where A is the area of contact between the magnet and the plate. The force can be rewritten in terms of the flux : B A, P F: . (4.169) 2 A This expression implies that the latching force can be increased by

4.4 MAGNETIC LATCHING

FIGURE 4.40

269

Contact latching: (a) single-pole contact; and (b) multipole contact.

reducing the contact area A and/or increasing the flux . This can be accomplished by tapering the poles of the magnet, or by adding tapered iron pole pieces to the magnet. However, there is a practical limit to this effect because Eq. (4.169) only applies when the material being held is well below saturation. As A decreases the material saturates and the force no longer increases. This is especially true in applications involving thin plates that have a relatively low saturation threshold due to their limited cross-sectional area. One way to achieve a higher latching force in such applications is to use a magnet with numerous closely spaced poles as shown in Fig. 4.40b. If there are N such poles, and each exerts a force F, then the total force is F : NF as long as the material being held is below saturation. We now consider noncontact latching. In noncontact latching the two attracted members are separated by a gap, and the magnet must project its field across the gap to exert the force of attraction. Thus, the poles of the magnet must be effectively separated so that the field will span the gap. Noncontact latching is often implemented using a magnetic circuit as opposed to a standalone magnet. There are numerous latching circuits, and most of these can be designed and optimized using magnetic circuit theory (Section 3.5). We demonstrate the design method via the analysis of two practical latching circuits.

270


Magnetic latching device: (a) magnet with two attached flux plates; and (b) latching circuit.

FIGURE 4.41

EXAMPLE 4.4.1 Consider the magnetic circuit of Fig. 4.41b. A magnet with attached flux plates is attracted to a wall made of soft magnetic material. Determine the force of attraction when a gap exists between the magnetic structure and the wall g 0. Assume that the flux plates and the wall have infinite permeability, and that the magnet has a linear second quadrant demagnetization curve of the form B :B ; H , K P K K

(4.170)

where : B /H . K P A SOLUTION 4.4.1 We use magnetic circuit theory (Section 3.5). As there are no currents we have w H · dl : 0, which gives H l ; 2H g : 0, (4.171) KK E where H is the field in the gap. The factor of 2 in Eq. (4.171) is due to the fact E that the path of` integration passes through the gap region twice as shown in Fig. 4.41b. In addition, notice that Eq. (4.171) ignores the path through the flux plates and the wall because H is negligible in these elements by assumption. As there is no fringing, w B · dS : 0, which implies that : . Thus, K E B A :B A (no fringing at gap), (4.172) K K E E

271

4.4 MAGNETIC LATCHING

where A : Lw is the area of the magnet, and A : Lw is the area of gap. K K E E We combine Eqs. (4.170), (4.171), and (4.172) and obtain A B P B (g) : K . E A (1 ; 2( / )(A /A )(g/l )) E K K E K Next, we determine the force using the Maxwell stress tensor (Section 3.2.3). We compute the force on the wall by applying Eq. (3.59) over an area bounding the wall (similar to the area indicated by the dotted line in Fig. 3.9 of Example 3.2.8). The only contributions to Eq. (3.59) come from the area in the gap regions. In these regions, B : B , and therefore Eq. (3.59) reduces to L E 1 F(g) : B(g)n ds L 2 1 1 (2B(g)A )n : E E 2 BA P K : n . (4.173) A (1 ; 2( / )(A /A )(g/l )) E K K E K The outward normal n of the wall points towards the magnet, and, therefore, the force is attractive. )

EXAMPLE 4.4.2 Determine the force of attraction for the latching circuit of Fig. 4.42b. Assume that the flux plate and the wall have infinite permeability and that the magnet has a linear second quadrant demagnetization curve of the form B :B ; H , K P K K

(4.174)

where : B /H . K P A SOLUTION 4.4.2 We use magnetic circuit theory (Section 3.5). First, apply w H · dl : 0 to the closed dotted line path on the right-hand side of the circuit. This gives H l ; 2H g : 0, (4.175) KK E where H is the field in the gap. The factor of 2 in Eq. (4.175) is due to the fact E that the path of integration passes through the gap region twice as shown. Notice that we have ignored contributions from the path through the flux plate and the wall as H is negligible in these elements. Next, because there is no fringing, : 2 and we have K E B A : 2B A (no fringing at gap), (4.176) K K E E

272


Magnetic latching circuit: (a) magnet with attached flux plate; (b) latching circuit.

FIGURE 4.42

where A : Lw is the area of the magnet, and A : Lw is the area of the gap. K K E E The factor of 2 in Eq. (4.176) accounts for the fact that the flux through the magnet splits evenly between the two return paths as indicated by the dotted lines in Fig. 4.42b. We combine Eqs. (4.174), (4.175), and (4.176) and find that A B P B (g) : K . E A 2(1 ; ( / )(A /A )(g/l )) E K K E K

273

4.5 MAGNETIC SUSPENSION

We use the Maxwell stress tensor to obtain the force (Section 3.2.3). Specifically, we apply Eq. (3.59) over an area bounding the wall (similar to the area indicated by the dotted line in Fig. 3.9 of Example 3.2.8). The only contributions to Eq. (3.59) come from the area in the gap regions. There are three such regions — a central region immediately beneath the magnet and two end regions. To apply Eq. (3.59) we need to know B (the field component normal to the wall). In the L central gap region B : B : 2(A /A )B , whereas in the end regions B : B . L K E K E L E Substituting these into Eq. (3.59) gives

1 B(g)n ds, L 2 1 1 : [2B(g)A ; B A ]n E K K E 2 BA A ; 2A P K K E n . : (4.177) 2 (1 ; ( / )(A /A )(g/l )) 2A K K E K E Because the outward normal at the wall n points towards the magnetic circuit, the force is attractive. ) F(g) :

4.5 MAGNETIC SUSPENSION In magnetic suspension, magnets are used to provide passive levitation and/or positioning of a structure. The magnets can be arranged in myriad configurations and substantial levitation force can be realized when modern high-strength materials such as NdFeB are used. The two most common suspension configurations are the radial- and axial bearing configurations shown in Fig. 4.43 [10]. Magnetic bearings are often used to provide low friction support for rotating mechanical structures as illustrated in Fig. 4.44. They enable substantially higher operating speeds relative to conventional mechanical bearings. It is well known that a passive magnetic field alone cannot provide stable support of a structure. Such stability can only be achieved with some form of active control [11, 12]. This is a consequence of Earnshaw’s theorem, which states that ‘‘A body with steady charges, magnetizations, or currents placed in a steady electric or magnetic field cannot rest in stable equilibrium under the action of electric and magnetic fields alone’’ [13]. Earnshaw’s theorem can be understood from the following heuristic argument [14]. Let Q denote the magnetic scalar potential due to a K permanent magnet support structure. Assume that the structure is

274


Cross-sectional view of axisymmetric magnetic bearings: (a) radial bearing configurations; and (b) axial bearing configurations.

FIGURE 4.43

designed to suspend a given magnet. Because the suspension point is outside the support structure, we have Q : 0. (4.178) K Now, the suspension force FQ is proportional to the field strength HQ : 9Q, K FQ . Q. K If the suspended magnet is to be in stable equilibrium, the net force must be zero, which implies that Q : 0. K

(4.179)


FIGURE 4.44

275

Magnetic suspension with lateral mechanical constraint.

However, Eq. (4.179) indicates a potential maximum or minimum of Q. K For stable equilibrium, the potential must be a minimum that occurs when Q 0. (4.180) K However, Eq. (4.180) contradicts Eq. (4.178), which implies that it is impossible to provide stable suspension with a passive magnetic field. Therefore, to implement magnetic suspension, magnets are usually used along with some form of mechanical structure that constrains the motion of the supported member. An example of this is shown in Fig. 4.44. Here, opposing disk magnets levitate a rotating member, while its lateral motion is constrained by a fixed support with a conventional mechanical bearing that enables rotation. The force between the disk magnets is derived in the following example. EXAMPLE 4.5.1 Derive an expression for the levitation force between two identical, axially polarized cylindrical disks as shown in Fig. 4.45 [15]. The disks have inner and outer radii R and R , thickness t , and are separated by K a distance h (Fig. 4.46). Assume that both magnets have a linear second quadrant demagnetization curve of the form B : (H < M z ). (4.181) Q where the < takes into account the opposite polarizations of the opposing magnets.

276


FIGURE 4.45

FIGURE 4.46

Levitated magnetic disk.

Cross section of disk geometry with reference frame.

277


SOLUTION 4.5.1 We use the charge model (Section 3.4). Specifically, we first reduce the magnets to equivalent charge distributions and then determine the force using the force formulas (3.120) to (3.123). The volume and surface charge densities and are given by Eq. (3.105). From Eq. (4.181) we have K K : 9 · M K : 9 · M z Q : 0, (4.182) and : M · n K : <M , (4.183) Q where the < sign takes into account the opposite polarity of the charge on opposing pole faces. Now, consider the force expressions (3.120) to (3.123) for a discretized magnet. Because : 0 in both magnets, the volume terms vanish K and the force between the magnets reduces to

(x ) (x )(x 9 x ) I A A . F: K N K I N (4.184) I N 4

x 9 x N I N I Instead of evaluating Eq. (4.184) directly, it is easier to derive the force in the following steps: First, obtain the force between two point sources; next, derive the force between opposing pole faces; and finally, obtain the total force by summing the contributions from all the surfaces. Point sources: We use cylindrical coordinates. Without loss of generality, one of the charges Q (x ) can be taken to be in the x-y plane at (r , , 0) while the K other Q (x ) is taken to be in a plane defined by z : h at (r , , h). The axial K force between these two point sources follows from Eq. (4.184): Q (r , , 0)Q (r , , h)h K K F : . (4.185) X 4 [r ; r 9 2r r cos( 9 ) ; h] Surfaces: The next step is to evaluate the axial force exerted by one pole face (surface) on an opposing pole face. Again, without loss of generality, one surface can be taken to lie on the x-y plane while the other can be taken to lie on the plane defined by z : h. The idea is to discretize each surface into a mesh of subsections (elements), define a point charge at the midpoint of each element in accordance with Eq. (3.110), and then compute the force exerted by one surface on the other by summing the contributions from the individual point charges. The area of each surface is (R 9 R). The first step is to divide this area into

278


a set of indexed elements. These elements are taken to have equal areas so that they contain an equal amount of surface charge. Because the area of the surface is spanned by the two coordinate variables r and , two indices (i, j) are required. Let N and N be the number of mesh points in these two variables, P respectively. Each element will have the same area A, (R 9 R) . N N P Choose each element so that it spans an angular measure A :

:

2 . N

(4.186)

(4.187)

Let (r , ) denote the midpoint of the element A , and let R(i) and R(i ; 1) G H GH denote its radial boundaries. Then r : R(i ; 1) 9 R(i), G R(i ; 1) ; R(i) r: (i : 1, 2, . . . , N ), G P 2

(4.188) (4.189)

and , : (j 9 1) ; H 2

(4.190)

for j : 1, 2, . . . , N. The end points are computed using the following equation:

(R 9 R) (i : 1, 2, . . . , N ), (4.191) P N P where R(1) : R (inner radius). Equation (4.191) is used to generate all the end point values from which all the midpoint values are obtained. Therefore, Eqs. (4.189) to (4.191) define a scheme for discretizing the surface into elements and assigning values to the midpoints of each element. Next, consider the magnetic interaction between an element of one surface and an element of an opposing surface. Let the coordinates of the midpoints of the two elements be (r , , 0) G H and (r , , h), respectively. The magnetic charge associated with each element GY HY is R(i ; 1) :

R(i) ;

Q (r , , 0) : (r , , 0)A K G H G H

(4.192)

Q (r , , h) : (r , , h)A, K GY HY GY HY

(4.193)

and

279


respectively. Substitute Eqs. (4.183) and (4.186) into these expressions and obtain Q : M K Q

(R 9 R) . N N P

(4.194)

The force is obtained by summing the contributions of all the elements in accordance with Eq. (4.185). By exploiting the symmetry of the problem the force reduces to

(R 9 R) F (h) : M X 4 Q N N P , , , h . (4.195) ;N P P [r ; r 9 2r r cos( ) ; h] GY G GY HY G GY HY G Force: There are two magnets to consider, which are labeled upper or lower in accordance with their respective positions. The upper magnet has its bottom surface in the plane defined by z : h and its top surface lies in the plane defined by z : h ; t (Fig. 4.46). The lower magnet has its top surface in the x-y plane K and its bottom surface in the plane defined by z : 9t . The axial force on the K top surface due to both the top and bottom surfaces of the upper magnet is F (h) : F (h) 9 F (h ; t ). X X X K Continuing in this fashion and taking into account the contributions from all the surfaces, the following formula is obtained for the total force between the magnets:

MN (R 9 R) F : Q X 4 N N P , , , (91)K h K K ; P P . [r ; r 9 2r r cos( ) ; h ] GY K K G GY HY G G GY HY where : (j 9 1) ; HY 2 :

2 N

(j : 1, 2, . . . , N)

(4.196)

280


and r: G

R(i ; 1) ; R(i) 2

r : GY

R(i ; 1) ; R(i) 2

(i : 1, 2, . . . , N ) P (i : 1, 2, . . . , N ) P

(R 9 R) (i : 1, 2, . . . , N ) P N P (R 9 R) R(i ; 1) : R(i) ; (i : 1, 2, . . . , N ) P N P R(1) : R (inner radius). The terms h and are given by K K h : h ; mt (m : 0, 1, 2) K K h : separation of magnets R(i ; 1) :

R(i) ;

t : thickness of magnets K and

1

(m : 0)

: 2 K 1

(m : 1) (m : 2).

Notice that Eq. (4.196) contains nested summations of elementary functions that are easy to program. However, it is important to note that when programming nested summations, a higher degree of accuracy can be achieved with fewer mesh points (lower values N and N) if the total of a single P summation over one index is evaluated and saved separately (as an intermediate step), and then the totals of the successive summations are added together to obtain the final sum. For example, Eq. (4.196) could be evaluated as follows:

MN (R 9 R) F : Q S (m), X 4 N N P K where , S (m) : P S (i, m), G


FIGURE 4.47

with

and

281

Levitation force vs separation distance (; : FEA).

, S (i, m) : P S (i, i, m), GY

, (91)K h K K S (i, i, m) : . [r ; r 9 2r r cos( ) ; h ] G GY K HY G GY HY Calculations: We demonstrate Eq. (4.196) with some sample calculations. We use the following parameters: M : 4.3 ; 10 A/m Q R : 2.0 cm R : 4.0 cm t : 1.0 cm. K Force values are computed for a series of separation distances with the mesh parameters set to N : 20 and N : 60. The computed data are compared with P corresponding FEA data in Fig. 4.47. )

282


4.6 MAGNETIC GEARS Magnetic gears can be used in place of mechanical gears to reduce undesired vibrations and for applications that require torque coupling between separated members. There are many gear configurations. One such configuration consists of two separated, radially polarized cylindrical magnets that are constrained to rotate about their respective axes (Fig. 4.48). The magnets are magnetically coupled to one another and when one magnet rotates it imparts a torque to the other, causing it to rotate [16]. The coupling between the magnets is a function of several variables including the number of poles, material properties, dimensions, and separation. Furthermore, substantial torque can be realized if modern, rare-earth materials such as NdFeB are used. We derive a formula for the torque in the following example. EXAMPLE 4.6.1 Derive an expression for the torque between two separated radially-polarized multipole cylinders (Fig. 4.48) [16]. Assume that the magnets have identical geometric dimensions, and that their length is much greater than their diameter. Let N denote the number of poles, and let R and R denote the inner and outer radii of each magnet. Assume that the magnets have different second quadrant demagnetization relations. The demagnetization curve for the magnets in the primed and unprimed coordinate systems are B : H ; M,

FIGURE 4.48

(4.197)

Magnetically coupled cylinders with reference frames.

283

4.6 MAGNETIC GEARS

and B : (H ; M),

(4.198)

respectively, where M : <M r . (4.199) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.6.1 The gear consists of two magnets. We refer to the magnet centered with respect to the primed coordinate system (x, y) as the source magnet. We refer to the magnet centered with respect to the unprimed coordinate system (x, y) as the drive magnet. The analysis proceeds in two steps. First, we use boundary-value theory to obtain a free-space field solution for the source magnet. Second, the drive magnet is reduced to an equivalent current distribution and the torque is computed by considering this current to be in the external field of the source magnet. Source magnet: As the length of the source magnet is much greater than its diameter, we can determine its field using a 2D analysis. The field due to the source magnet was derived in Example 4.2.5. Specifically, the field components in the primed coordinates are given by B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.200)

and (4.201) B (r, ) : ir\G>U (M , R , R , ) sin(i). G G G The coefficients U(M , R , R , ) are defined by Eq. (4.53), with M given by G G G Eq. (4.64). The field components (4.200) and (4.201) constitute the external field in the following analysis. Therefore, in what follows we adopt the notation (r, ). B(r, ) : B(r, ) and B (r, ) : B PY P Y Drive magnet: To determine the torque on the drive magnet we first reduce it to a distribution of equivalent volume and surface current densities J and j , K K respectively (Section 3.3). Once this is done, we compute the torque using Eq. (3.100), T:

4

;

r(r, ) ; (J (r, ) ; B (r, ))r dr d dz K

1

r(r, ) ; (j (r, ) ; B (r, )) da, K

(4.202)

284


where V and S denote the volume and surface of the magnet. Notice that B (r, ) is the field due to the source magnet represented in the unprimed coordinates. From Eq. (4.199) we have J : ; M : 0 and, therefore, the first K term in Eq. (4.202) is zero. As for the second term, there are N sectors, and each sector has two surfaces with current densities that contribute to the torque. These surfaces constitute the radial sides of the sector at angular positions and , respectively. If the magnet is rotated by an angle , then the surface current densities for the pth sector are R rR M z (p) : ; (1 ; 2p) Q N j (p, r, ) : (4.203) K R rR 9M z (p) : 9 (1 ; 2p), Q N where p : 0, 1, 2, . . . , N 9 1. Notice that : 0 occurs when the middle of the zeroth sector coincides with the x axis and its polarization is radially outward as shown in Fig. 4.48. It is assumed that the source magnet is held stationary with one of its sectors in a similar orientation with respect to the primed coordinate system and, therefore, is the relative angular offset of the two magnets. We substitute Eq. (4.203) into Eq. (4.202) and obtain

T() : 2M L Q

, \ 0 (91)N r[cos( (, p))B(r, (, p)) V 0 N ; sin( (, p))B(r, (, p))] dr, W (4.204)

where L is the length of the gear, and (, p) : ; (1 ; 2p). (4.205) N The coefficient 2M takes into account the fact that there are two surfaces at the Q interface between neighboring sectors (one for each sector). The integral in Eq. (4.204) can be evaluated numerically using Simpson’s method. The resulting torque formula is T() :

2M LR* , \ ,P Q (91)NS (q)r(q) P N P N O ;[cos( (, p))B(r(q), (, p)) V ;sin( (, p))B(r(q), (, p))], W

(4.206)

285

4.6 MAGNETIC GEARS

where N is the radial mesh coefficient (even), R* : R 9 R , and S (q) are the P P Simpson coefficient terms

S (q) : P

(q : 0) (q : 1, 3, 5, . . .) (q : 2, 4, 6, . . .)

(4.207)

(q : N ). P

The integration points are as follows: q r(q) : R ; (R 9 R ) N P

(q : 0, 1, 2, . . . , N ). P

(4.208)

To compute the torque we express the external field components B(r, ) PY and B(r, ) (Eqs. (4.200) and 4.201)) in terms of the unprimed coordinates, Y B(r, ) Y B(r(r, ), (r, )) cos((r, )) V PY 9 B (r(r, ), (r, )) sin((r, )) Y

(4.209)

and B(r, ) Y B(r(r, ), (r, )) sin((r, )) W PY ; B (r(r, ), (r, )) cos((r, )) Y

(4.210)

In Eqs. (4.209) and (4.210) we use the coordinate transformation relations r(r, ) : (r ; 2rd cos() ; d

(4.211)

and (r, ) : arctan

r sin() . r cos() ; d

(4.212)

The quantities B(r, (, p)) and B(r, (, p)) in Eq. (4.206) are V W evaluated using Eqs. (4.209) and (4.210). Calculations: We demonstrate the torque formula (4.206) via a practical example. To clarify the analysis, we distinguish the parameters for the source and drive magnets using superscripts s and d, respectively. For example, RQ and RQ refer to the inner and outer radii of the source magnet, and RB and RB refer to the inner and outer radii of the drive magnet. We use the following

286


parameters: MQ : 7.2 ; 10 A/m Q MB : 7.2 ; 10 A/m Q d : 80 mm RQ : 10 mm RQ : 20 mm RB : 15 mm RB : 30 mm N : 4 NB : 4. (4.213) The values of MQ and MB are characteristic of sintered NdFeB material. The Q Q torque per unit length is computed with the source magnet fixed and the drive magnet rotated through a series of angular values : 0, 5, 10, . . . , 90°. The analysis is performed with N : 20 and : in Eq. (4.197). The torque data P are compared with corresponding FEA data in Fig. 4.49. Notice that the peak torque occurs at : 45° when the drive magnet is rotated half the angular span of one pole. An analysis is also performed to determine the decrease of peak torque ( : 45°) as a function of separation distance d for a series of values d : 60, 70, . . . , 160 mm. A first set of data is computed with : in Eq. (4.197). These data are compared with FEA data in Fig. 4.50. An additional calculation is performed with : 2 . This represents a steeper slope in the demagnetization curve (4.197), which results in a lower external field B and a reduced torque as shown. )

4.7 MAGNETIC COUPLINGS Synchronous magnetic couplings are used to transmit torque between two separated members. Couplings are used for various purposes including the activation of devices in sealed containers and the transmission of instrument readings from vacuum or pressure vessels. The two most common coupling configurations are radial and axial couplings [17]. Radial couplings consist of two coaxial radially polarized cylindrical magnets (Fig. 4.51). Axial couplings consist of two axially polarized multipole disks that face one another (Fig. 4.52). For both configurations, the transmitted torque is a function of several variables

4.7 MAGNETIC COUPLINGS

FIGURE 4.49

FIGURE 4.50

287

Torque per unit length vs rotation angle (FEA : ;).

Torque per unit length vs separation distance d (FEA : ●).

288


FIGURE 4.51

Synchronous radial coupling.

FIGURE 4.52

Synchronous axial coupling.

289


including material properties, number of poles, dimensions, separation distance, and the relative angular offset of the magnets [18]. In this section, we derive formulas for the torque of radial and axial couplings. EXAMPLE 4.7.1 Derive an expression for the torque between two coaxial, radially polarized multipole cylinders (Fig. 4.51). Assume that the lengths of the magnets are much greater than their diameters. Further assume that the inner and outer magnets have linear second quadrant constitutive relations of the form B : H ; M,

(4.214)

B : (H ; M),

(4.215)

and

respectively, where M : <M r . (4.216) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.7.1 The coupling consists of an inner and outer magnet. We refer to the inner magnet as the source magnet and the outer magnet as the drive magnet. The analysis proceeds in two steps: First, we use boundary-value theory to obtain a free-space field solution for the inner magnet; and second, the outer magnet is reduced to a distribution of equivalent currents, and the torque is determined by considering the equivalent currents to be in the external field of the source magnet. Source magnet: Because the length of the source magnet is much greater than its diameter, we can determine its field using a 2D analysis. The field due to the inner magnet was derived in Example 4.2.5. The field components are given by B(r, ) : ir\G>U(M , R , R , ) cos(i) P G G G

(4.217)

and (4.218) B (r, ) : ir\G>U (M , R , R , ) sin(i). G G G The coefficients U(M , R , R , ) are defined by Eq. (4.53), with M given by G G G Eq. (4.64). In the following analysis the field components (4.217) and (4.218) constitute an external field. Therefore, we adopt the notation B(r, ) : P B(r, ) and B (r, ) : B (r, ). P Torque: We determine the torque on the outer magnet. Let N , M , and R Q and R denote the number of poles, magnetization, and inner and outer radii of

290


this magnet, respectively. As a first step, we reduce the magnet to a distribution of equivalent volume and surface current densities J and j , respectively K K (Section 3.3). Once this is done, we compute the torque using Eq. (3.100). Specifically, T:

r(r, ) ; (J (r, ) ; B(r, ))r dr d dz K

;

4

1

r(r, ) ; (j (r, ) ; B(r, )) da, K

(4.219)

where V and S denote the volume and surface of the magnet. From Eq. (4.216) we have J : ; M : 0. Therefore, the first term in Eq. (4.219) is zero. As K for the second term, there are N sectors to consider, and each sector has two surfaces with current densities that contribute to the torque. These surfaces constitute the radial sides of the sector at angular positions and , respectively. If the magnet is rotated by an angle , then the surface current densities for the pth sector are determined using

j (p, r, ) : K

M z Q

9M z Q

R rR (p) : ; (1 ; 2p) N

R rR (p) : 9 (1 ; 2p), N

(4.220)

where p : 0, 1, 2, . . . , N 9 1. Notice that : 0 occurs when the middle of the zeroth sector coincides with the x-axis and its polarization is radially outward as shown in Fig. 4.53. It it is assumed that the source magnet is held stationary with one of its sectors in a similar orientation; therefore, represents the relative angular offset of the two magnets. We substitute Eq. (4.220) into Eq. (4.219) and obtain

, \ 0 (91)N rB(r, (, p)) dr, P N 0 where L is the length of the coupling and T() : 2M L Q

(, p) : ; (1 ; 2p). N

(4.221)

(4.222)

The coefficient 2M takes into account the fact that there are two surfaces at the Q

291


FIGURE 4.53

Cross section of radial coupling with reference frame.

interface between neighboring sectors (one for each sector). The integral in Eq. (4.221) can be evaluated numerically using Simpson’s method. The resulting torque formula is 2M LR* , \ ,P Q (91)NS (q)r(q)B(r(q), (, p)), (4.223) P P N P N O where N is the radial mesh coefficient (even), R* : R 9 R , and S (q) are P P the Simpson coefficient terms T() :

S (q) : P The integration points are

(q : 0) (q : 1, 3, 5, . . . ) (q : 2, 4, 6, . . . )

(4.224)

(q : N ). P

q r(q) : R ; (R 9 R ) (q : 0, 1, 2, . . . , N ). N P P The field component B (r, ) is given by Eq. (4.217). P

(4.225)

Calculations: We demonstrate Eq. (4.223) with some sample calculations. To clarify the analysis, we distinguish the parameters for the inner (source) and outer (drive) magnets using superscripts in and out, respectively. For example, R and R refer to the inner and outer radii of the inner source magnet, and R and R refer to the inner and outer radii of the drive magnet. We consider

292


a coupling with the following parameters: M : 5.0;10 A/m Q M : 5.0;10 A/m Q R : 2.0 cm R : 5.0 cm R : 6.0 cm R : 9.0 cm N : 6.0. (4.226) We compute the torque per unit length for a series of angular values : 0, 5, 10, . . . , 60° (rotation from the center of one pole to the next). The computed data are compared with corresponding FEA data in Fig. 4.54. We also determine the number of poles that render the maximum torque (all other parameters held constant). These data are shown in Fig. 4.55. It is clear that the maximum torque occurs when N : 8. Finally, we compute the peak torque as a function of the aspect ratio L/D of the coupling, where L is the length of the coupling and D : 2R is the diameter of the outer magnet. The same

FIGURE 4.54

Torque per unit length vs rotation angle.

293


FIGURE 4.55

Torque per unit length vs number of poles.

parameters as in the preceding are used and L is varied from D to 10D. The percent difference between the scaled 2D data and a corresponding 3D analysis is shown in Fig. 4.56. According to this analysis, the scaled 2D value is within 7% of the actual value even when the aspect ratio is 1. ) EXAMPLE 4.7.2 Derive an expression for the torque between two coaxial radially polarized multipole cylinders (Fig. 4.51) [19]. Assume that both magnets have the same number of poles and linear second quadrant constitutive relations of the form B : (H ; M).

(4.227)

where M : <M r . (4.228) Q The < sign takes into account the alternating polarity of adjacent poles. SOLUTION 4.7.2 The coupling consists of an inner and outer magnet. We refer to the inner magnet as the source magnet and the outer magnet as the drive magnet. The analysis proceeds in two steps: First, we obtain a free-space field solution for the inner magnet; second, the outer magnet is reduced to a

294


FIGURE 4.56

Percent error of 2D model vs aspect ratio.

distribution of equivalent currents, and the torque is determined by considering these currents to be in the external field of the source magnet. Source Magnet: The field for the inner magnet was derived in Example 4.2.6. The only field component that contributes to the torque is the radial component, which is determined using M , B (r, , z) : Q (91)Q>>H>I P 4 Q H I , ; r P S (n)r (n) sin( 9 ( j)) Q Q P Q L ;I (r, , z; r (n), ( j), z (k)) Q Q Q , ; S (m)(z 9 z ( j )) cos( 9 (m)) Q Q Q K

;I (r, , z; r (k), (m), z ( j )) . Q Q Q

(4.229)

295


The various functions and parameters in Eq. (4.229) are defined in Example 4.2.6. In the following analysis this component is an external field. Therefore, we adopt the notation B (r, , z) : B (r, , z). P P Torque: We determine the torque on the outer (drive) magnet. Let N , M , Q R , R and z and z denote the number of poles (even), magnetization, inner and outer radii, and end positions of this magnet, respectively. As a first step, we reduce the magnet to a distribution of equivalent volume and surface current densities J and j , respectively (Section 3.3). Once this is done, we compute K K the torque using Eq. (3.100), T:

4

r(r, );(J (r, );B(r, ))r dr d dz K

r(r, );(j (r, );B(r, ))da, (4.230) K 1 where V and S denote the volume and surface of the magnet. From Eq. (4.228) we have J : ;M : 0 and, therefore, the first term in Eq. (4.230) is zero. As K for the second term, there are N sectors to consider, and each sector has two surfaces with current densities that contribute to the torque. Consider the sector that is centered with respect to the x axis as shown in Fig. 4.53. This sector has surface currents along its radial sides at angular positions : 9/N and : ;/N If the sector is rotated by degrees the contributing surface current densities are given by: ;

R rR M z : ; /N Q z zz j (r, , z) : (4.231) K R rR 9M z : 9 /N Q z zz . It is assumed that the inner magnet is held stationary with one of its sectors centered about the x-axis. Therefore, represents an angular offset between the two magnets We substitute Eq. (4.231) into Eq. (4.230) and solve for the axial component. This gives

X

0

rB r, ; , z dr dz, (4.232) P N X 0 where B (r, , z) is given by Eq. (4.229). The coefficient 2N reflects the fact P that each sector contains two surface currents that contribute equally to the T () : 2N M Q X

296


torque, and that the total axial torque is the same for each sector. The integrals in Eq. (4.232) can be evaluated using Simpson’s method. The resulting torque expression is M LR* ,X ,P Q S (p)S (q) X P NN P X N O ;r(q)B r(q), ; , z(p) , (4.233) P N where L : z 9 z , R* : R 9 R , and S (q) and S (p) are the Simpson P X coefficient terms T () : X

2N

and

S (q) : P

(q : 0) (q : 1, 3, 5, . . . ) (q : 2, 4, 6, . . . ) (q : N ) P

(p : 0) (p : 1, 3, 5, . . . ) S (p) : X (p : 2, 4, 6, . . . ) (p : N ) P The integration points are as follows:

and

q r(q) : R ; (R 9 R ) N P

(q : 0, 1, 2, . . . , N ), P

p z(p) : z ; (z 9 z ) (p : 0, 1, 2, . . . , N ). N X X Calculations: We demonstrate Eq. (4.233) via a practical design problem. Consider an application that requires a torque of T : 90 Nm to be transmitted through a 0.45-cm enclosure (i.e., the coaxial magnets must have a 0.45-cm radial gap between them). Assume that the magnets are made from bonded NdFeB with M : 4.3;10, and that they have the following dimensions: Q R : 1.0 cm (inner radius of inner magnet) R : 5.0 cm (inner radius of outer magnet) R : 8.0 cm (outer radius of outer magnet) L : 5.0 cm (length).

297


FIGURE 4.57

Torque vs number of poles.

We compute the peak torque as a function of the number of poles N for a series of values for the outer radius of the inner magnet, R : 3, 3.5, 4, and 4.5 cm (when R : 4.5 cm there is a 0.5-cm radial gap between the inner and outer magnet). The computed data are shown in Fig. 4.57. The analysis shows that the peak torque (which occurs at an angular offset of : /N ) occurs at 4, 4, 6, and 10 poles, respectively. Notice that only the R : 4.5-cm data has a torque of 90 Nm. Thus, a ten-pole coupling with R : 4.5 cm is optimal for this application. ) EXAMPLE 4.7.3 Derive an expression for the torque between two identical axially polarized multipole cylindrical disks as shown in Fig. 4.58 [20, 21]. The disks have inner and outer radii R and R , respectively, and a thickness t . Let K N denote the number of poles on the surface of each disk, and let h denote the spacing between disks. Assume that both magnets have a linear second quadrant demagnetization curve of the form B : (H < M z ), (4.234) Q where the < sign takes into account the alternating polarity of adjacent poles.

298


FIGURE 4.58

Synchronous axial coupling.

SOLUTION 4.7.3 We use the charge model (Section 3.4). Our approach is as follows: First, reduce the magnets to equivalent charge distributions; next, determine the force imparted by one magnet to the other using Eqs. (3.120) to (3.123); Last, determine the torque. The volume and surface charge densities K and are given by Eq. (3.105). From Eq. (4.234) we find that K : 9 · (<M z ) : 0, (4.235) K Q and that : M · n : <M . K Q

(4.236)

299


Now, consider the force formulas (3.120) to (3.123). Because : 0 in both K magnets, the volume terms vanish and the force between the magnets reduces to

(x ) (x )(x 9 x ) I A A . F: K N K I N (4.237) I N 4

x 9 x N I N I Instead of evaluating Eq. (4.237) directly it is easier to derive the force in the following steps. First, the force between two point sources is obtained. Next, the force between opposing pole faces (sectors) is derived. Third, an equation is derived for the force between a sector of one magnet and an entire surface of the opposing magnet. Last, the total force is obtained by summing the contributions from all of the sectors and surfaces. Because torque is obtained from the force, its derivation is presented as well. Point Sources: We use cylindrical coordinates with one of the charges Q (x ) K in the x-y plane at (r , , 0), and the other Q (x ) in a plane defined by z : h K at (r , , h). The axial force between Q (x ) and Q (x ) follows from Eq. K K (3.124), Q (r , , 0)Q (r , , h)h K K F : . (4.238) X 4 [r ; r 9 2r r cos( 9 ) ; h] The axial component of torque imparted to Q (x ) by Q (x ) is K K T : (r ;F ) · z , (4.239) X where F is the force imparted to Q (x ) by Q (x ), and r is the vector from K K the origin of the coordinate system to the first particle. The torque T is given by X Q (r , , 0)Q (r , , h)r r sin( 9 ) K . (4.240) T : K X 4 [r ; r 9 2r r cos( 9 ) ; h] Sectors: Next, we compute the axial force and torque imparted to one pole face (sector) by an opposing pole face. Again, without loss of generality, one sector is taken to lie on the x-y plane symmetrically positioned about the x-axis ( : 0), while the other is taken to lie on the plane defined by z : h, symmetrically positioned about an angular displacement : (Fig. 4.59). Q The idea is to discretize each sector into a mesh of subsections (elements), define a point charge at the midpoint of each element in accordance with Eq. (4.236), and then compute the force and torque imparted to one sector by the other by summing the contributions from the individual point charges. The area of each sector is area of sector :

(R 9 R) . N

(4.241)

300


FIGURE 4.59

Relative angular offset of two sectors.

In addition, each sector spans an angular measure 2 : . (4.242) N The first step is to discretize the area (4.241) into a set of indexed elements. These elements are chosen to have equal areas so that they contain an equal amount of surface charge. Because the area of the sector is spanned by the two coordinate variables r and , two indices (i, j) are required. Let N and N be P the number of mesh divisions in these two variables, respectively. Then each element will have the same area A given by (R 9 R) . (4.243) N N N P Furthermore, choose each element so that it spans an angular measure given by A :

:

. N

(4.244)

Let (r , ) denote the midpoint of the element A , and let R(i) and R(i ; 1) G H GH

301


denote its radial boundaries. Then r : R(i ; 1) 9 R(i), G

(4.245)

and r: G

R(i ; 1) ; R(i) 2

(i : 1, 2, . . . , N ). P

(4.246)

As the sector is centered about : , Q

: 9 ; (j 9 1) ; H Q N 2

(4.247)

for j : 1, 2, . . . , N . The endpoints are computed using R(i ; 1) :

R(i) ;

(R 9 R) N P

(i : 1, 2, . . . , N ), P

(4.248)

where R(1) : R (the inner radius). Equation (4.248) is used to generate the endpoint values from which the midpoint values are obtained. Therefore, Eqs. (4.246)—(4.248) define a scheme for discretizing the sector into elements and assigning values to the midpoints of each element. Next, consider the magnetic interaction between an element of one sector and an element of an opposing sector. Let the coordinates of the midpoints of the two elements be (r , , 0) and G H (r , , h), respectively. The magnetic charge associated with each element is GY HY Q (r , , 0) : (r , , 0)A K G H G H

(4.249)

Q (r , , h) : (r , , h)A, K GY HY GY HY

(4.250)

and

respectively. From Eqs. (4.236) and (4.243) these charges reduce to

(R 9 R) . Q : <M K Q N N N P

(4.251)

where, once again, the < term takes into account the polarity of oppositely charged sectors. The force and torque are obtained by summing the contributions of all of the elements in accordance with Eqs. (4.238) and (4.240). The resulting

302


expressions are

M (R 9 R) ,P , F (h, ) : Q X Q 4 N N N P G H , , h ; P , [r ; r 9 2r r cos( 9 ) ; h] GY GY HY G G GY H HY

(4.252)

and T

M (R 9 R) ,P , (h, ) : Q X Q 4 N N N P G H , , r r sin( 9 ) G GY H HY . ; P [r ; r 9 2r r cos( 9 ) ; h] GY G GY H HY GY HY G

(4.253)

Torque: Next, we determine the total axial force and torque taking into account the interaction of all the sectors of both magnets. There are two magnets that are labeled upper and lower in accordance with their relative positions. These magnets are identical and have a thickness t . Each magnet has two surfaces K (labeled top and bottom) and each surface contains N sectors. The upper magnet has its bottom surface in the plane defined by z : h and its top surface lies in the plane defined by z : h ; t . Assume that this magnet is rotated by K an angle as shown in Fig. 4.60. Under this condition, the angular positions

FIGURE 4.60

Rotated magnet surface.

303


of the centers of its sectors (, n) are Q 2 (, n) : ; (n 9 1) (n : 1, 2, . . . , N ). (4.254) Q N The lower magnet is centered about : 0, and has its top surface in the x-y plane, and its bottom surface in the plane defined by z : 9t . Consider the K sector of this magnet on the top surface centered about : 0. Based on the results of the previous section, the axial force on this sector due to all of the sectors on both the top and bottom surfaces of the upper magnet is , F (h, ) : (91)L[F (h, (, n)) X X Q L 9F (h ; t , (, n))]. X K Q Continuing in this fashion and taking into account the contributions from all the sectors and surfaces, the following formulas are obtained for the total force and torque,

N M (R 9 R) , ,P , F () : Q X N N N 4 P K L G H , , (91)L>K h K K , ; P [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K G GY H HY GY HY G and

(4.255)

N M (R 9 R) , ,P , T () : Q X 4 N N N P K L G H , , (91)L>K r r sin( 9 (, n)) K G GY H HY ; P , [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K GY HY G G GY H HY where

(4.256)

:9 ; (j 9 1) ; , H N 2 (, n) : (, n) 9 ; (j 9 1) ; , HY Q N 2 and (, n) is given by Eq. (4.254). The terms h and are given by Q K K h : h ; mt (m : 0, 1, 2) K K h : separation of magnets

t : thickness of magnets K

304


and

1 (m : 0)

: 2 (m : 1) K 1 (m : 2). The angle is the relative offset of the two magnets. The condition : 0 occurs when poles of one magnet align with the poles of opposite polarity on the other magnet (the torque is zero when : 0). Note that Eqs. (4.255) and (4.256) contain nested summations of elementary functions, which are easy to program. However, it is important to note that when programming nested summations, a higher degree of accuracy can be achieved with fewer mesh points (lower values N and N) if the total of a single summation over one index is evaluated and P saved separately (as an intermediate step), and then the totals of the successive summations are added together to obtain the final sum. For example, Eq. (4.256) can be evaluated as follows:

N M (R 9 R) , T () : Q S (, n, m), X N N N 4 P K L where , S (, n, m) : P S (, i, n, m), G with , S (, i, n, m) : S (, i, j, n, m), H and , S (, i, j, n, m) : P S (, i, i, j, n, m), GY and , (91)L>K r r sin( 9 (, n)) K G GY H HY S (, i, i, j, n, m) : . [r ; r 9 2r r cos( 9 (, n)) ; h ] GY K HY G G GY H HY Calculations: We demonstrate Eqs. (4.255) and (4.256) with a practical design. Consider an application that requires a torque of 1.0 Nm to be transmitted through a 10-mm enclosure. To begin the analysis, choose the

305


FIGURE 4.61

Torque vs number of poles.

following nominal magnet dimensions, R : 1.0 cm R : 4.0 cm t : 1.5 cm K

(inner radius) (outer radius) (thickness).

The coupling is to be made from bonded NdFeB with a 10% concentration of binder by volume which renders a bulk magnetization M : 4.3;10 A/m. The Q peak torque is computed as a function of the number of poles N for a series of separations h : 10, 12, 14, and 16 mm. The analysis is performed with N P and N set to 12. We find that the peak torque (which occurs at an angular offset of : /N ) occurs at 6, 4, 4, and 4 poles, respectively (Fig. 4.61). Notice that the magnets can be separated by as much as 14 mm and still render the desired torque. To complete the analysis, the torque and force are computed as a function of for the h : 14 mm separation (Fig. 4.62). The maximum force is approximately 60 N and is attractive (negative) at : 0° and repulsive (positive) at 90°. )

306


FIGURE 4.62

Torque and force vs .

4.8 MAGNETIC RESONANCE IMAGING Magnetic resonance imaging (MRI) is widely used for noninvasive diagnosis of the human body. In this section, we briefly review the principles of MRI, and the role that permanent magnets play in this technology. The MRI process itself is relatively complex and we present a brief and simplified description of it here. The interested reader can find more detailed discussions of MRI in numerous texts [22—24]. In MRI, radio frequency (rf) signals are used to create sectional images of the body. To obtain an image, a patient is first placed in a cavity (imaging region) that is permeated by a strong magnetic field. The nuclear magnetic moments within the patient’s tissue tend to align either parallel or antiparallel to the field, with the former having a lower energy than the latter (Fig. 4.63b). In a given specimen there are a large number of magnetic moments and most of the parallel moments are cancelled by the antiparallel moments leaving a relatively small number of unpaired parallel moments. The interaction between the magnetic

4.8 MAGNETIC RESONANCE IMAGING

307

moments and the field causes them to precess about the field direction at an angular frequency (Larmor frequency), : B , (4.257) where is the nuclear gyromagnetic ratio and B is the applied flux density. The magnetic moment of the hydrogen nucleus (proton) is used for MRI. Hydrogen is abundant in biological tissues, and the proton has a relatively high gyromagnetic ratio ( : 42.6 MHz/T). The Larmor frequency for hydrogen is 64 MHz when B : 1.5 T. Once the nuclear moments are aligned, the specimen is subjected to a pulsed rf signal that is tuned to the Larmor frequency. The rf signal induces transitions between the lower- and higher-energy proton spin populations (Fig. 4.63c). As the moments transition from the higher- to the lower-energy state they generate an rf signal that is sensed by a coil

Orientation of nuclear magnetic moments: (a) random orientation, (b) alignment in a B-field; (c) reorientation due to absorption of rf energy; and (d) reorientation due to reradiation.

FIGURE 4.63

308


and processed to create sectional 3D images (Fig. 4.63d). Spatial imaging can be accomplished by use of additional gradient field coils that have fields that superimpose with the primary field to produce a predefined variation in the bias field across the specimen. This gives rise to a spatial variation of the Larmor frequency, which in turn gives rise to a variation of the generated rf signal and which can be processed to form an image. There are essentially three different magnetic field systems used in MRI apparatus: 1. conventional electromagnets; 2. superconducting coils; and 3. permanent magnet structures. Conventional electromagnets consist of multiple turns of low gauge wire or conducting ribbons usually used in conjunction with soft magnetic cores and pole pieces. Electromagnet coils have a relatively high resistance and therefore they consume large amounts of energy and this causes heating. The heating sets a practical limit on the working current of such coils and, therefore, limits magnetic field strength. On the other hand, superconducting coils are capable of producing very stable and uniform fields of up to 4.0 T. However, they require liquid cryogens and cooling apparatus to suppress their resistance. Such coils are, therefore, expensive and require extensive preparation and maintenance. Permanent magnet systems are competitive with current-driven coils for MRI applications that require fields in the range of 0.1—0.5 T. They consume no energy and require little maintenance. When designing an MRI magnetic field system, the following parameters need to be considered: 1. 2. 3. 4. 5.

magnetic field strength; the homogeneity of the magnetic field; the temporal stability of the magnetic field; fringing fields; and patient access to the imaging volume.

A wide range of field strengths can be used for MRI. For example, whole body MRI has been achieved with fields ranging from 0.02 T (0.85 MHz) to 4.0 T (170 MHz). Higher field strengths are usually desirable because the number of transitions from high- to low-energy states scales approximately linearly with field strength. The three magnetic field systems can be ranked in terms of their field strength as follows: superconducting coil (3 T); conventional electromagnet (0.2—0.4 T); and permanent magnets (0.1—0.5 T).


309

A high degree of field uniformity is required for MRI. This is defined as the normalized maximum deviation in field strength between two points in the imaging volume B B 9 B

. : B B The field homogeneity is usually specified in ‘‘parts per million’’ for a given diameter of spherical volume (dsv). It can also be expressed as a ratio of frequencies using the Larmor relation (4.257). In this case it is given by f 9f f

: . f f Typical requirements for MRI field homogeneity are 10 to 100 ppm over a spherical imaging volume with a diameter of 50 cm (50 dsv). Superconducting coil systems have the highest degree of field uniformity (10 ppm, 50 dsv), followed by permanent magnet systems (10 ppm, 20 dsv), and then conventional electromagnet systems (30 ppm, 40 dsv). In MRI, data acquisition occurs over time. Therefore, it is important that the magnetic field have a high degree of temporal stability. Instability can be caused by variations in the field source itself, or by time dependent factors such as the movement of external field sources or ferromagnetic materials. Superconducting coils are the most stable field source. On the other hand, conventional electromagnets are susceptible to variations in their power source. Permanent magnets also exhibit field instability due to fluctuations in the ambient temperature. In this regard, SmCo magnets have an advantage over NdFeB due to their lower temperature coefficient (Section 1.20). The magnetic fringe field is another factor that must be considered when evaluating MRI magnet systems. A related issue is magnetic shielding. The fringe field can have undesired effects on medical equipment in proximity to the MR imager. Conversely, external field sources or the movement of ferromagnetic materials can degrade MR image quality if the system is not properly shielded. These effects can be reduced using active or passive shielding. In active shielding, additional coil sets are arranged inside the imager and activated to counter the fringe field. In passive shielding, a soft magnetic enclosure is used to confine the field and to prevent stray fields from penetrating the imaging cavity. The final factor we consider is patient access to the imaging region. An MRI magnet system must provide an imaging region that can

310


FIGURE 4.64

Comparison of MRI magnetic field systems [25].

accommodate the patient (or body part), a supporting structure (couch), and the rf coils. This space must be sufficient to provide minimal physical and psychological comfort for the patient. For example, in whole body MRI the magnet system cavity is cylindrical with a diameter of 1.0 m. However, the free access diameter for imaging is approximately 0.75 m. In pediatric MRI, the corresponding diameters are 0.5 and 0.39 m, respectively. For research applications, the magnet cavity has a diameter ranging from 0.2 to 0.4 m, and the imaging region has a diameter of 0.135 to 0.31 m. The characteristics of the three MRI magnetic field systems are summarized in Fig. 4.64. As already noted, permanent magnet systems can be used for MRI with field requirements that range from 0.1 to 0.5 T. The main advantage of these systems is that they consume no power. Moreover, when rareearth magnets are used the size and weight of such structures are greatly reduced relative to conventional materials. The MRI permanent magnet systems are classified as either yokeless, yoked, or hybrid. Yokeless systems consist of permanent magnets alone. The imaging field is confined and focused by the magnetization pattern within the magnets. The ring dipole arrangement shown in Fig. 4.65a is an example of a yokeless system [25]. This is composed of eight permanent magnet segments with magnetizations indicated by the solid


FIGURE 4.65

311

Two MRI magnet systems: (a) ring dipole; and (b) H-frame [25].

arrows. This assembly provides a transverse magnetic field across the imaging region as indicated by the dotted lines. The field generated by these systems is typically 0.3 T. In yoked systems a high-permeability material surrounds the permanent magnet structure. This enhances the field in the imaging region and minimizes the fringe field. The H-frame structure shown in Fig. 4.65b is an example of a yoked system [25]. In this structure, permanent magnet

312


blocks are supported in a ferromagnetic frame called a yoke. The yoke not only supports the magnets but also serves as a flux return path for the imaging field. Soft magnetic pole pieces are also used to enhance the strength and uniformity of the imaging field. The H-frame structure is the design of choice for MRI despite its weight and size. Hybrid structures contain yoked and yokeless sections. The design of MRI magnet structures is complicated. In the design of conventional magnet structures, the analyst solves the ‘‘direct’’ problem in which the geometry and magnetization are given, and the field distribution is determined. In MRI, the analyst is faced with the more difficult ‘‘inverse’’ problem in which the field strength and uniformity are specified across the imaging region and the geometry and magnetization of the structure need to be determined. There is no unique solution to such problems in that a specified field distribution within a closed region can be obtained using an infinite number of different structures [26]. One design approach is to select one of the many ideal closed structures that give the desired interior field distribution and then remove a portion of the material to allow access to the cavity. These openings give rise to undesired field nonuniformities that must be compensated for. Remedies for this include the modification of the geometry or magnetization, and/or the insertion of additional components such as high-permeability materials.

4.9 ELECTROPHOTOGRAPHY Electrophotography is one of the most widely used methods for image reproduction and printing [27, 28]. The central element of this process is the photoconductor (PC), which consists of a thin sheet (10—50 m) of photosensitive material that is coated onto a ground plane. The PC generates electron-hole pairs as it absorbs incident light. Once generated, the electrons and holes separate and move under the influence of an internal electric field with the electrons moving to ground. This process forms the bases of electrophotographic image reproduction. The electrophotographic process comprises six distinct steps: Charge; Expose; Develop; Transfer; Fuse; and Clean (Figs. 4.66 and 4.67). We briefly review these steps in the order in which they occur: 1. Charge: The first step in the electrophotographic process entails charging the PC. The PC (which is initially uncharged) passes through a corona-charging device. This device creates ionized gas

4.9 ELECTROPHOTOGRAPHY

FIGURE 4.66

313

Electrophotographic system [27].

molecules, which are absorbed by the PC and thereby create a uniform charge distribution on its surface. This charge distribution gives rise to an electric field within the PC (Fig. 4.67(1)). 2. Expose: To form an image, the charged surface of the PC is exposed to light that is reflected from an existing image or produced by a laser writer. A portion of the incident light is absorbed in the PC and electron-hole pairs are created in the exposed regions. These charged carriers separate and drift under the influence of the PC’s internal field. The electrons move to ground and the holes move to the charged surface where they discharge the exposed regions. This produces a charge pattern on the exposed surface of the PC that mirrors the image. This is called the latent image (Fig. 4.67(2)). 3. Develop: During this step, charged pigmented toner particles are transported proximal to the latent image and are electrostatically attracted to regions of the PC in accordance with the local surface charge (Fig. 4.67(3)). This transforms the latent image into a real image.

314


FIGURE 4.67

Electrophotographic process.

4. Transfer: In the transfer step, the toner on the PC is electrostatically transferred to paper. This is accomplished by depositing a uniform charge of opposite polarity to that of the toner on the back side of the paper prior to it contacting the PC (Fig. 4.67(4)).


315

5. Fuse: During this step, the image is permanently fixed to the paper by melting the toner onto the paper surface (Fig. 4.67(5)). 6. Clean: In this final step, the PC is discharged and cleaned of any excess toner using coronas, lamps, brushes, and/or scraper blades (Fig. 4.67(6)). Once this is done the PC is ready for the next imaging cycle. We are primarily interested in the development of the latent image since this is the most magnetically intensive step of the process. Most electrophotographic copiers and printers use ‘‘magnetic brush’’ technology to implement this step. While there are various implementations of this technology, they all perform essentially the same function, namely the transportation of developer material from a reservoir to the latent image. The developer material is usually either monocomponent or two component. In monocomponent systems, the toner particles are magnetic and on the order of 5—30 m in diameter. The magnetic transfer field couples directly to these particles. In two component materials, there are magnetic ‘‘carrier’’ particles and nonmagnetic toner particles. The carrier particles range from 70 to 250 m in diameter and are much larger than the toner particles. The magnetic transfer field couples to the carrier particles and several toner particles are transported by each carrier particle. A typical magnetic brush system is shown in Fig. 4.68. In this system the magnetic toner/carrier material is transported to the latent image by

FIGURE 4.68

Magnetic brush development station [28].

316


riding on the surface of a conductive cylindrical shell. The conductive shell encloses a rotating radially polarized permanent magnet cylinder. The rotating magnet gives rise to a time-varying magnetic field at the surface of the shell, which is typically a few tenths of a tesla or lower. This time-varying field provides a magnetic transport force that causes the toner particles to move in the opposite direction as shown. In magnetic brush systems the magnetic transport force dominates the centripetal, gravitational and electrostatic forces. When two magnetized toner/carrier particles approach one another they exert a dipole-dipole force on each another. This force is far greater than all the other ‘‘ambient’’ forces, including the magnetic force exerted by the magnetic brush. Specifically, the dipole-dipole force is typically more than an order of magnitude greater than the force due to the magnetic brush [27]. Consequently, coaligned neighboring toner/carrier particles tend to adhere to one another, forming chains of particles (Fig. 4.69. During transport, these chains behave as a single mechanical entity. They ride on the surface of the conductive shell and cartwheel end over end as they are transported. The dynamics of this are complicated and beyond the scope of this text. We simply derive here the force on an isolated toner particle in the field of the magnetic brush. For this, we need the field components for an infinite, radially polarized multipole

FIGURE 4.69

Monocomponent toner transport [28].

317


cylinder. These were derived in Example 4.2.5. The radial and azimuthal components are given by Eqs. (4.66) and (4.67), respectively, B (r, ) : ir\G>U (M , R , R , ) cos(i) P G G G

(4.258)

and B (r, ) : ir\G>U (M , R , R , ) sin(i), (4.259) G G G where U (M , R , R , ) is given by Eq. (4.65) with M given by (4.64) (the G G G superscript 2 has been dropped). In the following example we derive the force on a magnetic particle in the field of a magnetic brush. EXAMPLE 4.9.1 Determine the force on a spherical toner particle of radius R in the field of a cylindrical magnetic brush. Assume that the toner is made from a magnetically linear and isotropic material with permeability and susceptibility . K SOLUTION 4.9.1 The force on the toner particle is computed using Eq. (3.116). We assume that (M · )B is constant over the volume of the toner particle. This gives F:

(M · )B dv 4 (m · )B . where m is the magnetic dipole moment of the toner particle, and B is the field of the magnetic brush (multipole cylinder). Specifically, : B (r, )r ; B (r, ) , P where B (r, ) and B (r, ) are given by Eqs. (4.258) and (4.259), respectively. P From Example 1.8.1 we know that the dipole moment of the toner particle is B

4R( 9 ) H . ( ; 2 )

(4.260)

4 R( 9 ) (B · )B . ( ; 2 )

(4.261)

m: Thus, we find that F:

318


We evaluate Eq. (4.261) and obtain F(r, ) :

4R( 9 ) ( ; 2 ) ;

B (r, ) P

; B (r, ) P

B (r, ) 1 B (r, ) P P ; B (r, ) r r r

B (r, ) 1 B (r, ) ; B (r, ) . r r

(4.262)

Finally, we substitute Eqs. (4.258) and (4.259) into Eq. (4.262) and find that 4 R( 9 ) ( ; 2 ) ; ii(i ; 1)r\G>GY>U U cos(i) cos(i) G GY G GY ; iiYr\G>GY>U U sin(i) sin(i) r G GY G GY ; ii( i ; 1)r\G>GY>U U cos(i) sin(i) G GY G GY 9 iir\G>GY>U U sin(i) cos(i) , (4.263) G GY G GY where U (M , R , R , ) is given by Eq. (4.65) with M given by Eq. (4.64). G G G )

F(r, ) :9

4.10 Magnetooptical Recording Magnetooptical (M-O) recording is routinely used for digital data storage [29—31]. Just as in magnetic storage, the information is recorded along tracks on data disks. A track is a narrow annulus region a distance r from the center of the disk. Each track is divided into a number of smaller segments or sectors and each segment stores a single block of data of typically either 512 or 1024 bytes. In conventional M-O systems, data retrieval/storage is accomplished using a read/write head that is held stationary while the disk spins (Fig. 4.70). Typical disk rotation speeds are 1200 and 1800 rpm for slower drives, and 3600 rpm for high data-rate systems.

4.10 MAGNETOOPTICAL RECORDING

FIGURE 4.70

319

Magnetooptic recording system.

An M-O storage disk consists of a multilayer stack of thin films. The conventional quadrilayer structure is shown in Fig. 4.71. This structure is optically designed to provide near-optimum readback. The functional storage layer in this stack is a thin magnetic film (250 Å) that is currently made from a rare-earth transition-metal (RE-TM) material, preferably an alloy based on Tb, Fe, and Co. This material is characterized by a vertical magnetic anisotropy and a high magnetic coercivity on the order of several tenths of a tesla (several kilo-gauss at room temperature). Because of the high coercivity, the film is magnetically very stable and

FIGURE 4.71

An M-O recording thin film stack [31].

320


FIGURE 4.72

Magnetooptic recording process.

can retain a recorded magnetization pattern indefinitely under normal operation. However, this layer tends to be chemically unstable due to the high affinity of rare-earth elements for oxygen. In practice, it is sandwiched between two chemically stable dielectric layers as shown. Data is recorded in the RE-TM layer using a thermomagnetic process. In conventional systems, recording occurs with the media moving (linear velocity < 10 m/s) between a stationary laser source and a fixed bias field source (Fig. 4.72). The film is initially polarized in one direction (e.g., the down direction). To record information, the laser is pulsed (duration < 50 ns) and focused so as to heat a local spot of the film (diameter < 1m). As the temperature of the spot increases the coercivity within it decreases. Eventually, the spot reaches the Curie temperature (T < 473 K) at which point the local magnetization reverses A and aligns with the bias field, which is typically 0.02—0.04 T across the film. The temperature at the center of the spot can reach a value as high as 500 K, which, while sufficient to reverse the magnetization, is not high enough to damage the film. When the laser is turned off, the temperature decreases back to room temperature, and the magnetization within the spot remains reversed. Thus, the spot represents a recorded bit of information. By repeating this process, regions of the disk are selectively reversed-magnetized and the resulting magnetization pattern represents the recorded information. Data readout in M-O systems is based on the magnetooptic Kerr effect. Specifically, when linearly polarized light is normally incident on a perpendicular magnetic medium its plane of polarization undergoes a slight rotation upon reflection (Fig. 4.73). This rotation is known as the I polar Kerr effect and the direction of rotation depends on the orientation

4.10 MAGNETOOPTICAL RECORDING

321

Kerr effect: (a) reflection from a north pole; and (b) reflection from a south pole [30].

FIGURE 4.73

of the magnetization within the film. To read data, a polarized laser beam is applied to the disk as it rotates and the reflected light is collected and processed. The same laser source is usually used for both the read and write process. However, the output power is substantially reduced for readout so as to avoid erasure. To erase information, the bias field is reversed and the film is heated at a constant power level until an entire sector is erased. The disk makes at least one revolution between the erasure of an old block and the recording of a new one. Therefore, the bias field reversal required for the erase and rerecord sequence can be implemented over a relatively long

322


FIGURE 4.74

Cross section of a rotary actuator.

time frame (16 ms at 3600 rpm). It is convenient and common to use permanent magnet actuators for the bias field mechanism. In these actuators the magnet that provides the bias field is also a functional part of the actuator. A cross-sectional view of a rotary bias field actuator is shown in Fig. 4.74. This consists of a rotating bipolar cylindrical bias magnet surrounded by a stationary concentric conductive shell, and a drive coil. When the coil is energized it rotates the magnet to one of two equilibrium positions, thereby producing a bias field of desired polarity across the recording media. Rotation of the magnet gives rise to eddy currents in the conductive shell that impart a drag torque and reduce oscillations of the magnet about its equilibrium positions. We present an analysis of this actuator in Example 5.11.1. We work through a design of the bias magnet in the following example [32]. EXAMPLE 4.10.1 Determine the dimensions of a cylindrical bias field magnet for use in an M-O system (Fig. 4.10). The magnet must provide a bias field of between 0.02 and 0.04 T along a radial length of the disk which is taken to be 100 mm. Assume that the axis of the magnet is positioned 4 mm below the recording layer and aligned with the radial length of the disk. Assume that the magnet has a second quadrant constitutive equation B : (H ; M ). (4.264) Q Let R, z and z denote the radius and respective end points of the cylinder. SOLUTION 4.10.1 For the erase/rerecord sequence the magnet needs to be rotated between two equilibrium positions. We choose these positions to be when the magnetization is at S (n) cos((n)) P 2N L I ;[r 9 R cos( 9 (n))]I(r, , z; R, (n), z ), I where N is the Simpson’s method mesh number (even), (n) :

n 2 N

(4.265)

(n : 0, 1, 2, . . . , N),

and

(n : 0) (n : 1, 3, 5, . . . ) S (n) : (n : 2, 4, 6, . . . ) (n : N). The function I(r, , z; R, , z ) is given by I F(r, , z; r, , z ) I if

IY

where

r ; r 9 2rr cos( 9 ) " 0 or 91 2(z 9 z ) I if r : r, cos( 9 ) : 1, z " z , I

(z 9 z )g(r, , z; r, , z ) I I . F(r, , z; r, , z ) : I r ; r 9 2rr cos( 9 ) We want to evaluate the field along the y-axis. For a given observation point (0, y, z) on this axis the field is determined by substituting r : y, and : /2 9 in Eq. (4.265), that is, K B (y, , z) : B y, 9 , z , (4.266) W K P K 2

where is the angular position of the north pole of the magnet relative to the K

324


x-axis ( is positive in the counterclockwise direction). We use Eqs. (4.265) K and (4.266) to design the magnet. Calculations: For the design, we assume that the magnet is made from sintered NdFeB material with a uniform magnetization of M : 7.2;10 A/m Q (B : 9000 G). First, we determine a viable magnet radius R. The length of the P magnet is set to 100 mm (z : 950 mm, z : 50 mm) and Eq. (4.265) is evaluated for a range of R values, 0.5 mm R 1.5 mm with : 85°, r : 4 K mm, and z : 0. A plot of B vs R is shown in Fig. 4.75. From this plot it is P obvious that a magnet with R : 1 mm will suffice because its field is 0.028 T, which is within the desired range. Having determined a viable radius, we examine the angular dependency of the field. Specifically, we compute the field with R : 1 mm, r : 4 mm, and z : 0 for a series of angular offsets relative to the y-axis, 0 90° 9 20°. This corresponds to a magnet rotation K 70° 90°. These data are shown in Fig. 4.76. Note that the field is within K the desired operating range over this entire angular range. Thus, there is considerable tolerance in the angular position of the magnet. The next step is the determination of a viable length for the magnet. For this, we set R : 1 mm

FIGURE 4.75

The M-O bias field vs magnet radius R (h : 100 mm).

4.11 FREE-ELECTRON LASERS

325

The M-O bias field 4 mm above the magnet vs angular offset 90° 9 (z : 0). K

FIGURE 4.76

and vary the length h as follows: h : 100, 104, 108 mm (954 mm z 950 mm, and 50 mm z 54 mm). The bias field is computed with : 85° at K a distance r : 4 mm above the magnet along a series of points that span its length. A plot of B for h : 100, 104, 108 mm is shown in Fig. 4.77. Note that P the field data is plotted for positive values of z only (the data for negative z are the mirror image). We find that a magnet length of h : 108 mm will suffice because the field is virtually constant along the specified length of 100 mm (0 z 50 mm). In summary, the analysis shows that a sintered NdFeB magnet with a radius R : 1 mm and length h : 108 mm rotated an angle of > E · dl : 9 · dl Q \ \

source

where and > \

: 9( 9 ), > \ are the potentials at the positive and negative

340

CHAPTER 5 Electromechanical Devices

terminals of the source, respectively. The source voltage V is defined as Q the potential difference V : 9 and, therefore, Q > \ > V Y9 E · dl. (5.7) Q Q \ Thus, Eq. (5.6) becomes

V : Q

E · dl ;

resistor

E · dl ;

capacitor

E · dl ;

1

B · ds. t

(5.8)

inductor

In the resistor, J : E (Ohm’s law) and, therefore,

E · dl :

resistor

J · dl

resistor

: iR.

(5.9)

The same analysis applies to the inductor,

E · dl :

inductor

J

· dl

inductor

: iR , (5.10) where and R are the conductivity and resistance of the inductor. For the capacitor, we have

E · dl :

1 C

i dt.

(5.11)

capacitor

Substituting Eqs. (5.9), (5.10) and (5.11) into Eq. (5.8) yields 1 V : i(R ; R ) ; Q C

i dt ;

B · ds. t

(5.12)

1 Now, consider the last term in Eq. (5.12). This can be written as d dt

B · ds : 1

d" , dt

(5.13)

341

5.3 QUASI-STATIC FIELD THEORY

where " is called the flux linkage. That is, " is the total flux through (linking) the circuit, "Y

B · ds (flux linkage). 1 The term d"/dt represents the voltage induced in the circuit by a time rate of change of magnetic flux through it. For a simple RLC circuit (with no moving members), " is a function of the current only and, therefore, d" d"(i) di : . dt di dt

(5.14)

We define the circuit inductance L to be LY

d"(i) di

(inductance),

and then obtain d" di :L . dt dt

(5.15)

Finally, substitute Eq. (5.15) into Eq. (5.12) to obtain 1 V : i(R ; R ) ; Q C

i dt ; L

di . dt

(5.16)

This is Kirchhoff’s voltage law for an RLC circuit. In electromechanical devices, the flux linkage " is no longer a function of the current only. Instead, it is a function of both the current and the position of a moving member. Thus, for the analysis of electromechanical devices the preceding derivation needs to be generalized to include the effects of electromechanical coupling. We discuss this in detail in Section 5.4.1.

5.3.2 Moving Reference Frames In our study of electromechanical devices we will need to determine the voltage induced in a coil as it moves through a stationary magnetic field. This requires an understanding of quasi-static field theory as observed in two different inertial reference frames (in motion relative to one another). For our purposes, an inertial reference frame is one that moves with constant velocity. In this section, we derive the transformation laws

342


FIGURE 5.3

Inertial reference frames in relative motion along one axis.

for quasi-static fields as observed in two different inertial reference frames. Consider two inertial reference frames, a stationary ‘‘laboratory’’ reference frame O with coordinates (x, y, z, t), and a moving reference frame O with coordinates (x, y, z, t). Frame O moves with velocity u with respect to O as shown in Fig. 5.3. Observers in O and O measure different values for the electromagnetic fields. Rigorous comparisons of these measurements can be made if one knows how the fields transform from O to O or vice versa. The transformation laws follow from the theory of special relativity. A detailed presentation of this theory is beyond the scope of this text and is also unnecessary for our presentation. For our purposes, it suffices to state the underlying principles of the theory, and then simply write down the coordinate transform relations. The interested reader can find detailed presentations of special relativity in numerous excellent texts [6]. The basic postulates of special relativity are as follows: 1. The laws of physics are the same in all inertial reference frames. 2. Observers in inertial reference frames measure the same value for the speed of light. The first postulate implies that the quasi-static field equations must have the same form in both O and O as shown in Eqs. (5.17):

343


Quasi-Static Equations Reference Frame O

Reference Frame O

;H : J

;H : J

·B:0 B ;E : 9 t ·D:

· B : 0 B ;E : 9 t · D : .

(5.17)

Here, is the del operator with respect to the primed coordinate, :

x ; y ; z . x y z

Our goal is to determine the transformation laws for the fields going from O to O and vice versa. We can determine these if we know the transformation relations for the coordinate variables. These follow from the second postulate of special relativity and are given by t : (t 9 #x/c) x : (x 9 #ct) y : y

(5.18)

z : z, where :

1 (1 ; #

,

and u #: . c Here, c is the speed of light in a vacuum. We are interested in low-velocity electromechanical devices for which u c. Therefore, #0

(5.19)

1.

(5.20)

and

We substitute Eqs. (5.19) and Eq. (5.20) into Eq. (5.18), and find that for low velocities the Lorentz transformation relations Eq. (5.18) reduce to

344


FIGURE 5.4

Inertial reference frames in relative motion.

the Galilean transform relations, t : t x : x 9 ut y : y

(5.21)

z : z. This analysis generalizes to the case of an arbitrary velocity as shown in Fig. 5.4. For this case we find that t : t and x : x 9 ut, or x : x 9 u t V y : y 9 u t (5.22) W z : z 9 u t. X Now that we know the transform relations for the coordinates, we can determine the corresponding relations for the field equations. These will ultimately enable us to compare the fields in the two reference

345


frames. We start by considering a scalar-valued function f of the primed coordinates, f (x, y, z, t). By virtue of Eq. (5.22), f can be considered to be a function of the unprimed coordinates, f (x, y, z, t) : f (x(x, t), y(y, t), z(z, t), t(t)). We use the chain rule to determine the derivatives of f with respect to the unprimed coordinates. We consider spatial derivatives first and find that f f x : x x x :

f . x

It follows that f : f .

(5.23)

Next, we consider the time derivative. From the chain rule we have f f t f x f y f z : ; ; ; y t z t t t t x t :

f f f f 9 u 9 u 9 u . t x V y W z X

This can be rewritten as f f : 9 u · (f), t t or, alternatively, as f f : ; u · (f). t t

(5.24)

The relations (5.23) and (5.24) relate the derivatives in O and O for scalar-valued functions. A similar analysis applies to vector-valued functions. Let A(x, y, z, t) be a vector-valued function of the primed coordinate variables. This can also be considered to be a vector-valued function of the unprimed coordinates A(x, y, z, t) : A(x(x, t), y(y, t), z(z, t), t(t)). It is easy to check that · A : · A,

(5.25)

346


and that ;A : ;A.

(5.26)

Similarly, by taking the time derivative we find that A A : ; (u · )A. t t This can be rewritten as A A : ; u( · A) 9 ;(u;A), t t

(5.27)

where we have used the identity ;(A;B) : (B · )A 9 (A · )B ;A( · B) 9 B( · A). Finally, substitute Eqs. (5.25), (5.26), and (5.27) into the quasi-static equations for O in Eq. (5.17) and obtain ;H : J · B : 0 ;(E 9 u;B) : 9

B t

(5.28)

· J : 0. We have used · B : 0 in the third equation of Eq. (5.28). By comparing Eqs. (5.17) and (5.28) we find that the fields transform as follows: Transformation Relations H : H

(5.29)

B : B

(5.30)

E : E ; u;B

(5.31)

J : J.

(5.32)

In Eqs. (5.29)—(5.32) the variables (H, B, E, u, J) are measured in the stationary frame O, and the variables (H, B, E, J) are measured in the moving frame O [4]. For these relations to hold, the variables must be compared at the same physical point in space. For example, from Eqs. (5.22) and (5.31) we have E(x, y, z, t) : E(x 9 u t, y 9 u t, z 9 u t, t) V W X ;u;B(x 9 u t, y 9 u t, z 9 u t, t). V W X

347


Notice that we have assumed that u is a constant (independent of position). The relations (5.29)—(5.32) give the transformations for the fields and the source. However, we still need to know the constitutive relations in O and O. We apply a general principle that says that the constitutive relations for moving media are the same as for stationary media when they are written in terms of the fields defined in an inertial reference frame moving with the media [4]. For example, for a linear, homogeneous and isotropic media at rest in O we have J : E and B : H. However, in O (which is moving with respect to the media) the relations are modified by virtue of Eqs. (5.29)—(5.32). Specifically, we find that Constitutive Relations Reference Frame O

Reference Frame O

J : (E ; u;B)

J : E

B : H

B : H.

(5.33)

In our study of electromechanical devices we will need to determine the voltage induced in a coil as it moves through a stationary magnetic field. Therefore, we discuss this in some detail. Consider a reference frame O in which there is a static (but not necessarily uniform) B-field. Consider an open conductor moving with velocity u relative to O. Assume that the conductor has a conductivity (as measured by observers in both O and O). We want to determine the voltage induced across the terminals of the conductor as measured in O. We apply Faraday’s law to a stationary contour C that passes through the conductor at a given (but arbitrary) instant in time,

E · dl : 9

d dt

B · ds. (5.34) ! 1 We label the terminals a and b, and then decompose the left-hand side of Eq. (5.34) into integrals across the terminals and the conductor itself,

E · dl : !

?

E · dl ;

@

E · dl,

?

terminals

: 9V ;

@

conductor

@

E · dl,

?

conductor

(5.35)

348


where V is the terminal voltage (as measured in O). As for the remaining integral, from Eqs. (5.31) and (5.32) we have E : E 9 u;B and E : J/ : J/. Therefore, J E : 9 u;B

(moving conductor).

(5.36)

Notice that all the terms in Eq. (5.36) are measured in O (the conductor is moving with respect to O). As the conductor is open, there is no current through it (J : 0) and, therefore, E : 9u;B

(5.37)

Combining Eqs. (5.34), (5.35) and (5.37) gives us

V:9

@

(u;B) · dl ;

d dt

B · ds. (5.38) ? 1 Because B, C, S, and ds are constant in O, the last term in Eq. (5.38) is zero and the terminal voltage reduces to

V:9

@

(u;B) · dl. (5.39) ? It is important to note that if there were a time-varying current in the conductor, the flux through the circuit due to the current would add to the voltage via the last term in Eq. (5.38). We demonstrate this theory in the following examples. EXAMPLE 5.3.1 A conductive bar is in sliding contact with a pair of stationary conducting rails as shown in Fig. 5.5. The bar is moving through a constant uniform B-field with a constant velocity u relative to the rails. Determine the voltage across the rails. Assume that the bar and rails are good conductors. SOLUTION 5.3.1 We choose a reference frame O at rest with respect to the rails, and a stationary contour C (dotted line) that runs clockwise through the rails, and passes through the bar at a single instant in time (Fig. 5.5). Apply Faraday’s law (5.5) to C,

E · dl : 9

d dt

B · ds.

(5.40)

! 1 As B, C, S, and ds are constant in O we have d dt

1

B · ds : 0.

(5.41)

349


FIGURE 5.5

Conductive bar moving along conductive rails.

Substitute this into Eq. (5.40) and decompose the right-hand side into integrals over the various segments and obtain,

>

E · dl ;

\

terminals

E · dl ;

upper rail

E · dl ;

moving bar

E · dl : 0

(5.42)

lower rail

As the upper and lower rails are good conductors ( -) we have E 0 and, therefore,

E · dl :

upper rail

E · dl : 0.

(5.43)

lower rail

In the moving bar we have

E · dl :

(E 9 u;B) · dl,

moving bar

but E : J/ : J/ 0 and, therefore,

E · dl : 9 (u;B) · dl.

(5.44)

moving bar

To evaluate Eq. (5.44), we use u : ux , B : Bz , and dl : dyy (Fig. 5.5). Therefore, 9 (u;B) · dl : 9 9uB dy F : 9uBh,

350


or

E · dl : 9uBh.

(5.45)

moving bar

Finally, substitute Eqs. (5.43) and (5.45) into Eq. (5.42) and obtain V : 9uBh, where V is the terminal voltage,

V:9

>

E · dl.

)

\

EXAMPLE 5.3.2 Determine the lift and drag force on a wire carrying a current i and moving with velocity u relative to a conductive sheet of thickness as shown in Fig. 5.6a [7]. SOLUTION 5.3.2 We choose a reference frame at rest with respect to the wire (Fig. 5.6b). In this reference frame, the magnetic field B of the wire is constant (time invariant), and the sheet moves with velocity u : 9ux . The sheet experiences a time varying magnetic field. The current density induced at a point x in the sheet is J (x) : 9uBR(x), W X where BR(x) is the total magnetic field at x, that is, W BR(x) : B(x) ; BG(x). W W W Here, B (x) and BG (x) are the fields due to the current in the wire, and the W W induced current in the sheet, respectively. Specifically, we have i x B(x) : . W 2 x ; h and BG(x) : W 2

K ()) X d), x9) \

(5.46)

where K (x) : 9uBR(x) (5.47) W X is the current density per unit length ( is the thickness of the sheet). Notice

351


Wire moving above a conductive plate: (a) wire moving and plate stationary; and (b) wire stationary and plate moving [7].

FIGURE 5.6

that Eq. (5.46) contains a singularity. As K ()) is continuous, we evaluate the X integral using the Cauchy Principal value. Substituting Eq. (5.46) into Eq. (5.47) yields the following integral equation for K (x), X u K ()) 2u X K (x) : 9 d) 9 B(x), (5.48) X w x9) w W \ where w : 2/( ). We solve Eq. (5.48) using the Fourier transforms, eIVK (x) dx, F[K ] : X X \ and

F

K ()) 1 X d) : F[K ]F . X x9) x \

352


The second transform follows from convolution theory. In addition, notice that F

1 i k 0 : x 9i k 0.

We apply these to Eq. (5.48) and obtain

F[B] u 1 2u W 19 F . F[K ] : 9 X w x w 1 ; (u/w) Next, we apply an inverse transform to Eq. (5.49) and find that

(5.49)

2u 1 u B()) W K (x) : 9 B (x) 9 d) . (5.50) X w (1 ; u/w) W w )9x \ The last term in Eq. (5.50) can be rewritten in a more useful form using B : ;A. After some analysis we find that

2u 1 u B(x) ; B(x) , K (x) : 9 X w V w (1 ; u/w) W

(5.51)

where i h B(x) : , V 2 x ; h and i x . B(x) : W 2 x ; h Notice that Eq. (5.51) is not limited to a single moving current filament. Rather, it is valid for a system of moving field sources provided they are of limited extent, produce a one-dimensional (1D) current pattern, and travel at the same velocity. The lift and drag forces on the wire are equal and opposite to the forces on the sheet. The force density on the moving sheet is the Lorentz force density (2.23), f : K (x)z ;(B(x)x ; B(x)y ). V W X Therefore, f : 9K (x)B(x) W V X and f : K (x)B(x). V W X

353


We integrate over the sheet to obtain the total lift and drag force

>

K (x)B(x) dx V X \ ih > K (x) X dx : x ; h 2 \ i u : 4h w ; u

F : *

:F

u w ; u

(Lift force)

and w F : F " u *

(Drag force),

where i F : . 4h A normalized plot of these forces is shown in Fig. 5.7. Notice that the drag force reaches a maximum when u : w. )

FIGURE 5.7

Normalized plot of lift and drag forces vs normalized velocity [7].

354


5.4 ELECTRICAL EQUATIONS The electrical equations for an electromechanical device follow from circuit theory, which in turn is based on quasi-static field theory. In this section, we derive a generalized version of Kirchhoff’s voltage law that takes into account the effects of electromechanical coupling. We study two different types of circuits, stationary and moving coil. In a stationary circuit, all the circuit components are at rest with respect to one another. For these circuits, we apply quasi-static field theory in a reference frame fixed with respect to the circuit. The electromechanical coupling is accounted for by a flux linkage term that takes into account the voltage induced in the circuit by a time rate of change of magnetic flux through it. This change in flux is due to a change in the magnetic field that can be caused by a change in the current and/or by the movement of a ferromagnetic member that alters the field. Mechanical motion is taken into account by considering the flux to be a function of both the circuit current and mechanical displacement. In a moving coil circuit, a coil is moving with respect to an external magnetic field. To analyze this kind of circuit, we apply quasi-static field theory in a reference frame that is at rest with respect to the magnetic field. Mechanical motion is taken into account by the voltage induced in the coil as it moves through the field. Finally, it is important to note that the following analysis applies to magnetically linear devices.

5.4.1 Stationary Circuits In this section we study electromechanical devices with stationary electrical circuits. Such a device is shown in Fig. 5.8. It consists of a voltage source V , a resistor R, and a magnetic circuit with a fixed coil Q and a moving soft magnetic bar that alters the flux through the coil when it moves. We analyze the electrical circuit using quasi-static field theory. Specifically, we choose a reference frame at rest with respect to the circuit and a fixed integration contour C along the circuit. At any point along the circuit the electric field is given by E:9

A 9 , t

(5.52)

where 9 is due to the accumulation of charge along the circuit, and A/t is due to the time-varying current. We evaluate the line integral

355

5.4 ELECTRICAL EQUATIONS

Electromechanical actuator circuit.

FIGURE 5.8

of Eq. (5.52) clockwise along C and obtain

A · dl 9 t

E · dl : 9

· dl. ! ! ! The first term can be decomposed into integrals over the individual components

>

E · dl ;

\ source

E · dl : 9

E · dl ;

resistor

A · dl 9 t

· dl.

(5.53)

coil

Notice that w · dl : 0 because it entails the integration of a conservative vector field around a closed path. Also notice that although w A/t · dl is evaluated around the entire path, the major contribution to it is from the inductor. We assume that the source is ideal with no internal resistance and define the voltage impressed at the terminals in the usual way:

V Y9 Q

> \

E · dl. Q

(5.54)

Thus, Eq. (5.53) becomes V : Q

E · dl ;

resistor

E · dl ; coil

A · dl. t

(5.55)

356


In the resistor J : E (Ohm’s law). Therefore,

E · dl :

resistor

J · dl,

(5.56)

resistor

: iR. Similarly, in the coil J :

E, and, therefore,

E · dl : iR

.

(5.57)

coil

Substituting Eqs. (5.56) and (5.57) into Eq. (5.55) yields V : i(R ; R ) ; Q

A · dl. t

(5.58)

Now, consider the last term in Eq. (5.58). We can rewrite this as follows:

A · dl : t :

1

1 d : dt :

A ; · ds t

B · ds t B · ds

1

d" , dt

(5.59)

where "Y

B · ds (flux linkage). 1 The symbol " denotes the flux linkage, that is, the total flux through (linking) the circuit. To determine " we need to evaluate B · ds over an 1 area bounded by the circuit path in accordance with Stokes’ theorem. However, this integral can be difficult to evaluate depending on the topology of the circuit. For a simple single-turn current loop, the flux is denoted by and, therefore, " : . For a tightly wound N turn coil the flux linkage is " : N

(N turn coil),

where is the flux through each turn.

357


We digress briefly to discuss the second and third step in Eq. (5.59). Notice that we have replaced the partial time derivative inside the integral by a total time derivative outside the integral. We do this because inside the integral, B is a function of both time-independent coordinate variables and other time-dependent variables (including t itself). However, the integration effectively eliminates the time-independent variables, and leaves only the time-dependent variables. Therefore, the total time derivative is appropriate outside the integral. We return to the derivation. Substitute Eq. (5.59) into Eq. (5.58) and obtain V : i(R ; R ) ; Q

d" . dt

(5.60)

mechanical coupling

This is a generalization of Kirchhoff’s voltage law. The term d"/dt in Eq. (5.60) requires some discussion because it is key to the electromechanical coupling. This term represents the voltage induced in the circuit by a time rate of change of magnetic flux through the circuit. The change in flux can be due to a change in circuit current and/or the movement of a ferromagnetic member that alters the field. To account for this we write " as a function of both the circuit current and the position of the member. Let x and denote the position of the member when it executes linear and rotational motion, respectively. Then ":

"(i, x) (linear motion) "(i, ) (rotational motion).

Given this functional dependence, the total time derivative of " is

"(i, x) di "(i, x) dx ; (linear motion) i dt x dt

d" : dt "(i, ) di "(i, ) d ; (rotational motion) i dt dt or

d" : dt

L

di "(i, x) ; v dt x

di "(i, ) L ; dt

(linear motion) (5.61) (rotational motion)

358


where L is the inductance and which is defined by LY

" i

(inductance).

(5.62)

The variables v : dx/dt and : d/dt are the linear and angular velocities, respectively. It is important to note that the definition of inductance Eq. (5.62) applies only to magnetically linear devices. There is no simple definition of L for nonlinear devices [8]. Finally, substitute Eq. (5.61) into Eq. (5.60) and obtain

Electrical $ Equations

di(t) "(i, x) V (t) : i(t)(R ; R ) ; L ; v(t) Q dt x (linear motion)

di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt

.

(5.63)

(rotational motion)

These are the electrical equations of motion for stationary circuits with time varying currents and a moving ferromagnetic member.

5.4.2 Moving Coils In this section we derive Kirchhoff’s voltage law for circuits that have a coil moving through an external magnetic field B . All the other circuit components are stationary with respect to the field. Let O denote the stationary reference frame. The coil moves with velocity u through B , which is constant (but not necessarily uniform) with respect to O. Such a circuit is shown in Fig. 5.9. This figure shows a linear actuator consisting of a conductive bar of mass m in sliding contact with a pair of stationary conducting rails. The rails are connected to a voltage source V and a resistor R that limits the current. The bar is moving through a Q constant uniform B-field with a time-dependent velocity u(t) relative to the rails. This circuit is analyzed in Example 5.6.2. We derive Kirchhoff’s voltage law for a circuit with a moving coil by applying Faraday’s law to a stationary contour C that passes through the circuit and coincides with the coil for a given position of its travel (the dotted line in Fig. 5.9). Thus, C coincides with the entire circuit at a single instant in time. We evaluate

E · dl : 9 !

d dt

1

B · ds

(5.64)

359


Moving conductor actuator.

FIGURE 5.9

clockwise around the contour and obtain

>

E · dl ;

\

source

E · dl ;

resistor

E · dl : 9 coil

d dt

B · ds.

(5.65)

The integrals over the fixed components (source and resistor) follow from Eqs. (5.54)—(5.57). We obtain V : iR ; Q

E · dl ;

d dt

B · ds. (5.66) The coil is moving with velocity u through B . The E-field in it as measured in O is given by Eq. (5.36), J E : 9 u;B . Therefore,

where R and obtain

E · dl :

J 9 u;B · dl,

: iR 9

(u;B ) · dl, (5.67) is the resistance of the coil. Substitute Eq. (5.67) into Eq. (5.66)

V : i(R ; R ) 9 Q

d (u;B ) · dl ; dt

B · ds.

(5.68)

Now consider the last term in Eq. (5.68). The dominant contribution to this term comes from the coil. Moreover, as the system is magnetically linear we can write B as a superposition of two fields: the field B due G

360


to the current i and the external field B , B:B ;B . G Therefore, the last term in Eq. (5.68) can be written as d dt

B · ds : :

d dt

d B · ds ; G dt

d d " (i) ; dt dt

:L

di d ; dt dt

B · ds

B · ds

B · ds, (5.69) where " (i) : B · ds, and we have used the definition of inductance G " (i) L : . i Finally, we substitute Eq. (5.69) into Eq. (5.68) and find that di V : i(R ; R ) ; L 9 Q dt

d (u;B ) · dl ; dt

B · ds. (5.70) This is a generalization of Kirchhoff’s voltage law that takes into account the voltage induced in a coil as it moves through B . If B is static, d B · ds : 0 and Eq. (5.70) reduces to dt di V : i(R ; R ) ; L 9 (u;B ) · dl. (B static). (5.71) Q dt It is important to note that the line integral in Eq. (5.71) is evaluated as described in Section A.4 with the direction of integration in the direction of conventional current flow. The voltage equation (5.71) applies to a coil in linear motion. However, it can be generalized to the case of a coil that executes rotational motion. Specifically, if a segment dl of the coil is rotating with an angular velocity at a distance r with respect to a fixed reference frame, then its linear velocity is

u : ;r.

(5.72)

We substitute Eq. (5.72) into Eq. (5.71) and obtain the circuit equation for rotational motion, di V (t) : i(R ; R ) ; L 9 Q dt

[(;r);B ] · dl

(B static). (5.73)

361

5.5 MECHANICAL EQUATIONS

It is important to note that when evaluating Eq. (5.73), the velocity (;r) must be evaluated before the cross product with B is taken. If the rotation is with respect to a specific axis (which we label the z-axis), then : z , and Eq. (5.73) can be rewritten as di V (t) : i(t)(R ; R ) ; L 9 Q dt

[(;r );B ] · dl ,

(B static), (5.74)

where r is a vector from (and perpendicular to) the z-axis to the , segment dl.

5.5 MECHANICAL EQUATIONS In this section we review the mechanical equations of motion for electromechanical devices that execute linear or rotational motion with a single degree of freedom. Such linear or rotational motion is described by a single coordinate variable that we denote by x or , respectively. The equations for these variables follow from Newton’s second law of motion, m

dx : F(i, x) dt

(linear motion),

(5.75)

(rotational motion),

(5.76)

or d J : T(i, ) K dt

where m is the mass, J is the moment of inertia about the axis of K rotation, and F(i, x) and T(i, ) are the force and torque. The moment of inertia of a point mass m at a fixed distance r about the axis of rotation is J : mr. If the mass is distributed, then J is computed via integraK K tion, J : K

4

r dv,

where and V are the density and volume of the mass, respectively,

and r is the distance of the element dv from the axis of rotation. For

362


example, if a rigid body of density is rotating about the z-axis, its

moment of inertia is J : K

(x ; y) dv. (5.77)

4 Notice that in Eqs. (5.75) and (5.76) we have chosen the current i and position x or as the dependent variables. This is often a convenient choice, but other variables may be more convenient depending on the application. Regardless of the choice of variables, for lumped-parameter analysis we need to determine analytical expressions for F(i, x) and T(i, ). This is usually accomplished using one of the following methods: 1. Derive an analytical solution for the magnetic field and determine the force density using the Lorentz force formula or the Maxwell stress tensor. Then, perform the appropriate volume integration to obtain the total force or torque. This approach is most useful in the analysis of moving coil actuators. It is discussed in what follows. 2. Derive an analytical expression for the field energy (or coenergy) and then determine the force or torque via differentiation. This approach is especially useful in the analysis of magnetic circuit or moving magnet devices. It is discussed in Section 5.7. 3. Discretize the independent variables into a discrete set of values, use a numerical method such as FEA to obtain force or torque data for these values, and then fit the data to an analytical formula. This is called a hybrid approach and is discussed in Section 5.15. The first method is the most direct approach for determining the force or torque. It is especially useful in the analysis of moving coil actuators. The force and torque imparted to the coil are obtained using the Lorentz force formulas Eqs. (3.47) and (3.52) (Section 3.2.2). Specifically, if a coil carries a current i and moves with velocity u through an external field B , the force and torque imparted to it are F:i

(dl;B ),

(5.78)

and T:i

r;(dl;B ), (5.79) where r is the vector from the origin about which the torque is computed

5.6 ELECTROMECHANICAL EQUATIONS

363

and integration is over the length of the coil. When evaluating Eq. (5.79), the force (dl;B ) must be evaluated before the cross product with r is taken. Equations (5.78) and (5.79) are vector equations. We are interested in both the force in the x direction, and the torque about a given axis, which we choose to be in the z-axis. The magnitudes of force and torque in these directions are F :i V

(dl;B ) · x ,

(5.80)

[r ;(dl;B )] · z , ,

(5.81)

and T :i X

where r is a vector from (and perpendicular to) the z-axis to the , segment dl.

5.6 ELECTROMECHANICAL EQUATIONS In this section we summarize the equations of motion for electromechanical devices. We have seen that the electrical and mechanical equations are separately second-order equations in the independent variables i and x or , respectively. These equations are coupled and therefore represent a second-order system of equations. These coupled equations can sometimes be solved directly. However, for practical analysis it is often easier to reduce them to a first-order system of equations. Such systems lend themselves to various numerical timestepping solution procedures such as the Euler or Runge-Kutta methods. These are discussed in Appendix C.

5.6.1 Stationary Circuits In this section we summarize the equations of motion for electromechanical devices with stationary circuits. These include magnetic circuit and moving magnet actuators. The electrical and mechanical equations for these devices were derived in the foregoing text. We repeat

364


them here for convenience,


and

di(t) "(i, x) V (t) : i(t)(R ; R ) ; L ; v(t) Q dt x (linear motion)

di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt

.

(5.82)

(rotational motion)

dx(t) : F(i, x) (linear motion) dt Mechanical $ . d(t) Equations J : T(i, ) (rotational motion) K dt m

(5.83)

Since the same variables i and x or appear in both sets of equations, they need to be solved simultaneously. Thus, the analysis of an electromechanical device entails the solution of the following coupled second-order equations: Linear Motion di(t) "(i, x) ; v(t) V (t) : i(t)(R ; R ) ; L Q x dt dx(t) m : F(i, x) dt

(5.84)

or Rotational Motion di(t) "(i, ) V (t) : i(t)(R ; R ) ; L ; (t) Q dt d(t) J : T(i, ), K dt

(5.85)

For practical analysis is convenient to rewrite the second-order equations in terms of the following first-order systems,

365


Linear Motion di(t) 1 : dt L

V (t) 9 i(t)(R ; R ) 9 Q dv(t) 1 : F(i, x) dt m dx(t) : v(t) dt

(5.86)

(5.87)

"(i, x) v(t) x

and Rotational Motion di(t) 1 : dt L

V (t) 9 i(t)(R ; R ) 9 Q d (t) 1 : T(i, ) J dt K d(t) : (t). dt

"(i, ) (t)

Equations (5.86) and (5.87) constitute initial-value problems that need to be solved subject to the following initial conditions, i(t ) : i x(t ) : x v(t ) : v

(linear motion)

(5.88)

and i(t ) : i (t ) : (rotational motion) (5.89) (t ) : , respectively. Occasionally, the initial-value problems can be solved analytically. More often, their solution requires the application of a numerical time-stepping procedure such as the Euler or Runge Kutta method (Appendix C). EXAMPLE 5.6.1 Apply Euler’s method to Eq. (5.87). SOLUTION 5.6.1 We reduce Eq. (5.87) to a set of algebraic equations using Euler’s method as discussed in Appendix C. In Euler’s method, derivatives are

366


approximated by a simple difference quotient, for example, df f(t ; t) 9 f(t) . t dt

(5.90)

We apply Eq. (5.90) to Eq. (5.87) and obtain i(t ; t) : i(t) ;

t "(i, ) V (t) 9 i(t)(R ; R ) 9 (t) Q L

t T(i, ) J K (t ; t) : (t) ; t (t).

(t ; t) : (t) ;

(5.91)

To solve Eq. (5.91) we need expressions for "(i, ) , L:

"(i, ) , i

(5.92) (5.93)

and T(i, ).

(5.94)

Once these are known, Eq. (5.91) can be solved using the following iterative scheme: i(n ; 1) : i(n) ;

t "(i(n), (n)) V (n) 9 i(n)(R ; R ) 9 (n) Q L t T(i(n), (n)) J K (n ; 1) : (n) ; t (n).

(n ; 1) : (n) ;

(5.95)

The solution method is as follows: Choose a value for t. Then, given the initial conditions [i(0), (0), (0)] use Eq. (5.95) to obtain [i(1), (1), (1)], which gives the solution at t : t. Repeat this process to obtain [i(2), (2), (2)], [i(3), (3), (3)], . . . , [i(n), (n), (n)], [i(n ; 1), (n ; 1), (n ; 1)]. These are the solution values at t : 2t, 3t, . . . , nt, (n ; 1)t, . . . , respectively. )

5.6.2 Moving coils In this section we summarize the equations of motion for electromechanical devices with coils that move through an external field B

367


(moving coil actuators). The electrical and mechanical equations were derived in the preceding; we repeat them here for convenience:


di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt

(u ; B ) · dl

(linear motion) di(t) 9 V (t) : i(t)(R ; R ) ; L Q dt

(rotational motion)

[(;r);B ] · dl

(5.96)

and

Mechanical $ Equations

m

dx(t) :i dt

d(t) J :i K dt

(dl;B ) · x ; F (x)

(linear motion)

[r;(dl;B )] · z ; T ().

(rotational motion) (5.97)

In Eq. (5.97) the term T () represents the mechanical torque imposed

on the system. The electrical and mechanical equations are coupled and can be grouped into the following second-order systems of equations, Linear Motion di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt m

dx(t) : i(t) dt

(u ; B ) · dl

(dl ; B ) · x ; F (x)

(5.98)

368


and Rotational Motion di(t) V (t) : i(t)(R ; R ) ; L 9 Q dt d(t) J : i(t) K dt

[( ; r) ; B ] · dl

(5.99)

[r ; (dl ; B )] · z ; T ().

The line integrals in Eqs. (5.98) and (5.99) are evaluated as described in Section A.4. The direction of integration follows the current i. In addition, in Eq. (5.99) the velocity ( ; r) and force i(dl ; B ) are evaluated before the remaining cross products are taken. For conventional rotary devices, the analysis is usually performed using cylindrical coordinates with the rotation taken to be about the z-axis, : (t)z . The current flow is often taken to be along the z-axis with dl : dzz . When this is the case, Eq. (5.99) reduces to the following form: Rotational Motion di(t) V (t) : i(t)(R ; R ) ; L ; K (t) Q C dt J

(5.100)

d(t) : i(t)K ; T (),

K dt R

where K is called the electrical constant (or back emf constant) and K is C R called the torque constant. In the MKS system K : K . We derive Eq. C R (5.100) for two different rotational devices in Examples 5.6.3 and 5.11.1. These equations are also derived for a device with a different configuration in Example 5.9.1. For practical analysis, it is convenient to rewrite Eqs. (5.98) and (5.99) in terms of the following first-order systems: Linear Motion di(t) 1 : [V (t) 9 i(t)(R ; R ) ; dt L Q du(t) 1 : [i(t) dt m

(u ; B ) · dl]

(dl ; B ) · x ; F (x)]

dx(t) : u(t) dt

(5.101)

369


and Rotational Motion di(t) 1 : [V (t) 9 i(t)(R ; R ) ; L Q dt

[( ; r) ; B ] · dl]

(5.102)

d (t) 1 : [i(t) [r ; (dl ; B )] · z ; T ()]

J dt K d(t) : (t). dt

Equations (5.101) and (5.102) constitute initial-value problems that need to be solved subject to the initial conditions (5.88) and (5.89), respectively. We demonstrate the use of these equations in the following examples. EXAMPLE 5.6.2 Consider a linear actuator consisting of a conductive bar of mass m in sliding contact with a pair of stationary conducting rails as shown in Fig. 5.9. The rails are connected to a voltage source V , and a resistor R that Q limits the current. The bar is moving through a constant uniform B-field with a time-dependent velocity u(t) relative to the rails. Determine the equations of motion for the system. Assume that the rails are good conductors with negligible resistance. SOLUTION 5.6.2 This device has a moving conductor and is governed by the linear motion equations (5.101). In this case the coil is the moving bar, and there is no mechanical force (F (x) : 0). Therefore, Eq. (5.101) becomes

di(t) 1 : V (t) 9 i(t)(R ; R ) ; L Q dt du(t) i(t) : dt m

dx(t) : u(t), dt

(u ; B ) · dl

(dl ; B ) · x (5.103)

where L is the inductance of the complete circuit. We need to evaluate the integrals in Eq. (5.103). We use a Cartesian reference frame O at rest with respect to the rails as shown in Fig. 5.9.

370


Induced voltage: To evaluate the induced voltage, we need to determine (u ; B ) · dl along the bar. From Fig. 5.9 we have B : 9B z , (5.104) u(t) : u(t)x , (5.105) and dl : dyy . (5.106) Therefore (u ; B ) · dl : u(t)B dy. (5.107) We integrate Eq. (5.107) along the bar in the direction of current. This gives

(u ; B ) · dl :

F

(u ; B ) · dl

: B u(t)

dy

F : 9B hu(t). (5.108) Force: Next we evaluate the force. For this we need (dl ; B ) · x along the bar. From Eqs. (5.104) and (5.106) we have (dl ; B ) · x : 9B dy. We integrate this along the bar in the direction of current flow and obtain

(dl ; B ) · x :

F

(dl ; B ) · x

: 9B

dy

F : B h. (5.109) Therefore, the Lorentz force on the bar is F : iB h. Equations of motion: Substitute Eqs. (5.108) and (5.109) into Eq. (5.103), which gives di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 B hu(t)] L Q dt du(t) 1 : i(t)B h dt m dx(t) : u(t). dt


FIGURE 5.10

371

Rotating coil device: (a) perspective of device; (b) end view; and

(c) side view.

These equations need to be solved subject to initial conditions (5.88).

)

EXAMPLE 5.6.3 Consider the rotary device shown in Fig. 5.10a. This represents a basic direct-current (dc) motor. In this device, a magnetic field B (assumed constant) is produced between two pole pieces by a field current i through a D coil that is wrapped around the pole pieces. A rectangular coil is mounted between the pole pieces and is free to rotate about the z-axis. When a current i passes through the coil it experiences a torque that causes it to rotate as shown.

372


A split ring with brushes is connected to the terminals of the rotating coil so that the current through it reverses direction every half turn. This ensures that the torque will always be in the same direction, thereby producing continuous rotation. Write the equations of motion for this device. SOLUTION 5.6.3 This device has a moving coil and is governed by the rotational motion equations (5.102). There is no mechanical torque and Eqs. (5.102) reduce to di(t) 1 : [V (t) 9 i(t)(R ; R ) ; dt L Q

[( ; r) ; B ] · dl]

d (t) i(t) : [r ; (dl ; B )] · z J dt K d(t) : (t). dt

(5.110)

where L and R are the inductance and resistance of the coil, and J is the K moment of inertia of the coil about the z-axis. We need to evaluate the induced voltage and torque integrals in Eq. (5.110). We use a Cartesian reference frame at rest with respect to the pole pieces as shown in Fig. 5.10a. We choose a rotation angle , which is measured with respect to the y-axis in a counterclockwise sense about the z-axis as shown in Fig. 5.10b. Induced voltage: We determine the voltage induced at the terminals of the coil by summing the contributions from its various segments. The coil has four segments; relatively long left and right side segments, and two shorter end segments (Fig. 5.10b,c). We evaluate the side segments first. To evaluate the induced voltage we need [( ; r) ; B ] · dl along these segments. We know that, (t) : (t)z , and from Fig. 5.10b we find that

9b sin()x ; b cos()y b sin()x 9 b cos()y

r:


Therefore, ;r:

9 (t)b cos()x 9 (t)b sin()y (t)b cos()x ; (t)b sin()y

Note that B is in the x-direction, B : B x .


373


Therefore,

(t)bB sin()z ( ; r) ; B : 9 (t)bB sin()z


Because dl : dzz for both segments we have

(t)bB sin() dz [( ; r) ; B ] · dl : 9 (t)bB sin() dz


(5.111)

We evaluate Eq. (5.111) along the side segments with the integration in the direction of the current. This gives

[( ; r) ; B ] · dl :

(t)bB sin() dz

?

left side

: 9 (t)abB sin(),

(5.112)

and

[( ; r) ; B ] · dl : 9

?

(t)bB sin() dz

right side

: 9 (t)abB sin(). Next, consider the end segments. For these we have ( ; r) ; B . z , and dl . r. It follows that [( ; r) ; B ] · dl . z · r : 0

(end segments).

Therefore,

[( ; r) ; B ] · dl : 0. end segments

(5.113)

374


This shows that the end segments make no contribution to the induced voltage. The induced voltage is the sum of Eqs. (5.112) and (5.113). Specifically,

[( ; r) ; B ] · dl : 92 (t)abB sin()

: 9 (t)A B sin(), (5.114) where A : 2ab is the area of the loop. If there are n loops in the coil, then Eq. (5.114) is multiplied by n. Torque: We now determine the torque. Only the left- and right side segments of the coil contribute to the torque. We make a brief argument to show that the end segments make no contribution. Consider the torque on an element of an end segment, dT : r ; (dl ; B ) · z . X On these segments, dl dr and, therefore, dl ; B . dr ; B . z .

This gives

dT . (r ; z ) · z : 0. X Thus, the end segments generate no torque. Now consider the left- and right side segments. For these we have dl : dzz . Therefore, dl ; B : B dzy

It follows that

(side segments).

9bB sin() dz z (left side) (5.115) bB sin() dz z (right side). We evaluate the integral of Eq. (5.115) along the side segments, integrate in the direction of current flow, and obtain r ; (dl ; B ) :

[r ; (dl ; B )] · z : 9

?

bB sin() dz

left side

: abB sin(),

and

[r ; (dl ; B )] · z :

?

(5.116)

bB sin() dz

right side

: abB

sin().

(5.117)


375

The torque on the coil is the sum of Eqs. (5.116) and (5.117),

[r ; (dl ; B )] · z : 2abB sin() : A B sin().

(5.118)

If there are n loops in the coil then Eq. (5.118) is multiplied by n. Moment of inertia: Next, we determine the moment of inertial. We use Eq. (5.77), which we repeat here for convenience, J : K

(x ; y) dv.

4

The inertia of each side segment is J : K

?

bA dz

: abA : m b, where A and are the area and density of the wire, respectively, and

m : aA is the mass of each side segment. The inertia of each end segment is J : 2 K

@

A d

: bA : m b, where m : 2bA is the mass of each end segment. Summing the contributions from all segments gives the total moment of inertia, J : 2J ; 2J K K K : (2m ; m )b. If there are n loops in the coil then Eq. (5.119) is multiplied by n.

(5.119)

376


Equations of motion: We are finally ready to write the equations of motion. Substitute Eqs. (5.114) and (5.118) into Eq. (5.110) and obtain di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 A B sin((t)) (t)] dt L Q d (t) 1 : i(t)A B sin((t)) J dt K d(t) : (t). dt These equations are solved subject to the initial conditions (5.89). Notice that the electrical constant K equals the torque constant K (see Eq. (5.100)). C R Specifically, K : K : A B sin((t)). C R If the coil had n turns all at the same angle , then K and K would be C R multiplied by n. If the turns were distributed in an angular sense, then K and C K would be obtained by integrating the contributions of the individual turns R over the angular span of the coil. )

5.7 ENERGY ANALYSIS In this section we discuss an energy method for the analysis of electromechanical devices. This method, based on the conservation of energy, is used to determine the force or torque on a moving member. It is especially useful for the analysis of magnetic circuit actuators as we shall see. We start with the basic energy balance relation: electrical energy energy energy mechanical input at : dissipated by ; stored in ; energy device terminals device magnetic field output Apply this to a system that is subjected to an infinitesimal (virtual) displacement and obtain dW C

electrical energy input

: dW

energy dissipated

; dW

DJB

stored field energy

; dW

mechanical energy

(5.120)

377

5.7 ENERGY ANALYSIS

The term dW accounts for energy dissipation due to loss mechan isms such as ohmic heating and mechanical friction. For our purposes, we consider the system to be lossless and treat the loss mechanisms as external to the system. For example, a real inductor can be treated as a lossless coil in series with an external resistance. Thus, without loss of generality, we restrict our attention to lossless systems in which dW : 0. For these systems, energy balance equation (5.120) becomes dW : dW ; dW . C DJB

(5.121)

We consider the various terms in Eq. (5.121). Recall from Eq. (5.60) that mechanical motion couples to the electrical equations via flux linkage ". Moreover, the voltage e induced across the terminals of an electromechanical component such as a coil is e : d"/dt. Therefore, the electrical energy input into the terminals of the component during a time dt is d" dW : i dt C dt : d" i.

(5.122)

Substitute Eq. (5.122) into Eq. (5.121) and obtain dW

DJB

: d" i 9 dW

.

(5.123)

(linear motion) (rotational motion),

(5.124)

The mechanical energy output is given by dW

:

f dx DJB T d DJB

where f and T are the force and torque imparted to the mechanical DJB DJB member by the magnetic field. It is important to note that a positive force (or torque) is assumed to be in the same direction as a positive displacement dx (or d). From Eqs. (5.123) and (5.124) we have dW

DJB

:

id" 9 f dx DJB id" 9 T d DJB

(linear motion) (rotational motion).

(5.125)

We desire relations for f and T in terms of the field energy W . The DJB DJB DJB form of Eq. (5.125) suggests that we can obtain these if we choose " and

378


x or as the independent variables. With this choice, W is of the form DJB W (", x) (linear motion) DJB W : DJB W (", ) (rotational motion). DJB The total derivative of W is DJB W (", x) W (", x) DJB DJB d" ; dx (linear motion) " x (5.126) dW : DJB W (", ) W (", ) DJB DJB d" ; d (rotational motion). "

Now, we compare Eq. (5.125) with Eq. (5.126) and find that W

(", x) DJB x W (", ) DJB T :9 DJB f :9 DJB

(linear motion) (5.127) (rotational motion),

which are the desired results. Recall from Eq. (3.74) in Section 3.2.5 that for linear systems the energy stored in the magnetic field is W : Li (linear systems). DJB Moreover, from Eq. (3.78) we know that L:

" . i

Therefore, W

DJB

:

"i . 2

(5.128)

We use Eq. (5.128) to express W in terms of the independent variables DJB " and x or . This gives

"i(", x) " : 2 2L(x)

W : DJB "i(", ) " : 2 2L()

(linear motion) (5.129) (rotational motion).

379

5.7 ENERGY ANALYSIS

We can use Eq. (5.129) to determine f and T . First determine L(x) or DJB DJB L(), substitute these into Eq. (5.129), and then obtain f and T using DJB DJB Eq. (5.127). Instead of choosing flux linkage " and position (x or ) as the independent variables, it is often more natural and convenient to work with current i and position. When this is the case, the energy has a functional dependency of the form W (i, x) or W (i, ), and Eq. (5.125) DJB DJB becomes

id"(i, x) 9 f dx DJB dW : DJB id"(i, ) 9 T d DJB


(5.130)

where d"(i, x) :

"(i, x) "(i, x) di ; dx i x

(linear motion)

d"(i, ) :

"(i, ) "(i, ) di ; d i

(rotational motion). (5.131)

Similarly,

W (i, x) DJB W : DJB W (i, ) DJB


which gives

W (i, x) W (i, x) DJB DJB di ; dx i x

(linear motion)

dW : DJB W (i, ) W (i, ) DJB DJB di ; d i

(5.132) (rotational motion).

The functional form of these relations motivate the definition of the coenergy function W A , DJB W A (i, x) : i"(i, x) 9 W (i, x) DJB DJB (linear motion) (Coenergy)

(5.133)

W A (i, ) : i"(i, ) 9 W (i, ) DJB DJB (rotational motion). We can determine the force and torque directly from the coenergy. To

380


this end, first take the total derivative of W A (i, x). This gives DJB d(i"(i, x)) 9 dW (i, x) (linear motion) DJB dW A : DJB d(i"(i, )) 9 dW (i, ) (rotational motion) DJB or

"(i, x)di ; id"(i, x) 9 dW (i, x) DJB dW A : DJB "(i, )di ; id"(i, ) 9 dW (i, ) DJB


Substitute Eqs. (5.123) and (5.124) into Eq. (5.134) and obtain

"(i, x)di ; f dx (linear motion) DJB (5.135) dW A : DJB "(i, )di ; T d (rotational motion). DJB Now dW A is formally given by DJB W A (i, x) W A (i, x) DJB DJB di ; dx (linear motion) x i dW A : (5.136) DJB W A (i, ) W A (i, ) DJB DJB di ; d (rotational motion). i

We compare Eqs. (5.135) and (5.136) and find that ":

W A (i, x) DJB , i

(5.137)

and that f : DJB

W A (i, x) DJB x

W A (i, ) DJB T : DJB


The relations (5.138) are the desired expressions for force and torque. Last, we obtain an explicit expression for W A from Eqs. (5.129) and DJB (5.133),

"(i, x)i L(x)i : 2 2

WA : DJB "(i, )i L()i : 2 2


We can use Eq. (5.139) to determine both force and torque. First

381

5.7 ENERGY ANALYSIS

FIGURE 5.11

Electromechanical system: (a) actuator; and (b) energy conversion

diagram.

determine L(x) or L(), substitute these into Eq. (5.139), and then obtain f and T using Eq. (5.138). DJB DJB EXAMPLE 5.7.1 Set up the equations of motion for the actuator circuit of Fig. 5.11 [1]. SOLUTION 5.7.1 This is a magnetic circuit actuator. Choose a convention in which f (force imparted to the moving member by the field) is positive in the DJB direction of increasing displacement x as shown. For the equations of motion use the first-order system (5.86), which is written in terms of the independent variables (i, x),

1 "(i, x) di(t) : V (t) 9 i(t)(R ; R ) 9 v(t) L(x) Q x dt dv(t) 1 : F(i, x) dt m dx(t) : v(t). dt

(5.140)

382


To solve Eq. (5.140) we need expressions for the terms L(x), "(i, x)/x, and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). From Eq. (5.139) we have DJB L(x)i W A (i, x) : . (5.141) DJB 2 The magnetic circuit of Fig. 5.11 was analyzed in Example 3.5.2 from Section 3.5.1. The inductance L(x) was found to be NA E , (5.142) (A /A )( /)l ; 2x E A A where l is the path length in the core, and A and A are the cross-sectional A A E areas of the core and gap, respectively. Substitute Eq. (5.142) into Eq. (5.141) and obtain L(x) :

NA i E W A (i, x) : . (5.143) DJB 2((A /A )( /)l ; 2x) E A A Next, use Eqs. (5.137) and (5.138) to obtain " and f . We find that DJB W A (i, x) DJB "(i, x) : i :

NA i E , ((A /A )( /)l ; 2x) E A A

(5.144)

and that f (i, x) : DJB

W A (i, x) DJB x

NA i E . (5.145) ((A /A )( /)l ; 2x) E A A The minus sign in Eq. (5.145) implies that the direction of f is opposite to the DJB direction of increasing x (i.e., toward the actuator and opposite to the direction of increasing air gap). The total force F(i, x) on the moving member is a sum of the forces due to the field and spring :9

F(i, x) : f

(i, x) ; f (x), (5.146) DJB Q where f (x) is the force due to the spring. In its initial state, the actuator is Q unenergized with i(0) : 0, and the mass is at rest in an equilibrium position x . In this position the spring force is zero, f (x ) : 0. Thus, when the mass is Q

383

5.8 MAGNETIC CIRCUIT ACTUATORS

at a position x the spring force is f (x) : 9k(x 9 x ). (5.147) Q Notice that when x x (x x ) this force tends to move the mass towards (away from) the magnetic circuit. Substitute Eqs. (5.144), (5.145), (5.146), and (5.147) into Eq. (5.140) and obtain di(t) ((A /A )( /)l ; 2x) E A A : dt NA E

2 NA i E ; V (t)9i(t)(R;R ); v(t) ((A /A )( /)l ;2x) Q E A A dv(t) NA i 1 E :9 ; k(x 9 x ) dt m ((A /A )( /)l ; 2x) E A A dx(t) : v(t). (5.148) dt

This nonlinear first-order system has to be solved subject to the initial conditions x(0) : x , v(0) : v and i(0) : i . ) Finally, for linear systems the energy and coenergy are numerically equal, W : W A . This follows from the fact that " : Li for such DJB DJB systems, which gives " Li : . 2L 2 However, for nonlinear systems " is not proportional to i and W and DJB W A are not numerically equal. The difference between W and W A for DJB DJB DJB linear and nonlinear systems is illustrated in Fig. 5.12.

5.8 MAGNETIC CIRCUIT ACTUATORS In this section we study magnetic circuit actuators. In these actuators the moving member is part of the magnetic circuit, and its motion alters the flux through the electrical circuit. The energy method is especially useful for analyzing these actuators. We demonstrate this in the following examples. EXAMPLE 5.8.1 Consider the magnetic actuator circuit shown in Fig. 5.13a. Assume that the magnetic circuit operates in a linear region of its B-H curve

384


Graphical representation of energy and coenergy: (a) linear system; and (b) nonlinear system.

FIGURE 5.12

with , and that there is no flux leakage. Determine the force on the plunger [2]. SOLUTION 5.8.1 We determine an analytical expression for the force on the plunger using both energy and coenergy approaches. Choose a convention in which a positive f is in the direction of increasing displacement x as shown. DJB Energy: In the energy approach, " and x are the independent variables and the energy is determined using Eq. (5.129), " W : . DJB 2L(x)

(5.149)


385

Actuator with moving plunger: (a) actuator circuit; and (b) reference frame for plunger [2].

FIGURE 5.13

We need to determine the inductance L(x). We use the inductance relation (3.75). L:

1 i

B · H dv.

2 i

B · H dv,

(5.150)

4 As , the core and plunger have a high permeability and the H-field in these elements is negligible. Therefore, Eq. (5.150) reduces to an integration over the gap regions, L:

4g

(5.151)

386


where V : wg(d 9 x) is the volume of the gap. Factor 2 in Eq. (5.151) takes E into account the identical integrations over the gap regions above and below the plunger. To evaluate Eq. (5.151) we need to know the fields B and H in the E E gap. To this end, apply Eq. (3.141) to a path around the circuit through the plunger and obtain 2H g : ni, E which gives ni H : . E 2g Further, B : H . Substitute B and H into Eq. (5.151) and obtain E E E E nw L(x) : (d 9 x). (5.152) 2g Then, substitute this into Eq. (5.149), which gives g" W (", x) : . DJB nw(d 9 x) Finally, obtain the force using Eq. (5.127), f

DJB

:9

W (", x) DJB x

:9

g" . nw(d 9 x)

(5.153)

(5.154)

As 0 x d and " 0, the force is negative, which means that it acts to pull the plunger towards the core (in the direction of decreasing x). Formula (5.154) would be more useful if we had an explicit expression for ". For an n-turn coil we have " : n where is the flux through each turn of the coil. As there is no flux leakage, : : B A where A : w(d 9 x). Therefore E E E E ni ni : w(d 9 x) and " : w(d 9 x). 2g 2 g Coenergy: In the coenergy approach i and x are the independent variables and the coenergy is given by Eq. (5.139), WA : DJB

L(x)i . 2

(5.155)


387

Substitute Eq. (5.152) into Eq. (5.155) and obtain in WA : w(d 9 x). DJB 4g

(5.156)

The force follows from Eq. (5.138), inw f :9 . DJB 4g

(5.157)

Notice that the force is independent of x. The two force expressions (5.154) and (5.157) are equivalent. To see this, substitute Eq. (3.78) (i.e., " : L(x)i) into Eq. (5.154), which gives Eq. (5.157). ) EXAMPLE 5.8.2 Consider the cylindrical actuator shown in Fig. 5.14. Assume that the magnetic circuit is linear with , and that there is no flux leakage. Set up the equations of motion for the actuator [2]. SOLUTION 5.8.2 First, choose a convention in which a positive f is in the DJB direction of increasing x as shown. As the motion is linear, we use the first-order

FIGURE 5.14

Cross section of a cylindrical actuator circuit [2].

388


system of equations (5.86),

di(t) 1 "(i, x) : V (t) 9 i(t)(R ; R ) 9 v(t) dt L(x) Q x dv(t) 1 : F(i, x) m dt

(5.158)

dx(t) : v(t). dt To solve Eq. (5.158) we need expressions for L(x), "(i, x)/x, and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). DJB From Eq. (5.139) we have W A (i, x) : DJB

"(i, x)i 2

:

L(x)i . 2

(5.159)

Inductance: We determine the inductance using Eq. (3.75), L:

1 i

B · H dv. (5.160) 4 As the core and plunger have a high permeability, the H-field in these elements is negligible, and Eq. (5.160) reduces to an integration over the nonmagnetic sleeve regions, L:

1 i

B · H dv ;

1 i

B · H dv,

(5.161)

4 4 where the subscripts top and bot refer to the top and bottom sleeve regions, respectively. Notice that V dgh and V dg(h 9 x). We need to determine H and H . To this end, apply Eq. (3.141) to either of the dotted paths in Fig. 5.14. This gives g ; H g : ni. (5.162) In addition, as there is no flux leakage : , or B A :B A , (5.163) where A : dh and A : d(h 9 x). From Eq. (5.163) and the fact that H


B

: H and B : H , we find that h H :H . (h 9 x)

389

(5.164)

Substitute Eq. (5.164) into Eq. (5.162) and obtain H

:

ni (h 9 x) . g (2h 9 x)

(5.165)

Finally, substitute Eqs. (5.164) and (5.165) into Eq. (5.161), which gives the inductance ndh (h 9 x) . L(x) : g (2h 9 x)

(5.166)

Coenergy: Once we know the inductance, the coenergy is easily obtained from Eq. (5.159), indh (h 9 x) W A (i, x) : . DJB 2g (2h 9 x)

(5.167)

We use Eqs. (5.137) and (5.138) to obtain " and f , and find that DJB W A (i, x) DJB ": i indh (h 9 x) : , g (2h 9 x)

(5.168)

and f

DJB

:

W A (i, x) DJB x

ind h :9 . (2h 9 x) 2g

(5.169)

As x h, Eq. (5.169) is negative, which implies that the direction of f is DJB opposite to the direction of increasing x (i.e., opposite to the direction of increasing air gap). The total force F(i, x) is a sum of the forces due to the field and spring, F(i, x) : f where

DJB

(i, x) ; f (x). Q

(5.170)

f (x) : 9k(x 9 x ). (5.171) Q Here, x is the initial position of the plunger, which is assumed to be the equilibrium position for the spring. When x x (x x ), f (x) tends to move Q

390


the mass into (out of) the solenoid. Finally, substitute Eqs. (5.168), (5.169), (5.170), and (5.171) into Eq. (5.158) and obtain

di(t) (2h9x) ind h g : V (t)9i(t)(R;R ); v(t) Q dt ndh (h9x) g (2h9x) dv(t) 1 ind h :9 ; k(x 9 x ) m dt 2g (2h 9 x)

dx(t) : v(t). dt

(5.172)

This nonlinear first-order system is valid for 0 x h. It is solved subject to the initial conditions x(0) : x , v(0) : v and i(0) : i . ) EXAMPLE 5.8.3 Set up the equations of motion for the actuator shown in Fig. 5.15. Assume that all magnetic materials are linear with . The nonmagnetic sleeves of thickness s provide low friction support for the moving plunger [4]. SOLUTION 5.8.3 Set up a coordinate system with f positive in the direction DJB of increasing x as shown. As the motion is linear, we use the first-order system of equation (5.86)

di(t) 1 "(i, x) : V (t) 9 i(t)(R ; R ) 9 v(t) dt L(x) Q x dv(t) 1 : F(i, x) dt m dx(t) : v(t). dt

FIGURE 5.15

(5.173)

Cross section of an actuator circuit [4].

391


To solve Eq. (5.173) we need expressions for L(x), "(i, x)/x and F(i, x). As the independent variables are i and x, we work with the coenergy W A (i, x). From DJB Eq. (5.139) we have "(i, x)i W A (i, x) : DJB 2 :

L(x)i . 2

(5.174)

Inductance: We determine the inductance using Eq. (3.75), L:

1 i

B · H dv. (5.175) 4 As the core and plunger have a high permeability, H is negligible in these elements and Eq. (5.175) reduces to an integration over the gap region and the nonmagnetic sleeve regions, L:

1 i

B · H dv ;

2 i

B · H dv, (5.176) 4E 4Q where V and V are the volumes of the gap and sleeve regions, respectively. The E Q factor of 2 in the second integral takes into account the identical integrations over each sleeve region. Notice that V : A x and V : A s, where A and A E E Q Q E Q are the cross-sectional areas of the gap and sleeve, respectively. We need to determine H and H . To this end, apply Eq. (3.141) to the dotted path in Fig. E Q 5.15. This gives H x ; H s : ni. (5.177) E Q Further, as there is no flux leakage : 2 , or E Q B A : 2B A . (5.178) E E Q Q Combining Eqs. (5.177) and (5.178) with B : H and B : H gives E E Q Q ni H : , (5.179) E (x ; (A /2A )s) E Q and A ni H : E . (5.180) Q 2A (x ; (A /2A )s) Q E Q Next, substitute Eqs. (5.179) and (5.180) into Eq. (5.176), which gives the inductance L(x) :

nA E . (x ; (A /2A )s) E Q

(5.181)

392


Coenergy: Once we know the inductance, the coenergy is obtained easily from Eq. (5.159) inA E W A (i, x) : . DJB 2[x ; (A /2A )s] E Q Use Eqs. (5.137) and (5.138) to obtain " and f . We find that DJB W A (i, x) DJB ": i :

inA E , [x ; (A /2A )s] E Q

(5.182)

(5.183)

and f

DJB

:

W A (i, x) DJB x

inA E . (5.184) 2[x ; (A /2A )s] E Q Notice that f is negative, which implies that the direction of f is opposite to DJB DJB the direction of increasing x (i.e., opposite to the direction of increasing air gap). The total force F(i, x) is a sum of the forces due to the field and the spring, :9

F(i, x) : f

DJB

(i, x) ; f (x), Q

(5.185)

where f (x) : 9k(x 9 x ). (5.186) Q Here, x is the initial position of the plunger, which is assumed to be the equilibrium position for the spring. When x x (x x ), f (x) tends to move Q the mass into (out of) the circuit. Finally, substitute Eqs. (5.181), (5.184), (5.185), and (5.186) into Eq. (5.173) to obtain

di(t) (x ; (A /2A )s) inA E Q E : V (t) 9 i(t)(R ; R ) ; v(t) Q dt nA [x ; (A /2A )s] E E Q dv(t) 1 inA E :9 ; k(x 9 x ) dt m 2[x ; (A /2A )s] E Q dx(t) : v(t). (5.187) dt

This nonlinear first-order system is solved subject to the initial conditions x(0) : x , v(0) : v , and i(0) : i . )

393

5.9 AXIAL-FIELD ACTUATORS

5.9 AXIAL-FIELD ACTUATORS Axial-field actuators are moving coil actuators. These actuators are well suited and widely used for applications that require controllable bidirectional rotation. For example, they are routinely used for positioning read/write heads in computer disk drives. These actuators consist of pie-shaped magnets that are positioned above (and/or below) a coil that is free to rotate as shown in Fig. 5.16. When the coil is energized, it experiences a torque and rotates either clockwise or counterclockwise depending on the direction of current. A mechanical component such as a spring is often used to provide a restoring torque, and to fix the unenergized position of the coil (Fig. 5.17). Axial-field actuators can be designed and optimized prior to fabrication using lumped-parameter analysis [9, 10]. We develop a model for performing such analysis in the following example. EXAMPLE 5.9.1 Determine the equations of motion for the axial-field actuator shown in Fig. 5.16. Assume that the coil has n turns. SOLUTION 5.9.1 This device is a moving coil actuator and is governed by Eqs. (5.102),

di(t) 1 : V (t) 9 i(t)(R ; R ) ; dt L Q

[( ; r) ; B

· dl

1 d (t) : i(t) [r ; (dl ; B )] · z ; T ()

dt J K d(t) : (t). dt

FIGURE 5.16

Drive magnet and coil for an axial-field actuator.

(5.188)

394


FIGURE 5.17

Axial-field actuator with spring mechanism.

Here, L and R are the inductance and resistance of the coil, and J is its K moment of inertia above the z-axis. We need to evaluate the induced voltage and torque integrals in Eq. (5.188). We use a cylindrical coordinate system at rest with respect to magnets as shown in Fig. 5.16. Before we begin, we state some simplifying assumptions. Assumptions: The first assumption is that the magnetic field due to the magnet is essentially uniform and constant across the coil, that is,

9B z 0

K (5.189) B z 9 0,

K where 9 and ; define the angular positions of the radial edges of the K K B :

395


FIGURE 5.18

Electrical connection for a single turn of the coil.

magnet. In reality, the field distribution above the magnet varies with height, and from point to point at a fixed height. However, an average value of B

that is adequate for this analysis can be obtained using three-dimensional FEA, closed-form analysis [11], or measured data if the fabricated magnet is available. The second assumption is that each of the coil’s two radial edges rotates over a separate magnetic pole throughout the entire range of motion. Induced voltage: We determine the voltage induced at the terminals of the coil by summing the voltage induced across its various segments. For reference purposes, Fig. 5.18 shows the electrical connection of a single turn. The coil has four segments: an upper and a lower segment, and two side segments (Fig. 5.19). We consider a single turn of the coil and evaluate the side (radial) segments first. We need to determine ( ; r) ; B along these segments. First, we know that (t) : (t)z . Then from Fig. 5.19 we have r : rr . Thus, ; r : (t)r Note that B is in the z-direction, 9B z B : B z

(side segments).


396


FIGURE 5.19

Top view of the coil with the magnetic field.

Therefore, ( ; r) ; B

:

9 (t)rB r (t)rB r


(5.190)

We evaluate

[(;r);B ] · dl

along the left and right sides where dl : dr r

(side segments).

We obtain

[(;r);B ] · dl : 9 left side

0 (t)B r dr 0

R 9 R , : 9 (t)B 2

(5.191)

397


and

[( ; r) ; B ] · dl :

0 (t)B r dr 0

right side

R 9 R . : 9 (t)B 2

(5.192)

Now we consider the top and bottom segments. For these we have ( ; r) ; B r . Because dl , it follows that [( ; r) ; B ] · dl r · : 0

(top/bottom segments).

Therefore,

[( ; r) ; B ] · dl : 0, top/bottom segments

which shows that these segments do not contribute to the induced voltage. Finally, add Eqs. (5.191) and (5.192), and multiply by n (the number of turns) to obtain an expression for the induced voltage

[( ; r) ; B ] · dl : 9 (t)nB (R 9 R).

(5.193)

Torque: Only the side segments of the coil contribute to the torque. We make a brief argument to show that the upper and lower segments make no contribution. Consider the torque on an element of the upper or lower segment, dT : [r ; (dl ; B )] · z , X where r : rr and dl . . Notice that dl ; B

. ; B . r .

This gives dT . [r ; r ] · z : 0, X which shows that the upper and lower segments generate no torque. Now

398


consider the left- and right side segments. For these, we have dl : drr , and

B dr dl ; B : 9B dr


Therefore,

B r dr z r ; (dl ; B ) : 9B r dr z


(5.194)

We evaluate

[r ; (dl ; B )] · z

along the left- and right side segments with the integration in the direction of the current flow. This gives

[r ; (dl ; B )] · z :

0 B r dr 0

left side

:B and

(R 9 R) , 2

(5.195)

0 B r dr [r ; (dl ; B )] · z : 9 0 right side

:B

(R 9 R) . 2

(5.196)

The torque on the coil is ni times the sum of Eqs. (5.195) and (5.196),

[r ; (dl ; B )] · z : niB (R 9 R). (5.197) Equations of motion: We are finally ready to write the equations of motion. Substitute the integrals (5.193) and (5.197) into Eq. (5.188) and obtain di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 nB (R 9 R) (t)] dt L Q d (t) 1 : [i(t)nB (R 9 R) ; T ()]


(5.198)

399


These equations are solved subject to the initial conditions of Eq. (5.89). Finally, from Eq. (5.198) we see that the electrical constant K equals the torque constant C K (see Eq. (5.100)). Specifically, R K : K : nB (R 9 R). C R Calculations: We apply Eq. (5.198) to an actuator with the mechanical mechanism shown in Fig. 5.17. Here, the mechanical torque is supplied by the spring. To perform the analysis we need expressions for the coil inductance L and the mechanical restoring torque T ().

Inductance: The inductance of the coil can be estimated by considering a short circular coil with a height h equal to that of the actuator coil, and with A a radius r , A R 9R . r : A 2

(5.199)

To first order, the flux through the center of the coil (through each turn) is (i)

inr A . (h ; (2r )) A A

(5.200)

Therefore, the flux linkage is "(i) : n(i). Recall that the inductance is d"(i) , di

(5.201)

nr A . (h ; (2r )) A A

(5.202)

L: which gives L

This approximation is appropriate if the mean circumferential arc subtended by the coil is approximately equal to R 9 R , which is the case for many practical designs. However, if this is not the case the inductance can be determined empirically, or calculated using three-dimensional FEA. Mechanical Torque: The mechanical torque is provided by the spring mechanism shown in Fig. 5.17. The restoring force is F(Y) : k (Y 9 Y ) ; F , Q where k is the spring constant, Y is the initial spring length, F is the force Q

400


FIGURE 5.20

Applied voltage V (t). Q

when Y : Y , and Y is the stretched length. The length Y is defined by the angle #. From geometry we have Y : (X ; D 9 2XD cos(# ; ). The restoring torque is given by () : DF(Y()) cos(# ; ). (5.203)

Analysis: We substitute Eqs. (5.202) and (5.203) into Eq. (5.198) and perform the analysis. The applied voltage V (t) is shown in Fig. 5.20 and the remaining Q parameters are as follows: T

B : 0.5 T

R;R :3 n : 100 turns R : 10 mm R : 15 mm

401


FIGURE 5.21

Current i(t).

D : 10 mm # : 45° h : 1.5 mm A k : 1.5 N/mm Q J : 6 ; 10\ kg · m. K The moment of inertia is computed using Eq. (5.77), which gives J : (R 9 R) A ; (R 9 R) A , K where and A are the density and area of the wire and is the

angular span of the coil (left- to right side) in radians. The response of the actuator is computed for the initial conditions i(0) : 0A, (0) : 0°, and (0) : 0 rad/sec. The current profile i(t) is shown in Fig. 5.21. The rotation angle (t) and angular velocity (t) are shown in Fig. 5.22. Notice that the angular velocity (t) passes through zero and turns negative as the angular position (t) peaks at approximately 6.5 ms. )

402


FIGURE 5.22

Angular position (t) and velocity (t).

5.10 RESONANT ACTUATORS Resonant actuators are used as drive mechanisms for scanning elements in optical scanning systems [12]. An example of such an actuator is shown in Fig. 5.23. In this actuator a wedge-shaped magnet provides an axial bias field, and a coil rotates above the magnet once it is energized. A torsional pivot mechanism provides the restoring torque. When power consumption is an issue, the torsional mechanism can be designed to render the actuator resonant at the scanning frequency. When this is the case, the actuator can be brought to a resonant oscillation with a sequence of low-energy pulse excitations, thus minimizing power consumption. We demonstrate this in the following example. EXAMPLE 5.10.1 Develop a model for the design of the axial-field resonant actuator shown in Fig. 5.23. SOLUTION 5.10.1 The behavior of this device is governed by the equations for an axial-field actuator as derived in Example 5.9.1. We repeat them here for

403

5.10 RESONANT ACTUATORS

FIGURE 5.23

Resonant actuator.

convenience, di(t) 1 : [V (t) 9 i(t)(R ; R ) 9 nB (R 9 R) (t)] dt L Q d (t) 1 : [i(t)nB (R 9 R) ; T ()]


(5.204)

Mechanical torque: The restoring torque is provided by a torsional pivot mechanism T () : 9K . (5.205)

N Here, K is the spring constant of the pivot. This simple linear relation is N typically only valid for limited rotations, for example, 912° 12°. The

404


resonant frequency f for this mechanism is f:

1 2

K N. J K

(5.206)

Calculations: We demonstrate the use of Eq. (5.204) with some sample calculations. We numerically integrate these equations using the fourth-order Runge-Kutta method (Appendix C). The following parameters are used for the analysis: B

: 0.55 T

R;R :6 n : 100 turns R : 3.2 mm R : 11.5 mm L : 0.4 mH J : 0.3 ; 10\ kg · m K f : 100 Hz The field B : 0.55 T can be achieved using sintered NdFeB magnets above and below the coil with soft-magnetic flux plates attached to the outer surface of each magnet. The flux plates enhance the field across the coil. We apply a periodic step function voltage V (t) with a magnitude of L L L L L x t v : v ; F(i , x ) L> L m L L

x : x ; t v . (C.12) L> L L The solution procedure for Eq. (C.12) is similar to the preceding steps 1 through 4. Specifically, start with the initial conditions [i , v , x ] and use Eq. (C.12) to

502

APPENDIX C Systems of Equations

obtain [i , v , x ], which gives the solution at t : t. Repeat this process to obtain [i , v , x ], [i , v , x ], . . . , [i , v , x ], . . . , until solution values are L L L obtained for the entire interval (t , t ). )

C.2 IMPROVED EULER METHOD The accuracy of Euler’s method can be improved by using a more accurate estimate of the integral in Eq. (C.3). In the improved Euler method the integral is estimated using the trapezoidal rule as shown in Fig. C.4. This gives

RL>

t [f(y(t ), t ) ; f(y(t ), t )]. L> L> L L 2

(C.13)

t ) : y(t ) ; [f(y(t ), t ) ; f(y(t ), t )]. L> L L L L> L> 2

(C.14)

f(y, t) dt

RL We substitute Eq. (C.13) into Eq. (C.3) and obtain y(t

Notice that the unknown y(t ) appears on both sides of Eq. (C.14). L> Thus, it cannot be obtained explicitly. However, Eq. (C.14) can be transformed into an explicit relation for y(t ) by using an estimate on L> the right side. Specifically, y(t ) can be estimated using the Euler L> relation (C.5). Let y (t ) denote the estimate, L> y (t ) : y(t ) ; t f(y(t ), t ). L> L L L

FIGURE C.4

Integral estimate for the improved Euler method.

503

C.3 RUNGE-KUTTA METHODS

Substitute this in the right side of Eq. (C.14) and obtain t y(t ) : y(t ) ; [f(y(t ), t ) ; f(y (t ), t )]. L> L L L L> L> 2

(C.15)

This can be rewritten using index notation as follows: t y : y ; [f(y , t ) ; f(y , t )], L> L L L L> L> 2

(C.16)

where y : y ; t f(y , t ). (C.17) L> L L L Equations (C.16) and (C.17) can be solved iteratively as described in steps 1 through 4 given in the preceding text. A flow chart of the solution procedure is shown in Fig. C.5. The improved Euler method is more accurate and stable than the standard Euler method. Specifically, the error of this method in any one interval is proportional to t and global error is proportional to t.

C.3 RUNGE-KUTTA METHODS Runge-Kutta methods are widely used for integrating first-order equations. The fourth-order Runge-Kutta method is perhaps the most popular. In this method, the estimate of Eq. (C.3) is based on Simpson’s 1/3 rule [1]. This leads to the following set of formulas: k : t f(y , t ) L L

k t k : t f y ; , t ; L 2 L 2 k t k : t f y ; , t ; L 2 L 2 k : t f(y ; k , t ; t) L L y : y ; [k ; 2k ; 2k ; k ]. L> L

(C.18)

A flow chart of the solution procedure is shown in Fig. C.6. The fourth-order Runge-Kutta method is accurate to the fourth-order term in the Taylor expansion. Thus, the error in any interval is proportional to

504

FIGURE C.5


Flow chart for implementation of the improved Euler method.


FIGURE C.6

method.

505

Flow chart for implementation of the fourth-order Runge-Kutta

506


t. The following example demonstrates the use of this method for the solution of a first-order system of equations. EXAMPLE C.3.1 Apply the fourth-order Runge-Kutta method to the equations for rotational motion (5.87). SOLUTION C.3.1 The equations governing rotational motion are

di(t) 1 "(i, ) : V(t) 9 i(t)(R ; R ) 9 (t) , L dt

(C.19)

d(t) : (t), dt

(C.20)

and d (t) T(i, ) : . (C.21) J dt K Note that there are three dependent variables i(t), (t), and (t). We need a set of Runge-Kutta coefficients for each variable. We denote these coefficients by (k , k , k , k ), (m , m , m , m ), and (p , p , p , p ), respectively. To implement the Runge-Kutta method, it is convenient to rewrite Eqs. (C.19)—(C.21) as follows: di(t) : f(V(t), i(t), (t), (t)), dt d(t) : g( (t)), dt and d (t) : h(i(t), (t)), dt where f(V(t), i(t), (t), (t)) :

1 "(i(t), (t)) V(t) 9 i(t)(R ; R ) 9 (t) , L

g( (t)) : (t), and h(i(t), (t)) :

T(i(t), (t)) . J K

507


The Runge-Kutta coefficients are determined from the following relations: k : t f(V(n), i(n), (n), (n)) m : t g( (n)) p : t h(i(n), (n)), k m p k : t f V(n), i(n) ; , (n) ; , (n) ; 2 2 2

(C.22)

p m : t g (n) ; 2

k m p : t h i(n) ; , (n) ; , 2 2 k m p k : t f V(n), i(n) ; , (n) ; , (n) ; 2 2 2

(C.23)

p m : t g (n) ; 2

k m p : t h i(n) ; , (n) ; , 2 2

(C.24)

and k : t f(V(n), i(n) ; k , (n) ; m , (n) ; p ) m : t g( (n) ; p ) p : t h(i(n) ; k , (n) ; m ). (C.25) Once the coefficients have been computed, they are used to determine the subsequent solution values as follows: i : i ; [k ; 2k ; 2k ; k ] L> L : ; [m ; 2m ; 2m ; m ] L> L : ; [p ; 2p ; 2p ; p ]. (C.26) L> L The solution procedure for Eqs. (C.19)—(C.21) is similar to steps 1 through 4 of Euler’s method. Specifically, start with the initial conditions [i , , ] and use Eqs. (C.22)—(C.25) to determine the coefficients (k , k , k , k ), (m , m , m , m ), and (p , p , p , p ). Once these are known, use Eq. (C.26) to

508


obtain [i , , ], which gives the solution at t : t. Repeat this process to obtain [i , , ], [i , , ], . . . , [i , , ], . . . , until solution values are L L L obtained for the interval (t , t ). ) References 1. Nakamura, S. (1991). Applied Numerical Methods with Software, Englewood Cliffs, NJ: Prentice-Hall.

Appendix D: Units

In this section we discuss three systems of units that are commonly used in magnetism. These are the CGS or Gaussian system, and two SI systems (International System of Units or Système international d’unités). The SI systems are referred to as the Kennelly and Sommerfeld conventions, respectively. The CGS system is based on the fundamental units of centimeter, gram and second, and the SI systems are based on the MKSA units of meter, kilogram, second, and ampere. In this book we use SI system units in the Sommerfeld convention. The basic entities and units in the Sommerfeld and CGS systems are listed in Table D.1. Maxwell’s equations in these two systems are as follows: Maxwell’s Equation SI ;H:J;

CGS (Gaussian) D t

·B:0 B t ·D:

;E:9

4 1 D J; c c t ·B:0 1 B ;E:9 c t · D : 4.

;H:

(D.1)

509

510

APPENDIX D Units TABLE D.1

Fields and units in the SI (Sommerfeld) and CGS systems

Symbol

Description

SI

CGS

H B M

Magnetic field strength Flux density Magnetization Flux

A/m Tesla A/m Weber

Oe Gauss emu/cm Maxwell

The constitutive relations are as follows:

Constitutive Relations SI (Sommerfeld)

CGS (Gaussian)

B : (H ; M) D:+ E;P : 4 ; 10\ 10 + : 4c

B : H ; 4M

(D.2)

D : E ; 4P :1 + : 1.

To convert between the Sommerfeld and CGS systems we use Tables D.2 and D.3. For example, from Table D.1 we find that 1 A/m : 4/10 Oe and that 1 J/m : 40 GOe and so on. Similarly, from Table D.3 we find that 1 Oe : 10/4 A/m and that 1 MGOe : 10/4 kJ/m and so on.

TABLE D.2

Conversion from SI to CGS units

SI H in A/m B in Tesla M in A/m in Webers (BH) in J/m

(BH) in kJ/m

to ; ; ; ; ; ;

4/10 10 10\ 10 40 4/10

CGS : : : : : :

H in Oe B in Gauss M in emu/cm in Maxwells (BH) in GOe

(BH) in MGOe

511

UNITS TABLE D.3

Conversion from CGS to SI units

CGS H in Oe B in Gauss M in emu/cm in Maxwells (BH) in GOe

in MGOe (BH)

to ; 10/4 ; 10\ ; 10 ; 10\ ; 10\/4 ; 10/4

SI : : : : : :

H in A/m B in Tesla M in A/m in Webers (BH) in J/m

(BH) in kJ/m

Other useful relations are 1 T:1

N , A·m

and 1 Wb : 1 T m. The Kennelly and Sommerfeld conventions differ in the definition of magnetization. In the Kennelly convention the constitutive relation is B: H;J (Kennelly convention). Here, J is called the magnetic polarization or intensity of magnetization and is measured in tesla. It is also referred to as the intrinsic flux density and is denoted by B . Notice that J : M. The fundamental element in G magnetism is the magnetic dipole m. In the Sommerfeld convention m is measured in A · m. In the Kennelly and CGS systems it is measured in Wb · m and emu, respectively (1 emu : 4 ; 10\ Wb · m). If a magnetic dipole is subjected to an external B-field, it acquires an energy E : 9m · B,

(D.3)

T : m ; B.

(D.4)

and experiences a torque

In both the Kennelly and CGS systems these are given by E : 9m · H and T : m ; H. In the remainder of this section we give conversion factors for various mechanical units. There are three systems that are commonly used for mechanical analysis. These are the MKS, CGS and British systems. In the MKS system the fundamental units of length, mass, and

512

APPENDIX D Units TABLE D.4

Conversion of mechanical variables from MKS

to CGS units MKS

to ; ; ; ;

mass in kg force in newtons (N) torque in N · m energy in joules

10 10 10 10

CGS : : : :

mass in grams force in dynes torque in dyne · cm energy in ergs

time are the meter (m), kilogram (kg), and second. The unit of force is the newton (N), and the unit of energy is the joule (J). We use the MKS system throughout this text. In the CGS system, the fundamental units are the centimeter (cm), gram (g), and second. In this system the units of force and energy are the dyne and erg, respectively. Last, in the British system the units of mass, length, and time are the slug, foot, and second, respectively. For example, acceleration due to gravity is g : 32 ft/s, whereas in the MKS system g : 9.8 m/s. In the British system the unit of force is the pound (lb), and a commonly used unit of energy is the British thermal unit (Btu). Tables D.4 and D.5 give conversion factors for various mechanical units. Other useful relations are as follows: 1. Length: 1.0 in : 2.54 cm, 1.0 ft : 0.3048 m. 2. Mass density: 1.0 lb/ft : 16.02 kg/m. 3. Torque: 1.0 lb · in:0.113 N · m, 1.0 oz · in:72 gm · cm, 1.0 N · m : 1.019 ; 10 g · cm.

Conversion of mechanical variables from British to MKS units

TABLE D.5

British mass in slugs force in lbs torque in ft · lb energy in Btu

to ; ; ; ;

14.59 4.45 1.3558 1055

MKS : : : :

mass in kg force in newtons torque in N · m energy in joules

Index A Actuators axial-field, 393—401 linear, 413—421 magnetic circuit, 383—392 magnetooptical bias field, 406—413 moving magnet, 449 plunger, 383, 387, 390 resonant, 402—406 rotary bias field, 322 solenoid, 136 Alnico, 48—50 Ampere’s circuital law, 78, 88, 338 Angular momentum, orbital versus spin, 10—11 Anisotropic materials, 5, 77 Anisotropy magnetocrystalline, 27—31 shape, 31—33 Antiferromagnetic material, 6, 7 Atomic magnetic moments, 7—9 multielectron atoms, 14—17 single electron atoms, 9—14 Axial coupling, 286, 288, 297—306 Axial-field actuators, 393—401 Axial-field motors, 421—437 no-load speed, 424 stall-torque, 424 Axially polarized cylindrical shell, 245—253 Axially polarized multipole cylindrical disk magnet, 253—264 B Back emf constant, 368 Barium ferrite, 46 Barkhausen, 36 Bar magnet, 211—217 rectangular, 133—135 Bearing configurations, radial and axial, 273 Bessel function, 176 B-H hysteresis loop, 36—39 Bias field actuators, 406—413 magnetooptical recording and, 323 Bias magnets cylindrical, 218—264 rectangular, 208—218

Biot-Savart law, 104 Bipolar cylindrical magnet, 218—224 Bipolar sphere, 182—185 Bloch walls, 34—35 Bohr magneton, 11 Boltzmann’s constant, 17 Boltzmann statistics, 17 Bonded ferrite magnets, 54 Bonded magnets, 53—55 Boundary conditions, 79—81 Boundary-value problems, 161 Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 magnet between plates, 165—169 spherical coordinates, 180—185 Bound charge, 75 Br, 38 Brillouin function, 19, 22 Brushless dc motors. See Axial-field motors C Can-stack motors, 437—446 Carrier particles, 315 Cartesian coordinates, 469—473 boundary-value problem, 162—174 Laplace’s equation, 162—174 transformations, 491—494 Ceramic magnets, 46—47 CGS/Gaussian unit, 2—5, 509—512 Chain rule, 345 Charge bound, 75 free, 75 Charge density, free, 75 Charge model, 131—134 force, 135—140 torque, 141—143 Circuits current sources, 144—153 equivalent, 157—159 inductance, 151 magnet sources, 153—161 moving, 358—361, 366—376 no flux leakage, 145, 150 stationary, 354—358, 363—366 Coenergy, 379 Coercive force, 31

513

514 Coercivity, 31, 38, 39 intrinsic, 29, 32 Conduction current, 75 Constitutive relations, 75—77 magnetostatic, 99 Contact latching, 268—269 Continuity equation, 75 Convection current, 75 Conversion factors, 3, 509—512 Coordinate transformations, 491—494 Coulomb gauge condition, 86, 103, 127 Coupling, synchronous applications, 286 axial, 286, 288, 297—306 radial, 286, 288—297 Russell-Saunders, 14—15 spin-orbit, 14 Cross product, 471 Curie’s law, 19 Curie temperature, 20 Curl, 472 Current conduction, 75 convection, 75 electrolytic, 75 equivalent, 128 free, 75 line, 104—106 sources, 144—153 surface, 107 Current density displacement, 75 upper limit of, 58 Current model, 126—131 force, 131 torque, 131 Cylindrical coordinates, 473—476 boundary-value problem, 175—180 Laplace’s equation, 175—180 transformations, 491—494 Cylindrical magnets axially polarized multipole disk magnet, 253—264 axially polarized shell, 245—253 axial polarization, 128—130 bipolar, 218—223 radially polarized multipole, 224—245 D Del operator, 343, 472 Demagnetization curve, 44, 66—67 for ceramic, 67 for samarium-cobalt, 67 Demagnetization field, 23—27 Diamagnetic material, 6 Die-upsetting, 53 Differential Maxwell equations, 74, 79, 338

INDEX Dipole-dipole force, 140, 316 Dipole/dipole moment, 3, 108 Energy in external field, 4, 511 magnetic, 140 of sphere in external field, 26 torque in external field, 4, 511 Direct current motors. See Axial-field motors Dirichlet boundary-value problem, 162 Disk magnets, levitation, 273—281 Dislocations, 34 Displacement current density, 75 Divergence, 472 Divergence theorem, 78, 114, 487—489 Domains, 6, 20, 33—36 Domain walls, 34 motion of, 35—36 Dot product, 470 Drag torque, 423 Drive torque, 453 Dyadic, 113 E Earnshaw’s theorem, 273 Eddy currents, 350—353, 411—412 drag force, 352, 353 lift force, 352, 353 Electrical constant, 368 Electrical equations moving coils, 358—361 stationary circuits, 354—358 Electrolytic current, 75 Electromagnets, conventional, 308, 310 Electromechanical devices actuators, axial-field, 393—401 actuators, linear, 413—421 actuators, magnetic circuit, 383—392 actuators, magnetooptical bias field, 406—413 actuators, resonant, 402—406 basics, 335—337 electrical equations, 354—361 electromechanical equations, 363—376 energy analysis, 376—383 examples of, 335 hybrid analytical-FEM analysis, 446—455 mechanical equations, 361—363 micro, 455—465 motors, axial-field, 421—437 motors, stepper, 437—446 quasi-static field theory, 337—353 Electromechanical equations moving coils, 366—376 stationary circuits, 363—366 Electron spin, 10 Electrophotography, 312—318 Electrostatic theory, 89—91

515

INDEX emu, 3, 511 Energy analysis, 376—383 balance equation, 377 co-, 379 contours, 46 density, 117 exchange, 22 magnetostatic, 23, 116—117 point of maximum, 44, 153 product, 157 self-energy equation, 23, 117 Equivalent charge, 132 Equivalent circuit, 157—159 Equivalent currents, 128 Euler’s method, 363, 365—366, 418, 498—503 Exchange energy, 22 Exchange integral, 22 Expectation value, 10 F Faraday’s law of electromagnetic induction, 78, 85, 347 FEA. See Finite element analysis FEL. See Free-electron lasers FEM analysis, hybrid analytical, 446—455 Ferric oxide, 47 Ferrimagnetic material, 6, 7 Ferrites, 46—47 bonded, 54 Ferromagnetic material, 6 Ferromagnetism, 19—22 Finite bar magnet, 211—217 Finite difference method, 200—204 backward, 202 central, 202—203 forward, 202 Finite element analysis (FEA), 60, 190—200 Finite radially polarized multipole cylinder, 235—245 Flux density, 2, 3, 98 line, 62, 108—109 linkage, 119—120, 341 magnetic, 144 remanent, 31 vector field, 78 Force between cylindrical magnets, 139—140 between dipoles, 140 between two wires, 111 charge model, 135—140 coercive, 31 current model, 131 on dipole in external field, 317 Lorentz, 82, 110 magnet above plate, 188, 189

on steady currents, 110—112 on wire above plate, 188 Free charge density, 75 Free current density, 75 Free-electron lasers (FELs), 325—331 Fringe fields, 309 G Galilean transform, 344 Gamma ferric oxide, 47 Gauss’ law, 78, 90, 338 Gears, 282—286 Gradient, 472 Green’s function, free-space, 90, 103, 495—496 H Hall effect, 265, 422 Hard magnetic materials, 44—45 alnico, 48—50 bonded, 53—55 ferrites, 46—47 neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Helmholtz’s theorem, 489—491 Henry, 118 High field structures, 265—268 Homogeneity, magnetic resonance imaging and, 309 Homogeneous material, 5, 77 Hund’s rules, 15 Hybrid analytical-FEM analysis, 446—455 Hysteresis loop, 36—39 I Images, method of. See Method of images Impulse magnetizers, 58—59 Inclusions, 34 Inductance, 341 external, 119 internal, 119 magnetostatic analysis, 118—126 mutual, 120—121 parallel wires, 122—123 self, 118 solenoid, 123—124, 125—126 toroid, 124—125 wire, 121—122 Induced voltage, 370, 372 Inertial reference frame, 341—342 Infinitely long cylindrical magnet, 224—235 Infinite rectangular magnet, 208—211 Inhomogeneous material, 5, 77 Initial magnetization curve, 37 Integral Maxwell equations, 77—79, 338 Integrals, of vector functions, 479—485 Intrinsic coercivity, 29 shape anisotropy and, 32

516 Iron oxides, 47 Iron-silicon alloys, 39, 43 Irreversible changes, 67, 70 Isotropic materials, 5, 77 K Kennelly units, 2, 509—512 Kerr effect, 320—321 Kirchhoff’s laws current, 146, 338 voltage, 89, 146—147, 339, 341 voltage (moving coil), 358—361 voltage (stationary circuit), 354, 357 L Land g factor, 14 Langevin function, 18 Laplace operator, 472—473 Laplace’s equation Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 spherical coordinates, 180—185 Larmor frequency, 307, 308 Larmor relation, 309 Latching, magnetic, 268—273 Legendre polynomials, 182—185 Legendre’s equation, 181 Lenz’s law, 78 Levitation, passive, 273—281 Linear actuators, 413—421 Linear materials, 5, 77 Line integral, 479—482 Load line, 44—46, 155—156 stability, 66—70 Lodestone, 47 Lorentz force, 82, 110, 362 Lorentz gauge condition, 83 Lorentz transformation, 343—345 M Magic cylinder, 265, 266 Magic sphere, 265—268 Magnequench, 52—53 Magnetic bearing. See Magnetic levitation Magnetic brush technique, 315—318 Magnetic circuit actuators, 383—392 Magnetic circuit analysis, 144—161 current sources, 144—153 equivalent circuits, 157—159 magnet sources, 153—161 Magnetic coupling. See Coupling Magnetic dipole. See Dipole moment Magnetic field bar magnet, 211—217 bipolar sphere, 182—185 current loop, 108 cylindrical magnets. See Cylindrical magnets

INDEX line current, 104—107 magnet between plates, 165, 168 rectangular magnets. See Rectangular magnets solenoid, 100—101 strength, 2, 3, 98 structures of high, 265—268 surface current, 107 toroid, 101—102 wire, 99—100 Magnetic gears, 282—287 Magnetic latching, 268—273 Magnetic levitation, 273—281 Magnetic moments. See Atomic magnetic moments Magnetic quantum number, 10 Magnetic resonance imaging (MRI), 306—307 field strengths, 308 field systems for, 308, 310 fringe fields, 309 homogeneity, 309 hybrid, 310, 312 patient access, 309—310 stability, 309 yoked, 310, 311—312 yokeless, 310—311 Magnetic viscosity, 70 Magnetite, 47 Magnetization process, 55—66 Magnetizing field, 56 Magnetizing fixture design, 58—59 Magnetocrystalline anisotropy, 27—31 Magnetomotive force (mmf), 148 Magnetooptical bias field actuators, 406—413 Magnetooptical (M-O) recording, 318—325 Magnetostatic analysis Biot-Savart law, 104 energy, 116—117 field equations, 97—102 force and torque, 110—112 inductance, 118—126 Maxwell stress tensor, 112—116 vector potential, 102—109 Magnetostatic energy, 23 Magnetostatic theory, 87—89 Magnets equivalent circuit, 157—159 cylindrical. See Cylindrical magnets rectangular. See Rectangular magnets sources, 153—161 Magnets, structures of cylindrical, 218—264 rectangular, 208—218

517

INDEX Materials anisotropic, 5, 77 classification of, 6—7 hard, 44—55 homogeneous, 5, 77 inhomogeneous, 5, 77 isotropic, 5, 77 linear, 5, 77 nonlinear, 5, 77 soft, 39—44 Maxwell, James Clerk, 76 Maxwell’s equations, 74, 338 boundary conditions, 79—81 constitutive relations, 75—77 differential, 74, 79 force and torque, 82 integral, 77—79 Maxwell stress tensor, 112—116 force, 115—116, 362 Mechanical equations, 361—363 Melt spinning, 51, 52 MEMS (micro-electromechanical systems), 455—465 Method of images, 185—190 magnet on plate, 188—189 wire above plate, 188 Micro-electromechanical systems (MEMS), 455—465 Mixed boundary-value problem, 162 MKS system, 368, 509—512 Molecular field, 20 Moment of inertia, 361, 375 Motion, electromechanical equations moving coils, 366—376 stationary circuits, 363—366 Motion, mechanical equations, 361—363 linear, 361 rotational, 361 Motors axial-field, 421—437 stepper (can-stack), 437—446 Moving coils, 358—361, 366—376 Moving reference frames, 341—353 MRI. See Magnetic resonance imaging Multipolar cylinders, radially polarized, 235—245 Multipole disk magnet, axially polarized, 253—264 Mutual inductance, 120—121 N Neodymium-iron-boron, 51—53 Neumann boundary-value problem, 162 Neumann functions, 176 Newton’s second law of motion, 361 Nickel-iron alloys, 39, 43 Noncontact latching, 269

Nonlinear materials, 5, 77 O Ohm’s law, 356 Operating point, 153—161 Orbital angular momentum, 10—11 P Paramagnetic material, 6 Paramagnetism, 17—19 Pauli exclusion principle, 14, 15 Permanent magnet structures. See Magnets Permanent magnets, 44—46 alnico, 48—50 bonded, 53—55 ferrites, 46—47 fields of. See Magnetic field magnetic resonance imaging and, 308, 310 magnetization of. See Magnetization process neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Permeability absolute, 41 differential, 41 free space, 5, 76 initial, 41 maximum relative, 41 relative, 41 Permeance angular gap, 149 radial gap, 149 rectangular gap, 149 Permittivity, free space, 76 Planck’s constant, 10 Poisson’s equation, 200—201 Potential(s) difference, 90, 340 energy, 23 Maxwell’s equations and, 82—84 scalar, 82 vector, 82, 83, 102—109 Q Quadrupole magnet, 218 Quantum mechanics, 10 Quasi-static field theory, 85—86, 337 moving reference frames, 341—353 stationary reference frame, 338—341 R Radial coupling, 286, 288—297 Radially polarized magnets, 224—245 finite, 235—245 infinitely long, 224—235 Rapid quenching, 51, 52

518 Rare-earth magnets, 58 neodymium-iron-boron, 51—53 samarium-cobalt, 50—51 Rectangular magnets, 208—218 Reference frames moving, 341—353 stationary, 338—341 Relative permeability, 41 Reluctance, 145 torque, 453 Remanence, 31, 38 Remanent flux density, 31 Remanent induction, 38 Resonant actuators, 402—406 Reversible changes, 67 Reversible temperature coefficients, 67 Runge-Kutta method, 363, 418, 503—508 Russell-Saunders coupling, 14—15 S Samarium-cobalt, 50—51 Saturation induction, 38 Scalar-valued functions, 345, 469—473 Schrödinger’s equation, 9 Self-energy equation, 23, 117 Self-inductance, 118 Separation of variables Cartesian coordinates, 162—174 cylindrical coordinates, 175—180 spherical coordinates, 180—185 Shape anisotropy, 31—33 Shorted turn, 418—421 SI (Système international d’unités), 2—5, 509—512 Soft ferrites, 39, 43—44 Soft iron, 39, 42—43 Soft magnetic materials, 39—44 Solenoid B-field, 100—101 inductance, 123—124, 125—126 Sommerfeld units, 2—5, 509—512 Sphere in external field, 26 Spherical coordinates, 476—479 boundary-value problem, 180—185 Laplace’s equation, 180—185 transformations, 491—494 Spin, 10 Spin angular momentum, 10—11 Spin-orbit coupling, 14 Stability, 66—70 magnetic resonance imaging and, 309 Static theory electrostatic, 89—91 magnetostatic, 87—89 Stationary circuits, 354—358, 363—366 Stationary reference frame, 338—341 Stepper motors, 437—446

INDEX Stokes’ theorem, 77, 356, 485—487 Stoner and Wohlfarth model, 28 Strontium ferrite, 46 Structural changes, 67 Superconducting coils, 308, 310 Surface current, 107 Surface integral, 483—485 Susceptibilities absolute, 42 differential, 42 initial, 42 Suspension, 273—281 Synchronous coupling. See Coupling, synchronous T Tesla, 2—3, 510—511 Tesla per amp (TPA) ratio, 60 Toroid B-field, 101—102 inductance, 124—125 Torque, 82, 374 on bipolar cylinder, 142—143 charge model, 141—143 constant, 368 current model, 131 drag, 423 drive, 453 reluctance, 453 on steady currents, 110—112 U Undulator, 326 Units CGS/Gaussian, 2—5 conversion factors, 3 Kennelly, 2 SI (Sommerfeld), 2—5 summary of, 509—512 V Vector field flux, 78 Vector potential, 83, 102—109 line current, 104—106 surface current, 107—108 Vector-valued functions, 74, 345—346, 469— 473 Virtual displacement, 376—377 Volume integral, 483 W Webers, 2 Weiss molecular/mean field theory, 20—22 Wigglers, 326—328 Halbach, 329—330 helical field, 330 hybrid, 330—331

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