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0 over ~ sp ~:e 0 ~~nd (1) the magnitude of vector E as tric . with the pet:nntttvI y 8. he centre of the sphere and plot the function of the dl!~tanceEr(f)ro~d tm (r). (2) the surface and volume dencurves of the functions raT ' sitie5 of bound charge. d . E we shall use the Gauss theaSolution. (1) In .order to ketcr~h~e di~tribution of only extraneous rem for vector D smce we now
a
c~:
r Eon is the normal component of the vector Eo in a vacuum. Substituting these expressions into (1), we obtain
i.e. E < Eo. • 3.9. A point charge q is in a vacuum at a distanc\: 1 from the plane surface of a homogeneous dielectric filling the half-space below the plane. The dielectric's permittivity is e. Find (1) the surface density of the bound charge as a function of the distance r from the point charge '1 and analyse the obtained result; (2) the total hound charge on the surface of the dielectric. Solutioll. Let liS Use the continuity of the normal component vector D at the dielectric-vacuum interface (Fig. 3.17):
or
D 2n
=
1 - -q2 cos t}+--=e (J' --4n:f o r 2e o
0'
E 2n = eE1n
DIn' (
1 -cos q (J' ) -_ \'t--._ , 4:rte o r 2
,
~
e+ 1 2ltr 3
q'= -
e-1 q. e+1
, , n the plane separating a va.cu~I!1 • 3.~O: ~ p(),lIlt charge q dl'~I~ctric with the dil~lectric p~rmlttlvlfrom an fiomo~eneousf I D and E in the entIre space. F' mtimte d the magmtudes 0 vectors ty e. m . II from the continuity of the normal Solution. In this case, It fa o~ eE ,Only the surface cha.r~e .a' component of vector D thatf2 n n~;:t of veo;Lor E in the vlclmty willthe contribute to the n?rm a t"c~P~ence the above equality can be of point under consl d era 10 • written in the form 0'/2eo = e (-0'/2eo). We immediately und that ~; = ~. no bound surface charge (with the Thus, in the giv~n ca~e I~re:s 'ontact with the extraneous point exception o~ ~hc pomt~hl~ ~lh~e~le~tric licld iu the surrounding spa~e charge q). 'I hiS mea~s a , and b' depends only on the dlsis the field of the, pomt char~e ( ~eq ~harge q' is unknown, and hence tance r from thiS cha~ge'l u m for vector D. Taking fo~ the ~lose~ we musta use theofG~uds.s the centre at the point of locatIOn o. surface sphere ra 1U,St rlCo:'~h \\1 the charge q, we can write 2:rtr 2 D o
2e
o where the u'rm a' /2£0 is the component of the electric field created Dear the region of the plane under consideration, where the surface charge density is a'. From the last equality it follows that 0,=_e-1
- lIr As 1 -+- 0 the quantity th;inte~face , there is no surface
h t cos t}
Here we took into account. t ,a t t a' -+- 0, i.e. if the charge q IS JUS a
' O
• 3.8. Boundary conditions. In the vicinity of point A (Fig. 3.16) belonging to a dielectric-vacuum interface, the electric field intensity in vacuum is equal to Eo, and the vector Eo forms the angle ao with the normal to the interface at the given point. The dielectric permittiv~ ity is e. Find the ratio E/Eo, where E is the intensity of the field in~ side the dielectric in the Vicinity of point A.
E-r;
93
Problems
+ 2rr.r
2
J)
= q,
I II ""uit'ltll'S of H'l'tor D at the disLa,nce r o where Do hand J) are . t lC V"CUUIn lu 'I I'll tJ'e ";I'eleclric reS}lech"eiy, all( ,. • to . th fromBesides, the C arge · 0; if j # dI, then j 1 < O. Finally, we replace j 1 by liS, where I the current, is also an algebraic quantity (like j I)' Since' I is the same in all sections of the circuit when we are dealing with direct current, this quantity can be taken out of the integral. Cosequently, we get 2
2
fjdl=If
J 1
(J
~
(5.13)
.IPs· 1
The expression P dll S defines the resistance of the subcircuit of length dl, while the integral of this expression is the total resistance R of the circuit between cross sections 1 and 2. Let us now consider the right-hand side of (5.12). Here, '0 R the first integral is the poteoo---::.j~ . tial difference 'PI - 'Pz, while 1 2 the second integral is the electromotive force (e.mJ.) ~, act"'2 iog in the given subcircuit:
r£~
~
2
~t2 = )
E*dl.
(5.14)
1 2 1
Fig. 5.2 Like the current I, the electro.. motive force is an algebraic .. . quantity: If e.m.f. facIlItates the motion of positive carriers in a certain direction, then ifIll> 0; if, the e.m.f. hinders this motion t"12 < O. As a result of all the transformations described above (5.12) assumes the follOWing form: '
IRI='Pl-'P2+~t2'
125
5.3. Generalized Ohm's Law
5. Direct Current
Since the potential decreases over the segment R fr0l!! .left t? rig~t, O. This means that the current flows in the posItIve dIrectIOn (from 1 to 2). In the present case! Ipl < .lp2' but th~ curre~t .f1ows f~om point 1 to point 2, Le. towards .Ific~easII~g P?te~tlal. ThIs IS p~ssIb~e only becausp.an e.mJ. ,g is actIng Ifi this CIrcUIt from 1 to 2, I.e. m the' positive direction.
I
>
Let us consider Eq. (5.15). It follows from thi~ eq.uat~on that points 1 and 2 are identic.al for a . closed . cIrcUIt, I.e: «PI = «PI' In this case, the equatwn acqUIres a sImpler form. RI = ~, (5.16) where R is the total resistance of the closed cir~uit, and ~ is the algebraic sum of all the .e.m:f. 's in t?~ circuit. . Next we consider the subClrcUIt contallllllg the source of the ~.mJ. between terminals 1 and 2. Then, R in (5.15) will be the internal resistance of the e.m.f. source in t.he subcircuit under consideration, while 'PI - 'P2 the potential difference acro~ its terminals. If the source is disconnected, I '-= 0 and l= 'P2 - 'Pl' In other words, the e.m,f. of the source can be defmec! as the potential difference across its terminals in the open circuit. The potential difference across the terminals of an e.m.f. source connected to an external resistance is always less than its e.m.f. and depends on the load. Example. The external resistance of ~ circuit is TJ times higher th,an the internal resistance of the source. Fmd the ratIO of the potential dillerence across the terminals of the source to its e.m.f. Let R be the internal resistance of the source and R a the external resistance' of the circuit. According to (5, i5), '!P2 - !PI = ,g - R,! I while according to (5.16), (R j R a ) ! =,g. From these two equations, we get
+
1p2-Ip, = 1-
,
(5.15)
If
Ri! r!J
= 1-
Rj Rd-R a
Ra Ri+R a
=
TI 1+TI •
This equation is the integral form of Ohm's law for a n01tuniform subcircuit [ef. Eq. (5.11) which describes the same . law in the differential formJ.
It follows from here that the higher the value of TI, the closer the ~o tential difl'crence across the current terminals to its e,m.f., and VIce versa.
. Example. Co~sider the subcircuit shown in Fig. 5.2. The resistance IS nonz?ro. only. III the .segment H. The lower part of the tigure shows the variatIOn of potential !P over this region. Let us analyse the course of events in this circuit. .
Concludin a this section, let us consider a picture illustrating the 'how of direct current in a closed circuit. Figure 5.:3 shows the distribution of potential along a closed
126
5. Direct Current
circuit containing the e.m.f. source on segment AB. For the sake of clarity of representation, potential
0 f~r It > Ito and vice versa . • 5.8. The work of an e.mJ. source. A glass plate completely fills the gap between the plates of a parallel-plate capacitor whose capacitance is equal to C. when the plate is absent. The capacitor is connected to a source of permanent voltage U. Find the mechanical work which must be done against electric forces for extracting the plate out of the capacitor. Solution. According to the law of conservation of 'energy, we can write It can be seen that ql
Am
+ As =
~W,
(1)
where Am is the mechanical work accomplished by extraneous forces against electric forces, A s is the work of the voltage source in this process, and ~ W is the corresponding increment in the energy of the capacitor (we assume that contributions of other forms of energy to the change in the energy of the system is negligibly small). Let us find ~ Wand A s. It follows from the formula W = CU2/2 = = qU!2 for the energy of a capacitor that for U = const ~W
=
~C
U2/2
=
t!.q U/2.
(2)
Since the capacitance of the capacitor decreases upon the removal of the plate (~C < 0), the charge of the capacitor also decr~ases (t'..q < < 0). This means that the charge has passed through the source against the direction of the 1i.ction of extraneous forces, and the source has done negative work As = !'"q'U,
II1I 1'[' 1
1.
II 1
'I;
1
1'1
I'III
6.1. Lorentz Force. Field B
140
141
:S. Direct Current SoluUlIn. The required amount of heat is given by
Comparing formulas (3) and (2), we obtain
II
=
As
I
2.1 W.
Q=
Suh~t.itution of this expression into (1) gives
Am
=
-.1W
or A III
1 (e = 2"
"
1) Co1I 2 •
-
f Thus, extracting the plate out of th . o~ces) do a positivework(a<Ja't 1• .e capaCitor, we (extraneous Lrlc fdrces) .. The e.mJ. source in thIs case accomplishes a n~g:d~~ e tor decreases: wor , an the energy of the capaci-
('k
Am
>
: B],
I
(6.36)
where Pm is the magnetic moment of the current loop (Pm = B = ISn for a phme loop). * It is clear from ((j.36) that the IlWlnF. ent M of Ampere's forces acting on a current loop in 11 uniform magnetic field is perpendicular both to vector Pm and to vector B. The magnitude of vector 1\1 is M' = PmB sin ct, where ct is the angle between vecF tors Pm and B. When Pm tt B, M ~= 0 and it is not difficult to see that the Fig. 6.14 position of the loop is stable. When Pm H B, 1\'1 also eq uals zel'O but such a position of the loop is unstable: t he slightest deviation from it leads to the appearance of the moment of force that tends to deviate the loop from the ill itial po,'.:ition still further. .Example. Let liS verify the validity of formula (6.36) by using a Simple example of a rectangular current loop (Fig. 6.14). . It ca.n be seen from the figure that the fo~oes <Jcting on sides a are perpendicular to them and to vector B. Hence these forces <JfP directed along the hOrIzontal (they are not sh~':Vn in, the figure) and only strive to stretch (or compress) the loop. S,'1es 0 are perpendicular to H and hence each of them is acted upon by the force '
Considering that ab is the area bounded by the loop and Iba = Pm' we obtain M = PmB sin
a,
which in the vector form is written as (6.36).
Concluding this section, we note that expression (6.36) is valid for nonuniform magnetic fields as well. The only requirement is that the size of the current loop must be .sufftciently small. Then the effect of nonuniformity on the rotational moment (torque) M can be ignored. This precisely applies to the elementary current loop. An elementary current loop behaves in a nonuniform magneti(~ field in the same way as an electric ~dipole in an external nonuniform electric field: it will be rotated towards the position of stable equilibrium (for which Pm tt B) and, besides, under the action of the resultant force F it will be pulled into the region where induc~ion B {t' is larger. 6.8. Work Done upon'Displacement of Current Loop
When a current ioop is in an external magnetic field (we shall assume that this field is constant), certain elements of the loop are acted upon by Ampere's forces which hence will do work during the displacement of the loop. We shall show here that the work done by Ampere's forces upon an elementary displacement of the I loop with current I is given by 6A
F c= !hB. ~hese forces tend to rota te the' loop so tl,a t vedol' Pm becomes (hreeted lr ke vee tor B: Hence, the loop is acted npon by II couple of forces whose torque J~ t'9'ual to the p;'oduct of the arm Ii sin 1'1, of the couplr by the force P, I e.
M = lbBa 8i:'
Ct.
* If till' l00p .is ::ot plane, its ma:'ne,j,: rnO'lwnt = B dS are positive quantities (if the field B were dIrected towards us or the jumper were displaced to the left, d < 0). In any of these cases expression (6.38) can be represented in the form (6.37). 2. The obtained result is valid for an arbitrary direction of the fi.eld B as well. To prove this, let us decompose vector B Into thr~e components: B ~o Bn + Bz + B x . The compon~nt .Bz dIrected along, the jumper is parallel to the ~urrent In It and does not produce any force acting on the Jumper.. The component B x (along the displacement) is res~onsible for the force perpendicular to the displacement, whIch does not accomplish any work. The only remaining comJilonent is Bn which is normal to the plane in which the Jumper moves. Hence, in formula (6.38) instead of B we. must take only B n • But B n dS = d, and we again al'nve at (6.37). 3.. Let. us now ~onsider any current loop which is arbitrarIly dIsplaced I? a constant non uniform magnetic field (t~e contour of tIus loop may be arbitrarily deformed in thIS process). We mentally divide the given loop into infinitel~ small current elements and consider their infinitesimal ?Ispla~ements. Under these conditions, the magnetic field 10 whICh .each element of current is displaced can be assumed uOIform. For such a displacement, we can apply to each element of current the expression dA = I d'(p for the ele~entary work, where .d' expresses the contribution (If a gIven element of current to the increment of the flux through the contour. Summing up elementary. works for all the elements of the loop, we again obtain expression
2
A=
) 1 d.
(6.39)
1
!
If the current 1 is maintained constant during this displacement we obtain ': A = 1(2 - 1)' (6.40)
",
where 1 and 2 are magnetic fluxes through the contour in the initial and final positions. Thus, the work of Ampere's forces in this c;ase is equal to the product of the current by the incremei1t of the magnetic flux through the circuit. Expression (6.40) gives not only the magnitude but also the sign of the accompli~hed work. Example. A plane loop with current I is rotated in magnetic field B from the position in which n H B to the position in which n tt B, 0 being the normal to the loop (it spould be recalled that the direction of the normal is connected with the direction of the current through the right-hand screw rule). The area bounded by the loop is S. Find the work of Ampere's forces upon such a displacement, assuming that the current I is maintained constant. In accordance with (6.40), we have A = I [BS - (-BS)) = 2IBS. In the given casa, the work A > 0, while upon the reverse rotation A lI> = = lI>0. where ~ is the flux through the solenoid's cross section far from its cndface. Then we have
Fig. 6.20
Fig. 6.21
while B' is the magnetic induction f h . to the current flOWing through th 0 ttl' ~elt at the same point due be . cn removed in order to create th par ?f. t e conductor which has . Thus, the problem im lie fi e cavity. In~uction 8 inside the sO~id ~O:d~ all the ~alculation of magnetic ~Stng the theorem on circulation w~ or at ~ distance r from its axis.
0/
Fi;
~~~~ ~~I'jih;h~~c~~¥rf~~~~'can cb~n;:~~~s::[~ ;ir~~~~' ~~~n~f 1 B="2f!o[jxr).
Using now this formula for B o and B' w fi d' h' . , e n t elf dJilerence (1): 8_ lLO [' lLo· 1L - 2 J X r] - 2 [J X r']= 20 (j X (r-r')).
Figure 6.21 shoWs that r = I
+'r ,
1 B ="2
flo
h w ence r - r' = I, and . [J :< I) .
Thus, in Our case the magnetic field B . . If the current is flOWing towards us . In the cavity is uniform and plane of the lIgure and is dl'r t d (Fig. 6.21), the Held B lies i~ the ec e upwards.
11> ~ $0/2 = lLon!S /2.
By the way, we must pay attention to the following propertieof the field B at the endfacc of a long solenoid. 1. The lines of B are arranged as shown in Fig. 6.22. This can be easily shown with the help of superposition principle: if we place on the right one more solenoid, field B outside the thus formed solenoid must vanish, which is pcssible only with the field configuration shown in the figure. 2. From the principle of superposition, it follows that the normal component B n will he the same over the area of the end face, since when we form a composite solenoid, B n B n = Bo, where HI, is the field inside the solenoid away from its endfaces. At the centre of the elldface B = B n , and we obtain B = Bo/2. • 6.6. The field Gf 8 solenoid. The winding of a long solenoid of radius a is 11 thin conducting band of width h, wound in one layer practically without gaps. Direct current I flows along the band. Find magnetic field 8 inside and outside the solenoid as a funetion of distance r from its axis. Solution. The vector of linear current density i can be represented as the sum of two components:
+
i=i 1 + ill" The meaning of vectors i.L and ill is clear from Fig. 6.23b. In our case, the magnitudes of these vectors can be found with the help of
168
6. Magnetic Field in a Vacuum Problems
Fig. 6.23a by the formulas i.l =/Cosa=1 V1-sin 2 a=(I/h) V1-(h/2na)2. i lt = i sin a
= Il2na.
The magnetic induction B' .d h accordance with (6.20) by th mSI.e t .1.' sole~oid is determined in by , I . ' quantIty l.L' whIle outside the solen~id,
I,,:
B i = f!o/.l = (f!oI/h)
Vi -
(h/2na)2
B a = f!oilla/r = f!0I/2nr
(r
(r
>
Fe, and the wires repel each other. If, on the contrary, R > Ro, F m < Fe, and the wires attract each other. This can be observed experimentally. Thus, the statement that current-carrying wires attract each other is true only in the case when the electric component of the interaction can be neglected, Le. for a sufficiently small resistance R in the circuit shown in Fig. 6.24. Besides, by measuring the force of interaction between the wires (which is always resultant), we generally cannot determine current I. This should be borne in mind to avoid confusion. • 6.8. The moment of A-mpere's forces. A loop with current / i8 in the field of a long straight wire with current 1 0 (Fig. 6.25). The plane of the loop is perpendicular to the straight wire. Find the moment of Ampere's forces acting on this loop. The required dimensions of the system are given in the figure. Solution. Ampere's forces acting on curvilinear pllrts of the loop are equal to zero. On the other hand, the forces acting on the rectilinear parts form a couple of forces. We must calculate the torque of this , couple. Let us isolate two small elements of the loop (Fig. 6.26). It can be seen from the figure that the torque of the couple of forces correspond-
r
I, j
170
6. Magnetic Field in a Vacuum
dM = 2z tan cp dF, where the elementary Ampere's force is given by dF
=
(t)
I diB.
The dependence of the magnetic induction
(2)
X
I!'ig. 6.26 from the ~traight \ . b f circulation: vlre can I' ound with the help of the theorem on B = !-Lo[/2nr. (3) Let us now sub~titute (3) . t (2) h ing that· it = dr ;'1d x ~= r c~~ 0 . ' t en (2) into (1) and, consicleover r between a and b. This gi~~smtegrate the obtained expression M = (fto/n) II 0 (b -
. C1 • Pm, IS placed on the axis of a circul 'I avm"'f the magnetic moment current I is flowing. Find th f ce ar ~op 0 radius R, alo~ which F actlllg on the coil if its' distance from the centre of the loop in Fig. 6.27. s an d vector Pm is oriented as is shown
ft
.
0
.
h
, t e required force is defined as
F= Pm aB/an
where B is the magnetic indo ' . (t) at the Iscus of the coil. Let Us ~~~:n ,o!. ~~~. field cr~ate~f by the loop tor Pm' Then the projection of (1) t Zt atXII~ III t,he d.lrechon of the vecon 0 118. aXIs wIll be Fz
II
i I!
I I
III 'I
"
=
Pm
aB/az =
~here w~ took into account that B = (~rr~nt !II the loop. The magnet· .Z d (h.12), whenn'
.
IC III
Pm
ilB/az, .
B. for t~le given direction of IIctlOn B I~ defmcd by formula
aB __ 3 f.LoRz/1 az ----Z(lZ+RZ)6/Z.
• 6.10. Current I flows in a long thin-walled circular cylinder of radius R. Find the pressure exerted on the cylinder walls. Solution. Let us consider a surface current element idS, where i is the linear current density and dS is the surface element. We shall
~
'b./
,,' ./
08 Dr
B'
~'!m
dS
~B' Bi
l/
~ 6.9. A small coil with current h
(6 33)
It should be noted that if Pm (and hence Z-axis as well) were directed oppositely, then B z = -Band oBz/oz > O,and hence Fz>O and vector Fwould be directed to the right, Fig. 6.27 Le. again oppositely to the direction of Pm. HeBee, the obtained direction for F is valid for both orientations of Pm
~
a) sin lp,
vector M being- directed to the left (I!'ig. 6.26).
According t
Since aB/az < 0, the projection of the force Fz < 0, Le. vector F is directed towards the loop with current /. The obtained result can be represented in the vector form as follows:
B on the distance r
--
S alution
171
Problems
iug' to these elements is
Fig. 6.28
Fig. 6.29
lind the relation between the surface and volume elements of the current: j dV
=
j6h·6b dl
= i as.
The meaning of the lJuantities appearing in this relation is clarified in Fig. 6.28. In the vector form, we can write jdV=idS.
(1)
The Ampere's force acting on the surface current element" in thiS case is determined by the formula obtained from (6.28) with the help of substitution (1): (2) dF = Ii X D'] dS, where B' is the magnetic induction of the field at the point of lo'f" [I very low cylinder J et ,ICC Jt'tw('nll (wl)"lllaglletics as is shown
i.e. the normal component of vector B turns out to he the
-~-_.~---,
2 1
,,'t n'
Fig. 7.7
Fig. '7.8
,
same on both sides of the interface. This quantity does not have a discontinuity. Condition for Vector H. We shall assume, for higher generality that a surface conduction current with linear dens'iy i flows over the interfaeial surface of the magnetics. Let us apply the theorem on circulation of vector H to a very small rectangular contour whose height is negligibly small in comparison with its length l (the arrangement of the contour is shown in Fig. 7.8). Neglecting the contributions from smaller sides of the contoUl' to the circulation, we can write for the entire contour Il2~l Hi~,l = iNl,
+-
where iN is the projection of vector i onto the normal N to the ('oQutOllI' (vector N forms a right-handed systom with the direction of eontour circumvention). Taking the two projections of vector H onto the comm'on unit vec.tor of tlte tangent. T (in IUf\l\illlU 2), we obtain TT lt , ,== -lfl~' C·ancelling l Ollt of tlwpreviolls eqllation, we get (7.21)
184 7. Magnetic Field in a SUbstance'
i.e. the tangential co a discontinuity u on mpo~ent. ?f vector H general1 which is due to thPe prea tIansltlOn through the l'nteYI'f has If } sence of c d ' . Ilee, , IOwever, there are no o~ uctlOn Cllfl'ents. n f~ce between magnetics (i ~O~d) uctthlo currents Ilt the intero vector H turns 0 t . , e tangentilll co interface' u to be the same on botl 'd mponeJlt . I Sl es of the
1~2T=HIT·1
(7.22)
Thus, if there is no d . between two homo COli lICt,lOn Cllnent geneons magnetics, the at the intcrfare components B n B2T
I I
r---
1,1
f I a2 1
BIT
Fig. 7.9
FieJdB FieJdH Fig. 7.10
Ilnd H T Vll ry contlJl . lIOllSI ( . I . 1hrough this interface. 6n ~~tel~~~ Il Jump) upon tl tl'ullsition n d , Hn in this case have d' er ~lln~1 '. the components T . I s h ould be noted that IsContlnlllttes. 1I~ ;ect~r D, while vect::ci!rb ~ behav~s at the interface e raellon of Lines of B T . e laves lIke E. at the interface between 'twhe lInes of. B undergo refraction the case of dielectrics W h fl mllgnetlCs (Fig. 7 9) A. . of angles (Xl and (X2=' e s Il find the ratio of th~ t~ng:IJ~~
t
I I
(7
Ilomo~eneous
Magnetic
t85
:Taking these reilltio/ls into Ilecount. we obtllin the lllw of ~refraction of the lines of vector B (Ilnd hence of vector H " . weB) similllr to (2.25): tana 2 tan a l
---
~
(7.23)
III
Fignre 7.10 depicts the fields of vector's B HIIII H near the illterface between two mllgnctics (in the absellce of condut:tion currents). Here ~12 > ~ll' A compllrison of the densities of the lines shows that lJ 2 >R 1 • while H 2 < HI' The lines of B do not hllve a ~liscontinuity upon U transition through the interface, while the H lines do (due to the surfare magnetizlltion clIrrents). The refraction of magnetic field lines is used in magnetic protection, If, for example, a closed iron shell (layer) is introduced into an external ma~n~tic tield. the lield lim's will be concentrated (condensed) mainly on th~ shell itself. Inside th(' shell (in the cavity) the ma~netic field turns out to!d>e considerably weakened in comparison wi th the ~xtern:ll lield. In other words, the iron shell ael;; like a screen. This is used in order to protect sensitive devices from ('xtt'rnal magnetic fields. ,
2
W
7,/;, Firld in
tan a - B 2TIB2n ----.!. tana l -~/B • IT
In
and
B
e ~ la I confine ourselves to ~u~~)on .cllrre!!t at the interf:~: c~e w~e,n there is no con• ,lJI thiS Cllse we lillve . CCOr IJIg to (7.22) and
B 2T /1l2 = BfT/1l1
2n=
B
In'
7.5. Field ill a Homogeneous
~.lgneIic
It was mentioned in Sec. 7.1 thllt determination of the resultant magnetic field in the pl'esence of arhitral'y magnetics is generally a complicated problem. Indeed, for this purpose, in Ilccordance with (7.1), the fIeld B o of conduction currents mnst be supplemented by the macroscopic fwld B' createfl by magnetization c,ul'rents. The prohlem is that we do not know beforehllnrl the configuration of magnetization currents. We can only state that the distribution of these cnrrents depends on 'the nat1ll'e and configllration of the magnetic IlS well as on the configuration of the externlll lield B o, viz. the field of conduction C1ll'rents. And since we do not know the distribution of magnetization currents, we callnot calculate the fIeld B'. The only exception is the case when the entire space occnpiell by the fwld B is fdled by a homogeneolls isotropic dielectric. Let liS (",ollsider this ease in ~rp:ltel' dptail. Bllt first of all, we shall lIlIalyse the phenomenll obsPI'vpd whell COlHlllclion Cllrrent llows along a hOlllogenpolls conduetol'
186
187
7.5. Field in a Homogeneous Magnetic
in a vacuum. Since each conductor is a magnetic, magnetization currents also flow through it, viz. volume currents given by (7.18) and surface currents. Let us take a contour embracing our current-carrying conductor. In accordance with the theorem on circulation of vector J (7.5), the algebraic sum of magnetization (volume and surface) currents is equal to zero everywhere since J = 0 at all points of the contour, i.e. l' = I~ I~ = O. Hence I~ = -I~, i.e. the volume and surface magnetization currents are equal in magnitude and opposite in direction. Thus, it can be stated that in an ordinary case, when currents flow along sufficiently thin wires, the magnetic field in the surrounding space (in a vacuum) is determined only by conductioH currents, since magnetization currents c.ompensate each other (except fbI', perhaps, the points lying very close to the wire). Let us now fill the space surrounding the conductor by a homogeneous nonconducting magnetic (for the sake of definiteness, we assume that it is a paramagnetic, X > 0). At the interface between this magnetic and the wire, surface magnetization current l' will appear. It can be easily seen that this CUlTent has the same direction as the eonduction CUlTent I (when X >0). As a result, we shall have the conduction current 1, the surface and volume magnetization currents in the conductor (tho' magnetic fIelds of these currents compensate one another alld hence can be disregarded), and the surfac.e magnetizatioll current l' on the nonconducting magnetic. For sufficiently thin wires, the magnetic field B in the magnetic will be determined as the field of the eUI'l'ont I I'. Thus, the problem is reduced to finding the current 1'. For this pnrpose, we surround tho conductor by a contour arranged in the surface layer of the nonconducting magnetic. Let tlte plane of tho contour he perpendicular to the wire ax i,~, i.e. to tlte magnetization currents. ThOll, taking (7.7) ant! (7.11) into consideration, we can write
+
+
l' =
\~ i' dl = ~ .r dl = X
r
If II dl.
Acc(mlillg" 10 (7.12), it follows that = If. TrIO C()llji~lIl'ati()IlS of 1111' mag·llotizatioll current l' 'llld
I··
I
of the conduction current I pr~ctically h . coincide d tion (the B' ofwires .the .) . d I '''ce at all pomts t e In HC are thm ,an le...... ".1 s differs from the indueld f tl e magnetIzatIOll CII 1 " • fI~ BOfl ·tl field of the conuuction currents only In magtIOn de 0 the
1. F'In d th e magnetic . mduction . P ot' tJ\,' field in
. According to (6.20), in the b . mduction B in the solenoid is a ie~ce of mag'netJc the magnetic o t~e entIre space where t.he fi ld e~r'W 0 lonI. Since th.e magnetic fills ffi effects>, th!! magnet.ic induc~ion ~ ers (we larger: Ignore the edge mus trob e zero. /.L tImes
~ .. I I
. B = flP-onI. (7.28) In thIs cast', t.he of vJ~.1.or II remallls . ' , the S,;:!)n as III . the absence of the magnetic, i.e.field H = . The change in field B is caused b th current~ fl~)wing over the surface of the e appearance of magnetization magnetJ~ IU the same direction as the c~nd!Jcholl currents in the solenoid wlDdl.ngw~en /.L > 1. If, however, P-< 1 the dl~ctIons of these currents will b~ opposIte. h The obtained results are also valid w en the magnetic has the form of a ve!y long rod arranged inside the soleno~d ,so th~t it is parallel to the solea r nOId saxIs. Exa"!ple 2. The field of straight. CUneDt III the presenee of a magnetic Su~pose that a magnetic fills a long cyl~nder of radius a along whose axis Fig. 7.11 a.g.lven current I flows. The permeablhty of the magnetic 11 > 1 Find th magnet' 'd uc t'IOn B .as. a. function . e the distance r from the cylinder aXi~~ In of
sinc';;~~:;::~t~~ ~~e directly the theorem on circulation of vector B is very helpful~ t~nc~~::;i::.s ar? ud~now!1' ~n thlis situation, vector II rents, For a circle of radiusIOn IS he errn me , on y byhconduction curr, we ave 2nrH = I, w ence
I
B = llP- oH = IJ.lloIl2rrr. Upon a t T magnetic i~'d~a~~l IOn Bthroul~ the magnetic-vacuum interface, the (Fig, 7,11). c Jon ,un Jke H. undergoes a discontinuity . . c~used by surJace . increase in tionAn currents ThB insid . e the m~gn~tIc!s magnetizain the wire o~ h. ese. ~.urrents cOIUClde ,,, d:rect.ion with the current I On the other \l:ndxl~~is~~e sYhtem a.nll hence "amplify" this current. current is liirectl'd' oppos\t:lyt e cil~~dder the surface magnetization .' ' .b11 I oes not produce ally effect on field B in th ' of both current: cmomagpneent!Ct' OutSIde thhe magnetic, the magnetic fields . " sa e one anot er.
7.6. Ferromagnetism beFerromagneties. In, Il.w g netic ,respecl, all suh~tanc(>s call divided into weakly magnetic (para magnetics and dia-
F rf' ')m.:-a::cg"..:n~e:..::.t~is~m~
1_8~!)
,
magnetics) and strongly Illilgnetic (fel'l'olllagnetics). It is well known that in the absence of magnetic lield, paraand diamagnetics are not magnetized and are characterized by a olle-to-one cOITespol\(lence between magnetization J and vector H. Ferrumagl/etics are the subst;lllces (solids) that Illay possess spontaneous magnetization, Le. which arc Illagnetizell even J
B
H ,,'Fig. 7.12
Fig. 7.13
in the absence of eKtemal magnetic field. Typical ferromagnetics are iron, eobalt, allil many of their alloys. . Basic Magnetization Curve. A typical featlll'e of felTomagnetics is the complex nonlinear dependence J (H) or B (H). Figure 7.12 shows the magnetization curve for a ferromagnetic whose magnetization for H ~= 0 is also zero. This curve is called the Imsic maKnetization curve. Even at comparatively small values of H, magnetization .I attains saturation (.I"). Magnetic induction B = [lo (ll I) also increases with H. After attaining saturation, lJ continues ll to grow with II accorcing to the linear law B= ilo const, where canst = [loIs. Figure 7.13 represents the basic magnetization cllrve on the lJ-H diagram. In view of the nonlinear dependence B (II), the permeability [.t for ferromagnetics cannot be denned as a constant clwracterizing tlw magnetic properties of J specific felTomagnetic. However, as before, ills aswmed that j.l= BllloH, bllt here ~l is a function of II (Fig. 7.14), The value i1m:.x of permeability for ferromagnetics may lIe very large. For example, for pure iron pmax ~ 5,000. wilde for sHpennalloy [.tmax = bOO/lOO. it should bl~ lIoted that the concept of pCl"Iueability is
+
+
+
too
__?-
Magnetic Field in a Substance
applieahlt) only tu will ()I) SI'OWII, Lite
basic lllili[lIetizatioll curve since, as depcndcncc is nOlltllliq Ill'. Magnetic Hysteresis. Besides the lIonlinear d~ependence B (Il) or.J ([j), fCiTnmn!~'netics also exh ibit magnetic hysteresis: tltr rrlal iOIl hd,\Vef>11 f] and II or J and II turns out to till'
n (ll)
7.6. Ferromagnettsm
191
The val lies of lJ r and II c for II ifferen t fl'lTomagnetics var.y over oroad ranges. Fot' transformel' steel, the hysteresis loop is narrow (He is small), while for fe.rro.magl.tetics use.d for manufacturing permanent magnets it IS Wide (He IS large). For eXilmple, for alnico alloy, He = ;')0,000 Aim and B T = 0.9 T. These peculiarities of magnetization curves are u.sed in a conven!ent practical method for demagnetizing fei:romagnetlcs. ~ magnetlze.d sample is placed into a coil through whIch an alternatIng cu~ent IS passed, its amplitude being gradually ~educed to zero. In thIS c~se. the ferromagnetic is subjected to multIple cyclIc reverse J.llagnehzations in which hysteresis loops gradually decrease, contractIng to the point where magn~tization is zero. .
1
,'1,I
Fig. 7,14
Fig. 7.15
be ambiguous and is determined by the history of the ferromagnetic's. magnetization. If we magnetize an initially nomnaglletlZed ferromagnetic by increasing II from zero to tho valuo at which saturation sets in (point 1 in Fig. 7.15), and then reduce H from +H 1 to --HI' the ma; B ~ ~recstem the charge will move to t ~Po in the svst"1ll I\. . III thIS sy . ' 'n occur in crossL',ld cB' (or E' /dt. ueSl es" . < " its -,vanatlOll, ,
;y
"'ll
218
9. Electromagnetic Induction
9,1. Faraday', lAw. Len.', Law
of the del'ivative d<J.>/dt I d "direction" of f . ea s to a change in the sign or i Faraday found out that th . d ated in two different wa s e In. u~e~ current may be generwhich shows the coil C ~iih ThIs IS Illustrated in Fig. 9.1, magnetic field) and the loa clrrent I (the coil creates the meter G which indicates th: ind con3ected to the galvanoThe first way consists in th ~ce current. (or its separate parts) in the ~e1~s~~acement of t~e loop L In the second case the I . the fixed call C. field varies either due to the ::~f 0 IS ¥xed b?t the magnetic of a change in the current I . I.~ 0 the coIl C or as a result In all these cases the gal In I , or due to both reasons ence of induced current in :hanolmetelr G indicates the pres~ Len' L . e oop z saw. The direction of th . hence the sign of the induce i f )~ Induced. current (and law: the induced current is ~.e.m .. IS determIned by Lenz's cause generating it. In oth:;e~~d;o that i.t counteracts the creates a magnetic flux h' t r s, the Induced current magnetic flux generatin~ ~~ 1 ~re;ents the variation of the If, for example the 100 e Ill( ~ced,e.rn.f. L coil C, the magne'tic flux FIg. 9.1) is.drawn to the c~rre~t induced in the loop i~o~~~ the I?o~ Illcreases. The directIOn (if we look at th 1 'f .case IS In the clockwise creates til(' magnetic fJuxe,,~?p ~od~ the right). This current vents an increa!';e in It Irec e. ~o the left, which pre. Ie magnetIc flux, which causes the induced current'. 'f . The same situation takes I· in the coil C, keeping fixed ihace \ we Increase the current other hand, if we decrca~e the e COl a~d the loop L. On the rent induced in the loop L 'lfu~rent Ill. the .coil C, the curnow be directed countercloc~~is~ei~6I'seIlts dl~~ctioll (it wi11 Lenz's law expresses an im or we oo~ from the right). tendency of a system to ~ tant phYSIcal fact, viz. the (electromagnetic inertia) Coun eract the change in its state Faraday's Law of Ele~trorn . to this law, the c.m.f i d :g~ehC Ind~ction. According formula . n uce III a loop IS defined by the J.
&
(!).1)
219
regardless of the cause of the variation of the magnetic flux embraced by the closed conducting loop. The minus sign in this equation is connected with a certain sign rule. The sign of the magnetic flux <J.> is determined by the choice of the normal to the surface S bounded by the loop under consideration, while the sign of ~i depends on the choice of the positive direction of circumvention of the loop. As before we assume that the direction of the normal n to the surfa~e Sand the positive direction of the loop cir-, cumvention are connected through the right-hand screw rule· (Fig. 9.2). Consequently, when we choose . (arbitrarily) the direction of the normal, we denne the sign of the flux <J.> as well as the sign (and hence the "direction") of the induced e.mJ. ll' + With such a choice of positive directions Fig. 9.2 (in accordance with the right-hand screw rule), the qua9tities ~l and d<J.>ldt have opposite signs.. The unit of the magnetic flux is the weber (Wb). If the rate of variation of the magnetic flux is 1 Wh/s, the e. m.f. induced in the loop is equal to 1 V [see (9.1)1. Total Magnetic Flux (Magnetic-flux Linkage). If a closed loop in which an c.m.f. is illlillced contains not one but N turns (like H coil), 'ii will be eqnal to the sum of the e.m.f.s induced in each turn. And if the magnetic flux embraced by each tUl'll is the same and equal to /dt, where d is the increment of the magnetic flux through the area of the loop (in our case, d > 0). Thus, ~l
=
-d/dt. (9.5) It can be proved in the general form that law (9.1) is valid for any loop moving arbitrarily in a permanent nonuniform magnetic field (see Problem 9.2). Thus, the excitation of induced e.m.f. during the motion of a loop in a permanent magnetic field is explained by the action of the magnetic force proportional to [v X BI, which appears during the motion of the conductor. It should be noted by the way that the idea of the circuit shown in Fig. 9.3 forms the basis of all induction generators in which a rofOr with a winding rotates in an external magnetic field. . A Loop Is at Rest in -a Varying Magnetic Field. In this case also, the induced current is an evidence of the fact that the magnetic field varying with time creates extraneous forces in the loop. But what are these forces? What is their origin? Clearly, they cannot be magnetic forces proportional to [v X HI, since these forces cannot set in motion the charges that have been at rest (v = 0). But there are no other forces besides qE and q [v X Bl! This leaves the only conclusion that the induced current is due to the appearance of an electric field E in the wire. It is this field which is responsible for the induced e.m.f. in a fixed loop placed into a magnetic field varying with time.. Maxwell assumed that a magnetic field varying with time leads to the appearance in space of an electric field regardless of the presence of a conducting loop. The latter just allows us to reveal the existence of this electric field due to the current induced in the loop. Thus, according to Maxwell, a magnetic field varying with time generates an electric field. Circulation of vect9r E of this field around any fixed loop is defined as
~ E dl = - ~; .
(9.6)
9.2. Origin 0/ Electromagnetic Inductton
222
223
9. Electromagnetic induction
Here the symbol of parti I d ' . (a/at) emphasizes the fa:t t~~~v~~vel with respect t h e oop and the ~ retc ed on it are fixed. Since the flux tIl = ~ B Integration is performed aro y an a. . ed over the loop we are int~~eds t ed In, !b)ltrar sur ace we have
to time surface dS (the stretch-
a i r aB
at J BdS= J at dS , In this equalitv, we exchan d I ~ith respect to'time and int:; a t': Ie or(\er' of differentiation IS possible since the loop anJ t~on ov~r the slll'face, which
Eq. (9.6) can be represented in theef:~~ ace are fixed. Then •
,{"Edl= _ ';Y
Jr ~ at dS.
(9.7)
This equation has the same stru t becto~ j being played by the vec~r~:~/~q· ~6.17)
thl:' role of the e written in the differential form th e same as t. Eq. on.sequently, (6.26), i.e. it can
V
X E = -aB/at.
Th' . . ~~ IS equatIOn expresses a local relat' fields: the time variation ot ma neUe IOn hetween e~ectric and magnetic the c~rl 0/ vector E at the same ~otnt !!;~d ~t gwen point determines zero IS an evidence of the presence ~f thee aC V X itself. E differs from t ~t field . eIectnc
r
h
The fact that the circulation of th I . by a varying magnetic field d'ff e e ectflc field ind uced ~h~s electric field is not a pote:ti:~sfifl~mL~r;o indica~es that It IS a vortex field. Thus, electric fiel e . 1 e. magnetic field, field (in electrostatics) or a vorte~ c~~l:e eIther a potential In the general case electric field E . electrostatic field a~d the fi ld . dm~y be the sum of the field varying in time Since the e. In :lCed by a magnetic field is equal to ze;o, Eqs.· {9~~)~(~18\lOn of t?e elec~rostat~c for the general case as well when the' Ii l~u~n. cut to ne vahd of these two ftehls.' e IS the vector sum
radius (see Problem 9.5). Since the electriC field is a vortexlfieldi the direction of the force acting on the electrons always coincides witli the direction of motion; and the electrons continuously increase their energy. During the time when the magnetic field increases (-- 1 ros), the electrons manage to make about a million turns and acquire an energy up to 400 MeV (at such energies, the electrons' velocity is almost equal to the velocity of light c in vacuum). Thp inductioll arcelerat.or (bptatron) resembles a transformer in which the role of the secondary winding consisting of a single Fig. \J.4 turn is played by an electron beam.
Conclusion. Thus, the Jaw of electromagnetic induction (9.1) is valid when the magnetic flux through a loop varies either due to the motion of the loop or lIw tillle v,ll'iatioll of the maglletic ticl!l (01' due [0 both reasons). Uowever, we had to fcsprt to two quite different phenOi nena for explaining the f- 00). EquatIOn (9.22) shows that the rate of stabilization of the currentis ?etermin.ed by the same constant T. The curve I(t) characterizing the IDcrease m the current with time is shown in Fig. 9.8 (curve 2).
I.
On the Conservation of Magnetic Flux. Let a current loop move and be deformed in an arbitrary external magnetIC field (permanent or varying). The current induced in the loop in this case is given by 1= 'jgt+~. R
= __ 1 R
231
9.4. Mutual Induction
9. Electromagnettc Induction
d$
dt'
If the resistance of the circuit R = 0, d$/dt must also be equal to zero, since the current I cannot be infinitely large. Hence it follows that cI> = const. Thus, when a superconducting loop moves in a magnetic field, the magnetic flux through its contour remains constant. This conservation of the flux is ensured by induced currents which, according to Lenz's law, prevent any change in the magnetic flux through the contour. The tendency to conserve the magnetic flux through 11 coutour' always exists but is exhibited in the clearest form in the circll~ts of supercoilductors. . .Examp.le. A superconducting ring of radius a and inductance L' IS m .a u~Iform magnetic field B. In the initial position, the plane of the rmg IS parallel to vector B, and the current in the ring is equal t~ zero. The ring is turned to the position perpendicular to vector B. Fmd the current in the ring in the final position and the magnetic induction at its centre. . The magne~ic flux through the ring does not change upon its rotatIOn and remams equal to zero. This means that the magnetic fluxes
of the field of the induced current and of the. ext!,!rn~l current throu~ the ring are equal in magnitude and opposite m SIgn. Hence LI = na'B, whence 1= na2 B/L. This current in accordance with (6.13), creates a field B I = 1tl1oaB/2~ at the cent;e of the ring. The resultant magnetic induction at this point is given by B = B- B I = B (1 - 1tl1oa/2L). res
9.4. Mutual Induction Mutual Inductance. Let us consider two fixed loops 1 and 2 (Fig. 9.9) arranged sufficiently close to each other. If current I flows in loop 1, it creates through loop 2 the total m~gnetic flux $2 proportional (in the absence of ferromagnetics) to the current II: $2 = .{i21Il'
(9.23)
Similarly, if current 12 flows in loop 2, it creates through the contour 1 the total magnetic flux (9.24) 9.9
The proportionality factors L 12 and L 21 are called mutual indue. tances of the loops. Clearly, mutual' inductance IS numerically equal to the magnetic flux through one of the loops created by a unit current in the other loop. The coefficients L 21 and L 12 depend on the shape, size, an~ mutual arrangement of the loops, as well as on the magnetic permeability of the medium surrounding. the loops: These coefficients are measured in the same umts as the lllducta?ce L. Reciprocity Theorem. Calculations show (an~ expenments confrrm) that in the absence of fel'romagnetlcs, the coefficients L 12 and L 21 are equal: (9.25) This remarkable property of mutual inductance is usually called the reciprocity theorem. Owing to this theorem\ we
....,...
232
9. Electromagnetic Induction
do not have to distinguish between L 12 and L.J.l and can simply speak of the mutual inductance ·of two circuits. The meaning of equality (9.25) is that in any case the magnetic flux 1 through loop 1, created by current I in loop 2, is equal to the magnetic flux 2 through loop 2, created by the same current lin loop 1. This circumstance often alIows us to considerably simplify the calculation, for example, of magnetic fluxes. Here are two examples. I'
Example f. Two circular loops 1 and 2 whose centres coincide lie in a plane (Fig. 9.10). The radii of the loops are 4 1 and Current I flows in loop 1. Find tile magnetic flux 4>2 embraced by loop 2, if 41 0) depends only on the mutual ?flentation of lhe currents themselves anft is independent of the chOice of the positive direction~ of circumventi?n of t~e loops. We recall that the sign of the quantity La was conSidered m Sec. 9.4. I
Field Treatment of Energy (9.34). There are some more important problems that can be solved by calculating the magnetic energy of two loops in a different way, viz. from the point of view of localization of energy in the field. Let B 1 be the magnetic field of current. /1' and ~2. the field of current /2' Then, in accordance WIth the pnnciple of superposition, the field at each point is B~-= B 1 -t- B 2 , and according to (9.31), the field energy of this system of currents is W = ~ (H2/2~lflo) dV. Substituting into this formula .13 2 B~ + B; + 2B 1 B 2 , we obtain nO
~
•
.
I
W=
-=.9-,-._
i-')~
1_ _ ~l.lJ.lo
Electromagnetic Induction
dV+
r 2LdV+ I' -!Jl·B 2l.lf o J l.ll.lo
J
t
2
dV
I
•I
(9.35)
II 12
r ]l1 ·B2 av.
J
l.lfto
into account that
+ L 121 2 ,
} = L}1}
he correspondence between individual terms in formulas 9.35) and (9.34) is beyolHl doubt. Formulas (9.:3'1) and (9.35) lead to the following important sequences. 1. The. magnetic energy of a system of two (or more) CUl'wnts IS an essentially positi~e quantity (W > 0). This rOllo\~s from tile fact that tv ex: \ Jr2 dF, wllOre the integrand contall1S positive quantities. • 2. The ~nergy of currents is a nonadditive quantity (due to, tll,n, ex lsten~e of m 1I1llal energy). . .3. I he last lJItogTHI .in (~).;3:l) is proportional to the prol!t,Ctt' 1 1/ 2I . of Cllrren t, _ L ~ dt
Hence the required
dt •
arnollnt of electricity (charge) is
_r 1 q-- I I dt= - - •
• \ (dIll
I
R.'
I dl) 1 -=-7f(M>[-!.M).
~ince the frame has been sto I, f . . valllshes, and hence A/- 0 a Ppel ,! tcr rotatIOn, the current in it of the flux Art> through 'tlll: fra~~n(r~~lJ f~r ~s to find the increment et us choose the normal n to th I 2 - 11>1)' rl that in the lin.al Positior; n is dire~t~/b~h~f ~he frame, for instance;' ah?t g . B). Then It can be easily seen that' ihelPlane of the figure II' I e III the initial positio~ ' ,JII e ma position 11>2 > 0 and A$ turns out to be' I I < 0 (the normal is opposite to B)' boumled by the jj~al aneti~~·~'leqU > O. According to Lenz'" law, the induced eurrent I causes the Ampere force counteracting the motion, directed to the left. Having chosen the X-axis to the right, we write the equation of motion of the jumper
b+a b:a
q='
ance of the circuit (the situation would be oifferent if the circuit were
m dvldt = /lB,
(2)
where the right-hand side is the projection of the Ampere force onto the X-axis (this quantity is negative, but we omit the minus sign sina, as can be seen from (1), current I < 0). Elinlinating' I from l':qs. (1) allli (2), we obtain dvlv
=
-a dt.
a = IJ2/2/mR.
The integration of this expression, taking into account the initial condition, gives In (171170) = -at, v = vee- t
251 Problems
250
9. Electromagnetic
Fig. 9.23 the f inductan~es L e.:..
Induc~ion
. • 9.8. The role of t ranslent . processes In th . . I!f of the source, its iniernal e .clrcUlt shown in currents establish:d I;z :::Iktance R the S w e cOlIs after key coils K h b nown. Fwd the and
~1sup~rconducting
~L:~on. Let us use .
~nd
Kirchhoff's laws
l RI ='f,-L 1 dl dt'
for:~ec :~n rIC closed. circuits
~Ll
(erroneous) result, namely, insteal\ of 1J1i, 1/2 is obtained in the parentheses. The thinner the central wire, "i.e. the larger the ratio bla, the smaller the relative diflerence in the results of calculation by these two methods, viz. through the energy and through the flux. • 9.10. Mutual indudion. A long straight wire is arranged along the symmetry axis of a toroidal coil of rectangular crosS section, whose B
RI = 'fb - L __ d/ z 2 dt
the expresshlOns . while A comparison for stabilized ofcurre~t s we ave shows that L t d1 1 = L zdJ z , Lt /
Besides,
10
=
(1)
LzJ zO'
+
(2) J 10 1 20 = 1 0 = ~ I R . From these eq ua t'IOns, we find LZ 1 10 ='fi,6 L R L 1 L 2 ' J 20 = -R L LL 1 . f'. I', 2 . • 9.9. Calculati mternal solid conduc on 0 ~n~uetanee. A coaxial· . aU C?nSlsts. of tube. of radius b. a and extemal thi consIdering that th t e lllductance of a unit I n canductmg internal conductor distribution over the eng. of cable, everywhere . IS umform. The permeabTt sectIOn the I I Yc~oss IS equal to of unity
+
Fi~dr l! r~d1Us e.curr~nt
(
c~b~e
ili ~he
t~e3~)nergyrath::~ha~et~~o~:~athce
~bi d:~e:~~~~dc~~dt~~tor
notSolution. thin and Inh the Case under considerati tl' should iSr WIth 1 ,we can write e magnetIc nux. In acco r dancems.o
Lu =
1
J2
r.\b -;;BZ 2rtr dr
ii"'o
(1 )
'
where r iswethemu dist fince from the cable axis In d integral, -circulation, we hstavend the dependence B(·r). Using or er to theevaluate theoremthis on
Th f
B r
~o\enoitl
10.1 Displacement Current
252
-------~~
fJ. EleclrtJlnll!:netic Induction
-------
area of the solenoid, and B 2 = f-tJ.1onS/2' Hence formula (1) can be written (after cancelling /2) as follows: I L I2 1 =f-tf-ton2NtS=llllonln2V,
r'" f-tflon(V YflflonW= yr
• 9.12. Reciprocity them·em. A small cylindrical magnet M is placed at the centre of a thin coil of rarlius a, containing N tnrns (Fig. 9.26). The coil is connect.ed to a ballistic galvanometer. The resistalKeof the circuit is R. After the magnet had been rapidly removed from the coil, a chaq;e q passed through the galvanometer. Find the magnetic moment of the magnet. Solution. In the process of removal of the magnet, the total magnetic flux through the coil was changing, which resulted in the emergence of induced current defined by the following equation: R / = __ d(D _ L d/ dt dt .
We multiply both sides of this equation by dt and take into acconnt that / dt = dq. This gives R dq
=
-d, sincl' all directions perpendicular to the radial dirrclion arc absolutely I'quivull'nt. It remains for us to conclude that magnetic flCld is equal to zero everywhere. The absence of magnetic field in the presence of electric current of density j indicates that in addition to conduction current j, displacement current jd is also present in the system. This current is such that the total current is equal to zero eYcrywhere, Le. jd = - j at each point. This means that .
.
I
Jd = J = 4nrz
where we took into account that Gauss theorem.
iJD
q c=
4nrz
J)
=
= 7ft '
q/4nr z in accordance with the
10.1. Maxwell's Equations
Maxwell's Equations in the Integral Form. The introduction of displacemtilUt current brilliantly completed the macroscopic theory of electromagnetic field. The discovery of displacement current (aD/at) allowed Maxwell to create a unified theory of electric and magnetic. phenomena. Maxwell's theory not only explained all individual phenomena of electricity and magnetism from a single point of view, but also predicted a number of new phenomena whose existence was subsequently confmned. . , Hitherto, we considered separate parts of this theory. Now we can represent the entire theory in the form of a system of fundamental equations in electrodynamic.s, called l'vfaxwell's equations for stationary media. In all, theN are four Maxwell's equations (we are already acquainted with each of these equations separately from previous sections; here we simply gather them together). In the integral form, the system of Maxwell's equations is written as follows:
r ~Edl= I'~H 1
dl =
-)
~~ dS,
~ (j +- ~~_ ) dS,
~ DdS= ~
t" BdS=O,
pdV,
(10.10) (10.11)
where p is the volume density of extraneous charges and j is the density of conduction ClllTent. 17-0181
•
258
_-.!():
J1!a.r'Nll's
Equalion.~.
EleclromaKnclic Field Enerr:y
These et)uations express in .. concise f01'l1l all our knowlOlL:,,> 0, id tI v, and vice .vers~. '. ht solenoid with the • 10.2. A current n?wm~ III a long str~~{1netic field inside the radius R of cross sec~ion I~ varIed sOrtha\~h~he l~w B = ~t2, where ~ solenoid increaFs~ (Iensity as a function of is a constant. 'Illl IW~~~ lt~pla~~~;~:~~rrent .. the distance r from the solenoId aXIS.
272
10. Maxwell', Equation,. Electromagnetic Field Ener,,,
273
Problem' Solution. In order to find the displacement current density, we must, in accordance with (to.5), first find the electric fiehl strength (here it will be a vortex field). Using Maxwell's equation for circulation of vector' E, we write 2nrE = nr2 aBlat, E = r~t (r < R); 2nrE = nR2 aBlat, E = R2~tlr (r> R).
Now, using the formula id = eo iJElat, we can find the displacement current density: id = eo~r (r < R); id = eo~R2/r (r > R).
Let us transform the expression in the parentheses to cosine. For
thi~
~~1~h:~7:t:d~~~I:n~1:t~~~~;~i~h~xF;::~I:SbJi/:: c~ 6 te~~iiwL 2
,
=sin 6. This gives
H = ~ rEm
va +(eeoco 2
)2 cos (cot+6).
A point charge q moves in a vacuum l;lniformly ~l,d
rec~i~:a'~y with a nonrelalivistic velocity v. Usmg Maxwe
8
The plot of the dependence ; d (r) is shown in Fig. to.13•
• to.3. A parallel-plate capacitor is formed by two discs the space between which is filled with a homogeneous, poorly conducting medium. The capacitor was charged and then disconnected from a power source. Ignoring edge effects, show that the magnetic fIeld inside the capacitor is absent. Solution. Magnetic field will be absent since the total current (conduction current plus displacement current) is equal to zero. Let us prove this. We consider the current density. Suppose that at a certain instant the density of conduction current is j. Obviously,j ex D and D = an, where a is the surface charge density on the positively charged plate and n is the normal (Fig. 10.14). The presence of conduction current leads to a decrease in the surface charge density a, and hence in D as well. This means that conduction current will be accompanied by the displacement current whose density is jd = aD/at = (aalat) n = -jn = -j. Hence it follows that, indeed
it =
j
ot ! ~D
2nrH=(i n + ee o Taking into account Ohm's law jn
=
8it) nr
q~~_V
Fig. 10.15
Fig. 10.14
• . o'f t H obtain the expression 'for H equati0!l for thhe CIrcuI~tt~~~ rel::i~~rto the charge is characterized by at a pOlUt P w ose POSI I radius vector r (Fig. 10.15). -Solution. It tour arohund
iShcl:~r f~om ls~~~e;?v~~~g:d~t:~~~dthba:~~~e~~o:;
whi~cle wel'~hcc~:t:~ 0 (its trace is'shown in Fig. 10.16
must c oose a clr by the dashed line). Then
+ jd = O.
2nRH=..!at
• 10.4. The space between the plates of a parallel-plate cafacitor in the form of circular discs is filled with a homogeneous poor y conducting medium with a conductivity a and permittivity e. Ignoring edge effects, find the maci'litude of vector H between the plates at a distance r from their axeb, if the electric field strength between the plates varies with time in accordance with the law E = = Em cos wt. Solution. From Maxwell's equation for circulation of vector H. it follows that
r (E H ="2
o
l
eeo aE n ) rEnt . n+oiJ't =-r(acoswt-eeows!nwt).
(1)
wheLre~tR iSfit~e tr:ed~~xo~fte:c~~r~l~. through a surface boun~ed by this
reus n f' 1"t e shall take a spherIcal surface 16) Then the flux of D through circle. For t~e safke 0 Stlmp IC(£.lg' • • • . with the radIUS 0 curva ure r an elementary ring of this spherical surface IS q . , :1_' q. , da.' D ds=--2nr SlD a ·r ..... ='2 SlUa , 4nrll
fO
while the total flux through the selected surface is
.
aE n (t). we obtain
Jr D n dS,
~ DdS=~ (i-cos a).
(2)
Now, in accordance with (1), we differentiate (2) with respect to time: 8\ q. cia. (3} 8t J D dS = 2: SlD a dt . 18-0181
275
10. MGZlIIelt, Equation•. Electromagnetic Field Energy
Problem.
For the displacement of the charge from point 1 to point 2 (Fig. 10.17) over the distanee v dt, we have 1.1 dt ·sin a. = r da., whence da. 1.1 sin a. Cit r (4)
SolutiorL l"igure 10.19 shows that S tt v. Let us find the magnitude of S: S = EH, where E and H depend on r. According to the GaUS8 theonm, we have 2rtrE = 'AJeo, where A is the charge per unit length of the beam. Besides, it follows from the theorem on circulation of vector H that 2rtrH = f. Having deWnnined E and H from the last two equations and taking into aee:oant. that f = 1.1.1, we obtain S = EH = f 8/4n 8 eol.lr2.
274
SubstitutiDg(4) into (3) and then (3) into (1), we obtain H = qvr sin a./4m,a,
(5)
where we took into account that R = r sin a.. Relation (5) in vector form can be written as follows: H=_LJvXr) 4n
r3
e 10.8. A current flowing through the winding of a long straight
Thus we see that expression (6.3) Which we have postulated earlier is a corollary of Maxwell's equations.
dB
'k-
el0.6.CuriofE.Acertain :re2ion ofan inertial system of reference contains a magnetic field of m'1g11itude B = const, rotating at an
\
l'
H
S
I \
C;;2:'$Z;::::~=Z&-~~ \ /
\J
p
1 Fig. 10.16
2 Fig. 10,17
o
angular velocity roo Find V X E in this region as a function of vectors and B.
Fig. 10.18
solenoid is being increased. Show that the rate of increase in the energy of the magnetic field in the solenoid is equal to the flux of Poynting's vector through its lateral surface. Solution. As the current increases, the magnetic field in the !'olenoid also increases, and hence a vortex electric field appears, Suppose that the radius of the solenoid cross section is equal to a. Then the strength of the vortex electric field near the lateral surface of the solenoid can be determined with the help of Maxwell's equation that expresses the law of electromagnetic induction:
fI}
Solution. It follows from the equation V X E = - aB/at that vector V X E is directed oppositely to vector dB. The magnitude of this vector can be calculated with the help of Fig. 10.18:
I dB I = Bro dt,
I dB/dt I = Bro.
Henee
V
X E = -fro X . BI.
e 10.7. Poynting's vector. Protons haVing the same velocity v form a beam of a circular cross section with current f. Find the direction and magnitude of Poynting's vector S outside the beam at a distanee r from its axia.
l"ig. 10.19
2rtaE = rta
8
aB
at '
a aB E =2"
at·
The energy flux through the lateral surface of the solenoid can be represented as follows: a ( B8 ) 2 cD=EH·2nal=rta / at 2f.1o ' where I is the solenoid length and rta2 l is its volume. Thus. we see that the energy flux through the lateral surface of the solenoid (the flux of vector S) is equal to the rate of variation of the magnetic energy inside the solenoid: cD = S ·2rtal = ow/at. 18*
11.1. Equation of an Olctllatory Circuit 216
plates be h. Then the electric energy of the capacitor is equal to
10. Mazwtll's Equations. Eltctromagnttic Fldtl Elltr"
II
• 10.9. The energy from a source of constant voltage U is transmitted to a consumer via a long coaxial cable with a negligibly small resistance. The curr~nt in the c!lble is f. Find the energy nux through the cable cross sectIOn, assumIng that the outer conducting shell of the cable has thin walls. E Solution. The required ener~ gy llux is defined by the formula
t
dfr-:=-H=
S
e---o
:=:;
j
S·2nr dr,
(1)
a
Fig. to.20 where S=EH is the nux density, . 2'lr dr is the elementary ring of width dr within which the value of S is constant, and a and b are the radii of the internal wire and of the outer shell of the cable (Fig. 10.20). In order to evaluate this integral, we must know the dependence S (r), or E (r) and H (r). Using the Gauss theorem, we obtain 2'lrE = '}Jllo, (2) where Ais the charge per unit length of the wire. Further, by the theorem on the circulation' we have 2'lrH
=
(3)
f.
After substituting E and H from formulas (2) and (3) into expression (1) and integrating, we get ,
AI b =--ln 2'lll o
U=
)
E
b A dr=--ln-. 2'lllo
a
EI
II
'la l
We=+'la2h=~ UfucoslCllt.
(1)
The maguetic energy can be determined through the formula Wm =
\
•
:1
110
dV.
(2)
The quantity B required for ~valuating this integral can be found from the theorem on the circulation of vector H: 2'lrH = 'lr2 aDlat. Hence, considering that H = BII10 and aDlat = -'-eo (Umlh) CIl sin CIlt, we obtain 1 rCllU m • (3) B=2'llolto-h- \ SlDCIlt I· It remains for us to substitute (3) into (2), where for dV we must take an elemeutary volume in the form of a ring for which dV = = 2nr dr·". All a result of integration, we obtain 'l J.LollifalSa6Ufu '1 W m=16 h sIn CIlt.
(4)
The ratio of th.e maximum values of magnetic energy (4) and electric energy (f) is given by
,For example, for 5 X to-D.
a
=
WmU:-ax W emax 6 cm and
1 1 1 8 l1011oiJ cIl • cIl
= toDD
S-I,
this ratio is equal to
(4)
a
The values of A, a, and b are not given in the problem. Instead, we know U. Let us find the relation between these quantities: b
277
a
II. Electric Oscillations H.t. Equation of an Oscillatory Circuit
(5)
A comparison of (4) and (5) gives III = Uf.
This coincides with the value of power liberated in the load. • 10.tO. A parallel-plate air capacitor whose plates are made in the form of disks of radius a are connected to a source of varying harmonic voltage of frequency CIl. Find the ratio of the maximum values of magnetic and electric energy inside the capacitor. Solution. Let the voltage across the capacitor vary in accordance with the law U = Urn cos rot and the distance between the capacitor
Quasi-steady Conditions. When electric oscillations occur, the current in a circuit varieswith time and, generally speaking, turns out to be different at each instant of time in different sections of the circuit (due to the fact that electromagnetic perturbations propagate although at a very high but still finite velocity). There are, however, many cases when instantaneous values of current prove to be practically the same in all sections of the circuit (such a current is called quasistationary). In this case, all time variations should occur SO slowly that the propagation of electromagnetic pefturba-
278
11.1. Equation of an Oscillatory Circuit
11. Electric Oscillations
tio~s c~uld
be c.onsidere~ instantaneous. If 1 is the length of a c~rcUlt, the time reqUlred for an electromagnetic perturbatIOn to cover the distance 1 is of the order of 1: = lie. F~r a periodically varying current, quasi-steady condition WIll be observed if 't
=
lIc~T,
where T is the period of variations. For e:ample, for a circuit of length 1 = 3 m, the time 't = 10 8 s, and the current can be assumed to be quasistaR
L
(b)
(a)
Fig. 11.1
6
Fig. 11.2
tionary down to frequencies of 106 Hz (which corresponds to T = 10-6 s). In this chapter, we shall assume everywhere that in the cases under consideration quasi-steady conditions are observed and currents are quasistationary. This will allow us to use formulas obtained for static fields. In particular we shall b.e using the fact that instantaneous values of 'quasistatIOnary currents obey Ohm's law. Oscillatory Circuit. In a circuit including a coil of induc~ance L and a capacitor of capacitance G, electric oscillatIOns may appear. For this reason, such a circuit is called an oscillatory circuit. Let us find out how electric oscillations emerge and are sustained in an oscillatory circuit. . Suppose tha~ .initially, the upper plate of the capacitor IS charged p~sltlvely and t~e lower plate, negatively (Fig. 11: 1a). In tIus case, the entIre energy of the oscillatory cirCUlt IS co~centrated in the capacitor. Let us close key K. T~e capacItor star!s to discharge, and a current flows through cOlI L. The electrIc energy of the capacitor is converted into the magnetic energy of the,coil. This process terminates when the capacitor is discharged completely, while current
279
in the circuit attains its maximum value (Fig. 11.1b). Starting from this moment, the current begins to decrease, retaining its direction. It will not, however, cease immediately since it will he sustained by self-induced e.m.f. The current recharges the capacitor, and the appearing electric field will tend to reduce the current. Finally, the current ceases, while the charge on the capacitor attains its maximum value. From this moment, the capacitor starts to discharge again, the current flows in the opposite direction, and the process . is repeated. If the conductors constituting the oscillatory circuit have no resistance, strictly periodic oscillations will be observed in the circuit. In the course of the process, the charge on the capacitor plates, the voltage across the capacitor and the current in the induction coil vary periodically. The oscillations are accompanied by mutual conversion of the energy of electric and magnetic fields. If, however.... the resistance of conductors R =1= 0, then, in addition to the process described above, electromagnetic energy will be transfor~ed into Joule's heat. Equation of an Oscillatory Ci~uit. Let us derive theequation describing oscillations in a circuit containing seriesconnected capacitor G, induction coil L, resistor R, and varying external e.m.f. liJ (Fig. 11.2). First, we choose the positive direction of circumvention, e.g. clockwise. We denote by q the charge on the capacitor plate the ~iirection from which to the other plate coincides with the chosen direction of circumvention. Then current in the circuit is defined as (11.1) I = dqldt. Consequently, if I > 0, then dq > 0 as well, and vice versa (the sign of I coincides with that of dq). In accordance with Ohm's law for section lRL2 of the circuit, we have (11.2) RI = CPl - CP2 ~8 'IS,
+
+
where 'fe s is the self-induced e.m.f. In the case under consideration,
ts = -
L dIldt and CP2 - CPl.= giG
:1" ........
280
11. Electric Oscillations
(the sign of q must coincide with th . difference (JJ2 - (JJ since G >0) He sIgnE,of the potential written in the f~rm . ence q. (H.2) can be L dI
q
7t+RI+c=~, or, taking into account (H.t), L
aSq
dq
(iii"+R 7t
+ c1
q=~.
(H.3)
I
(11.4)
This is the equation of an 'llat . . near second-order nonho OSCl o,:y clrcU.lt, which is a Hconstant coefficients U~i~ggenteho.us differ~ntiafl equation with IS equatIOn or calculat' t) . q ( ,we can easily obtain the voltage across th .mg e capaCitor as U c = (JJ2 - (JJI = qlG and current I b en;~~r,::::uation of an oscillatory circuit ca:b~o;~~~aa~i~:L
l·q+2Pq+ro:q=~/L,
(11.5)
where the following notation is introduced:
. 2p = RIL, ro: = lILG. (11.6) TIle quantity ro' II d h and p is the dar:;, I~nca e t e natural f!equency of the circuit will be explaine~ b~/actor. The meanmg of these quantities If _ . ow: tions ~The~' t~ll ~cllladatlOns are usually called free oscillaR' WI e un mped for R = 0 and dam ed fI =F O. Let us consider consecutively all these casts. or 11.2. Free Electric Oscillations Free Undamped Oscill a t· If . . IIal e.m.f. ~ and if 't lon~. a CIrCUit contains no exter'11 t' . I S resistance R -. 0 the 111 such a circuit will be free d d- , OSCI a Ions '11" an un amped. The equation describin th (O~, an aperiodic discharge of the capacitor will occur instead of oscillations. The resistance of the circuit for which the aperiodic process'sets in is called the critical resistance:
R cr =2YL/C.
(11.24)
Let us consider two examples. Example t. An oscillatory circuit has a capacitance C inductance L, and resistance R. Find the number of oscillations a lter which the
,
11~
' g tbl'S time Ne oscillations will Durm 'h
V (W
)2
.!£-1.
illations in a CllCUlt WIth a g of damped OSCIllatIOns IS equ osc. . 't'a1 value if the frequency of 1ts lUl 1 ~ , • • h to roo . . . e- IlI , the time to durmg wl1!-c Since the current amphtu~e of Tj is determined by the equatIOn the amplitude decreases by a ac 0 1\ = ept•• Hence Ill. to = (In 1'\) p. ,
;mrex:
On the other •
•
Elimmatmg
bftDd the Q-factor is also related to ~: 'Q A
I'
= Ttl;, = n/~T = ro/2~.
f m the laa\ two equations, we obtain ro 2Q to = c;) In Tj.
11.3. Forced electric Oscillations . . Let us return to equations (11.3) Steady-state OscJlla~lOns. . 't and consider the case 1 4) for an OSCillatory Clf~Ul t al emf ~ whose an d (1 . . ' 1 d a varymg ex ern .,. when the circUlt ,mc ~ es reased by the harmonic law: (11.25) dependence on tlUle IS dec. ~ = Om cos (Ot. . .al lace owing to the proper t'les of This law OCCUPI~S a .Sp~Cl If lo retain a harmonic form of osthe oscillatory CIrCUlt ItS~. of external harmonic c.m.!. dIlations under the aCt' lonfor a~ oscillatory circuit IS In this case, the equa Ion written in the form (11 26)
L ~+RI+...L=~m·cos(Ot,
amplitude of cunent decreases to fie of its initial value.
The eunent amplitude (/m ex: e -Pt) decreases to fie of its initial
-
value during the time : -d f damped oscillations, t en If T is the perlo 0 occur. 1 't 1.{~ =_ _0 -1. Nc=-= r~ 2n ~ T 2n/V ro6-P 2 = 1./ LC and l> = R {2L, we obtain ---c._Considering tllat roo _1._ ~ / N e = 2n VCR' . . hi Ii tbe amylitude of current Example 2. Fin~ t~e ti?1e duni~~nwQ-factor wH. de~rea~ to 1~
dt
or
C
.
(11.27)
11.3. Forced Itkctrlc O.ctUatio..
11. ItZectrlc OBctli4ttoIU
As is known from mathematics, the solution of this equation is the sum of the general solution of the homogeneous equation (wiih the zero right-hand side) and a particular solution of the nonhomogeneous equation. We shall be interested only in steady-state oscillations, i.e. in the particular solution of this equation (the general solution of the homogeneous equation exponentially attenuates and after the elapse of a certain time it virtually vanishes). It can be easily seen that this solution has the form q = qrn cos (rot -'Ii')~ (11.28) where qrn is the amplitude of charge on the capacitor and 'Ii' is the phase shift -between oscillations of the charge and of the external e.m.f. ~ (11.25).• It will be shown that qrn and", depend only on the properties of the circuit itself and the driving e.m.f. ~. It turns out that 'Ii' > 0, and hence q always lags behind ~ in phase. In order to determine the constants qrn and 'Ii', we must substitme (11.28) into the initial equation (11.27) and transform the result. However, for the sake of simplicity, we shall pr~ in a different way: first find current I and then substitute its expression into (11.26). By the way, we shall solve the problem of determining the constants qrn and 'Ii'. Differentiation of (11.28) with respect to time t gives 1= - roqrn sin (rot -'\jJ) = roqrn cos (rot -'\jJ + n/2). Let us write this expression as follows: I = I rn cos (rot - cp), (11.29) where I rn is the current amplitude and cp is the phase shift between the current and external e.m.f. ~, (11.30) I rn = roqrn, cp = 'Ii' - n/2.
We aim at finding I rn and cpo For this purpose, we proceed as follows. Let us represent the initial equation (11.26) in the fonn UL + Un + Ue = ~rn cos rot, (11.31) where the left-hand side is the sum of voltages across induction coil L, capacitor C and resistor R. Thus, we see that at each instant of time, the sum of these voltages is equal
·1· f '£! Taking into account relation (11.30), totheexterna e.m.. ",. we write (11.32) U R = RI = RI m cos (rot - cp), q --.J!!!. cos (rot _ '\jJ) = I Tn cos (rot - cp U e=c- c we
UL
~
) , (11.33)
= L ~~ = - roLlrn sin (rot - cp) =
ro LIm cos trot - cp + T •
11: )
(11.34)
.
tor Diagnun The last three formulas show that U i:;hase with cu~rent I, U e lags behind I by n/2, and U
V is
R l.
Im
o Fig. 11.5
Fig. H.4
~his
v:~ugal;~er:::;f~~~~ w;jh vt~~
leads I by n/2. candbe. help of a vector dtagram, epIC m
tages
- Rf U em = 1 m/roC, U Lm = roLl rn Uam m, and their vector sum which, according to (11.31), is equal to vector ~m (Fig. 11 .4). . d' can Considering the right triangle of. thIS Iagram,;,e easily obtain the following expressIons for I m and cpo ~m (11.35) 1m = l!R'+«(a}L-1/wC)Z' wL-1/(a}(: (11.36) tancp= R Th the problem is solved. . d b It :~uld be noted that the vector diagram obtame a ove
288
289
11. Electric Oscillations
11.3. Forced Electric Oscillattons
proves to be vel'y con . f problems. It permits a vi:~~lent 01' solving .many specific various situations. ' easy, and rapId analysis of .Resonance Curves. This is the . dependences of the folloWing q n:-~e gIven to the plots of of external e.m.f. ~: current I uahn lIes on the fre~uency (J) , c arge q on a capaCitor, and
The maximum of this function or, which is the same, the minimum of the radicand can be found by equating to zerO the derivative of the radicand with respect to cu. This gives the resonance frequency (11.38). Let us now see how the amplitudes of voltages U R' U e, and U L are redistributed depending on the frequency W of the external e.m.f. This pattern is depicted in Fig. 11.7. The resonance frequencies for U R' U e and U L are determined by the following formulas:
--'-'-------
c6
Fig. 11.6
Fig. H.7
voltages UR Ue and U d fi . (11.34). " L e ned by formulas (11.32)Resonance curves for current I ( ) '. U.5. Expression (11.35) shows th m cu are shown m Fig. has the maximum value for cuL _ ~t the~urrent amplitude the resonance frequency for current lcur: -: O. C?nsequentlys tural frequency of the oscl'll t .col~cldes with the naa ory CIrCUIt: . WI reB = Wo = lIV LC. (11.37) The maXImum at resonance i th h' the smaller the damping facto: A. ~ ~~~er and the sharper Resonance curves for the cha p L. .shown in Fig. 11.6 (resonance rge qm (cu) On a capacitor are curves for voltage Uem across the capacitor have the charge amptitude is attaine~a~~ slhlape). The maximum of . t e resonance frequency . Wqre8=.Vw~-2~2, (11.38) which comes closer and closer to . In order to obtain (11 38) W o WIth decreasing p. ance with (11.30), i~ the ~~rmust represent qm, in accordm qm = I mlw where 1 m is defined by (11.35). Then
qm =
- _ ~=/L==-_ V(W3-WI)2+4~IWI
(11.39)
= WoY 1- 2 (~/(l)O)2,
(J}e
res
(J}L
res =
(11.40)
WoY 1- 2 (~/WO)2.
The smaller the value of ~, the closer are the resonance frequencies for all quantities to the value Woo Res6n3nce UII'Ves and Q-factor. The shape of resonance curves is connected in a certain way with the Q-factor of an oscillatory circuit. This ~onnection has the simplest form for the case of weak damping, I.e. for ~2 «w~. In this case, Ueres/cem = Q (11.41) (Fig. 11.7). Indeed, for ~2« cu~, the quantity w res ~ W o and, according to (11.33) and (11.35), U em res = I m/cuoC = = tm/woCR, Or Uemres!'tm = YLCICR = (1IR) YLIC which is, in accordance with formula (11.22), just the Qfactor. Thus, the Q-factor of a circuit (for ~2« w~) shows how many times the maximum value of the voltage amplitude across the capacitor (and induction coil) exceeds the amplitude of the external e.m.f. The Q-factor of a circuit is also connected with another important characteristic of the resonance curve, viz. its width. It turns out that for ~2« w~ Q = cuolfJw, (11.42) where CU o is the resonance frequency and fJw is the width of the resonance curve at a "height" equal to 0.7 of the peak height, I.e. at resonance. Resonance. Resonance in the case under consideration is 1{2
19-0181
1111
III! 11:1
I!II
Iii
II
I II
290
11. Electric Oscillations 11.4, Alternating Current
the excitation of strong oscillations at the frequency of external e.m.f. Or voltage. This frequency is equal to the natura~ fre~\lency of an oscillatory circuit. Resonance is used for slug-hng out a required component from a composite voltage. The entire radio reception technique is based on resonan~e. In order to receive with a given radio receiver the station we are interested in, the receiver must be tuned. In other words, by varying C and L of the oscillatory circuit we must attain the cOincid~nce between its natural {requeuc; and. the frequency of radIo wayes emitted bv the radio statlop. . he phenomenon of resonance is also associated with a certam danger: the external e.m.f. Or voltage may be small but t:le,v?ltages.acros~in(li,:i~Ju(d elem~nts of the circuit'(the capaCItor or lllductlOn coIl} may attalll the values utlnaerous for people. This should always be remembered. "
:r
11.4. Alternating Current
~otal. Hesistance (Impedance). Steady-state forced electric oscillatIOns can he treated as an altel'llatinrr current nowing in ~.circuit having a capacitance, inducta~ce, and resistance. Lnder the action of external voltage (which plays the role of external e.m. f.) U =" U m cos
(11.43)
tl)t,
the current in the circuit varies according to the law I = 1 m cos (cut '- cp),
(11.44)
where I
Um m-R2+(roL_1/roC)2'
tancp=
roL-1/roc R
(11.45)
The problem is reduced to determining the current amplitude and the phase shift of the current relative to U. The obtained ~xpression for the current amplitude 1 (ro) m can be formally Interpreted as Ohm's law for the amplitude values of current and voltage. The quantity in the denominatOr of this expression, which has the dimension of resistance, is denoted by Z and is called the total resistance , Or
291
impedance:
Z = V' R2 + (roL -1/coC)2. (11.46) It can be seen that for ro = roo = 1/V' LC, the impedance has the minimum value and is equal to the resistance R. The quantity appearing in the parentheses in formula (11.46) is denoted by X and is called the reactance: X = ro[, - 1/roC. (11.47) Here the quantity roL is called the inductive reactance, while the quantity VroC is called the capacitive reactance. They are denoted XL and Xc respectively. Thus, XL=roL, X c =1/roC, X=XL-X c ,
z= V R2+X~".
(11.48)
It should be noted that the inductive reactance grows with the frequency",w, while t~e. cap~citive rea~tan.ce decreases with increasing ro, When It IS saId that a CircUlt .has no capacitance, this must be t;nderstood so that there IS no c~pac itive reactance which is equal to 1/roC and hence valllsh.es if C -lo- 00, (when a capacitor is replaced by a short-CIrcuited section), ., Finally, although the reac~anc~ is I:Ue~suredII~ t~e same units as the resistance, they dIffer 1Il prmclple. ThIS difference consists in that only the resistance determines irreversi~le processes in a circuit such as, for example, the converSIon of electromagnetic energy into Joule's heat, Power Liberated in an A.C. Circuit. The instantaneous value of power is equal to the product of instantaneous values of voltage and current: P {t) = VI =-c= Vml m cos rot cos (rot - cp). (11.49) Using the formula cos (rot - (f') = cos rot cos cp sin rot sin if, we transform (11.49) as follows: P (t) = U ml m (C08 2 wt cos cp + sin rot cos rot sin cp),
+
+
Of praeticnl importance is the value of power averaged over a period of oscillation. ConsJdering that (cos 2 cut) = =--= 1/2 and (sin wt cos wt) = 0, we obtain (P) = 19*
U I
m m
2
cos cpo
(11.50)
292
11. Electric Oscillations
Problems
Th~s expl'ession can be written' l"ff . take lIlto account that 'IS foIl It a {I erent form If we ws (see Fig. 11.'1) U 0" _ nOI ram the vector diagram , m C S lp m' Hence, 1 {P)--2 RI:". (11.51)
the concept of electric resistance becomes wider since in addition to the conversion of electric. energy into Joule's heat, other types of energy transformations are also possible. For example, a part of electric energy can be converted into mechanical work (electric motors).
TI,lis power.i~ equal to that of the direct current I' _ ,T' Ie quantities - I m/V2".
I = I ml V2, V = V mlV2 (11.52) are called effective (root-mea ' t IHsquare) values of current alld voltaO'e All a '"' . mme ers 'md VO It t r.m.s. values of curl"ellt' . d meel'S are graduated for '('1 an vo l tage . . Ie expression for the 'Ivera . r.m.s. \'alues of CUlTent ayajct Fu_uol 01 Physiu . A Prob*n Book in M~ AnlIyals A Q>U.coon 01 Ounu_ tond ProbItomf .. Phyojgo ~Idbot 01 E~ MlIllIInMio;$ 00 VDol Know et.niJ,.., I l/loI9anic CIMomosUy Vol. I & 11 au.&iwiw CIwnouI s.mimicr........lysi1
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