Livro - Basic Laws of Electromagnetism por Irodov - Mir Publisher (1986)

where

28

1. Electrostattc Field in

tJ

2U

Vacuum

called the field potential. A comparison of (1.23) with the expression for the work done by the forces of the potential field (the work being equal to the decrease in the potential energy of a particle in the field) leads to the conclusion that the potential is the quantity numerically equal to the potential energy of a unit positive charge at a given point of the field. We can conditionally ascribe to an arbitrary point 0 of the field any value CPo of the potential. Then the potentials of all other points of the field will be unambiguously determined by formula (1.23). If we change CPo by a certain value Acp, the potentials of all other points of the field will change by the same value. Thus, potential cp is determined to within an arbitrary additive constant. The value of this constant does not play any role, since all electric phenomena depend only on the eleetric field strength. It is determined, as .will be"'shown later, not by the potentia] at a given point but by the·potential difference between neighbouring points of the field. The unit of potential is the volt (V). Potential of the Field of a Point Charge. Formula (1.23) contains, in addition to the definition of potential cp, the method of finding this function. For this purpose, it is sufficient to evaluate the integral ~ E dl over any path between two points and than represent the obtained result as a decrease in a certain function which is just

field of a lixed point charge: • 1

q

= -d

Ii r

IJ ,l.'l:'o

E.dt -- - (' r ·die /trrf,o ,.2

(_1_ _.'L _i-const) 4Jlco r I

r:!.

,

where we took into account that cr dt =-= 1· (dl}r= dr, sillee the projection of dl onto en ~nd hence ?ll 1', is ~qual to ~he increment of the magnitude of vector r, Le. dr. 'I he quantlty appearing in the parentheses under the diffe~enti~l is exactly i:p (1'). Since the additive constant contamed III ~he fo~'­ mula- does not play any physical role, it is usually omitted ill order to simplify the expression for (p. Thus, the potential of the field of a point charge is given by

I~

= __ 1 }L II _ _ 4n~_'_

(1.2:»

The absence'6f an additive constant in this expression indicates· that we conventionally assume that the potential is equal to zero at infinity (for r -+ 00). Potential of the Field ola System of Charges. Let a system consist of fixed point charges ql' q2' .... In accordance with the principle of superposition, the field intensity at any point of the field is given by E = E I + E 2 + ..., wh~re E l is the field intensity from the charge ql' etc. By llsmg formula (1.24)1 we can then write E·dl = (E l + E 2 + ...) dl = EI·dl + E 2 ·dl + ... = -dcpl - dCP2 - ... = --dcpl where cp = 2jcpf, Le. the principle of superposition t~rns out to be valid for potential as well. Thus, the potential of a system of fixed point charges is given by

.- _1_ '" !!i- -I I cp - 4nco L..J r 'I 1.

(1.26)

_i

where ri is the distance froID the point charge qj 10 the point under consideration. Here we also omitted an arbitrary constant. This is in complete agreement witll the faet that any real system of charges is Lounded in space, and honce its potential can bo taken equal to zero at inflIlity.

1.6. Relation Between Potential and Vector I!:

30 1. Electrostatic Field in a Vacuum -----------_._---.-'-'-'=";~--If the charges forming the system are distributed cUlItinuously, then, as hefore, we assume that each volume element dV contains 3 "point" charge p dV, where p is the charge . density in the volume dV. Taking this into consideration, we call write formula (1.26) in a different form: ( 1.27)

where the integration is performed either over the entire space or over its part containing the charges. If the charges are located only on the surface S, we can write
=

1 ~ adS 4neo -r-'

(1.28)

where (J is the surface charge density and dS is the element of the surface S. A similar expression corresponds to the case when the charges have a linear distribution. Thus, if we know the charge distribution (discrete or continuous), we can, in principle, fmd the potential of any system. 1.6. Relation Between Potential and Vector E

It is known that electric fwld is completely described by vector function E (r). Knowing this function, we can find the force acting on a charge under investigation at any point of the neld, calculate the work of fwld forces for any displacement of the charge, and so on. And what do we get by introducing potential? First of all, it turns out that if we know the potential cp (r) of a given electric fIeld, we can reconstruct the fwld E (r) quite easily. Let us consider this question in greater detail. The relation between rp and E can be established with tho help of Eq. (1.24). Let the displacement dl be parallel to the X-axis; then dl = i dx, where i is the unit vector along the X-axis and dx is the increment of the coordinate x. In this case E·dl = E·i dx = Ex dx, where Ex is the projectioll of vector E onto the ullit vector i (and not on the displacement dll). A comparison of this ex-

3t

pression with formula (1.24) gives Ex

= - orp/vx,

(1.29)

wher~ the symbol of partial.derivative emphasizes that the functIOn rp ~x, y, z) must he differentiated only with respect to x, assummg that y and z are constant in this case. . In a similar wa~, w.e can obtain the corresponding expres810.ns for the prOJectIOns E y and E z. Having determined E", Ell' and E z, we can easily fmd vector E itself:

E=

_ ( ocp ox

i

+ ~j + ~ k) iJy oz .

(1.30)

T~e quantity in the par~Iltheses is the gradient of the potentzaZ cp (grad cp or Vcp). We shall be using the latter more convenient notation and will formally consider VCP 'as the product of a symbolic vector V and the scalar cpo Then Eq. (1.30) can be represented in the form

(1.31) L~. the field inte~sity E'is equal to the potential gradient With the miflUS SIgn. This is exactly the formula that can be used for reconstructing the field E if we know the function

f

+

Let ~s derive one more useful formula. We write the righthand sl~e of (1.24) in the form E·dl c=o R I dZ, where dl = . Id1 l IS an elementary displacement and E I is the projectIOn of vector E onto the displacement dl. Hence

IE

l

= -ocpI8Z,

I

(1.32)

Le. the projection of vector E onto the direction of the dis-

1.6. llelation Between Potential and Vector E

--------------~_. ----------~

g2

On the Advantages o~ Potential. 1l was noted earlier that . electrostatic field is completely cltaracterizeu by vector function E (r). Thea what is the use of introducing potential? There are several sound reasons for doing that. The concept of potential is indeed very useful, and it is not by chance that this concept is widely used not only in physics but in engineering as well.. it. If we know the potential

1. Electrostatic Feeld in a Vacuum

placement dl is equal to the d' • . . p.otential (this is emphasized b;r~~~onal telflvfative .of the rivative). sym 0 0 partial deEquipotential Surfaces Let . d equipotential surface, viz. 'the s:;a~~t:~ :l~e t~etco~cept. of potentiallp has the same value. We shall sho~o~~ ~ 0 whIch each point o.f the surface is directed along thea e eqUipotentIal surface and. towards the decrease in the potential. Indeed, it follows from formula (1.32) that the projection of vector E onto any direction tangent to the eq,!ipo~entialsurface at a given pomt 18 equal to zero. This means that vector E is normal to the given surface. Further, let us take a displaC&ment dl along the normal to the surface, towards decreasing Fig. 1.13 p. Then Olp < 0, and accordmg to (1 32) E > 0 . v,ector E is. directed towards decreasing ~, ~r i~ the tIon ?pposite. to ~hat of the vector Vlp. It IS expedIent,.to draw equipotential surfaces in such a way that the potential difference between two neighbourin surfaces b~ t~e same. Then the density of e ui otentiJ s~aces WI.ll vIsually indicate the magnitudes fieYd intensttIes a~ dIfferent points. Field intensity. will be higher . t e re~lOns ~h~re equipotential surfaces are denser '"th~ pot~ntial rehef IS steeper"). \ Smce vector E ~s normal to an equipotential surface eve ry where, the field hnes are orthogonal to these surfaces.

:~

n::::.~r~

where
d;;;'

uating the integral ~ q'E dl in this case. This problem can be easily solved with the help of potential. Indeed, since all elements of the ring are at the same distance a from the centre of the ring, the potential of this point is Ipo ~= q/4Jv ou. And we know that cp = 0 at intinity. Consequently, the work A = q'cpo = q'q/4nlO oa.

of

Th

l

F~guhe df 3

shows a two-dimensional

patt~rn of

an electric field

h liDes eorres~o~d to equipoten~ial rlUrfaces, while thetiled MUd I. t. e mes of E. Such a representatIOn ean be easil v· i~~~:t:yd\:tiYghhows dthe hdirec.t!o.n of vector E, the regiorrs wi::e field er an were lower as we'l's tIl . "\h

lin:s ::

~

10 13

~~~~r steepness of the putential relief. SU(~ri a ;ai'tern ~~?':sed'to

d. t' qilal~~i~lve answers to a number of .~lU. ,.stiOJl8 sU~.h as ';In'wha.t lrec ,IOn WI "charge placed at a certain point m~veP Whe I> • iliagiutude o! the potential gradient At which 1 e orce actmg on the charge be greater~" et.c.

hi~her?

U:in~.;t.e

A

33

------

"

2. It turns out in many cases that in order to find electric field intensity E, it is easier lirst to calculate the potential lp and then take its gradient than to calculate the value of E directly. This is a considerable advantage of potential. Indeed, for calculating
r

I

34

1. Electrostatic Field in a Vacuum

----------'--'--'-.:-:..'-.:-:..~....:..::..:..::......:::..:......:=-:..=::.::::=-------

1.7. Electric Dipole

There are some other advantages in using potential which will be discussed later.

from the negative to the positive charge: p = ql,

1.1. Electric Dipole

where q > 0 and I is the vector directed as p. It can be seen from formula (1.34) that the dipole field depends on its electric mom~nt p.. It will be shown below that the behaviour of the dIpole III an external field .al~o depends on p. Consequently, p is an important characterIstIc ' . of the dipole. It should also be noted that the potential of the dI~)Qle field decreases with the distance r faste~ than the ~otentlal of the field of a point charge (in proportIOn to 1frz Illstead of 1fr). !In order to find the dipole field, we shall use formula (1.32) a~d calculate the projections of vector E onto two mutually perpendicular directions along the unit vectors er and ~ (Fig. 1. 14b):

The Field of a Dipole. The electric dipole is a system of two g and - g, separated equal in magnitude unlike charges by a certain distance l. When the dipole field is considered . ,

+

p

(b)

eb +q

/ q

P

E

E

it is assumed that the dipole itself is pointlike, i.e. the distance r from the dipole to the points under consideration is assumed to be much greater than l. The dipole field is axisymmetric. Therefore, in any plane passing through the dipole axis the pattern of the field is the same, vector E lying in this plane. L~t us first find the potential of the dipole field and then its illtensity. According to (1.25), the potential of the dipole field at the point P (Fig. 1.14a) is defined as 1 4itl:: o

'1) = 7..- - -,::

(lJ

1 4ne

aq>

"",' r

Fig. 1.14

<.p ,-

o

(1.35)

t)

= - --ar = = -

aq> r'~

=

2pcos~

1

4nll

o

-,-.- ,

1 p sin ~ 4neo - - , . -

(1.36)

Hence, the modulus of vector E will be z-tt. E=VE~+E6= 'tJ,80 J.: -?V-1-+-3-c-o-s r

(1.37)

In particular, for tt = 0 and tt = n/2 we obtain the expressions for the field intensity on the dipole axis (E u) and on the normal to it (E.d:

q(r_-r.) r.r_ •

Ell

= ~80 2~,

EJ.

= 4~80 ~,

(1.38)

Le. for the same r the intensity E u is twice as high as EJ.' The Force Acting on a Dipole. Let us place a dipole into a nonuniform electric field. Suppose that E+ and E_ are the intensities of the external field at the points where the positive and negative dipole charges are located. Then the resultant force F acting on the dipole is (Fig. 1.i5a): F = qE+ -qE_ = q (E+ - E_).

Since r» l, it call 1e seen fro.:n Fig. 1.14a that r__ r + = = 1 cos \} and r +r_ = r 2 , where r is the distance from the point P to the dipole (it is pointlikel). Taking this into accoulJt, we get (1.34) where p = ql is the electric moment oj the dipole. This quantity corresponds to a vector directed along the dipole axis

The difference E+ - E_ is the increment LiE of vector E on the segment equal to the dipole length 1 in the direction 3*

J

1. Electro$tatlc Field In a Vacuum f

I'

of vector I. Since the length of this segment is small, we can write ~E

~E=E+-E_=-. l

=

aE az l.

Substituting this expression into the formula for F, we find that the force acting on the d'ipole is equal to

I

F=p"*,

(1.39)

where p = ql is the dipole electric moment. The derivative appearing in this expression is called the directional derivative of the vector. The symbol of partial derivative indicates that it is taken with respect to a certain direction, viz. the direction coinciding with vector I or p. Unfortunately, the simplicity of formula (1.39) is delusive: taking the derivative fJE/fJl is a rather com~q~plicated mathematical operation. We shall not discuss this question
(.~

...

~:-q~

z:s.-

Figure 1.16 shows the directions of the force F acting on a dipole in the field of a point charge q for three different dipole orientations. We suggest that the reader prove independently that it is really so.

lf we are interested in the projection of force F onto a certain direction Xi, it is sufficient to write equation (1.39) in terms of the projections onto this direction, and we get

Fx = P

a:zx

,

37

1.7. Electric Dipole

where fJE ~/az is the derivative of the corresponding projection of vector E again onto the direction of vector I or p. Let a dipole with moment p be oriented along the symmetry axis of a certain nonuniform field E. We take the positive direction of the X-axis, for example, F as shown in Fig. 1.17. Since the increqe <= p. ment of the projection E ~ in the direc~F tion of vector p will be negative, F:r < O. and hence vector F is directed to the left, i.e. towards increasing field intensity. If we rotate vector p shown in FI the figure through 90° so that the dipole q-----~ centre coincides with the symmetry axis of the field, it can be easily seen Fig. 1.16 that in this position F:II: = O. The Moment of Forces Acting on 8 Dipole. Let<4Is consider behaviour of a dipole in an external electric field in its centre-of-mass system and find out whether the dipole will rotate or not. For this purpose, we must find the moment. of external forces with respect to the dipole centre of mass*. ~By definition, the moment of forces F + = qE+ and F _ = = -qE_ with respect to the centre of mass C (Fig. 1.18) is equal to M = [r+ X F+I [r_ X FJ=[r+ X qE+l -[r_ X qEJ,

q...-----tp

+

+

where r + and r _ are the radius vectors of the charges q and - q relative to the point C. For a sufficiently small dipole length, E+ ~ E_ and M = [(r+ - r_) X qEl. It remains for us to take into account that r + - r _ = I and qI = p, which gives

IM=

[p X

EJ.!

(1.41)

This moment of force tends to rotate the dipole so that its electric moment p is oriented along the external field E. Such a position of the dipole is stable. Thus, in a nonuniform electric field a dipole behaves as

(1.40) • We take the moment with respect to the centre of mass in order

to eliminate the moment of inertial forces.

38

1. Electroltattc Field in a Vacuum

39

Problem'

follo,,":s: under the action of the moment of force (1.41), the. dIpole tends to. get oriented along the field (p tt E), whIle under the action of the resultant force (1.39) it is

position, the moment of external forces will return the dipole to the equilibrium position.. Problems

• t.t. A very thin disc is uniformly charged with sut'fa?e char~ density lJ > O. Find the electric field intensity E on the axiS of thIS disc at the point from which the disc is seen at an angle .Q.

+q

E -q

Fig. 1.17

Fig. 1.18

£.

displaced towards the region where the field E has larger magnitude. Both these motions are simultaneous. The Energy of a Dipole in an External Field. We know that the energy of a point charge q in an external field is W. qcp, where cp is the fIeld potential at the point of locatIOn of the charge q. A dipole is the system of two charges and hence its energy in an external field is ' W = q+cp+ + q_cp_ = q (cp+ _ cp_), where CP+ and cp_ are the potentials of the external field at the points of location of the charges +q and -q. To within a quantity of the second order of smallness, we can write

Solu.io1l. It is clear from symmetry considerations th.at o~ the disc axis vector E must coincide with the direction of this aXIS (FIg. 1.19). Hence, it is safWentto find the component d!£z from the charg; of the area e1emeBt tiS at the point A and th~n mtegrate the o.btamed expression over the entire surface of lthe dISC. It can be easIly seen that' 1 lJdS dE z =-4- - I cos 'It. (1' JtEo

alp

cp+-cp-= az 1, ,,:here acp/Ol is the derivative of the potential in the directIOn of the vector l. According to (1.32), acp/al = -E It and henen '" + - '"_ ~ L E ,I from which we Il"l

Ell.

IW = -p.E.

Fig. 1.20

Fig. Lt9

r

In our case (dS cos 'It)/rz = dO is the solid angle at which the area element dS is seen from the point A, and expression (1) can be written as 1 dE z = -4- (J dO. JtEo

Hence, the required quantity is

(1.42)

It follows from this formula that the dipole has the minimUt;D. energy (Wmin = -pE) in the position p tt E (the positIOn of stable equilibrium). If it is displaced from this

1

E=-4- aQ. Jtto

It should he noted that at large distances from the disc, Q = SIrS, where S is the area of the disc and E = ql4nEors just as the field of the point charge q = lJS. In the immediate vicinity of the point O,the solid angle g = 2a and E = a/2eo'

40

Problems

1. Electrostatic Field in a Vacuum

• f .2. A thin nonconducting ring of radius R is charged with a linear density A = Ao cos cp, where '-0 is a positive eoDStant and cp is the azimuth angle. Find the electric field intensity E at the centre of the ring. Solution. The given charge distribution is shown in Fig. t,20. The symmetry of this distribution implies that vector E at the point 0 is directed to the right, and its magnitude is equal to the sum of the projections onto the direction of E of vectors dE from elementary charges dq. The projection of vector dE onto vector B Is

1 dq dE cos cp = -4- -R' cos cp,

ne o

This gives dEy = (A cos a da)/4ne oy and E y = A/4.tteoy·

We have obtained an interesting result: Ex = E y independently of y,

(1)

where dq = AR dlp = AoR cos cp dcp. IntegTating (1) over cp between 0 and 2n, we find the magnitude of the vector E:

"-0 R 4ne o

E

.=

a

2n

Jr cost

d

cp cp =

o

"-0

4e R • o

It should be noted that this Integral is evaluated in the most simple way if we take into account that (cost cp) = 1/2. Then 2n

~

Fig. 1.21 i.e. vector E is oriented at the angle of 45° to the filament. The modulus of vector E is E=

cos' cp dcp=(cos' cp) 2n=n.

o

• 1.3. A semi-infinite straight uniformly charged filament has a charge'" per unit length. Find the magnitude and the direction of the field Intensity at the point separated from the filament by a distanC'-t! y and lying on the normal to the filament, passing through its end. Solution. The prohlem is reduced to finding Ex and E., viz. the projections of vector E (Fig. L21, where it is assumed fliat '" > 0). Let us start with Ex. The contribution to Ex from the-charge element of the segment dz Is . 1 Adx. dE x =-4- -t-sma. (1) neo r j Let us reduce this expression to the form convenient for integration. In our case, dx = r dalcos a, T = ylcos a. Then "

V E~+E~=j,. Y2i4rt8 oY •

• 1.4. The Gauss theorem. The intensity of an electric field depends only on the coordinates x and y as follows: E = a (xi yj)/(x 2 y2),

+

q=80

dE x = - A 4 sin ex da.

neoy

Integrating this expression over a between 0 and n/2, we find Ex

= "'/4neoy·

In order to find the projectiou E y , it is sufficient to recall that dEli diffeq from dE~ in that sin a"ln (1) is simply replaced by C08 lx.

+

where a is a constant, and i and j are the unit vectors of the X-and Y;-axes. Find the charge within a sphere of radius' R with the centre at the origin. Solution. In accordance with the Gauss theorem, the required charge is equal to the flux of E through this sphere, divided by &0' In our case, we can determine the flux as follows. Since the field E is axisymmetric (as the field of a uniformly charged filament), we arrive at the conclusion that the flux through the sphere of radim(R is equal to the flux through the lateral surface of a cylinder having the same radius and the height 2R. and arranged as shown in Fig. 1.22. Then

where Er

=

~ E dS=8 oE rS,

aIR and S = 2nR·2R = 4nR2. Finally, we get q = 4n8 oaR.

• 1.5. A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density p = aIr, where a is a positive constant and r is the distance from

42

43

1. Electroirtattc Field in a Vacuum

Problem.

the centre of the sphere F' d th h electric field intensity E; o~~ide ~hc arghe of. th~ dsphere for which the the value of E. e sp ere IS In ependent of r. Find

of the spheres and of the distance between their centres. In particular, it is valid when one sphere is completely within the other or, in other words, when there is a spherical cavity in a sphere (Fig. 1.25). • 1.7. Using the solution of the previous problem, find the field intensity E inside the sphere over which a charge is distributed with

Solution. Let the sought charge of the sphere be q. Then, Using the

,I II· 1

I

lie

Iii

Iii I

Fig. 1.22 Fig. 1.23 Gauss theorem we can writ th f II . surf~ of radi~s r (outside :he :ph~r~~i~t ~h~r~h~ir~~ ~r a spherical r

E~ ·4n r 2 - _ q .t- 1 a - \. -4nr2dr. 80

£0. R

r

After integration, we transform this equation to 2 E ·4nr = (q ~ 2naR2)leo 4nar 2/28 0 • • The intensity E does t d d theses is equal to ze~~. J!rt~~ on r when the expression in the paren-

+

=

2naR2 and E = a/280 ' ~ f .6. Find the electric fi ld . t · . sectIon of two spheres unifo In enslty .Ii: In the region of intervolume densities p and _p th c3~rged by unlike charges with the spheres is determined by v~ctor le(F::~af.r:3)~etween the centres of the Solution. Using the Gauss th electric field intensity Within eo~efm, lwe can easilY.lshow that the a UD! orm Y charged spflere is E = (p/38 0 ) r, where r is the radius vector r I t' . can consider the field in the r e.a Ive .to the c~ntre of the sphere. We sUperposition of the fields of et~on of.;n\ersecthlOn of the spheres as the an arbitrary point A (Fig 1 24)0 \lfntlh~r my.c arged spheres. Then at " 0 IS regIOn we have q

WI

E = E+

+ JL =

P (r+ -

r_V3e n = pI/3e . n

Thus, in the region of intersectio f th h form. This conclusion is valid retpardle o f etse sp ~rebs the field is uni" ss 0 he ratio etween the radii

Fig. 1.25

Fig. 1.24

the surface density CT = CTo cos {Jo, where 0"0 is a constant and {Jo is the polar angle...' Solution. Let us consider two spheres of the same radius, having uniformly distributed volume charges with the densities p and -po Suppose that the centres of the spheres are separated by the distance 1 (Fig. 1.26). Then, in accordance with the solution of the previous problem, the field in the region of intersection of these spheres will be uniform: E = (p/380 ) 1. (1)

tz

In our case, the volume charge differs from zero only in the surface layer. For a very small l, we shall arrive at the concept of the surface charge density on the sphere. The thickness of the charged layer at the points determined by angle -& (Fig. 1.26) is equal to 1 cos {Jo. Hence, in this region the charge per unit area is (J = pl cos {Jo = 0"0 cos a, where °0 = pl, and expression (1) can be represented in the form E = -(0"0/380) k,

Fig. 1.26

where k is the unit vector of the Z-axis from which the angle it ii'J measured. • 1.8. Potential. Tho potential of a certain electric field has the form Ip = ex (xy - z2). Find the projection of vector E onto thp. direction of the vector a = i 3k at the point M (2, 1, -3). Solution. Let us first find vector E: E = -Vip = -a (yl xj - 2zk).

+

+

45

2.1. Field in a Substance

1. Electrodatlc Field In a Vacuam

where dq is the charge contained between the spheres of radii rand r + dr. Hence

The sought projection is E a =E . ..!..= -0: (yl-zj- 2lk)(i+3k). . a }f1+31 At the point M we have

E.

-0:~1t18)

1 , d r, r:dE r +2rE.r dr=-pr

-0: (y-61)

Eo

JfiO

p

• f.9. Find the potential lp at the edge of a thin df!e of l'Jdius R with a charge uniformly distributed over olle of its sides with the .... face density o. . S~lut~on. ~y definition, the potential in the case of a surface charge dl8trl~ution IS defined by integral (1.28). In order to simplify in. tegratlOD, we shall choose the area element dS in the fonn of a part of the ring of radius rand width!,dr (Fig. 1.27). Then dS = 2'&r dr, r = 2R cos '6-, and dr = -2R sin '6- t:H}. After substituting these expressions into integral (1.28), we obtain the expression for cp at &be point 0:

neo

Jo ~sin'6'd{;. n/2

We integrate by parts, denoting and sin t}- dt}- = dv:

~

=•

) 'It sin t}- dt}- = - '6- cos '6-

+ \ cos'6-dtJo=-'6-cos{}+sinil'.

Fig. 1.27

01

which gives :-1 after substituting the limits of integration. As a result, we obtam ~/ lp = oRfneo'

• 1.10. T.he potential of the field inside a charged sphere depends ?nly on the dI.stance r from its. Cl'ntre to the point under consideration I~ the foll?wI~g 'Yay: lp = ar' + b, where a and b are constants. Fmd the dIstrIbution of the volume charge p (r) within the 'sphere. Solution. Let us first find the field intensity. According to (1.32) we have • E r = -oq>/ar = -2ar. (1) I

Ii

I

T~en we US? th~ Gauss theorem: 4nr'Er = q/eo • The differential of thiS expressIOn IS 1 i 4n d(rIE r) = - dq = - p·4nr' dr, 66

eJ

ar

+2. E_' r r

__

...£... eo

Substituting (1) into the last equation, we obtain

J~o 0:.

lp=- oR

aE r

=-"

-flEoa,

i.e. the charge is distributed uniformly within the sphere. • 10ft. Dipole. Find the force of interaction between two point dipoles with moments PI and P2, if the vectors PI anti P2 are directed along the straight line connecting the dipoles and the distance between the dipoles is l. Solution. According to (1.39), we have F = PI I aElal I• where E is the field intensity from the dipole P2' determined by the first of formulas (1.38): E=_1_ 2P2 ...' 4nEo za Taking the derivative of this expression with respect to I and substituting it into the formull\. for fl, we obtain 1 _ 6PIP2 F __ - 41tEo l4' It should be noted that the dipoles will be attracted when PI t t P2 and repulsed when PI H P2'

2. A Conductor in an Electrostatic Field 2.1. Field in a Substance Micro- and Macroscopic Fields. The real electric field in any substance (which is called the microscopic field) varies abruptly both in space and in time. It is different at different points of atoms and in the interstices. In order to find the intensity E of a real field at a certain point at a given instant, we should sum up the intensities of the fields of all individual charged particles of the substance, viz. electrons and nuclei. The solution of this problem is obviously not feasible. In any case, the result would be so complicated

2.2. Fields Inside and Outside a Conductor

46

2. A

Co..a~ctor·tn

an Electrodatic Pield

that i~ would ~e impossibl~ to use it. Moreover, the knowledge of tillS field IS not reqUIred for the solution of macroscopic problems. In many cases it is sufficient to have a simpler and rougher description which we shall be using henceforth. Under the electric field E in a substance (which is called the rruu:roscopic field) we shall understand the microscopic field averaged over space (in this case time averaging is superfluous). This averaging is performed over what is called a physically infinitesimal volume, viz. the volume containing a large number of atoms and having the dimensions that are many times smaller than the dli!tances over which the macroscopic field noticeably changes. The averaging over such vo~umes smooth~ns all i.rregular and rapidly varying fluctuatlOns of the mIcroscopIC field over the distances of the order of atomic ones, but retains smooth variations of the macroscopic field over macroscopic distances. Thus, the field in the substance is E=Emacro=(Emlcro). (2.f) The Inftuence of a Substance on a Field. If any suhstance is introduced into an electric field, the positive and negative charges (nuclei and electrons) are displaced, which in turn leads to a parti.al separation· of these charges. In certain regions of the suhstance, uncompensated charges of different signs appear. This phenomenon is called the electrostatic . / induction, while the charges appearing as a result of separation are called induced charges. Induced charges create an additional electric field which in combination with the initial (external) field forms the resultant field. Knowing the external field and the distribution of induced charges, we can forget about the presence of the substance itself while calculating the reSultant iield since the role of the substance has already been taken int~ account with the help of induced dlarges Thus, the resultant field in the presence of a substance is determined simply as the superposition of the external Bald and the field of .wauced charges. However, in many cases the situation is complicated by the fact that we do not kaow beforehand how all these charges are distributed in space, and the problem turns out to be not as simple as it eooJd seem at first sight. It will be shown later that the distri-

41

bution of induced charges is mainly determined by the properties of the substance, i.e. its physical nature and the shape of the bodies. We shall have to consider these questions in greater detail. 2.2. Fields Inside and Outside a Conductor Inside a Conductor E = O. Let us place a metallic conductor into an external electrostatic field or impart a certain charge to it. In both cases, the electric field will act on all the ch2.l.'ges of the conductor, and as a result all the negative charges (electrons) will be displaced in the direction against the field. This displacement (current) will continue until (this practically takes a small fraction of a second) a certain charge distribution sets in, at which the electric fi.eld at a~l the points inside the conductor vanishes. Thus, in the statIC case the electric field inside a conductor is absent (E = 0). Fu.rther, siw;e E = 0 everyWhere in the conductor, the densIty of excess (uncompensated) charges inside the conductor is· also equal to zero at all points (p = 0). This can be easily explained with the help of the Gauss theorem. Indeed since inside the conductor E = 0, the flux of E through an; closed surface inside the conductor is also equal to zero. And this means that there are no excess charges inside the conductor. Excess charges appear only on the conductor surface with a certain density (J which is generally different for different points of the surface. It should he noted that the excess s~face charge is located in a very thin surface layer (whose thIckness amounts to one or two interatomic distances). The absence of a field inside a conductor indicates in accordance with (1.31), that potential cp in the cond~r has the same value for all its points, i.e. any conductor in an electrostatic field is an equipotential region, its surface being an equipotential surface. The fact that the surface of a conductor is equipotential implies that in the immediate vicinity of this surface the field E at each point is directed along the normal to the surface. If the opposite were true, the tangential component of E would make the charges move over the surface of the

2. A Condllctor in an Electrostatic Field

-------

49

conductor, Le. charge equilibrium would be impossible.

2.3. Forces Acting on the Surface of a Conductor

Example. Find the potential of an undlarged condudiug sphere provided that a point charge q is located at a distance r from its centre (Fig. 2.1). PotentialljJ is the same fur all points of the sphere. Thus we can calculate its value at the centre 0 of the sphere, r because only for this point it can be calculated in the most simple way:

shall take a small cylinder and arrange it as is shown in Fig. 2.3. Then the flux of E through this surface will be equal only to the flux through the "outer" endface of the cylinder- (the fluxes through the lateral surf~ce and the inner endface are equal to zero). Thus we obtam En /18 =

Ip=

1

q.,

4nco' r~t-fjl·

(1)

where the tJrst term is the potential of the charge q, while the second is the potential of the charges induced on the surface of the sphere.Dut since aU induced charges are at the same distance a from the point 0 and the total induced charge is equal to zero, q;'=O as well. Thus, in this case the putential \if the sphere will be determined only by the Erst term ill (1). Fig. 2.1

l"igure 2.2 shows the field and the charge distributions for a system consisting of two conducting spheres one of which (left) is charged. As a result of electric induction, the ch?rges of the opposite sign appear on the surface of the right uncharged sphere. The field _of these.eharges will in turn cause a redistribution of charges on the surface of the left sphere, and their surface distribution will become nonuniform. The solid lines in the figure are the lines of E, while the dashed lines show the intersection of tlquipotential surfa~s with the plane of the figure. As we move away from this...system, the equipotential. surfaces become closer_and closer to spherical, and the field lines become closer to radial. The field itself in this case resembles more and more the field of a point charge q, viz. the total charge of the given system. The Field Near aConductor Surface. We shall show that the electric fwld intensity in the immediate vicinity of the surface of a conductor is connected with the local charge density at the conductor surface through a simple relation. This relation can be established with the help of the Gauss theorem . . Suppose that the region of the conductor surface we are interested in borders on a vacuum. The field lines are normal to the conductor surface. Hence for a closed surface we

n

4'

Fig. 2.2 Fig. 2.3 = 0/18/80' where En' is ..the projection of (vector E onto the outward normal n (with respect to the conductor), /18 is"9the cross-sectional area of the cylinder and 0 is the local s~face charge density of the conductor. Cancelling both sides of this expression by /18. we get

I

En = a/so·

I

(2.2)

If 0 > 0, then En > 0, Le. vector E is directed from the conductor surface (coincides in direction with the normal n). If 0 < 0, then En < 0, and vector E is directed towards the conductor surface. Relation (2.2) may lead to the erroneous conclusion that the field E in the vicinity of a conductor depends only on the local charge density 0'. This is not so. The intensity E is determined by all the charges of the system under consideration as well as the value of a itself. 2.3. Forces Acting on the Surface of a Conductor Let us consider the case when a charged region of the surface of a conductor borders on a vacuum. The force acting on a small area· ~S of the conductor surface is ~F = at.S .E o, (2.3)

\ 50

2.4,., Properties

2. A Conductor in an Electrostatic Field

where 0 f!S ;is the charge of this element and Eo is the field created by all the other charges of the system "in the region where the charge 0 f!S is located. It should b~ noted at the very outset that Eo is not equal to the field intensity E in the vicinity of the given surface element of the cojductor, although there exists ~a certain relation between th~m. Let us find this relation, Le. express Eo through Be' Let E a be the intensity of the field created by the charge on. the area element f!S at the points that are very close to thIS element. In this region, it behaves as an infinite uniformly charged plane. Then, in accordance with (1.10), E a = = 0/ 28 0' The resultant field both inside and outside the conductor (near the area element f!S) is the superposition of the fields Eo and Eo. On both sides of the area element f!S the field E is p~actically ~he same, while the field Eo has opposite di~ ractIOns (see FIg. 2.4 where (for the sake of definiteness it is assumed that C1 > 0). From 'the condition E = 0 inside the conductor, it follows that Eo = Eo, and then outside the conductor, near its surface, E = Eo Eo = 2Eo. Thus, Eo = E/2, (2.4) and Eq. (2.3) becomes

+ 1

AF=T aAS·E. AS

(2.5)

Dividing both sides of this equation by f!S, we obtain the expression for the force acting on unit surface of a conductor: 1 Fu-T.aE.

(2.6)

Fig. 2.4 :-

We can write this expression in a different form since the quantities a and E appearing in it are interconnected. Indeed, in accordance with (2.2), En = a/eo, or E = (a/eo) n, where n is the outward normal to the surface element at a given point of the conductor. Hence

(2.7)

of a Closed Conducting Shell

5i 2

where we took into account that 0 = eoEn \ and E; = E • The quantity F u is called the surface density of force. Equation (2.7) shows that regardless of the sign of 0, and hence of the direction of E, the force Fu is always directed outside the conductor, tending to stretch it: Example. Find the expression for the electric force acting in a vacuum on a conductor as a whole, assuming that the field intensities E are known at all points in the vicinity of the conductor surface. Multiplying (2.7) by dS, we obtain the expression for the force dF acting on the surface element dS:

dF=+ eo£2 dS, where dS = n dS. The resultant force acting on the entire conductor can be found by integrating this equation over the entire conductor lIUliace:

2.4. Properties of a Closed Conduding Shell

It was sho.wn that in equilibrium there.are no excess charges inside a conductor, viz. the material inside the conductor is electrically neutral. Consequently ~ if the substance is removed from a certain volume inside a conductor (a closed cavity is created), this does not change the field anywhere, Le. does not affect the equilibrium distribution of charges. This means that the excess charge is distributed on a conductor with a cavity in the same way as on a uniform conductor, viz. on its outer surface. Thus, in the absence of electric charges within the cavity the electric field is equal to zero in it. External charges, including the charges on the outer surface of the conductor, do not create any electric field in the cavity inside the conductor.

This forms the basis of electrostatic shielding, Le. the screening of bodies, e.g. measuring instruments, from the influence of external electrostatic fields. In practice, a solid conducting shell can be replaced by a sufficiently dense metallic grating. That there is no electric field inside an empty cavity can be proved in a different way. Let us take a closed surface S enveloping the cavity and lying completely in the material 4*

52

2. A Conductor in an Electrostatic Field

2.4. Properties Df a Closed Conducting Shell

of the conductor. Since the field E is equal to zero inside the conductor, the flux of E through the turface S is also equal to zero. Hence, in accordance with the Gauss theorem, the total charge inside S is equal to zero as well. This does not exclude the situation depicted in Fig 2.5, when the surf/tge

space outside the cavity, the field created by the charges located inside the cavity. Since the conducting medium is electrically neutral everywhere it does not influence the electric field in any way. Therefore, if we remove the medium, leavi.ng only a conducting shell around the cavity, the field WIll not be ~hanged anywhere and will remain equal to zero beyond thIS shell. Thus the field of the charges surrounded by a conducting shell a~d of the charges induced on the surface of the cavity (on the inner surface of the shell) is equal to zero in the entire outer space. We arrive at the following important conclusion: a closed conducting shell divides the entire space into the inner .and outer parts which are completely independent oj one another in respect of electric fields. This must be i~te:preted as follows: any arbitrary displacement of charges InSIde the shell does not introduce any change in the field of the outer space, and hence the ch.arge distribution on the outer surface of the shell remains unchanged. The same refers to the field inside the cavity (if it contains cbarges) and to the ~istribution ?f charges induced on the cavity walls. They WIll also remam unchanged upon the displacement of charges outside the shell. Naturally, the above arguments are applicable only in the framework of electrostatics.

p

r

Fig. 2.5

Fig. 2.6

of the cavity itself contains equal quantities of positive and negative charges. However, this assumption is prohibited by another theorem, viz. the theorem on circulation· of vector E. Indeed, let the contour r cross the cavity along one of the lines of E and be closed in the conductor material. It is clear that the line integral of vector E along this contour differs from zero, which is in contradiction with the theorem on circulation. Let us now consider the case when the cavity is not empty but contains a certain electric charge g (or several charges). Suppose also that the entire external space is filled by a conducting medium. In equilibrium, the field in this medium is equal to zero, which means that the medium is electrically neutral and contains no excess charges. Since E = 0 inside the conductor, the field flux through a closed surface surrounding the cavity is also equal to zero. According to the Gauss theorem, this means that the algebraic sum of the charges within this closed surface is equal to zero as well. Thus, the algebraic sum of the charges induced on the cavity surface is equal in magnitude and opposite in sign to the algebraic sum of the charges inside the cavity. In equilibrium the charges induced on the surface of the cavity are\arranged so as to compensate compleLely, in the

53

Example. A point charge q is within an electrically neutral sh~ll whose outer surface has spherical shape (Fig. 2.6). Find the potential cp at the point P lying outside the shell at a distance r from the centre 0 of the outer surface. The field at the point P is determined only by charges induced on . the outer spherical surface since, a~ was shown ab«;lve, the field of the point charge q and of the charges mdu.ced on the. mner surf~ee ?f the sphere is equal to zero everywhere outsIde the caVIty. Next, III VIew of symmetry, the charge on the outer surface of the shell is distributed uniformly, and hence 1 q qJ =

4n: eo -,:-.

An infinite conducting plane is a special case of a closed conducting shell. The -space on one side of this plane is .electrically independent of the space on its other side. We shall repeatedly use this property of a closed conducting shell.

2.5. General Problem of Electrostatilcs

2.5. General Problem of Electrostatics.

static case the charge is distributed over thlle surface of a conductor in a unique way as well. Indeed, th'Iere is a one-toone correspondence (2.2) between the charges; on the conductor and the electric field in the vicinity of itts surface: (J =

Image Method

d . ~r~~ue~tlY~ we must solve problems in which the charge t~:.rl ~tJon IS . unknown but the potentials of conductors Jr s ape an d relative arrangement are given We must find th~ pot~ntial cp (r) at any point of the field between the ~~a~d uc ors. t should be recalled that if we know the potend th (r),. the field E (r) itself can be easily reconstructed an f en Its value in the immediate vicinity of the condu~tor s'!r a~es .can be used for determining the surface ch dlstrlbutlOll for the conductors. arge The Poisson and I aplac t· L . equation for the functi e equa .lOns. et ~s derIve the differential into the left-hand sid~~( /r~t)n~hal). For t~IS p¥rpose: we substitute i.e. E = -v
.

ell

e

I

Olsson equation:

V2
7thh:~ r~eifo~ Lapl~ce

(2.8)

operator (Laplacian). In Cartesian coordinates

V2=~ 2

1:\

55

2. A Conductor in an El~ctrostatic Field

ax

a +~ ay + az' 2

2

'

i.e. the scalar product v·v [see (1.19)). is tran~f~:ma:d ~~tcharg~s beltween th!! cond.uctors (p = 0), Eq. (2.8) o a SImp er equatIOn, VIZ. the Laplace equation:

I

V'

O.

I

(2.9)

Eq.S:(2~h8)t~~(4~~ f~\h~t~~ii;;:s~a~~t ~:t~:e~u~h~i~~n~~~ :~~sfies

qUIres t e given values
It can ~e prove? theoretically that this problem has a unique solutIOn. ThiS statement is called the uniqueness theorerr;. Fro~ the physical point of view, this conclusion is q'!Ite obvIOUS: If there are more than one solution there wIll be .several potential "reliefs", and hence the field E at each pom~ generally has not a single value. Thus we arrive at a phYSically absurd conclusion. Usm~ the uniqueness theorem, we can state that in a

(a)

(b)

(e) l

Fig. 2.7 = EllEn. Hence it immediately follows that the uniqueness of the field E determines the uniqueness of the charge distribution over the conductor surface.

The solution of Eqs. (2.8) aM (2.9) in the generail case is a complicated and laborious ·problem. The analytic solutions i of these equations were obtained only for a few particular cases. As ffor the uniqueness theorem, it simplifies the solution of a number of «electrostatic problems. If a solution of the problem satisfies the Lalplace (or Poisson) equation and the boundary conditions, we can stat6e that it is correct and unique regardless of the methods by which itt was obtained (if only by guess). Example. Prove that in an empty cavity of a cconductor the field is absent. The potential

d that makes it possible to calculate in a simple way the ,electric field in some (unfortunately few) cases. Let us consiider this method by using a simple example of a point charge ~q near an infinite conducting plane (Fig. 2.7a). , The idea of this method lies in that we mmst find another problem which can be easily solved and whwse solution or a part of it can be used in our problem. In ourr case such a simple problem is the problem about two changes: q and -g.

---56

2. A Conductor in an Electrostatic Field

2.6. C.pacUance. Capacitors

57

The field of this system is well known (its equipotential surfaces and field lines are shown in Fig. 2.7b). Let us make the conducting plane coincide with the middle equipotential surface (its potential qJ = 0) and remove the charge --g. According to the uniqueness theorem, the field in the upper half-space will remain unchanged. Indeed, qJ = 0 on the conducting plane and everywhere at infinity. The point charge g can be considered to be the limiting case of a small spherical conductor whose radius tends to zero and potential to infinity. Thus, the boundary conditions for the potential in the upper half-space remain the same, and hence the field in this region is also the same (Fig. 2.7c). It should he noted that we can arrive at this conclusion proceeding from the properties of a closed conducting shell [see Sec.2.1J. since hoth half-spaces separated by the conducting plane are electrically independent of one another, and the removal of the charge -g will not affect the field in the upper half-space. Thus, in the case under consideration the field differs from zero only in the upper half-space. In order to calculate this field, it is sufficient to introduce a fictitious image charge g' = -g, opposite in sign to the charge q, by placing it on the other side of the conducting plane at the same distance as the distance from g to the plane. The fictitious charge g' creates in the upper half-space the same field as that of tlie charges induced on the plane. This is precisely what is meant when we say that the fictitious charge produces the same "effect" as all the induced charges. We must only bear in mind that the "effect" of the fictitious charge extends only to the half-space where the real charge q is located,. In another half-space the field is absent. Summing up, we can say that the image method is essentially based on driving the potential to the boundary conditions, I.e. we strive to find another problem (eonfignration of charges) in which the field configuration in the region of space we are interested in is the same. The image method proves to be very effective if this·can be done with the help of sufficiently simple configurations. Let liS consider one more example.

charges whose action on the charge q,is eq~ivalent to the action of all

Example. A point charge q is placed between two mutually perpendicular half-planes (Fig. 2.8a). Find the location of fictitious point

It can he seen from formula (2.10) that for this jlurpose. we must mentally chargre the conductor by a charge q and calculate Its potell-

cha~~\\~~deu~~dii~3:~~~etc~;l~l~~r::'chargesfor which the eqnipoten-

tial ~ul'face~ with cp = 0 wo~J!d (a) (~) coinCide With the cond~c!Illg half-planes. One or two fI~tltIO~StL'q f charges are insuflicient III tillS I case' there should be three of -q I q the~ (Fig. 2.8b). Only with such a configuration of the ~:ystem of fOl~r I charges we can realize the ft.--l--quired "trimming"', i.e. ensure 7, j that the potential on the conI dueling" half-plane~ ~e equal to q b---lf---<>- q zero. These three hctl tlOllS cl.Jarges create just the same held within the "right angle" as the Fig. 2.8 field of the charges induced on

con~ucting plah.es. n alion of point chal'gl's {anolher proble~n), HavIll\r.found I IS con 19l1b f tlH'r questions, for example, lmd ili'e ~~~e~t~~{a~ldsfi~f/i~~~~~i~; ~n ~ny point within the "right angle" or determine the fw-ce acting Oil the charge q. the

2.6. CapaCitance. Capacitors Capacitance of an Isolated Conductor. Let I~S fconSid~r a solitarv conductor, i.e. the conductor .remove( r~m 0 101' . 1)0 d'Ies". and charge" show that the conductors, , , , . Expenments . . 1 't cl·r.r e of this conductor is directly propor~lOJ.la to I S p~~e~tia\

58

2. A Conductor in an Electro8tattc Field

2.6. Capacitance. Capacitors

tial
J'


00

1 q 1 q Erdr=---2 dr = - - - . 4nfo . r 4n8 o R R R

I

Substituting this result into (2.10), we find

C

=

4m oR.

. the charges on the plates must be equal in magnitude I.e. ( d ) and opposite in sign ~ :.n ~apacitor is its capacitance. The basic charactens lC 0 . d onductor the capaciUnlike the capacitance of an Isolate c t' f i'ts charge to 'tor is denned as t lle ra 10 0 • tance 0 f a. cap~cl b t n the plates (lhis difference 18 the potentIal difference e wee called the voltage):

f!

f

00

59

(2.U)

The unit 'ofcapacitance is the capacitance of a conductor whose potential changes by 1 V when a charge of 1 C is supplied to it. This unit of capacitance is called the farad (F). The farad is a very large quantity. It corresponds to the capacitance of an isolated sphere 9 X 106 km in radius, which is 1500 times the radius of the Earth (the capacitance of the Earth is 0.7 mF). In actual practice, we encounter capacitances between 1 ~F and 1 pF. Capacitors. If a conductor is not isolated, its capacitance will considerably increase as other bodies approach it. This is due to the fact Jhat the field of the given conductor causes a redistribution of charges on the surrounding bodies, Le. induces charges on them. Let the charge of the conductor be g > O. Then negative induced charges will be nearer to the conductor than the positive charges. For this reason, the potential of the conductor, which is the algebraic sum of the potentials of its own charge and of the charges induced on other bodies will decrease when other uncharged bodies approach it. This means that its capacitance increases. This circumstance made it possible to create the system of conductors, which has a considerably higher capacitance than that of an isolated conductor. Moreover, the capacitance of this system does not depend on surrounding bodies. Such a system is called a capacitor. The simplest capacitor consists' of two eonductors (plates) separated by a small distance. In order to exclude the effect of external bodies on the capacitance of a eapacitor, its plates are arranged with respeet to one another in such a way that the field created by the charges accumulated on them is concentrated almost completely inside the capaeHor. This means that the lines of E emerging on one plate must terminate on the other,

(2.12) C=qlU. \ \ The charge g ofa capacitor is the charge of its positively charged' plate. ., f a capacitor is also measured Naturally, the capaCitance 0 , in farads. " t depends on its O'eometry The capacitance of a capaCl or a between th~ plates, (size and shape. of i~S pl~\~s)th~leC;p~citor. Let us derive and the m~teqalf t ~~e capacitances of some capacitors the expreSSIOns or. cuum between their plates. assuming that there is a I I te Capacitor. This capacitor Capacitance of a Para e -p a d b a ap of width h. consists of two parallel pl~ttes ~epar~~:n :cco~ding to (1.11), If the charge of the capaCi or IS g, . '1 . E _ ale . h fi Id between Its pates IS 0' the intensity of t e .e h f each plate. Consequentwhere a = glS and S IS t e area 0 . ly, the voltage between the plates IS

rr

U

=

Eh

=

ghlfoS.

Substituting this expression into (2.12), we obtain (2.13) C = foS Ih . d 'thout taking into account This calculation was rna e WI lates (edge effects). field disto~tions n~ar thel e~~~e °1a~~~lor is determined ~Y The capaCItance 0 a rea p tely the smaller the gap h III this formula the more. accura. nsions of the plates. compari~on with th; l~ne~r t~~acitor. Let the radii of thE CapaCItance of a p. enca es be a and b respectively. II inner and outer capaclt?r pl.at field intensity between thE the charge of the capaCItor. IS g, . plates is determined by the Gauss theorem. 1

Er = into

q

rt·

\'

60

2. A Conductor in an Electrostatic Field Problems

The voltage of the capacitor is charges on the outer surface. It follows from this expression that 1 _ _q_ cP - 4n80 r a b

u = J"r E r dr = _q_ ( ..!.. __ 1 ). 4n80 a b a

(..i.. __ +..!..).

It can be easily seen that the capacitance of a spherical capacitor is given by C=4Jt£o ~ b-a .

(2.14)

It is interesting that when the gap between the plates is small, Le. when b - a «a (or b), this expression is reduced to (2.13), viz. the expression for the capacitance of a parallel-plate capacitor. Capacitance of a Cylindrical Capacitor. By using the same line of reasoning as in the case of a spherical capacitor, we obtain ol c-·- In2n8 (b/a)

(2.15)

,

where l is the capacitor's length, a and b are the radii of the inner and outer cylindrical plates. Like in the previous case, the obtained expression is reduced to (2.13) when the gap between the plates is small. The influence of the medium on the capacitance of a capacitor will be discussed in Sec. 3.7. Problems • 2.1. On the determination or potential. A point charge q is at a distance r from the centre 0 of an uncharged spherical conducting layer, whose inner and outer radii are equal to a and b respectively. Find the potential at the point 0 if r < a. Solution. As a result of electrostatic induction, say, negative charges will be induced on the inner surface of the layer and positive charges on its out{~r surface (Fig. 2.9). According to the principle of superposition, the sought potential at the point 0 can be represented in the form q-J - -1 -4rtE o

(q- - -t- \.~' - -dST, '1) - -dS- ) 0_

r

•

0+

a

:.

b

'

where the first integ-ral is taken over all the charges induced on the inner surface of the layer, while the second integral, over all the

a

It should be noted that the potential.can. be found in such ~f~e form only at the poin~ 0 s~nce all the. like mduced charges are a e same distance from this pomt and theIr distribution (which is unknown to us) does .not play any role.

• 2.2. A system consists of two concentric spheres, the inner sphere of radius R l having a charge ql' What charge q must be placed onto the outer spher~ of radius R 2 to make the potential of the inner sphere equal to zero? What will be the dependence of potential cP on the distance r from the centre of the system? Plot schematica~ly the graph of this dependence, assummg that ql < o. 4.' 'Fig. 2.9 Solution. We write the expressions. . h for potentials outside the slstem (CPn) and In the regIOn between t e spheres (CPI): 1 ql +q2 1 ql CPn=-- - - - , CPI=-4-- - - CPo, 4n80 r n80 r

+

where CPo is a c~r~ain constant. Its val~ can be easily found from the boundary conditIOn: for r = R 2 , CPn - CPl, Hence lfu = q2/4ncoR2' From the condition
ql

1-R 2/R J r

,CPr

=.3..L 4n80

(_1r

=

-ql R 2 IR l •

~).

R1

• 2.3. The force acting on a surface charge. An. uncharged metallic sphere of radius R is placed into an external umform. field'Js a result of which an induced charge appears on the sphere Wit~ su ice density = 00 cos 'ft, where 00 is a positive consta.nt an~ ~ IS ~ po ar angle. Find the magnitude of the resultant electnc force actmg on like charges. Solution. According to (2.5), the force acting on the area element dS is

°

dF=-taEdS. .

'"

(1)

It follows from symmetry considerations that t~e resultant force f is directed along the Z-axis (Fig. 2.11), and henc0 it can be represented

2. A Conductor in an Electrostatic field

Problem.

as the sum (integral) of the projections of elementary forces (1) onto the Z-axis: dF z = dF cos {}. (2)

where the minus sign indicates that the induced charge is opposite to sign to the point charge q.

It is expedient to take for the area element dS a spherical zone dS =

• 2.5. A point charge q is at a distance 1 from an infinite conducting plane. Find the work of the electric force acting on the charge q done upon its slow removal to a very large distance from the plane. f

1

o F

-q

Eq

I

,

o Fig. 2.10

= 2nR sin t}·R dt}. Considering that E = a/eo, we transform (2) as follows: clFz = (na I R2/ eo ) sin t} cos {} d{} = - (na~R2/po) cos3 {} d (cos t}). Integrating this expression over the half-sphere (i.e. with respect to cos if between 1 and 0), we obtain F = na8R2/4eo' • 2.4. Image method. A point charge q is at a distance 1 from an infinite conducting plane. Find the density of surface charges induced on the plane as a function of the distance r from the base of the perpendicular dropped from the charge q onto the plane. ~olutton. Accord.ing to (2.2), the :surface charge:density on a conductor IS connected .wIth the electric field near its surface (in vacuum) through the relatIOn a = eoEn' Consequently, the problem is reduced to determining the field E in the vicinity of the conducting plane. Using the image method, we find that the field at the point P (Fig. 2.12) which is at a distance r from the point 0 is

Hence

dx

x

Fig. 2.13

Fig. 2.12 Fig. 2.11

q 1 E=2Eqcosa=2 - - - - 4neox2 x '

x

q

Solution. By definition, the work of this force done upon an elementary displacement dx (F;.ig. 2.13) is given by ql 6A = F % dx = - 4neo (2%)1 dx,

where the expression for the force is obtained with the help of the image method. Integrating this equation over x between land 00, we find

r

00

ql

A= - 16neo

dx

J 7= I

ql

16ne ol •

Remark. An attempt to solve this problem in a different way (through potential) leads to an erroneous result ~h!ch differs from w~at was obtained by us by a factor of two. ThIS IS because the re.latIOn A = q (IJ>l -
• 2.6. A thin conducting ring of radius R, having a charge q, is arranged 80 that it is parallel to an infinite conducting plane at a ~is­ tance I, from it. Find (1) the surface charge density at a point of the plane, which is symmetric with respect to the ring and (2) the electric field potential at the centre of the ring. . Solutton. It can be easily seen that in accordance with the image method, a fictitious charge -q must be located on a similar ring but

65

Problems

2. A Conductor in an Electrostatic Field

on the other side of the conducting plane (Fig. 2.14). Indeed, only in this case the potential of the midplane between these rings is equal to zero, Le. it coincides with the potential of the conducting plane. Let us now use the formulas we already know. (1) In order to find {J at the point 0, we must, according to (2.2), find the field E at this point (Fig. 2.14). The expression for E on the R

the required force (see Fig. 2.15b) 2

V 2-

1

q2

F =c F 2 - F 1= ---'74- 2a 2 neo

,

and answer the second question. • 2.8. Capacitance of parallel wir~s. :rwo long straight wires with the same cross section are arranged In aIr parallel to one another.

1 2

1I

I

A

-qC---l;---~

........ _----

(8)

B

(b) Fig. 2.15

Fig. 2.14

Fig. 2.17

Fig. 2.16 axis of a ring was obtained in Example 1 (see p. 14). In our case, this expression must he doubled. As a result, we obtain ql {J=

21t (RI+12)'/1

(2) The potential at the centre of the ring is equal to the algebraic

sum of the potentials at this point created by the charges q and -q: ljl=

1 41t8o

(q if

q) yR2+411 •

• 2.7. Three unlike point charges are arranged as shown in Fig. 2.15a, where AOB is the right angle formed by two conducting half-planes. The magnitude of each of the charges is I q I· and the distances between them are shown in the figure. Find (1) the total charge induced on the conducting half-planes and (2) the force acting on the charge -q. Solution. The half-planeD forming the angle AGE go to infinity, and hence their potential q; = O. It can be easily seen that a ·~Ylltem haVing equipotential surfaces with ljl = 0 coinciding with the conducting half-planes has the form sho1inl in Fig. 2.15b. Hence the action of the charges induced on the conducting hali-planes Is equivalent to the action of the fictitious charge -q placed in the lower left comer of the dashed sauare. Thus we have already answered the first question: -qe By reducing the syst0m to iour point ch:trges, we can easily find

The distance between the wires is 11 times larger th~n the radi~s of the wires' cross section. Find the capacitance of the WIres per umt length provided that 11:;;}> 1. Solution. Let us mentally charge the two wires by charg~s of the same magnitude and opposite signs so that t~e charge per um.t length is equal to "'. Then, by defmition, the reqUIred capaCitance is Cu

=

",IU,

(1)

and it remains for us to find the potential difference between the wi~es. It follows from Fig. 2.16 showing the dependences of the potentl~ls lp + and Ip_ on the distance between the plates that the sought potential difference is

u= ILlcp+1 + ILlCP-l =2jLl(jl+I·

(2)

The intensity of the electric field created by one of the wires at a distance x from its axis can be easily found wi th ,the help of the Gauss theorem: E = "'/2ne ox. Then b-a

I 8(jl+ I =

J\

'" b-a E dx = 2neo In --a- ,

(3 )

a

where a is the radius of the wires' cross section and b is the separation between the axes of the wires. 5-0181

.,

!i 'I'

A Conductur ------tili

'I' 1,1., .

'I' i I

!

!

2.

3.1. Polarization

in an Electroltatlc Field

difference between them is equal to zero. Consequently,

It follows from (1), (2) and (3) .nat

E 1X l 1

Cu = neo/ln 1').

where we took into account that b :it> a. • 2.9. Four identical metallic plates are arranged in air at the same distance h from each other. The outer plates are connected by a conductor. The area of each plate is equal to S. Find the capacitance of this system (between points 1 and 2, Fig. 2.17). Solution. Let us charge the plates 1 and 2 by charges qo and -qo. Under the action of the dissipation field appearing between these plates (edge effect), a charge will move in the connecting wire, after which the plate A will be charged negatively while the plate B will acquire a positive charge. An electric field appears in the gaps between the plates, accompanied by the corresponding distribution of potential cp (Fig. 2.18). It should be noted that as follows from the symmetry of the system, the potentials at the Imiddle of the system as well as on its outer plates are equal to zero. By defini tion, the capacitance of the system in this case is C = qo/V

E = 2E'.

E 2x (l -

= 0,

11)

where E 1x and E x are the projections of ve.ctor E onto the X-axis to the left and to tte right of the plane P (Fig. 2.19b). On the other hand, it is clear that a

=

-(a1

+ a.),

where, in accordance with (2.2), a 1 = eOE1n1 = eOE 1X and a2 = e E 2 = -eOE 2X (the minus sign indicates that the normal o n, . .) is directed oppositely to the umt vector 0 f t.he X,-axIs. . Eliminating E!X and E 2x from these equatIOns, we obtam a1

=

-a (I -

11)/1,

a2

=

= U2

-al 1/l·

The formulas for charges q1 and q2 in terms ~f q have a si~ilar form. It would be difficult, however, to solve thiS problem WIth the help p

-

r------

r-r-A

-=01" _02

B

x

~

r-1 r-

a

;r

(2)

And since E ex a, we can state that according to (2) the charge qo on plate 1 is divided into two parts: qo/3 on the left side of the plate 1 and 2qo/3 on its right side. Hence V = Eh = ahlE o = 2qoh/3e oS, and the capacitance of the system (between points 1 and 2) is 2h

+

. (1)

where U is the required potential difference between the points 1 and 2. Figure 2.18 shows that the putential difference V between the inner plates is twice as large as the potential difference between the outside pair of plates (both on the right and on the left). This also refers to the field intensity:

C= 3eoS

67

0/ Dielectrics

(b)

(a)

Fig. 2.1st=

Fig. 2.19

of the image method, since it would require an infmite series of fIctitious charges arranged on both sides of the charge q, and to find the field of such a system is a complicated problem.

•

• 2.10. Distribution 01 an induced charge. A point charge q is placed between two large parallel conducting plates 1 and 2 separated by a distance 1. Find the tutal charges q1 and q2 induced on each plate, if the plates are connected by a wire and the charge q is located at a distance 11 from the left plate 1 (Fig. 2.19a). Solution. Let us use the superposition principle. We mentally place somewhere on a plane P the same charge q. Clearly, this will double the surface charge on ~ach plate. If we now distribute uniformly on the surface P a certain cnarge with surface density a, the electric field can be easily calculated (Fig. 2.19b). The plates are connected by the wire, and hence the potential

3. Electric Field in Dielectrics 3.1. Polarization of Dielectrics Dielectrics. Dielectrics (or insulators) are substances that practically do not conduct electric current.. This ~neans that in contrast, for example, to conductors dielectriCS do not contain charcrcs that can move over considerable distances t> and create elcctric current.

~

'

,"~i)

69

3.1. Polarization of Dielectrics

68

3. Electric Field in Dielectrics

When even a neutral dielectric is introduced into an external electric field, appreciable changes are observed in the field and in the dielectric itself. This is because the dielec~ric is acted up?n. by a force, the capacitance of a capacitor Increases when It IS filled by a dielectric, and so on. In order to understand the nature of these phenomena we must take into consideration that dielectrics consist either of neutral molecules or of charged ions located at the sites . of a crystal lattice (ionic crystals, for example, of the NaCI type). The molecules can be either polar or nonpolar. In a polar molecule, the centre of "mass" of the negative charge is displaced relative to the centre of "mass" of the positive charge. As a result, the molecule acquires an intrinsic dipole moment p. Nonpolar molecules do not have intrinsic dipole mo~ents, since the "centres of mass" of the positive and negatIve charges in them coincide. Polarization. Under the action of an external electric field, dielectric is polarized. This phenomenon consists in the following. If a dielectric is made up by nonpolar molecules, the positive cll~rge. in each molecule is shifted along the field and the negatIve, In the opposite direction. If a dielectric consists of polar molecules, then in the absence of the field their d~pole moments are oriented at random (due to thermal motIOn). L nder the action of an external field the dipole moments acquire predominant orientation in' the direction of the external field. Finally, in dielectric crystals ?f the NaCI ty~e, an external field displaces all the positive IOns along the field and the negative ions, against the field. * Thus, the m~chanism of polarization depends on the structure of a dielectric. For further discussion it is only important that regardless of the polarization mechanism all the positive ~harges duri~g this process are displaced aiong the field, whIle the negatIve charges, against the field. It should be noted that under normal conditions the displacements of charges are very small even in comparison with the dimensions of the molecules. This is due to the fact that the intensity of the external field acting on the dielectric is consider* There exist ionic crystals polarized even in the absence of an external field. This property is inherent in dielectrics which are called electrets (they resemble permanent magnets).

ably lower than the intensities of internal electric fields in the molecules. Bulk and Surface Bound Charges. As a result of polarization, uncompensated charges appear on the dielectric surf!"ce as well as in its bulk. To understand better the mechamsm £=0 p~

7~

.!.-

~I

I

I

I I I

I I

p~

,

It I p,+ I

I:

I

I

I. I.

X

X

X

(a)

p~

(b)

(c)

Fig. 3.1 A.~

of emergence of these charges (and especially bulk charges), let us consider the following model: Suppose that we have a plate made of a neutral inhomogeneous dielectric (Fig. 3.~a) whose density increases with the coordinate x accordIng to a certain law. We denote by p~ and p~ the magnitudes of the volume densities of the positive and negative charges in the material (these charges are associated with:nuclei and electrons). In the absence of an external field, p~ = p~ at each point of the dielectric, since the dielectric is electrically neutral. However, p~ as well as p~ increase with x due to inhomog~­ neity of the dielectric (Fig. 3.1b). This figure shows that III , the absence of external field, these two distributions exactly coincide (the distribution of p~(x) is shown by the solid line, while that of p~ (x), by the dashed line). Switching on of the external field leads to a displacement of the positive charges along the field and of the negative charges against the field, and the two distributions will be shifted relative to one another (Fig. 3.1c). As a result, uncompensated charges will appear on the dielectric surface as well as in the bulk (in Fig. 3.1 an uncompensated negative charge appears in the bulk). It should be noted that the re-

3.2. Polarization 70

3. Electric Field in Dielectric,

version of the field direction changes the sign of all these charges. It can be easily seen that in the case of a plate made of a homogeneous dielectric, the distributions p' (x) and p~ (x) would be IT-shaped, and only uncompensated surface charges would appear upon their relative displacement in the field E. Uncompensated charges appearing as a result of polarization of a dielectric are called polarization, or bound, charges. The latter term emphasizes that the displacements of these charges are limited. They can move only within electrically n~utr~l m?lecules. We shall denote bound charges by a prime (q , p , (J). Thus, in the general case the polarization of a dielectric !ea~s to the appearance of surface and bulk bound charges III It.

We shall call the charges that do not constitute dielectric molecules the extraneous charges. * These charges may be located both inside and outside the dielectric. The Field in a Dielectric. The field E in a dielectric is the term applied to the superposition of the field E of extraneous charges and the field E' of bound charges: 0 E = Eo + E', (3.1) where Eo and E' are macroscopic fields, Le. the microscopic fields of extraneous and bound charges, averaged over a physically infinitesimal volume. Clearly, the field E in the dielectric defined in this way is also a macroscopic field. 3.2. Polarization

Definition. It is natural to describe polarization of a dielectric with the help of the dipole moment of a unit volume. If an .ext~rnal field or a dielectric (or both) are nonuniform, polarl.zatlOn. turns out to be different at different points of t~e dlel~ctrlc. In order to characterize the polarization at a given POIllt, we must mentally isolate an infinitesimal volume !'1 V containing this point and then find the vector sum of the • ~xtraneous c~arg~s are frequently called free charges, but this term IS not convement In some cases since extraneous charges may be not free.

,

I

'1 1,1

'1'

i i!

','II

7f

dipole moments of the molecules in this volume and write the ratio

(3.2) Vector P defined in this way is called the polarization of a dielectric. This vector is numerically equal to the dipole moment of a unit volume of the substance. There are two more useful representations of vector P. Let a volume !'1 V contain !'1N dipoles. We multiply and divide the right-hand side of (3.2) by !'1N. Then we can write P = n (p), (3.3) where n = !'1NI!'1V is the concentration of molecules (their number in a unit volume) and (p) = ('T.PI)!!'1N is the mean dipole moment of a molecule. Another expression for P corresponds to the model of a dielectric as a mixture of positive and negative "fluids". Let us isolate a very small volume !'1 V inside the dielectric. Upon polarization, the positive charge p+!'1 V contai~ed in this volume will he displaced relative to the negative charge by a distance I, and these charges will acquire the dipole moment !'1p = p~ !'1 V·1. Dividing both sides of this formula by !'1 V, we obtain the expression for the dipole moment of a unit volume, i.e. vector P: P = pi.!. (3.4) The unit of polarization P is the coulomb per square meter (elm'). Relation Between P and E. Experiments show that for a large number of dielectrics and a broad class of phenomena, 'Polarization P linearly depends on the field E in a dielectric. For an isotropic dielectric and for not very large E, there exists a relation P = 'X8 oE J (3.5) where 'X is a dimensionless quantity called the dielectric susceptibility of a substance. This quantity is independent of E and characterizes the properties of the dielectric itself. 'X is alwavs greater than zero. Hencef~rth, if the opposite is not stipulated, we shall

73

3. Electric Field in Dielectrics

3.3. Properties 0/ the Field 0/ P

consider only isotropic dielectrics for which relation (3.5) is valid. H~wever, there exist dielectrics for which (3.5) is not applIcable. These are some ionic crystals (see footnote on page oR) and ferroeZectrics. The relation between P and E for ferr~electr~cs .is nonlinear and depends on the history of the dIeI?ctnc, l.e. on the previous values of (this'phenomenon IS called hysteresIs).

of the positive and negative bound charges as a result of polarization. Then it is clear that the positive charge p:Z+dS cos a inclosed in the "inner"part of the oblique cylinder will pass through the area element dS from the surface S outwards (Fig. 3.2b). Besides, the negative charge p:tdS cos a enclosed in the "outer" part of the oblique cylinder will enter the surface S through the area element dS. But be know that the transport of a negative charge in a certain direction is equivalent to the transport of the positive charge in the opposite direction. Taking this into account, we can write the expression for the total bound charge passing through the. area element dS of the surfa~e S in the outward direction: dq' = p:Z+ dS cos a + Ip:IL dS cos a.

72

E;

3.3. Properties of the Field of P The Gauss Theorem for th£' Field of P. We shall show that the field of P has the following remarkable and important property. It turns out that the flux of P through an

Since

p

I r: I =

dq'

=

p~ we have

P+ (l+

+ L) dS

cos a

=

p+l dS cos a,

(3.7) where l = l+ L is the relative displacement of positive and negative bound chaJ;ges in the dielectric during polarization. Next, according to (3.4), p:l = P and dq' = P dS cos at or

"+

(a)

( b) Fig. 3.2

ar~itrary closed sur~ace S is equal to the excess bound charge

(wIth the reverse sIgn) of the dielectric in the volume en. closed by the surface S, i.e.

dq'

=

P n dS

=

P·dS.

(3.8)

Integrating this expression over the entire closed surface

S, we find the total charge that left the volume enclosed by the surface S upon polarization. This charge is equal to

~ PdS=

-qint.

(3.6)

This equation expresses the Gauss theorem for vector P. Proof of the theorem. Let an arbitrary closed surface S envelope a part of a dielectric (Fig. 3.2a, the dielectric is hatched). When an external electric field is switched on the ~ielectric is polarized-its positive charges are displaced relative to the negative charges. Let us find the charge which passes through an element dS of the closed surface S in the outward direction (Fig. 3.2b). Let 1+ and C be vectors characterizing the displacement

~ P.dS. As a result, a certain excess bound charge

?'

will be left inside the surface S. Clearly, the charge leavlllg the volume must be equal to the excess bound charge remaining within the surface S, taken with the opposite sign. Thus, we arrive at (3.6). Differential form of Eq. (3.6). Equation (3.6), viz. the Gauss theorem for the field of vector P, can be written in the differential form as follows:

I

V·P=-p',

(3.9)

c:

74

3.3. Properties of the Field of P 3. Electric Field in Dielectrics

i.e. the divergence of the field of vector P is equal to the volume density of the excess bound charge at the same point, but taken with the oppOsite sign. This equation can be obtained from (3.6) in the same manner as the similar expression for vector E was obtained (see p. 24). For this purpose, it is sufficient to replace E by P and p by p' ..

When Is p' Equal to Zero in a Dielectric? We shall show that the volume density of excess bound charges in a dielectric is equal to zero if two conditions are simultaneously satisfied: (1) the dielectric is homogeneous and (2) there are no extraneous charges within it (p = 0). Indeed, it follows from the main property (3.6) of the field of vector P that in the case of a homogeneous dielectric, we can substitute xeoE for P in accordance with (3.5), take x out of the integral, and write x

&eoE·dS .. -q'. oJ

The remaining integral is just the algebraic sum of all the charges-extraneous and bound-inside the closed surface S under consideration, Le. it is equal to q + q'. Hence, x (q + q') = - q', from which we obtain q' =

- 1~X

q.

(3.10)

This relation between the excess bound charge q' and the extraneous charge q is valid for any volume inside the dielectric, in particular, for a physically infinitesimal volume, when q' __ dq' = p'dV and q -+- dq = pl1V. Then, after cancelling out dV, Eq. (3.10) becomes , p =

-

x 1+x p.

• (J.H)

Hence it follows that in a homogeneous dielectric p' = 0 when p = O. Thus, if we place a homogeneous isotropic dielectric of any shape into an arbitrary electric field, we can be sure that its polarization will give rise only to the surface bound charge, while the bulk excess bound charge will be zero at all points of such a dielectric, Boundary Conditions for Vector P. Let us consider the behaviour of vector P at the interface between two homogeneous isotropic dielectrics.. We have just shown that in

75

such a dielectric there is no excess bound bulk1ch~rge tn~ only a surface bound charge appears as a resu t 0 po arl-

zat~~·us

find the relation between polarizatio? P a nd ~he 'ty a' of the bound charge at the lOterf ace ed surf ace ensl . 1 t s use property tween the dielectrics. For this purpose, e u ' (3.6) of the field of vector~. We D choose the closed surface lO the form of a flat cylinder whose end2~J~~~t--faces are on different sides of the 1 interface (Fig. 3.3). We shall asI sume that the height of the cyl• n' inder is negligibly small and the F' 3 3 area !1S of each endface is so small Ig. . that vector P is the same at all . d' . ts of each endface (thiS also refers to the surface ensl~Y ~?I~f the bound charge). Let 11 be the common normal to t e interface at jl. given point. We shall always draw vector n froDm. dielecd~riC ~h~o g~~e~}ri~ ~hrOugh the lateral surface d 'th (3 6)' Isregar Illg ., of the cylinder, we ca'h write, in accor ance WI P !1S + Pin' t1S = - a' t1S, 2n h P and P , are the projections of vector P in d ielecw. er; on2to the n~~al n and in dielectric 1 onto the normal t~I(p' 3 3) Considering that the projection of. vector Ponto n Ig. . 1 ~, is equal to the projection of thIS vector o~to . t h e norma. ( m n) normal n taken with the opposIte t~e o~pos~te ~m ~ we can ~ite the previous equation sIgn, l.e. tn' - In' 11' t1S)' in the following form (after cance mg . P2n - Pin = - a'. (3.12) This means that at the interface bet~een ~iel~ctrics the normal component of vector P ha~ a dlsc.ofntIDdu!ty, ;~ose . d d d a' In partIcular, I me mID IS a magmtu e epen s on . d d't' (3 12) acquires a vacuum, then P 2n = 0, an con I Ion . simpler form: (3.13) where P n is the projection of vector ~ onto th~ outward normal to the surface of a given dielectrIC. The SIgn of the pro-

('

76

3. Electric Field in Dielectrics

jection P n determines the sign of the surface bound charge a' at a given point. Formula (3.13) can be written in a diHerent form. In accordance with (3.5), we can write (3.14) where En is the projection of vector E (inside the dielectric and in the vicinity of its surface) onto the outward normal. Here, too, the sign of En determines the sign of a'. A Remark about th(' Field of Vector P. Relations (3.6) and (3.13) may lead to the erroneous conclusion that the field of vector P depends only on the hound charge. Actually. this is not true. The field of vector P, as well as the field of E, depends on all the charges, both hou~d and extraneous. This can he proved if only by the fact that vectors P and E are connected through the relation P = xeoE. The hound charge determines the flux of vector P through a closed surface S rather than the field of P. Moreover, this flux is determined not hy the whole hound charge hut by its part enclosed hy the surface S.

3.4. Vedor D The Gauss Theorem for Field D. Since the sources of an electric field E are all the electric charges-extraneous and hound, we can write the Gauss theorem for the field E in the following form:

~ eoE dS =

(q + q')lnt,

(3.15)

where q and q' are the extraneous and hound charges enclosed by the surface S. The appearance of the hound charge q' complicates the analysis, and formula (3.15) turns out to be of little use for finding the field E in a dielectric even in the case of a "sufficiently good" symmetry. Indeed, this formula expresses the properties of unknown field E in terms of the bound charge q' which in turn is determined hy unknown field E. This difficulty, however, can he overcome hy expressing the charge q' in terms of the flux of P hy formula (3.6). Then

a.4.

77

Vector D

expression (3.15) C an he transformed as follows:,

~ (eoE + P) dS = qlnt. The quantity in the parentheses in t~~ integrand by D. Thus, we have defined an auxIlIary vector

ID=eoE+P, I

(3.1.6) ~

denoted (3.17)

through an arbitrary dosed surface is equal.to the h fl w ose uxsum of extraneous c1larges enclosed hy this sur'algebraic face:

I~ DdS =

(3.18)

qlnt.\

.- This statement is called the ~auss thearer- for fiel~e~ I t should h~. n ote? ~h~t ve~or D ISpth~~~~~ist~Za~~~P it is d ly different quantities. eo a~. have an' deep inde~d an aux~liary veft~~rW~~: ;::~e~t~ of the leld of physICal meamng. Howe 't. (3 18) justifies the introvdectt~r Do'f etxhf:~:~~or~~ne~ua~~o~ase~ it ~onsiderably simpliuc IOn . d' I t ' * fies the ~naly(s3iS107f) thedfle(idH~n) a:: ~can~sfor any dielectric, '. , RelatIOns . an

both isotr?pic (~nf7)anS\~~~~P;~~t the dimensions o.f vect~r D as those of vector P. The 2quantlty D IS measured in coulombs per square metre (elm ).

~~~r~:~lOs~me'

Differential form of Eq. (3.18) is

I V·D=p, I

(3.19)

. equal to the volume densi ty of an Le. the divergence 0 f th e field D is oint extran~ous ch~rge .at the obat~i~fJ fro~ (3.18) in the same way as it can Ee,\see p. 24\" It suffices to replace E by D and wasThis doneequatiOn for the field take into account only extraneous charges. . '. ftPJ] called dielectric displacement! or electro* The ql;'antity this D term, however, III order to static inductton. We Dh~~llo s a no i b e u~ing ~ .

emphasize the auxiliary nature of

\~etor

.

78

79

3.4. Vector lJ

3. Electric field in Dielectrics

At the points where the divergence of vector D is positive we have the sources of the field D (p > 0), while at the points where the divergence is negative, the sinks of the lield D (p < 0).

Relation Between Vectors D and .E. In the case of isotropic dielectrics, polarization P = xeoE. Substituting this expression into (3.17), we obtain D = 8 0 (1 x) E, or

Let us illustrate what was said above by several examples. Example t. An extrallPUUS 'poil~t charge '1. is loc.atc(~ at th~ cenl~e of a sphere of radius a, made 01 an IsotropIC dIelectrrc WIth a dlelectrrc

+

I

D=eoeE,

I

(3.20)

where e is the dielectric constant of a substance:

8=1+x.

(3.21)

The dielectric constant e (as well as x) is the basic electric characteristic of a dielectric. For materials 8> 1, while for vacuum 8 = 1. The value of 8 depends on the nature of the dielectric and varies between the values slightly differing from unity (for gases) and several thousands (for some ceramics). The value of e for water is rather high (8 = 81). Formula (3.20) shows that in isotropic dielectrics vector D is collinear to vector E. For anisotropic dielectrics, these vectors are generally noncollinear. The field D can be graphically represented by the lines of vector D, whose direction and density are determined in the same way as for vector E. The lines of E may emerge and terminate on extraneous as well as bound charges. We say that any charges may be the sources and sinks of vector E. The sIJurces and sinks of field D, however, are only extraneous charges, since only on these charges the lines of D emerge and terminate. The lines of D pass without discontinuities through the regions of the field containing bound charges. A Remark about the Field of Vector D. The field of vector D generally depends on extraneous as well as bound charges (just as the field of vector E). This follows if only from the relation D = 8 oeE. .However, in certain cases the field of vector D is determined only by extraneous charges. It is just the cases for which vector D is especially useful. At the same time, this may lead to the erroneous conclusion that vector D always depends only on extraneous charges and to au incorrect interpretation of the laws (3.18) and (a.1!J). These laws express only a certain property of field D but do not determine this field proper.

r

Fig.

Fig. ;LS

~L4

constant e. Find"'ihe projectiun 1,', of Iield intensity E as a functiun of the distance r frolll the centre of this sphere. The syIllllletry uf the system allows us to use the Gauss th~orem for vector D for sol ving the 'problem (we cannut usc here the srn.l1lar theorem for vector E, since the bound char~e is unlwo.wn to us). l' or a sphere of radius r with the centre at the POlllt of locatIOn of the charge q we can write the following relation: 4;r.r"lJ r . = q. Hence we can lind lj, and then, usillg formula (3.20), the required quantity t r : 1

E r (r < a) = 4ne";

'/ <:/:2'

E

r (r > a

)

1

q

=="4ne o ~.

Figure 3.4 shows the curves lJ (r) and E (r). Example 2. Suppose that a system consists of apoint ~harg.e q >.0 and an arbitrary sample 01 a homogelleou~. IsotropIc dielectrrc (Fig. 3.5), where S isa certain closed surface. h~d ?ut what Will happen to the ileIds of vectors E and D (and to theIr fluxes through the surface .) if the dielectric is removed. The field E at any point of space is determined by the charge q and by bound charges of the polarized diele~tr.ic. Since in o.ur case D = e eE this refers to the lield D as well: It IS also determIned by the exrra~eous charge q and by the bound char~es of the dielectric. The removal of the dielectric will change the held E, and hence the lield D. The flux' of vector E through the surface S will also cha~ge, since negative bound charges will vanish from inside. this s~rface. However the flux of vector D through the surface SWill remalll the same in ;pite of the change in the lield D. . . . Example 3. Let us consider a system C?ntallllllg no extraneo.us charges and having only bound charges. :::;uc~ a ~ystem can, for example, be a sphere made 01 an electret (see p. 6b). 1, I~ure 3.?a shows the field E of this sy::;tem. What can we say about the fIeld Dt

RO

3, Electric Field in Dielectrics

First of all, t~e abs~nee of extraneous charges means that the field has ~o sou:(;~s, the Imes of D d.o IIOt emerge or terminate anywhE're. lJowever, the 11I~ld J) PXISts 'md.IS.sllown I'n I"I'g . 3 .u. "b 'fl}' _ ( Ie Illes 0 l' LL' J)

81

3.5. Boundary Conditions

where the projections of vector E are taken on the direction of the circumvention of the contour, shown by arrows in the figure. If in the lower region of the contour we take the n n ..

Lines ofE

Lines of D

(a)

Cb)

Fig, 3.G and ~ coi~cid~ outsi?e the sphere, but inside the sphere they have 01.PdPoslte dlfectlOns, smce here the relation D = 2 eE is no longer val and D = FoE P. 0 '

+

!------I ---:' I

c

2 [ ~ ~ '"'r'

I~

J

•

Fig. 3.7

Fig. 3.8

projection of vector E not onto the unit vector 't' bu~ onto the common unit vector 't, then En' = - E H , and It follows from the above equation that Ii:

(3.22)

3.5. Boundary Conditions Let ~s first consider the behaviour of vectors E and D at. the mterface between two homogeneous isotropic dielectrics. Suppose that, for greater generality, an extraneous su:face charge exists at tl16 interface between these dielectrICS. The required conditions cari be easily obtained with the help of two theorems: the theorem on circulation of vector E and the Gauss theorem for vector D:

~E dI = 0

and

~ n dS =--= glnt-

. Boundary Condition for Veclor E. Let the field near the lDterface be E 1 in dielectric 1 and E 2 in dielectric 2. We choose ~ sm~ll ?l(~_ngateu rectangular contour and orient it as ~hown In Fig, .L'. The sides of the contour parallel to the lllterfac~ must have slIch a length that the fwld E over this !,eng.th ,,1Tl each dielectric can be assumed constant. The height of th~ contour must be negligibly small. Then, in accordance With the tllf'orem on circulation of vector E we IJave '

I.e. the tangential component of vector E turns out to be the same on both sides of the interface (it does no.t have a discontinuity). . Boundary Condition for Vector D. ~et us ta~e a cylInder of a very small height and arrange it at the lll~erface between two dielectrics (Fig. 3.8). The cross sectIOn of the cylinder must be such that vector D is the same within each of its endfaces. Then, in accordance'with the Gauss theorem for vector D, we have D 2n .!1S DIn' ·!1S = (JAS,

+

where (J is the surface density of the extraneous charge at the interface. Taking both projections of vector D onto the common normal n (which is directed from dielectric 1 to dielectric 2), we obtain Din' = - DIn' and the previous equation can be reduced to the form \

D 2n - Din = a.

(3.23)

It follows from this relation that the normal component of vector D generally has a discontinuity when passing 6-0181

82

throngh tllO inlt'l'[a(;('. If, however, tltel'O aro charges at the interface (u = 0), wo ohtain

I

Din

/10

extraneous

E H /E 2n E u /E l1l

tan Cl2 tan (;(,

LJ 1

E2

Et

FieldE Fig 3.9

FieldD

Fig. 3.1U

We see that in the case under consideration, the lines of E are

refracted and undergo discontinuities (due to the presence of bound

charges), whileihe lines of D are only refracted, WI ~hou t thscon ttnuities (since there are no extraneous charges at the Illterlace).

Boundary Condition 'on the Conductor-Dielectric Interface. If medium 1 is a conductor and medium 2 is a dielectric (see Fig. 3.8), it follows from formula (3.23) that (3.26)

(;3.2!'i)

This means that lines of D and E will form a larger angle with the normal to the interface in the dielectric with a larger value of 8 (in Fig. ;3.9, 8 2 > 8 1), Example. Let us represent graphically the fields E and D at the i II ["ri'ace hetween two homogeneous dielectrics 1 and 2, assuming that

awl llli~re is 110 extraneous charge on this surface. Since 1'2 > ;'n in accordance with (3.25) Cl 2 > at (Fig. 3.10). Considering that the tangential component of vector E remains unchanged anrl using Fig. ;·1.9, We can easily show that F 2 < Hl in magnitude, i.e. the lines of E ill dielectric 1 must be denser than in dielectric 2, as is shown in Fig. 3.10. The fact t~at. the normal corn:.'2> 1-'1

>

•

Taking into aceount the above conditions, we obtain the law of refraction of lines E, and hence of lines D: . ~

pouents o[ vectors D are equal leads to the conclusion that LJ 2 in magnitude, Le. the lines of D must be denser in dielectric 2.

fl~1l" I

In this case the normal components of vector l) do not have a discontinuity and tUI'll out to he tlte same on different sides of the interfaGo. Thus, in the absence of extraneous charges at the interface hetween two homogeneous isotropic dielectrics, the components E, anfl J)n vary continuollsly dnring a transition through this interface, while the components En and D, have discontinuities. Hdl'action o[ E and D Lines. The boundary conditions which we obtained for the components of vectors E and D at the interface between two dielectrics indicate (as will be shown later) that these vectors have a break at this interface, Le. are refracted (Fig. 3.9). Let us flIld the relation between the angles a l and a 2 • In the absence of extraneous charges at the interface, we have, in accordancp- with (3.22) and (3.24), E 2 , = E l , and 8 2 E 2n = 8 r E ln • Figure 3.9 shows that tan Cl2 tan Cll

83

9.5. Boundary Conditions

3. Electric Field in Dielectrics

where n is the conductor's outward normal (we omitted the subscript 2 :;ince it is inessential in the given case). Let us verify formula (~).26). In equilibriulll, the electric fteld inside a conductor is E =,c 0, and hence the polarization P = O. This means, according to (;3.17), that vector D =~ 0 inside the conductor, Le. in the notations of formula (:L~3) D1 = 0 and DIn = O. IIenceD 2n = a. Bound Charge at the Conductor Surface. If a homogeneous dielectric adjoins a charged region of the surface of a COIlductor, bound charges of a certain r1en:;ity a' appeal' at the conductor-dielectric interface (recall that the volume density of bound charges p' = 0 for a homogeneous dielectric). Let us now apply the Gauss theorem to vector E in the same way as it was done while deriving formula (2.2). Considering that there are both bound and extraneous charges (a 0*

3.6. Field in 84

3. Electric Field in Dielectrics

' I

I

(3.27)

I t can be seen that the surface density G' of the bound charge in the dielectric is unambiguously connected with the surface density G of the extraneous charge on the conductor, the signs of these charges being opposite. 3.6. Field in a Homogeneous Dielectric

III II' I :1

II

1

I'

Homogeneous Dielectric

85

a' are unambiguously connected with the extraneous charges

and G') at the conductor-dielectric interface we arrive at the following expression: E" = (G + a')le o' On the other hantl, according to (3.2G) E" = Dn/ee,o = alee o' Combining these two equations, we obtain Gle = G + a', whence 0 = - -8-1 8- 0 .

/l

It was noted in Sec. 2.1 that the determination of the resultant field E in a substance is associated with considerable difficulties, since the distribution of induced charges in the substance is not known beforehand. It is only clear that the distribution of these charges depends on tho nature and shape of the substance as well as on the configuration of the external field Eo. Consequently, in the general case, while solving the problem about the resultant field E in a dielectric, we encounter serious difficulties: determination of the macroscopic field E' of bound charges in each specific case is generally a complicated independent problem, since unfortunately there is no universal formula for finding E'. An exception is the case when the entire space where there is a field Eo is filled by a homogeneous isotropic dielectric. Let us consider this case in greater detail. Suppose that we have a charged conductor (or several conductors) in a vacuum. Normally, extraneous charges are located on conductors. As we already know, in equilibrium the field E inside the conductor is zero, which corresponds to a certain unique distribution of the surface charge G. Let the fIeld created in the space surrounding the conductor be Eo. Let us now fill the entire space of the field with a homogeneous dielectric. As a result of polarization, only surface bound charges G will appear in this dielectric at the interface with the conductor. According to (3.27) the charges

a on the surface of the conductor. As before, the,re will be no field inside the conductor (E = = 0). This means that the distribution of surface charges (extraneous charges a and bound charges a') at the conductordielectric interface will be similar to the previous distribution of extraneous charges (G), and the configuration of the resultant field E in the dielectric will remain the same as in the absence of the dielectric. Only the magnitude of the field at each point will be different. In accordance with the Gauss theorem, a + G' = eDEn, where En = Dnlee o = a/eeo' and hence G + a' = Gle. (3.28) But if the charges creating the field have decreased by a factor of e everywhere at the interface, the field E itself has become less than the field Eo by the same factor: (3.29) E = Eo/e. Multiplying both sides of this equation by eeo, we obtain • D = Do, (3.30) Le. the field of vector D does not change in this case. It turns out that formulas (3.29) and (3.30) are also valid in a more general case when a homogeneous dielectric fills the volume enclosed between the equipotential sur1aces of the field Eo of extraneous charges (or of an external field). In this caBe also E = Eo/e and D = Do inside the dielectric. In the eases indicated above, the intensity E of the field of bound charges is connected by a simple relation with the polarization P of the dielectric, namely, E'

=

-Ple{}.

(3.3t)

This relation can be easily obtained from the formula E = = Eo + E' ifl we take into account diat Eo = eE and

P = xeoE.

.

As was mentioned above, in other cases the situation is much more complicated, and forml1las~ (3.29)-(3.3t) are inapplicable. CornU.Fie. Thus, if a homogeneous dielectric fills the entire space oeeupied by a field, the intensity E of the field

87

Problems 9. Electric Field in Dtl!lectrics

will be lower than the intensity Eo of the field of the same extraneous charges, but in the absence of dielectric, by a factor of e. Hence it follows that potential cp at all points will also decrease by a factor of e: cp = cpo/e, (3.32) where CPo is the field potential in the ahrnce 01. the dielectric. The same applies to the potential difference: U = Uoh, (3.33) where U0 is the potential difference in a vacuum, in the absence of dielectric. In the simplest case, when a homogeneous dielectric fills the entire space between the plates of a capacitor, the potential difference U between its plates will he by a factor of e less than that in the absence of dielectric (naturally, at the same magnitude of the charge q on the plates). And since it is so, the capacitance e = ql U of the capacitor filled by dielectric will increase e times e' = ee, (3.34) where e is the capacitance of the capacitor in. the absence of dielectric. It should be noted that this formula is valid when the entire space between the plates is filled and edge effects are ignored.

Problems

. .3.1. Polarization of a dielectric and the bound charge. An extraneous point charge q is at the centre of a spherical layer of a heterogeneous isotropic dielectric whose dielectric constant varies only in the radial direction as 8 = air, where a is a constant and r is the distance from the centre of the system. Find the volume density p' of a bound charge as a fuuction of r within the layer. Solution. We shall use Eq. (3.6), taking a sphere of radius r as the closed surface, the centre of the sphere coinciding with the centre of the system. Then 4Jtr2 • P r = -q' (r),

dq'.

+

(1)

between the spheres with a;:erp'4Jtr2 dr, we transform (1) , 2

r 2 dP r +2rP r dr=-pr

d

r,

whence (2)

In the case under consideration we have e-l 8-1 Pr=xeoE r =

-e D r =-;-

q 4n:r 2

. (2) will have the form and after certain transformation expressIOn 1 q p'=~

?"

which is just the required result.

f vector D An infmitely large pla~e • 3.2. The Gauss thedo~i t~~c with th~ dielectric constant e IS made of a hBtnogeneous Ie ee 1 uniformly charged by an. extraneous Ex I{J charge with volume densIt~ P > O. The thickness of the platt! IS 2a. 1~h1lii-l~ Find the magnitude of vector E an. l;; E?, the potential cp as functions of the dIstance 1 from the middl~ of. the pl a~e (assume that the poten~Ial IS ~ero n the middle of the plate), ChOOSlll I' X-axis perpendicular to the pal'. Plot schematic curves for the pro. . n E (x) of vector E and the po~~~t:fal ~(x). (2) Find the surface and volume densities of the bound charge. / Solution. (1) From symmetry co~. siderations it is clear that ~'I= ~;ii FIg. 3.11 the middle of the plate, w 1 I' a d-' other points vectors E f a~h pef P:: In order to determine E, w,e s~bIl icular to the surface 0 I' P a . D (since we know the dIstrr u'use the Gauss theorem for vector W take for the closed su~face a tion of only extran~ous charges)! hose endfaces coincid~s WIth the right cylinder of helghthl, one 0 ~'onal area of this cylInder be S. midplane (x = 0). Let t I' cross-sec 1 Then

(ll

h

y\

DS

where q' (r) is the bound charge inside the sphere. Let us take the differential of this expression: 4Jt d(rl·p r) = -

. h' I Here dq' is the bound c~arg~ ill th \ ~n, radii rand r dr. Considermg a q as follows:

DS

=

pSI,

=

pSa,

D

D

= =

pi, pu,

E

=

pl/ee:o (1 ~ a),

E = paleo (l ~ a).

88

The graphs of the funtions Ex (x) and

,

89

Problems

3. Electric Field in Dielectrics

where P is the volume density of the extraneous charge. Hence IJ p r3 -a' E-- -=-- --,- eEo 38 08 r The corresponding curves for E (r) and cp (r) are shown in Fig.· 3.12b.

e-1 =Pn="X8 0 E n =(e-1)pa/e=--. pa>O. e

This result is valid for Il0th sides of the plate. Thus, if the extraneous charge p > 0, the bound charges appearing on both surfaces of the plate are also positive. In order to find the volume density of the bound charge, we use Eq. (3.9) which in our case will have a simpler form:

p'=_ ap x ax

=_.!.( ax

o\---'---!---~, a b

e-1 px)=-- e-1 p. e e

(b)

(a)

It can be seen that the hound charge is uniformly distributed over

Fig. 3.12

the bulk and has the sign opposite to that of the extraneous charge.

• 3.3. A homogeneous dielectric has the shape of a spherical layer whose inner and outer radii are a and b. Plot schematically the curves of intensity E and potential

a) = q/4nr 2 •

The required intensity is E (r < a) = 0,

E (r

>

a) = D/eeo.

The curve for E (r) is shown in Fig. 3.12a. The curve for

a

0

. ·f I d· tributed with the volumc .3.4. Extraneous ~argerUd·lu~r:\:ad~of a homogeneous dielecdeMity p > 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
4

D

,

4nr3IJ=Tnr p, 2 4, 4nr D =3 na p,

D

p

P r

E=-= - r, 388

D_ Pa3 - 3r' '

E=!!-=pa3~.

=7'

The curves for the functions E (r) and (2) The surface density of the boun

0

8100

eo

3eo

r

~ (r~

e-1 fJ'=Pn=-e-

are ~hown in Fig. 3.13. c arge IS pa

T·

d ·t of bound charge it is sufficient to In order to find the volume enSI y f I (3 11) ~nd we get repeat the reasoning that led us to ormu a . , p' =c

8--1 -

--;;-

fl·

(1)

'rh••• __It can be obtaincd ill a different way, viZ· . b Y 1I~(i.llg n:;:KU E d x doc~ not depend on coor(I rna t es mEq. (3.9). Since P = "XfO an

90

3. Electric Field in Dielectrics

91

Problems

side the sphere), we obtain where we assume that the charge of the inner plate is q > O. Let us find E with the help of the Gauss theorem for vector D: D 1 q 1 q 2 E = EEo = 4nEo ~ ~ 4'1Eo r;;:. . 4nr D = q;

-V'p = -XEllV.E H' ' . ence p = -x (p p'), which gives for-

p' =

where EoV'E = P mula (1).

+ p'

+

• 3.5. Capacitance of a cond t . spherical conductor of radius a, su~~o~~de~lIbd the capacitance of a y a layer of a homoge-

Substituting the latter expression into (1) and integrating, we find q b 4ne oa U = 4neoa In a' C = In (bla) •

• 3.7. The Ga_ theorem and the principle of SU,ertlOSitiOD. Suppose that we have a dielectric sphere which retains polarization

a

r

·Fig. 3.13

Fig. 3.14

Ileous diplectric and hav· tI stant f. Plot approximat~r:fur~:sf~~Pf radius b and the dielectric contan ce from the centre of the sphere Y)t~nd
Fig. 3.15

after an external electric field is switched off. If the sphere is. polarized uniformly, the field intensity inside it is E' = - P/SEo, where P is the polarization. (1) Derive this formula assuming that the sphere is polarized as a result of a small displacement of all positive charges of the dielectric with respect to all its negative charges. (2) Using this formula, find the intensity Eo of the field in the spherical cavity inside an infinite homogeneous dielectric with the permittivity f if the field intensity in the dielectric away from the cavity is E. Solution. (1) Let us represent this sphere as a combination of two spheres of the same radii, bearing uniformly distributed charges with volume densities p and -po Suppose that as a result of a small shift, the centres of the spheres are displaced relative to one another by a distance I (Fig. 3.15). Then at an arbitrary point A inside the sphere we have

r

0

I

ntcgrating this expression, we obtain

h


4nE oEa 1-c,~--"'-,--_ -HE-1) a!b Ill' curves for E (r) and ()
p

a

,L- =

b

E' =E~+E~=-3P(r+-r_)= __ .J:!.,

So/ution. In accordance w'th th d .. !U) th I ~ efillltlOnofthecapacitanceof

eo

capacltor(C=

potential

djffere~ce 'u f~lrob~em

jreduced to determination of th: a gIven c large q: IS

J

Edr,

a

3eo

where we used the fact that field intensity inside a uniformly charged sphere is E = pr/3E o, which directly follows from the Gauss theorem. It remains for us to take into account that in accordance with (3.4), pI ~ P. , (2) The creation of a spherical cavity in a dielectric is equivalent to the removal from a sphere of a ball made of a polarized material.

b

u=

Fig. 3.16

(1)

~ ..

i' ,

,

-;;-'-'-.'-. '.

" .•

m ,.

IIl II ',I II'.1.[

92

3. Electric Field in Dielectrics

1

Consequently, in a('cordance with the principle of superposition, the field E inside the dielectric can he represented as the sum E = E' + + Eo· Hence

1. '/

Ii

1'1

Eo

Considering that P

'II

=

=

E -

E'

=

+ P/3E

E

(e:- 1) EoE, we obtain Eo

=

+

(2

e) E/3.

Solution. The field intensity inside the dielectric is

' f a CI' a thin rinO' with the centre at charge on the plane. (2) Let us imagine at the lllt~h t the inner a;' outer radii of this the point 0 (Fig. 3.17). Suppose ,a are r' and r' + dr'. The surface rmg . th' . g is bound charge WI. thm ,IS rID dq'=o' • 2:rtr' dr'. It can be ~: n from the figure that r 2 = 12 r • whence r dr = r' dr', an~ the exp.re~~ sion for dq' combined WIth (1) glV~S e-1 dr

+

dq'=-e'+1 qlrz·

R= V -E~-I-E~.

(1)

Integrating this equatio~ over r between land 00, we obtam

UsilJJ{ we fmdconditions (3.22) and (3.21) at the interface between dielectrics,

=

Eo sin a o, En

=

IJnlce o

=

Eonle

=

Fig. 3.17

Eo cos aole,

when> 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 · <. 'tvJ of lhe tanrrential component of vee r from qlHI t Ie conet WUI 5 E it follows that J) = d'o.

(1) '

Combining these two conditions, we find q

Do ~ 2lt (1+e) r 2

eq

'

j)oc=

2lt

<1+1;) r

2

•

94

4. Energy 01 Electric Field

and the ele&tric field intensity in the entire space is

E = .!.!J!. = 80

4.1. Electric Energy

de.

2n (t +e)

95

Then the corresponding work of these forces is

'I

6/1 1• z = FI·rll l + F 2 ·dl z·

for 2

It ean ~ seen that for 8 = 1 these formulas are reduced to the already famlhar expressions for D and E of the point charge in a vacuum.

0/ a System of Charges

Considering that F 2 = - 1'\ (according to the N.ewton third law), we can write this expression in the form

The obtained results are represented graphically in Fig. 3.18. It

61'11.2= F I (dll-dl l )·

The quantity in the parentheses is the dis~la~ement .of charge 1 relative to charge 2. In other words, It IS ,the ~IS­ placement of charge 1 in the system of. referen.ce whIch is rigidly lixell to charge 2 and accomplishes wIth It a trallSlational motion relative to the initial reference system K. Indeed the displacement dl l of charge 1 in system K can be rep;esented as the displacement dl 2 of .system K' plus the displacement dli of charge 1 relatIve to system K': dl l = dl z +dlj. Hence, dl l - dl z = dli, and

1! '

. Field E

FieldD Fig. 3.18

should be noled that the field n in this case is not determined by the extranC?us charge alone (otherwise it would have the form of the field of a pOlllt charge).

4. Energy of Electric Field 4.1. Electric Energy of a System of Charges

~nergy .Approach to Interaction. The energy approach to lllteractlOn between electric charges is, as will be shown rather fruitful in respect of its applications. Besides, thi~ approach makes it possible to consider the electric field from a different point of view. First of all, let us find out how we can arrive at the concept of the energy of interaction in a system of charges. 1. Let us first consider a system of two point charges 1 and 2. We shall [lOd the algebraic sum of the elementary works of the forces F I and F 2 of interaction between the charges. Suppose" that ill a certain system of reference K the charges were displaced by dl l and dl 2 during the time

I{

6At,2==Ft.dl~.

Thus, it turns out that, the sum of the elementary works done by two charges in an arbitrary system of reference K is always equal to the elementary work done by the~or~e acting on one charge in another reference system (/\ ) III which the other charge is at rest. In other words, the work 6A l.Z does not depend on the choice of the initial system of reference. The force F acting on charge 1 from charge 2 is conservative (as a cellt~al force). Consequently, the work of the given force in the displacement dli can be represented as the decrease in the potential energy of interaction between the pair of charges under consideration:

6.111. 2 - dW l2 , where W l2 is the quantity depending only on the distance between these charges. 2. Let us now go over to a system of three point charges (the result obtained for this case can be easily g~neralizerl for a system of an arbitrary nllmber oE charges). fhe war\.. performed by all forces oE interaction during elementary displacements of all the charges can be represented as the sum of the works of three pairs of interactions, i.e. 6.1 = l'lA1. 2 + 6A1.3 + 6.1 2 ,3' But as it has just been shown,

4J. Electric Energy of a System of Charges

96

4. Energy of Electric Field

for each pair of interactions 6A i, 6A = -

d (W12

= -

k

+ W13 + W 23 )

dW ik , and hence =

dW,

-

97

obviously docs 1I0t depend on the number ~f charg?s constituting the system. Thus, the energy of mteractlOn for a system of point charges is

t4.2)

where W is the energy of interaction for the given system of charges:

=

W

W 12

+ W13 + W 23 •

Each term of this sum depends on the distance betwoon corresponding charges, and hence the energy W of the given system of charges is a function of its configuration. Similar arguments are obviously valid for a system of any number of charges. Consequently, we can state that to each configuration of an arbitrary system of charges, there corresponds a certain value of energy W, and the work of all the forces of interaction upon a change in this configuration is equal to the decrease in the energy W 6A

= -

dW.

(4.1)

Considering that WI =c= qi(Pio where qi is the ith cha:ge of tho system and (ei is the potential create~ ~t the pomt of location of the ith charge by all the remaunn~ charg~s, we obtain the final expression for the energy of mteractlOn of the system of point charges: II! --- -- - 21~

(4.3)

g.(D· I riO

Example. Four similar point ch,~rges q ar~local('d at th: ve:tice~ of a tet.rdbedron w.ith an edge a (I'lg. 4.1). I·tnd the energ) of tnter ac:tinll of charges in this syst.em.

Energy of Interaction. Let us find the expression for the energy W. We again first consider a system of three point charges, for which we have found that W = W 12 + WllI + W 23 • This sum can be transformed as follows. We represent each term W ik in a symmetric form: Wilt =

The energy of interaction fn.. each pair of charges of t.his system i~ t.he same and equal to 1V I = q2 /4m oa. It. can be seen from the figure that. the total number of interacting pairs is six, a~d hence the energy of interaction of all point charges of the system IS

= ~

Another approach to the solution of this problem is based on formula. (4.3). The p,0t.ential q> at the point of 10catlOII of one of the charges. created by the Held of all the other charges. is q> = 3q/4ne oa. lienee

+

(W lk

+ W kl ),

since W ik

=

W k !. Then

W = ; (W t2 + W21 + W 13 + W31 + W 23 + W 32 ).

W = 6 WI = 6 q2/4ne oa.

q

q

Let us group the terms with similar first indices: W

=+

q

[(W 12 + W (3) + (TV21 + W23 ) + (W:11 + W32 )].

Each sum in the parentheses is the energy WI of interaction between the ith charge with all the remaining charges. Hence the latter expression can be written in the form

=+ ~ 3

W

= ;

(W t

+ W 2 + W s)

Fig. 4.1

Total Energy of Interaction. If the charges are arranged continuously, then, representing the system of charges. as a combination of elementary charges dq = p dV and gOlQg over in (4.:1) from summation to integration, we obtain

Wj'

w· -~- ~

i=l

This expression can be generalized to a system of an ubitrary number of charges, since the above line of reasoning

q

7-0181

plpdV,

I

98

4.2, Charged Conductor and Capacitor Energies

4. Energy of Electric Field

where (jJ is the potential created by all the charges of the system in the volume element dV. A similar expression can be written, for example, for a surface distribution of charges. For this pmpose, we must replace in (4.4) p by 0' and dV hy dS.

Expression (4.4) can be erroneously interpreted (and this often leads to misunderstanding) as a modification of expression (4.3), corresponding to the replacement of the concept of point charges by that of a continuously distributed charge. Actually, this is not so since the two expressions differ essentially. The origin of this difference is in different meanings of the potential (jJ appearing in these expressions. Let us explain this difference with the help of the following example.j Suppose a system consists of two small balls having charges qi and q2' Thedistance between the balls is considerably larger than their dimensions, hence qi and t}z can he assumed to be point charges. Let us find the energy W of the given system by using both formulas. According to (4.3), we have 1

W ="'2 (qt(jJt

+ q2(jJ2) = qt(jJl = q2(jJ2,

where
II 'I

Iii I!!

=

WI

+ W 2 + W 12 ,

corresponds only to the energy W I2 , ~hile c~lcllla.tion b~ formula (4.4) gives the total energy o( lllteractwn: III a~dl­ tion to W 12 , it gives intrinsic energies WI and W 2 • DIsregard of this circumstance is frequently a cause of gross errors. We shall return to this question in Sec. 4.4. Now, we shall use formula (itA) for obtaining several important results.

4.2. Energies of a Charged Conductor and a Charged Capacitor

Energy of an Isolated Conduetor. Let a conductor. have a charge q and a potential (jJ. Si?ce the value of (jJ IS the same at all points where charge IS located,w,e .can. take (jJ in formula {4.4),:outof the integral. The remallllllg !ntegral is just the charge q on the conductor, and we oLtam

~

. W ..

-*

=

0f -f-. \ =

(1t.G)

These three expressions are written assuming that C = q/(jJ. Energy of a Capacitor. Let q and (jJ + be the charge. and potential of the positively charged plate. o~ a capaCitor,. According to (4.4), the integral can be splIt lllto two parts (for the two plates). Then W Since q_

(4.5)

where WI is the energy of interaction of the charge elements of the first ball with each other, W 2 the same but for the second ball, and W 12 the energy of interaction between the cjJ.arge elements of the first ball and the charge elements of the second ball. The energies WI and W 2 are called the intrinsic energies of charges ql and q2' while W 12 is the energy of interaction between charge ql and charge q2' Thus, we see that the energy W calculated by formula (4.3)

!lO

=

-

=

+

(q+(jJ+

q+, we have 1

W

=2 q+

«(jJ+

+ q-(jJ-). 1

-
where q = q+ is the charge of the capacitor ~nd .U is the potential difference across its plates. Consldermg that C = q/ U, we obtain the following expression for the energy of the capacitor:

----------\

I w~5'{-~!f7*

f,.

(4.7)

100

4. Energy oj Electric Field

J I :,;hulIJd Le, I~uled he~e thai these formulas determine the lutal cllergy 01 lIltel'aCllOn, viz. not only the energy of interactIOII Lot \\ een the charges of one plate and those of the ut.he~ plate, but also the energy of interaction of charges wlthlll each plate. And Wh~t if We Have a Dielectric? We shall show that formula,s (4. fJ) and (4.7) are valid in the presence of a dielectric. For t!llS purpose, we consider the process of charging a capacitor a" a transport of small portions of charge (d ') from one plate to the other. q . Th~ elen~entary .\I'ork performed against the forces of the f1eld 111 th IS case IS bLl == l./' dq' = (q'IC) dq',

where l./' i::> the potential tltfference between the plates at the moment whelJ the next portion of charge dq' is being transferred. Integruting this expression over q' between 0 and q, we obtain A

= q2/2C,

which ~oilJcides with the expression for the total energy of a capacitor. ~onsequel~tly, the work done against the forces of the eleclnc lleld IS completely spent for accumulating t~Je en~rg~ W of the charged capacitor. Moreover, the expression o.DtalD~d for the work A is also valid in the case when there IS a dIelectric between the plates of a capacitor. Thus, w.e hav~ proved the validity of (4.7) in the presence of a dlOlectl'lc. Obviously, all this applies to (4.6) as well. 4.3. Energy of Electric Field

On Loc?lizatioll of Energy. Formula (4.4) defines electric energy n' of any system in terms of charge and potential. It turns out, however, t?at energy W can also be expressed t}~rough another quantity characterizing the field itself Vl~. throu~h field intensity E. Let us at first demonstrat~ thIS .by u~lDg the simple example of a parallel-plate capacitor, Igno.nn g field distortions near the edges of the plates (edge effect). Substituting into the formula W = CU 2/2

4.3. Energy of Electric Field

iOt

= eeoS/h, we obtain cUt --'- tlloSU' _ £Ilo W -22 --2 h

the expression C

(Y )2 Sh •

And since U1h = E and Sh capacitor plates), we get

=

V (the volume between the

W =+eoeE2V.

(4.8)

This formula is valid for the uniform field which fills the volume V. In the general theory, it is p~oved that the en~rgy can be expressed in terms of E (m the case of an £SOtrop~c dielectric) through the formula

':v

W= ~ llo~E' dV=

j

E;D dV.

(4~9)

4/

The integrand in this equation has the meaning of the ?nergy contained in the volume dV. This leads us to a very Important and fruitful physic'al idea about localization ot energy in the field. This assumption was confirmed in experiments with fields varying in time. It is the domain where we encounter phenomena that can be explained with the help. of the notion of energy localization in the field. These vary~ng fields mav exist independently of electric charges WhICh have generated them and may propaga~e in space in the form of electromagnetic waves. Expenments show that electromagnetic waves carry energy. This circumstance confirms the idea that the field itself is a carrier of energy. The last two formulas show that the electric energy is distributed in space with the volume density footE'

E·D

W=-2-=-2-

(4.10)

It should be noted that this formula is valid only in the case of isotropic dielectrics for which the relation P = x€oE holds. l' For anisotropic dielectrics the situation is more comp lcated.

f

102

4. Energy 01 Electric Field

A~other substantiation for fo ~hr:gl of an i~olated charged cond~:~~~s(~9~ It is known that the energ;r{:;ufhe Ifield rrect , proceeding from th;J~~2~fel·otcuaSI~hOt~ thaft L ' Iza IOn 0 et us consider an arbitrar .. denbtallly isolate a tube of infinit~ .PosItIvely charged conductor. We I' y mes of E (Fig. 4.2), and tak::'i~.cross section, which is bound=dS dt T~·an ellementary volume dV= . IS vo ume contains the energy dt ED DdS . "2 dSdt =~ Edt. Let us now find th th~d in the entire isol:te~n~~?elo~al­ IS purpose, .we !ntegrate the l~st I' o~ pressIOn, conSldermg that th d x D dS is the sa . I I' pro uct m of the tube andeh a 1. cross sections out of the i~tegral~nce It can be taken

D dS ~ dW=-2- \ •.

Fig. 4.2

Edt=~ 2 cp,

A

wh ere A is a point at the b . . '. It remains for us to make ~1lllDlng of t~e Imagmary tube. e tep fix£ression over all the tubes and Ja~t th , I.e. integrate the obtained d. Considering that pote~tial n. the energy localized in the entire t ~ tubes (since they originate ~h esrme at the end faces of all write I' sur ace of the conductor), we

h

0:

W=

i~

DdS,

w~ere the integration is performed ov lth one of the equipotential surf h fr a closed surlace coinciding t eorem, this integral is equal to ~chs. hn accordance with the Gauss

we finally get

Q.E.D.

I'

W

=

c arge q of the conductor, and

qrp/2,

Let us consider two exam l ' I l . we get by using the ide~ of ~n~r 1 ulstral~Ing. the. advantages gy oca lZatlOn In the field. Example 1. A point char I' . !"'lade of a homogeneous diel:ctr1 IS ~t the ce?tre of a spherical layer lI~ner and outer radii of the la eCr ~~th the dielectric constant E. The e Fmd the electric energy contaiKed in thlJuadl. tlo a ~nd b respectively. IS Ie ectrIC layer. We mentally isolate in the d' 1 . ical layer of radius from r to Ie +ect~lc ~hvery thin concentric spherr r. I' energy localized in this

4.3. Energy of Electric Field

103

layer is given by BOllEI 4 2d dW=-Zitl" r,

where E = q2/4nB oBr 2, Integrating this expression over r between a and b, we obtain q2

W= 8nE oB

(1' 1) . a -T

. Example 2. Find the work that must be done against the electric forces in order to remove a dielectric plate from a parallel-plate char~d capacitor. I t is assumed that the charge q of the capacitor remains unchanged and that the dielectric fills the entire space between the capacitor plates. The capacitance of the capacitor in the abseace of the dielectric is C. The work against the electric forces in this system.is equal to the increment of the electric energy of the system: A =

6.W =

W2

-

WlJ

where WI is the energy of the field between the capacitor plates in the presence of thi!'Qielectric and W 2 is the same quantity in the absence of the dielectric. Bearing in mind that the magnitude of vector D will not change as a result of the removal of the plate, 1.1'. that D 2 = = D1 = (J, we can write • A=W2 -W I =

\/~ __ ~) 2B . 2B B o

o

V =

~ (1-~), 2C B

where V = Sh and C= BoS/h, S being the area of each plate and h the distance bl'tween them.

The Work of the Field During Polarization 01 a Dielectric. An analysis of formula (,'1.10) for the volume energy density reveals that for the same value of E, the quantity w is E times greater in the presence of a dielectric than when it is absent. At first glance this may seem strange: field intensity in both cases is maintained the same. As a matter of fact. when a field is induced i'l a dielectric, it does an additional work associated with polarizatioll. Therefore, under the energy of the field in the dielectric we must understand the -sum of the electric energy proper and an additional work which is accomplished during polarization of the dielectric. In order to prove this, let lIS substitute into (4.10) the quantity EoE + P for D, which gives l'oE2+ E . P -2 2'

w=--

(4.11)

104

4. Energy of Electric Field

The first term on the ri ht-ha d ' '. energy density of the fiel~ E . n side comcldes with the us to verify that the "addit'o I~" a vacuufI.l. It remains for with polarization I na energy E·P/2 is associated Le~ us calculate' the work done b . polarIzation of a unit volume of th YI ~hle el~tr~e .field for e ( Ie ectnc, I.e. for the dJ -

-

J-

.....- - - - 1 = 0

:

I

+

..-: -

===~==: 1-1.- L

dl

4.4. ..1 S,/stem of -------_...::......-

Suppose that we have a system of two clwrged bodies in a vacuum. Let one body create in the surrounding space the field E I , while the other body, the field E 2 • The resultant field is E = E) -I- E 2 , and the square of this quantity is E2=E;+E~

+

6..1 = p: (dl+ - dL). E ~= p~ dl. E, where dl = dl + - dl is h . . ~he positive charges relative\ e t~ddItlon~1 displacement of mg to (3.4) p' dl _ dP d o e negatIve charges. Aceord, + an we get E.dP.

E = xe o , we have

oA =E'xe o dE= d

(4.12)

()(e~E2) =

d

(E~P).

Hence, the total work f unit volume of a dielectric ~~ent or the polarization of a

. .

A

=

E·P/2,

(413)

wlndl coincides w'th th . Thus, the volu~e enee:econd t~rm in formula (4.11). the energy eoE 2 /2 of the densIty w = E· D/2 includes associated with the 'p I . eld. proper and the energy E.P/2 o aflzatlOn of the substance.

lY

j

·E2 •

Therefore, according ,to (4.9), the total energy W of the given system is equal to the sum of three integrals: lV --

=

+ 2E

+

disRlacements of char es' , . agamst the field upon g p+ and ~- I'espectIvely along and E to E dE. Neglect~~ I~~rease m the field intensity from g e second-order terms we write 6A=p~E.dl++r'E.dJ ' where dJ + and dl are add' . -. increase in the fi""eld b dIEtlO(nFa.1 dIsplacements due to an Ig . 4 . 3) . ConSI' d ' P"- = -- P+, we obtain Y ermg that

6A

10;:;

----._------

4.4. A System of Two Charged Bodies

Fig. 4.3

Since P

Tu'o Charged Bod/.-s

r

Il"

.1 -.,-- (r FoFt

) -2Eof~ dl .,I '

JeoE

j'

E ~ c.il'. ,

(4.14)

which coincides with formula (4.5) and reveals the field meaning of the terms appearing in this sum. The first two integrals in (4.14) are the intrinsic energies of the first and second chaJitred bodies (WI and W 2 ), while the last integral is the energy of their interaction (W I2 ). The following impprtant circumstances should be mentioned in connection with formula (4.14). 1. The intrinsic energy of each charged body is an essentially positive quantity. The total energy (1.9) is also always positive. This can be readily seen from the fact that the in-tegrand contains essentially positive quantities. However, the energy of interaction can be either positive or ne~ative. 2. The intrinsic energy of hod ies remains constant upon all possible displacements that do not. change the configuration of charges on each body, and consequently this energy can be assumed to he an additive constant in the expression for the total energy W. In such cases, the changes in Ware completely determined only by the changes in the interaction energy W)2' In particular, this is just the mode of behaviour of the energy of a system consisting of two point charges upon a change in the distance between them. 3. Unlike vector E, the energy of the electric field is not an additive quantity, Le. the energy of a field E which is the sum of fields E) and E 2 is generally not equal to the sum of the energies of these fields in view of the presence of the interaction energy lV12 . In particular, if E increases n times everywhere, the energy of the field increases n2 times.

~

I;:

106

4. Energy oj Electric Field

4.5, Forces A cUng in a Dielectric

4.5. Forces ~cting in a Dielectric

expense of a decrease in the electric energy W of the system (or its field). Here we assume that these displacements do not cause the transformation of electric energy into other kinds of energy. To be mOre precise, it is assumed that such transformations are negligibly small. Thus, for infinitesimal displacements we can w.rite ~A= -dW\q, (4.15)

Electr~striction. E~periments show that a dielectric in an electnc field expenences the action of forces (sometimes these forces are called ponderomotive). These forces appear when the diel.ectric is neutral as a whole. Ponderomotive forces appa.ar In the lo.ng run due to the action of a nonunifo:m ~lec.tnc field on dIpole molecules of the polarized dielectnc ~It IS known that a dipole in a nonuniform electric field IS acte~l upon by a force direcled towards the increasing field). In thIS case, the forces Ilre cansed by the nonuniformi~YI of not only ~he ~acroscopic fIeld hut the microscopic e d as well, whIch IS created mainly by the nearest molecules of the polarized dielectric. . Unde~ t~e action of these electric forces, the polarized dI~le~tnc IS deformed. This phenomenon is called electrostnctwn: As a r~sult of electrostriction, mechanical stresses appear In the dIelectric. ( 0,wing to electrostriction, not only the electric force WhlC~l depe;'1ds on the charges) acts on a conductor in a polanzed dIelectric, but also an additional mechanical f~rce c.aused .by the dielectric. In the general case, the effect o a dIelectrIc on. the resultant force acting on a conductor cannot be taken Into account by any simple relations and the ~I:oblem of calcul.ating the forces with simulta~eous aJlaJ~~Is of th,e mechamsm of their appearance is, as a rule, rather complIcated. However, in many cases these forces can .be clllcnlat:ed in a sufficienLly simple way without a ~et~I1ed :malysls of their origin by using the law of conservatIOn of energy. Energy Method for Calculating Forces. This method is ~,he most general. It allows us to take into account automat~cally' all force interactions (both electric and mechanical) IgiorIng thei~ origin, and hence leads to a correct result. I e~ us consIder the essence of the energy method for calc~ atmg forces. The simplest case corresponds to a situation when charged conductors are disconnected from the powe~ supply. In this case, the charges on the conductors rem?lll unchanged, and we may state that the work A of all mternnl forces of tII:e syst~m upon slow displacements of the conductors and dIelectncs is done completely at the

107

where the symbol q emphasizes that the decrease in the energy of the system must be calculated ,when charges. on the conductors are constant. Equation (4.15) is the initial equation for determining the forces acting on conductors and dielectrics in the electric field. This can be done as follows. Suppose that we are interested in the force acting on a given body (a conductor or a dielectric). Let us displace this body by an infinitesimal distance dx in the direction X we are interested in. Then the work of the required force F over the distance dx is ~A = F xdx, -Where F x is the projection of the force F onto the positive direction of the X-axis, Substituting this expression for ~A into (4. t'5) and dividing both parts of (4.15) by dx, we obtain (4.16) We must pay attention to the following circumstance. It is well known that the force depends only on the position

of bodies and on the distribution of charges at a given instant. It cannot depend on how the energy process will develope if the system starts to move under the action of forces. And this means that in order to calculate F x by formula (4.16), we do not have to select conditions un~er which all the charges of the conductor are necessarIly constant (q = const). We must simply find the increment dW under the condition that q = const, which is a purely mathematical operation. It should be noted that if a displacement is performed at constant potential on the conductors, the corresponding calculation leads to another:expression for the force: Fx = = aw/ax 11jO' However (and it is important!) the result

+

,; !.~ ".' ".'

~ . . .'

... :..L

•.'

...

.;'

i08

4. Energy of Electric Field

4.5. Forces Acting in a Dielectric

of the. calculation of F:c with the help of this formula or (4.16) IS the same, as should be expected. Therefore, henceforth we shall confine ourselves to the application of only formula (4.16) and will use it for any conditions, including those where q =1= const upon small displacements. We must not be confus.ed: the derivative aw/ox will be cakulated at q = const 10 such cases as well. Example: Fin? the !orce acting on one of the plates of a II 1plate ~apaCltor III a .hquid dielectric, jf the distance het~a the ~lateJ h , tre capac~tance. of !he capacitor under given conditions is an vo tage U IS malOtamed across its plates. II?- this case, if we mentally move the plates ap1l.rl the volta e U

:te

full:~~f:::h~~el~~~: ~h~c;~f~I~ ~;i::~/tli:~i~~sh:iin::;j:~ ~e ~orce u;nder t.he assumption that

X

-

x'+dx

-q---- ----x

q

=

const, I.e. WIth the help of formula (1..16). Here the mo.st convenient expresSion for the energy of the capacitor is

~ 2ee oS ' Fig. 4.4 w~ere ~ is the permittivity of the dielectriC, ~ IS the area of each plate, and :r is th~ .dlsta;llce ~etween them (x = h). . ~ext, let us cho~se the posItIve dIrection of the X -axis as is sho\vn In Fthlg. 4.4. :'-CCO~dlllg to (4.16), the force acting on the upper plate of e capaci tor IS c

_ ~

F

W-

+q----
Q

F,,=- oW ax

-n-

"toG

CC= _ _ :r

I=--~

(1)

2ee oS •

q

The minus sign i!I this formula indicates that vector F is directed

~owards the neg~tIv~ values on the X-axis, Le. the force is attractive

y nature. Consldenng that q = we transform (1) to

as

=

Ds =

ee ES and E = Ulh

0 ,

F" = -GUt/2h.

Forces in a Liquid Dieleetric. Formula (1) of the last example shows that the force of interaction between the plates of a parallel-plate capacitor in a liquid dielectric IS smaller t~an the corresponding force in a vacuum by a factor of e (Ill vacuum e = 1). Experiment shows that this resul~ can be gener~liz~d: if the entire space occupied by a ~eld IS ?lled by a lIqUId or gaseous dielectric, the forces of mteractlOn between charged conductors (at constant charaes

'"

109

on them) decrease by a factor of e: F

=

Fo/e.

(4.17)

Hence it follows that two point charges ql and q2 separated by a distance r and placed into an infinite liquid or gaseous dielectric interact with the force F --- _1 -~, 4ne o er 2

(4.18)

which is smaller than the force in a vacuum by a factor of e. This formula expresses the Coulomb law for point charges in an infinite dielectric. We must pay a special attention to the fact that point charges in this law are extraneous charges concentrated on macroscopic bodies whose dimensions are small in comparison with the distance between them. Th'.1s, the law (4.18) has, unlike the Coulomb law in vacuum, a very narrow field of applicationi!<-the dielectric must be homogeneous, infinite, liquid or gaseous, while the interacting bodies must be pointlike in the macros~opic sense. I t is interesting to note that the electric field intensity E as well as the· forceF acting on a point charge q in a h~mogeneous liquid or gaseous dielectric filling the entire space of the field, are by a factor of e smaller than the values of Eo and F 0 in the absence of dielectric. This means that the force F acting on the point charge q in this case is determined by the same formula as in vacuum: F = qE, (4.19) where E is the field intensity in the dielectric at the point where the extraneous charge q is placed. Only in this case formula (4.19) makes it possible to determine the field E in the dielectric from the known force F. It should be noted that Gnother field differing from the field in the dielectric will be acting on the extraneous charge itself (which is located on some small body). Nevertheless, formula (4.19) gives the correct result, strange as it may seem.. Surface Density of a Force. We shall be speakmg of the force acting on a unit surface area of a charged conductor in a liquid or gaseous dielectric. For this. pu.rpos?, let .us consider a parallel-plate capacitor in a hqUld dielectnc.

110

111

4. Energy of Electric Field

Problems

Suppose that the capacitor is charged and then disconnected from the power supply to maintain the charge and the field E of the capacitor constant when the plates are moved apart. Let us consider once again Fig. 4.4. The energy of the capacitor is the energy of the field within it. In accordance with (4.9), this energy is W = (1/2) EDSx, where S is the surface area of each plate and x is the distance between them (Sx is the volume occupied by the field). By formula (4.16), the force acting on the upper plate is

S l t' In accordance with (4.6), the intrinsic energy of each h It u e/qo:~l to qm/2 where
4..-1:£0

2

2R I- : ; '

The energy of interaction between the charged shells is equhl tt: the char e of one shell multiplied by the po.tential R), we have

1 Fx=c -8W/8x l q = -"2 EDS,

W 12 ~~ qt

whence the surface density of the force is

1 4n£0

q2 R z =--=

1

q I q2

~.

4)1£0

The total electric energy of the system is 1

W=WI+Wz-j-It'IZ= 4ne

(4.20)

o

(qI ' -2H t· t

'1~ 28

: qlqz) ' 2

R

z

•

• 4 3 Two small metallic balls of radii R I and H 2 are in vacuum at a di~t~nce cl'1hsiderably exceeding their dimensions and ha;e I a certain total charge. Find the ratio ql /q2. bet:-v~en the chafl~es o. t l~ balls at which the energy of.the sys.tem ~s. mlmmaI. What IS the po tential difference between the balls III thiS case? Solutiun. The elecLric energy of this system is

We have obtained an interesting and important result of a general nature (for a liquid or gaseous dielectric). It turns out that the surface density of the force acting on a conductor is equal to the volume density of the electric energy near the surface. This force is directed always outward along the normal to the surface of the conductor (tending to stretch it) regardless of the sign of the surface charge.

1 W=WI-IWz-!-W IZ = 4ne

o

(qI -iL1_qlqZ) 2R -+- 2H 2 ' l ' I

where Wand Ware the intrinsic electric energies of the b~IIs ~qljJ~~~, W is th~ energ) of their interaction (ql
IU h

Problems • 4.1. Energy of interaction. A point charge q is at a distance l from an infinite conducting plane. Find the energy W of interaction between this charge and the charges induced on the plane.

W _ _1_ _. 4neo

Solution. Let us mentally "freeze" the charge distributed over the plane and displace under these conditions the point charge q to infinity. In this case the charge q will move in the potential field which is 'equivalent to the field of a fixed fictitious point charge __q, located at a fixed distance l on the other side of the plane. We can write straightaway

[-..1L +- (q_q,)2 -+2R 2H I

2

J.

ql (qL-qtl

l

The energy W is minimal when aWlaql ,= O. IIence

R

I q ~q I~ Rt+R z

1 q2 W = - -_ _ 4..-1:80

q:.

1

W 12 =;

Rz and qz""'q RI.J.R I

z

'

where we took into account that R I and R 2 are consideraIlly smaller than land

'U'

• 4.2. Intrinsic, mutual, and total energies. A system consists of two concentric metallic shells of radii R I and R 2 with charges ql and q2 respectively. Find intrinsic energies WI and It' 2 of each shell, the energy W of interaction between the shells, and the total electric energy 0 the system W, if R 2 > R I •

q\lq2

= HI/Hzo

The potential of each hall (lIwy can he. considered isob ted) ,is (p ~ q!,R: Hence it follows (rolll tlip ahove I'dallOll that '.I' . : qe, I.e, lhe I'0ll n tial difference is ('qllal to zp['o lor SIlI·,II a d,sll'lh"llOlI. ,

l2

I',

112

4. Energy of Electric Field

• 4.4. Energy localization in the field. A charge q is uniformly distrtbuCed inside a sphere of radius R. Assuming that the dielectric constant is equal to unity everywhere, find the intrinsic electric energy of the sphere and the ratio of the energy Wt localized inside the sphere to the energy W z in the surrounding space. Solution. Let us first find the fields inside and outside the sphere with the help of the Gauss theorem:

Et

=

4 q

neoR

3

r (r

~ R),

113

Problems

As a result of integration, we obtain ..1-- q(qo+q/2)

--

4JH'o

(_1

RI

1_)

R z .•

Remark. If we try to calculate the work in terms o[ the potential as A = q (
q 1 E z =-4---(r;?R). neo r Z

.

We can now calculate the intrinsic electric energy of the sphere:

a W=Wt+Wz=)

00

e~Ef 4nrz dr +) e~E~ 4nr 3 dr= 8n~:R (--}+1).

o

a

Hence, it follows that

Fig. 4.6

Fig. 4.5

1 3q z W=---4neo 5R ' It is interesting to note that the ratio W/W z does not depend on

the radios of the sphere.

• 4.5. A spherical shell is uniformly charged by a charge q. A point charge qp is at its centre. Find the work of electric forces upon the expansion of the shell from radius R t to R z• Solution. The work of electric forces ia equal to the decrease in

the electric energy of the system: A

=

Wt -

W2•

In order to find the difference WI - Wz, we note that upon the expansion of the shell (Fig. 4.5), the electric field, and hence the energy localized in it, changed only in the hatched spherical layer. Consequently,

rJ T Z

W t - W z=

eo

(Ei- En 4m z dr,

and incorrect. T~ is due to the fact that this approac~ does not take into account the additional work pNformerl by electnc forces upon the change in the configuratiOI) of the charge q located on the l'Xpanding shell. • ~.6. A point charge q is at the centr~. of a sphlfical unch~rged conducting layer whose inner .and oute~ radl.l are a and b respectl vcl~ . Find the work done by electnc forces III thiS system upon the remo\al of the charge q from its original position through a small hole (Fig. 4.6) to a very large distance from the spherical layer. Solution. We shall proceed from the fact that the work of electric forces is equal to the decrease in. the .electric ene.rgy of the system. As is well known, the latter is locahzed III the field.ltself. ',fhus, the problem is reduced to determination of the change III the lleld as a result of this process. . . It can be easily seen that the field around the charge q wtll change only within the spherical layer with the inner radius a and the outer radius b. Indeed, in the initial positiono.f .the charg~; there ~as no field in this region, while in the fmal. POSI twn there IS a certalll field (since the conducting spherical layer IS far from the charge q). Consequently, the required work is b

at

A=O-Wl= -

where E t and E z are the field intensities (in the hatched region at a distance r from the centre of the system) before and after the expansion of the shell. By using the Gauss theorem, we find E =_1_ q+qo I 4ne o rZ

~ _ _1_ k 2-

4neo

r2 '

V J\' -EoEz 2- d . a

Considering that R = q/4nEor2 and dV = 4m 2 dr, and integrating, we obtain qZ a-b , 1 = - - - -
H4 . •

4_,_Energy 01 Electric Field

4.7. The work done upon moving capacitor plates apart

A parallel-plate air capacitor has the plates of area S eacb. Find th~ work A' against the electric forces, done to increase the di8tance bct,we~n the plates from x 1.to ~2' if (1) the charge q of the capacitor and (2) It~ voltage U ,fire mallltallled constant. Find the increml'nt.."l of thl;

elpclnc energy or the capacitor in ,he two cases. Snlution. (1) The required work is

pl~l'manent voltage U. F.inu the force /' acting on a unit surface of the

plate from the dielectrIC. Sol~tion. The resultant force / acting per unit area of each plate can be represented as

1=/o-/',

whpl'l', A\ is th~ intensity of the l1eld createJ by one plate (E = u/2£.0). ,It !S In tillS, fJeld that the charge located on the other plate moves. fillS work IS completely spent for increasing the f'lectric energy'

AW-- A', . (2) III this case, the force acting on each capacitor pJat~ will de-

(lpnd on the, oistanc~ between the plat!'s. Let us write the elementary work 01 the force actlllg on a pla1.l~ during its displacement over a distane(' d.1' relative to the other plate:

where we took into account that q

Aft,,!, integration, we obtain

=

eu, £1 =

'

U!2;r, and C

c=

eoS/x.

The ill('J'eIlH'lIt of thp el"('lrie energy of the capacitor is

= lole,

/' =

10 (1 -

=

-A',

(a~amst the e!eetI'lc forces), The energy of the rapacitor deaeases in tillS case. III ordrl' to understand this, we musl ronsider a SOJ'rce main!?i!ling the potentwl diff?rence of the capacitor at a constant value. J hiS source also accomplIshes the work A S' According to the law of conserva,tion ,')f energy, A A' = AlV, whence As = A W _ A' = = ---2.'1 ~__ 0,

+

(It !j .8.

ro~ccs ~ct~!I~ betwee~ conductors in a dieledric. A paral111 the horizontal position, into a liq-

l~l-pl~te capacJ~or IS Im!!H'rsed,

Uid !helectr1c WJth the flIelectric constant 8, filling the gap of width h between the plates. Then thf' capacitor is f'onnected to a SOur('(' of

=

uE

=

f.12/2eo,

(2)

lie)

=

10 (10 -

1) f.oU2/2h 2•

For example, for U = 500 V, h = 1.0 mm and 10 = 81 (water), we get = 7 kPa (0.07 atm). • 4.9. The force ~cti~g on a dielect~ic. A cylindri~,al,layer of a homogeneous di~lectrlc. WIth the dielectrIC c~nstant c IS wtro,duced into a cylindridH capaCItor so that the layer hll~ the gap of WIdth d between the plates. The mean radius of the plates IS Il sHch that R ~ cl. The capacitor is connect~d tp ~ s~ur~e of a perman.ent voltage U. Fwu the force pulling the dlelectl'lc wSlde the capaCItor. Solution. Using the formula W = q2!2C for thf) energy of a capacitor, we fmd that, in accordance with (4.10), the required force is

l'

-

a;: Iq = ~

O~2()X =~~ ~~ .

(1)

Since d ,,<:: R the capacitance of the given capacitor can b~ calcul~ted by the formula for a paraIJel-plate capacitor.. Therefore, .If the dielectric is introducer! to a depth x ann the capacitor length IS l, we have Cc= f.f.ox~2nR

~hus, by movi,ng the plates apart, we perform a positive work

10

where E is the field intensity in the region occu~ied.by one plate, created by the charges of the other plate. ConsJderwg that a c= D = = eeoUlh and substituting (2) into (1), we obtain

Fx =

It should be noted that AW

(1 )

where /0 is the electric force acting per,unit area of a plate from the other plate (it is just the force per umt area when the dielectric is absent). In our case, we have

I

IiA'= qE1.dx= eoSU~~ 2 x2

H5

Problems

+

Eo

(l-;>.2nR

eo.~nR (ex-j-l-x),

(2)

Substituting (2) into (1), we obtain Fx

=

eo (e -

1) nRU2!d.

• 4.10. A capacitor consists of two fixed plates in the form of a semicirck of radius R and a movable plate of thiekness h made of a dielectric with the dielectric constant 10, placed between them. The latter plate can freely rotate about. th(' axis 0 (Fig. 4.7) and practica~­ ly fills the entire gap between the l1xed plat!~s. A constant voltage f! IS maintained between the plates. Find the moment M about the aXIs 0 of forces ading 011 the movable pIa te wllPn it is placed as shown in the figure. SolutitJrt. The work performed by the moment of forces M upon 8*

116

5. Direct Current

5.1. Current Density. Continuity Equation

thc rotation of thc platc through an ". I ., I .' tit'crcnsc in the electric cnergy of the ..gtc"e ctlllcnt da IS equal to thc . sys em a q = const [see (4.16»): _'\;/ z da= -dWl q , where W = q2 /2C. Hence

M z = - oW /. =L~ iJa q 2 C2 • (1) In the case under consideration, C = C + C h dr:l~~frfcap~~tances of the parts ?f the cap~citor ~ith a~d ;;\h~~~ tf: . e area 0 a sector wl~.h an angle a is determined as S = = aR2'12, and hence C =~ EoaR2/2h + EEo (n - a) R2/2h. Differentiating with respect to a we find ac,liJa = (EoR2/2h) (1-E). SUbstitutin t~IS ~xpression into formula (1) and con~ sIderlllg that C = qlU, we obtain U2 eoRI M Z =-2-

Fig. 4.7

2h (1-e)

= -(8-1) 8oR2U2

'4h

<

0

.

The ncgative sign of M indicat th h acting clockwise (oppositelyZ to the eps 't~t t de. mOI?ent of the force is see Fig 4 7) Th' , OSI Ive IrectlOn of the angle a' pacitor: " IS moment tends to pull the dielectric inside the ca~

~n :~;i~fb;i~~~ ~~:~ ~I~

fh:n~ependent of~he angl~

AJz a .. However, IS due to the fact that for smail vaIu oment M Z - o..ThIS dIscrepancy a as was done in the solut:on of tb' es of we cannot Ignore edge eflects , ' I S pro b l em. 5. Direct Current 5.1. Current Density. ContinUity Equation

Electric Current. In this chapter we sb~ll co ~ selves to l' f .,. nlme ourd' . an a~a YSI~ 0 conduct JOn CUl'l'ent in a conducting me HIm, espeCIally m metals. I t is well known that ele t . ~ur(rent iSh the transfe~ of charge through a certain sU~f:~~ say, t e cro~s sectJOn of a conductor). In a. conductmg medium, current can be c . d b trons/Ulhmetals), ions (in electrolytes), or sOr:;~fl:the/p:~~~= ~l~as~ti~ ~ \ ~bsenc~ of electric field, curren t carriers perform f' 0. JOn all( on average the some number of carriers ~ur eIther sign passes tllrough each side of any imaginal' face S. Thus, the current passing through S in this 'cas~

1t7

is zero. However, when an electric field is applied, an ordered motion with a certain average veloeity u is imposed on the chaotic motion of the carriers. and a current flows through the surface S. Thus, an electric current is essentially an ordered transfer of electric charges. The quantitative measure of electric current is intensity I, defined as the charge transferred across the surface S in a unit time: 1= dQ/dt. The unit of current is the ampere (A). Current Density. Electric. current may be distributed nonuniformly over the surface through which it passes. Hence, in order to characterize the current in greater detail, current density 'vector j is introduced. The magnitude of this vector is equal to the ratio of the current dI through a surface element perpen,dicular to the direction of motion of charge carriers to th€" area dS J of this element: j =ceo dI/dS J. For the direction of vector j we take the direction of velocity vector u of the ordered'motion of positive carriers (or the direction opposite to that of the velocity vector of the ordered motion of negative carriers). If the current consists of positive and negative charges, its density is derIDed by the formula (5.1) i = P+U+ p_u_,

+

where p+ and p_ are the volume densities of positive and negative charge carriers respectively, and u + and u_ are the velocities of their ordered motion. In conductors,where the charge is carried only by electrons (p_ < 0 and" u+ = 0), the current density is given by j = p_u_. (5.2) The field of vector i can be graphically represented wnh the help of lines of current (lines of vector j), which are drawn in the same way as for vector E. Knowing the current density vector at each point of the surface S under consideration, we can also fLlld the C'Irrent passing through th is fiurfflce f1S the f1nx of vector j: I

=

J j·dS,

(d)

118

5. Direct Current 5.2. Ohm's Law for a Homogeneous Condnctor

-. Current I is a scalar al ebra" from formula (5.a) that b~side;co{luanttty . It can be seen the current is determined b th 1 I~r factors, the sign of, the normal at each point 0 YtJ e cflolce of the direction of of tlI e d"ll"eetlOn of vectorsn dSlie Sur i e: b Y tl Ie ch' If hace S. " Olce vectors dS are reversed th . t e dIrectIOns of all the Continuitv Equation' L e curr~nt I changes sign. . .J' • e t us Imagine a I se rl III ? conllucting medium throu ,h wI . c.o . sUl'face S It IS customary to choose the ~ut nch a. cUl~ent IS passing. I h ward dIrectIOn for vectors normal to closed surfaces , aD( ence for ve t dS . sequently, the integralh . d .,' COl'S. Conthe volume V (' I J'] S gl\ es the charge leaving A enve oped by the su f S) . ccording to the law of 'conse t' I' ace per unit time. IS ?qu~l to the decrease in tl rva ~o~ of .ch?rge, this integral unIt tIme: 18 c alge IllSHle volume V per

I~idL -*.) 'C

dS= O.

(5.5)

This means that in this . start or terminate any' I case, the lInes of vector j tlo not to have no sources in tl:-'" ~ere. The. fteld of vector j is said ,e case of dIrect current. 5 ' Differential form or the conti . t ( JI) and (5.5) in the diITt'n'ntial forn~ul~ y ell~~ation. Let us write Eq~.

j'

.

. or t liS purpose, we express the ch Ulo'e q u.s p d V J I ' ... . aUf tit' right··hand ~ide of Eq (5 4)" d __ JiJ ' . as --at P dl ~, -- -P d I' J . 1ii'. "lere, we have taken the partial time derivative of SineI' r can dpPl'llll 011 lim. , . P. C .IS wl'I1 as on coordinatp< TI c HIS,

J .

'-0

(r,;Y j ciS -

_

I' J

-.± (,Iii at .

Procl'edin~

in the same way as for the flux of vector E in Sec. 1.4, we lind that the divergence of vector j at a certain point is equal to the decrease in the charge density per unit time at the same point:

r V·j= -ap/at. I This lcatls to thc steady-state condition (when iJp/DI

(5.6)

=

0):

I V·j=o.j

(5.7)

This Illeam; that Ior dircct current the field of vector j does not have

sources.

5.2. Ohm's Law for a Homogeneous Conductor

Obm's law, which was discovered experimentally, states: the current passing through a lwmogeneous conductor is proportional to the potential ditJerence across its terminals (0[' to the l.101taf'!.' .ftfr T

(.5.4)

Tllis relation is called the '. pression for the law of . .contl~ulty equation. I t is an exIn the case of a t dcollservatI~n of charge. s ea v-state (du'CCt) · t 'b d IS n lItion in Sp'lce . 't ." current, the charge W I' I mu~ remalll unch I I . Of( S, 011 t IS right-hand side of E allge(. II other sequently, for direct C\lr' t I q. (5.4) dq/dt = O. COIl, " Tell we lave.

~j

1 t!J

,.

l)

JI=UiR,

l

(5.8)

where Jl is the electric resistance of the conductor. The unit of resistance is the ohm (Q). Resistance R depends on the shape and size of the conductor, on its material and temperature, and above all on the confIguration (distribution) of the current through the conductor. The meaning of resistance is quite clear when we are dealing with a wire. In a more general case of the volume distribution of current, it is meaningless to speak of resistance until we know either the position of the leads attached to the eonductor or the current configuration. In the simplest case of a homogeneous eylindrical eOllductor, resistance is given by l

R = Ps'

(:i.!l)

where l is the length of the conductor, S is its cross-sectional area, and p is the resistivity, which depends on the material of the conductor and its temperature. Resistivity is measured in ohm-metres (Q·m). Resistivity of the best conductors like copper and aluminium is oC_the order of 10-8 Q·m at room temperature.

5.2. Ohm's Law for a Homogeneous Conductor

5. Direct Current

120

Differential Form of Ohl:!.'s Law. Let us establish a relation between the current density j and the field E at a point of a conducting medium. We shall confine ourselves to an isotropic conductor in which the directions of vectors j and E coincide. Le~ us me~tally iso!ate. around a certain point in a conductIng medIUm a cylIndrICal volume element with .generatrices parallel to vector j, and hence to vector E. If the cylinder has a cr?ss-sectional area dS and length dl, we can write on the basIs of (5.8) and (5.9) the following expression for such cylindrical element:

]'dS =

E dl pdlldS •

After cancelling out common terms we obtain the following equation in vector form:

Ij~fE~.E,

wh?re a

=

I

(5.10)

1/p is the conductivity of the medium. The unit an ohm is called a mho (Siemens). Consequently, the UOlt .for the measurement of a is mho per metre. EquatIOn (5.10) is the differential form' of Ohm's law It does not .contai~ any diff~rentials (derivatives), but i~ called the dIfferentIal form SInce it establishes a relation between quantities corresponding to the same point in a conductor. In other words, Eq. (5.10) is an expression for Ohm's law at a point. Methods for Calculating Resistance (R). There are several methods for measuring resistance and all of them are ultimately ba~ed on the application of relations (5.8)-(5.10). The expedIence of using a particular method is determined by the formulation of the problem and by its symmetry. The practical applications of these methods are described in Problems 5.1-5.3 and 5.6. .Charge in a Current-carrying Conductor. If the current is dlrect (constant) the excess charge inside a homogeneous ~Qndu~tor is ~qual to zero everywhere. Indeed, Eq. (5.5) IS valId for dIrect current. Taking into account Ohm's law reClpro~al to

121

in the form (5.10) we can rewrite (5.5) as follows:

l~ aEdS=O, u

where the integration is performed over any closed surface S inside the conductor. For a homogeneous conductor, the quantity a can be taken out of the integral:

a~ EdS=O. ' According to the Gauss theorem, the remaining i?te.gral is proportional to the algebraic sum of the charges mSlde the closed surface S, Le. proportional to the excess charge in this surface. HowEn ever it can be directly seen from the last' equation that this integral is equal to zero (since a =1= 0). This means that the excejEl charge is also equal to . zero. Since the surface S is chosen arbitrarily, the excess char,ge 'under th~se conditions is always equal to zero mside a conductor. , Fi~. 5.1 The excess charge can appear only at the surface of a homogeneous conductor at places where it comes in C,I}lildlL with other conductors, as well as in the regions wher8 the conductor has inhomogeneities. Electric Field of a Current-carryillgConductor. Thus, when current flows, an excess charge appears on the surf~ce of a conductor (inhomogeneity region). In accordance WIth (2.2), this means that the normal component of vector E appears at the outer surface of the conductor .. Further, from the continuity of the tangential component ?f vector E, we conclude that a tangential component of thIS vector also. exists near the surface of the conductor. Thus, near the. surface of the conductor, vector E forms (in the ~resence ofthe current) a nonzero angle (J. with the vertical (FIg. 5.1). When the current is absent a = O. . . . If the current flows in a steady state, the dlstl'lbutlOn of electric eharges in the conducting medium (generally speaking, inhomogeneous) does not change in time, although

122

5. Direct Current

the charges are in motion: at each point, new charges continuously replace the departing ones. These moving charges create a Coulomb field which is identieal to tbe one created by stationary charges of the same configuration. This means that the electric fIeld of stationary currents is a potential field. At the same time, electric fIeld in the case of stationary currents differs sig-nineantly from electrostatic, or Coulomb, field of fixed charges. If charges are in equilibrium, the electrostatic fwld inside a conductor is always equal to zero. The electric field of stationary currents is also an electrostatic (Coulomb) field, although the charges inducing this field are in motion. Hence, the field E of steady-state currents exists also inside a current-carrying conductor. 5.3. Generalized Ohm's Law

Exlrant'ous Forces. If all the forces acting all charge curriers wcre reduced to electrostatic forces, the positive carriers would move under the action of these forces from regions with a higher potential to those with a lower potential, while the negative carriers would move in the opposite direction. TIds would lead to the equalization of potentials, and the potentials of all mutually connected conductors would become the same, thus terminating the current. In other words, lmder the action of Coulomb forces alone, a stationary licld must be a static field. In order to prevent this, a direct CUl'l'cut circuit must contain, besides the subcircuits in whieh positive carriers move in the direction of deereasing potential q;, subcircuits ill whieh positive carriers move ill the direction of increasing If' or against the electric field forces. In these subcircuits, the transfer of carriers is possible only through forces that are not electrostatic. vVe shall call such forces extraneous. Thus, a direct current can be sustained only witlt the help of extraneous forces whicu are applied either to certain parts of the cirellit, or to the entire eircuit. The physieal naturo of oxl I'illH'OliS forcos may be ql1ite diverse. Snell fOl'ecs may bo ('!"pated by physieal or chemical inhomo,gclloili(~s ill a ('olldllclor, rot· O\,llllplc, roreps l'esllltin,g from a cO/llact ol" rondllctors or different types (galvanic

5.3. Generalized Oh!!'.m~·s~L~a~w~

-----------

-_-1_2~3

cells, accumulators) or eondllctors at different temperatures (thermocouples). . · d Ohm's Law In order to deSCrIbe extraneous GIenera Il z e · f e fJ el d forces quantitatively, the concepts of extrane?us orc. I . and its strength E* are introduced. The lat~er IS .n~mencallY equal to the extraneous foree acting on a umt posltl\:e C~I~~~ Let us now turn to current density. If an ~lec~r~ e it generates in a conductor a current of d.enslty l - oE, E is obvious that under the combined actlOn of the, field '11 and the extraneous force field E*, the current denSIty WI be given by

t

j=o(E+E*). \

(;).11)

This equation generalizes the law ~5.10) to, the ~ase ~; inhomogeneolls regions of a conductlllg medI~lm, and called the generalized Ohm's law in the ~iffe~entIaI fOfl~l. Ohm's La,,' for a Nonun iform Subcll'cml.. Nonumform subcil'cuits are characterized by the presence of extraneOIlS ,., t t forces in t h e m , ' Let us consider a special, but practICally qUIte.lmpor an case when a current flows along a thin wire. In thIS ~ase, t~Ie d'il'ec.tion of the current will coincide with the aXIs of t 10 wire and the current density j can be assum~d to be th~ sa~~ at all points on the cross section of the WIre, Suppose th .. ~he cross-sectional area of the wire is equal t~ 8 (8 need not be the same over the entire length of the WIre): _1- . Dividing- both sides of Eq. (5.1~) by ~, formIIlg a SC,l HI '. prodnc.t of the resulting expresSIOn ":Ith the element. dl of the wire along its axis from erosS s~c.tIOn .~ to ~ross sect I?" 2 (this direction is taken as the P?SItlve dIrectIOn) and r~~~ tegrating over the length of the WIre between the two c c' sections, we obtain 2

2

2

j

1

1

\1:1=~Edl+~E*dI.

Let lIS (J re.p]oc(' .. ~ . of vee tor that jt is

(;).12)

transform the integrand in the, fl.rst integr,al: .we hv iff) and J. ill by jidl, where JI IS the pl'OJectIOll. .' " . . II F·t! we IIOt(' j onto the (hrcdlOII of vecLor {, III \Cr, .' .~ an algebraic quantity and depends OIl the Ollenta

124

tion of vector j relative to dl: if j tt dl then j 1 > 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

III order to pl'OVl' Ihi~ law. il i~ ~ul'l'J('il'ul to eOllsidl'" the C,lse when lhe circuil cOllsists of three su!lcireuils (Fig. :i.:l)_ LeI liS assunH' Ihe dirl'dioll of cil'CllmVl'lIlioll to he clockwise,

Fig-. 5.;>

Fig. 5.4

as shown ill lhe figure. Applying Ohm's Jaw (;j.1S) of the three suhcircuits. we obtain 11 R I = qJ2-- cps

+ 'f l ,

5.4. Branched Circuits. Kirchhotrs Laws

12 Hz = cps -

CfJI

-t- 'f 2 ,

Calculations of branched circuits, for example, determination of current in individual branches, can be considerably simplified by using the following two Kirchhoff's laws. Kirchhoff's Fircit Law pertains to the junctions, Le. branch points in a circuit, and states that the algebraic sum of the currents meeting at a junction is equal to zero:

I sRI. = qJl -

IP2

+ '(s·

~Ik

=

O.

(5.17)

~fere, currents converging at a junction and diverging from It are supposed to have opposite signs; for example, we can assu~e the forme~ t? ~e positi~e and .the latter. to be negative (or VIce versa, tillS IS ImmaterIal): When applIed to Fig. 5.4, Eq. (5.17) assumes the form II _.- 1 2 + 1 3 0-== O. Equation (5.17) is a consequence of the steady-state condition (5.7); otherwise, the charge at junction would cllilnge and the currents would not be in a steady state. Kirchhoff's Second Law. This law is applicahle to any closed contour in a brandied eirenit dnd states that the algebraic sum of the products of cwnfit and resistance in each part of a network is equal tv the algebraic sum of e.m.f. 's in the circuit: '

a

(5.18)

10

cach

Adding these equations and cancelling all potentials, we arrive at formula (5.18), i.e. Kirchhoff's second law. Thus, Eq. (5.18) is a consequence of Ohm's law for 1l011lIniforIll subcireuils. Setting up of a System of Equations. In each specifIC case, 1\ irchhoff's laws lead 10 a complete syslem of alg-cbrait: equaliollswhich call he used, say, for l'ilHlilig all the llllknown currents in the circllit. The number of equations of the form (:'>.17) and (5.18) Illust he equal to the Ilumber of unknown quan! ilies. Care should be taken to ensure that Ilone of the equiltions is iI corollary of allY olher equalioll ill lhe system: (1) if a branchell circuit has N jllnctiolls, illdejwllt!I'nt equations of lype (5.17) can he set up only for N 1 JUIII'lions, and the equation for lhe last jUliet ion will he iI r'ornllary of tile preceding equations; (:2) if a D!'i111clied cirelli! containC' seve!'al ell'S!"! Inops. tlw indepc'nl[pill eqllil1ions o[ t~'pe C,.iS) Coli! 1)(' ~e! up only fol' loops which eal!lIo( Ill' "htainpd hy !'u!lL'rilililosillg" the 100Jls con:,idl'l'ed heron'_ 1"01' p\amplp, "111'11 r-ljll,lI iO/lS will Lt, independl'lil for loops 7.'24 and 234 oj' the

Iii

ii

'Ii

129

5.5. loule'lJ Law

128

5. Direct Current

circuit shown in Fig. j,G. Equation fot' tho loop 1234 will follow from the two preceding olles. It is possible to set up autonomous equations for two other loops, say, 124 and 1234. Theil the equation for contour 284 will be a 3

r----~4

1

Fig. 5.6

Fig. 5.7

consequence of these two equations. The number of autonomous eql..\ations of type (5.18) will be equal to the smallest number of discontinuities which must be created in a circuit . in order to break all the loops. This number is equal to the number of subcircuits bounded by conductors if the circuit can be drawn in a plane without intersections. I:or example, it is necessary to set up three equations of type (5.17) and three of type (5.18) for a circuit (Fig. 5.7) containing four junctions. This is so because the number of discontinuities (marked by crosses in the circuit) breaking all the loops is three (the number of subcircuits 'is also equal to three in this case). If we assume the currents to be unknown the number of discontinuities will be six in accordance with the number of subcircuits between junctions, which corresponds to the number of independent equations. The following procedure should be adopted while setting up equations of type (5.17) and (5.18). 1. The directions of currents are marked hypothetically by arrows, without 'caring for the direction of these arrows. . If a certain current turns out to be positive as a result of calculations, this means that the direction chosen for it is correct. If, however, the current is found to be negative, its actual direction will be opposite to the one pointed by the arrow. 2. Having selected an arbitrary dosed loop, we circumvent its sections in one direction, say l clockwise. If the assumed direction for any current coincides with the

direction of circumvention, the:corresponding Iterm IR .in (5.t8) should be taken with the positive sign; in the OPPosl~e case, the minus sign should be used. The sam~ pr?cedure ~s applicable to lJ: if an e.m.f. increases po.tentIaI 1Il th~.dl­ rection of circumvention, it should be assIgned the pOSItive sign; the negative sign should be used in the opposite case. Example. Find the ma~nit~de and d.irec~ion of the cur~ent passing through resistor R in the CIrCUIt shown m Fig. 5.8. All reSIstances and e.m.f.'s are assumed to be known. R There are three subcircuits, and hence, three unknown currents I, I).. and 1 2 in this circuit. We mark (arbitrarily) the supposed directions of these currents by arrows (at the right jU!lction). . . I The circuit contams N = 2 JunctIOns. 12 This means that there is only one independent equation of type (5.17): I

+ II + 1

2

= O.

Let us now ,Met up equations of type lS2 (5.18). According to the number of subcircuits there should be two of them. Let F' 58 us con;ider the loop co.taining Rand Ig. . R I and the loop w~th R an~ R 2 • Taki.ng the clockwise directIOn for clrcumventmg each of these loops, we can write -IR

+ I 1R 1 =

-1!!1>

-IR

-+ I 2R 2 =

1!!2'

We can verify that the corresponding equation fo.r the lo~p with Rand R can be obtained from these two equatIOns. Solvmg the I • . we 0 btam system of 2three equatIOns, 1- -R I I!!2+ R 21!!1 - R I R 2+RR I +RR 2 '

If we fmd that as a result of substitution of numerical .valu~s into this equation I > 0 the current actually flows as shown III Fig. 5.8. If I < 0, the cllrre~t flows in the opposite direction.

5.5. Joule's Law

The passage of current through a conductor having a resistance is invariably accompanied by liberation of heat (heating of conductors). Our task is to find the quanti~y of heat liberated per unit time in a certain part of the mr\I-OIM I

5. Direct Current 130._-------=:.:..=..::..:..::.:...::..::::..:.;:.:::..--------

cuit. We shall consider two cases which are possible, viz. uniform and nonuniform subcircuits. The problem is considered on the basis of the law of conservation of energy and Ohm's law. Uniform Subcircuit, Suppose that we are interested in the region between cross sections 1 and 2 of a conductor (Fig. 5.9). Let us find the work 2 done, by the field during a time interval dt in the region 12. If the cuuent through the conductor is equal to I, a charge dq = I dt will pass through each . cross 'section of the conductor . FIg, 5.9 during the time dt. In particular, this charge dq will enter cross s~ction 1 and the tsame charge will leave cross section 2. ~lDce ~harge distribution in the conductor remains unchanged ~n th~s case (the current is dire~t), the whole process IS eqUIvalent to a transfer of charge dq from section 1 to section 2 with potentials CPI and CPI respectively. Hence the work done by the field in such a charge transfer is 6A =dq (CPI - CPI) = I (CPI - CPa) dt. According to the law of cons9l'vation of energy, an equivalent amount of energy must be liberated in another form. If the conductor is stationary and no chemical transformations take place in it, this energy must be liberated in the form of internal (thermal) energy.. Consequently, the conductor gets heated. The mechanism of this transformation is quite si~ple: as a result of the work done by the field, charge . c~rrlers . (fo~example, electrons in metals) acquire an ~ddItI.onal.kmetIcenergy which is then spent on exciting lattIce VIbratIOns due to collisions of the carriers with the lattice sites' atoms. Thus, in accordance with the law of conservation of energy, the elementary work 6A = 'Q dt, where Q is the heat liberated per unit time (thermal power). A comparison of this equation with the preceding one gives

------------_.

5.5. Joule's Law

131

~~:.::.::::.::-.::...:.=~---------

Since in accordanee with Ohm's law CPI - lJl2 = RI, we get (5.19) This is the expression for the well-known Joule's law. We shall now derive an expression for the differential form of this law, eharacterizing the liberation of heat at different parts of a conducting medium. For this purpose, we isolate a volume element of this medium in the form of a cylinder with its generatrices parallel to vector j, viz. the current density at the given point. Let dS and .dllbe the cross-sectional area and length of the small cylmder respectively. The amount of~eat liberatedin .this volu~e durin~ the time dt will be given, m accordance WIth Joule s law, by

fJQ =;,RI2 dt = ~~l (j dS)2 dt = pj2 dV dt, where dV = dS dl is the volume of the small cylinder. Dividing the last equation' by dV dt, ':Ie .obta.in a fo 7mula for the amonnt of heat liberated per umt tIme m a umt volume of the conducting medium, or the thermal power density of the currtmt:

(5.20) This formula expresses Joule's law in the differential form: the thermal power density of current at any point is p'rop~r­ tional to the square of the current density and to the res£stw£ty of the medium at that point. .

Equation (5.20) is the most. general. form, of l.oule's law, applicable to all conductors -IrrespectIve of thel~ sh~pe or homogeneity, as well as the nature of .forces WhIC~ mduce the electric current. If the charge earners are subJected to electric forces only, we can write, on the basis of Ohm's law (5.10), Qd=j.E=aE2.

(5.21)

This equation is of a less general n~tur~ than (~.20). Nonuniform Subcircuit. If a subClrcUIt contams a source

.5,6. Transient Processes in a Capacitor Circllit

5. Direct C', rent

132

----

pJ'2•,,-~• Q"d (see (5.20». 1Il an mllOffiogeneous medium can then he written in the form

of c.m.f., the charge carriers will be subjected not only to electric forces, but to extraneous forces as wel1. In accordance with the law of cOllservatioD of energy, the amount of heat liberated in this case will be equal to the algebraic sum of the works done by the electric and extraneous forces. The same applies to the corresponding powers: the thermal power must be equal to the algebraic sum of the powers due to electric and extraneous forces. This can be easily verified by multiplying (5.15) by /: R/2

=

(
'P2) /

+ W,/.

.

. Ior tl 13 t lJ,."

COIlSH

c.c-=

1/p

311d

(5.24)

5.6. Transient Processes in a Capacitor Circuit

0

(5.22)

Transient Processes. These are th~ processes i?vol.ving a transition from one stationary regIme of the CircUlt to I another. An example of such a processis the charging and dischar~ing of.a capacitor, which will be conSIdered III detail in this section. . C q± R So far, we have considered only (hrect current#It turns out, h~wev~r, that , K most of the laws obtained lfl thIS case are also applic~ble to .alterna~ing c~lrFig. 5.10 rents. This applIes to aU cases lD Wlll.ch the variation ofeurrent is not too rapId. . )I The 'instantaneous value of current will then practlca y be the same at all cross sections of the eircuit. SUC~I cll~Tents . ted' W'\·tli aIllI el1e ld s aSSOCIa ' them . are called quaslstatzonary . d (a more exact eriterion for quasi~tationary currents an fields is given in Sec. 11.1). . I • I ws Quasistationary currents can be deSCribed. by t 10 a , obtained 'for constant currents by applying these laws to instantaneous values of quan~ities... . . r f a Let us now caIlsidet' the dlschargmg and ChalgIIlg 0 . processes that t h e 'curren t s in these ' capacitor, assumwgI I are quasistationan. Discharging of ~ Capacitor. If the plates of a c larg~( c,apacitor wilh eapacitance C are connected t~lI'ougILa r~~Hs­ tor R a current will flow through the resistOl'. et ,q and U be the instantaneous values ~f th~ current, ~harge of the positive plate, ana the potential lhfference bet.ween the plates (voltage). . ' .\ ' Assuming that the CIIlTent I is p~sitIve whel~. It from the positive plate to the negatIve plate (F Ig. . ,

The left-hand side of this expression is the t,hermal power Q liberated in the subcircuit. In the presence of extraneous forces, the value of Qis determined by the same formula (5.19) which is applied to a uniform subcircuit. The last term on the right-hand side is th..! power generated by extraneous forces in the subcircuit uNIHf consideration. It should also be noted that the last H'1'ill (ll) is an algebraic quantity and, unlike R/2, it roverSt~;~ its sign wh!m the di. rection of the current / is revt)f<;ed. Thus, (5.22) indicates that tho heat liberated in the region between points 1 and 2 is equnl, to the algebraic sum of the electric and extraneous powers. The sum of these powers, Le. the right-hand side of this equation, is called the electric power developed in the subcircuit. It can then be stated that for a stationary subcircuit, the thermal power evolved is equal to the electric power developed in the subcircuit by the current. Applying (5.22) to the entire unbranched circuit ('PI = =
,

134

5. Direct Current

C ·t· Circuit 135 5.6. Transient Processes i.n a apacl or .~----

---------

we can write I = - dqldt. According to Ohm's law, we obtain the following expression for the external subcircuit having resistance R: .

RI

dq

+

If

RC ~oc

= giG,

O.

where go is the initial charge on the capacitor and constant having dimensions of time:

= RG.

(5.25)

T

o

is a

dq

Fig. 5.12

Fig. 5.11

U

that I = dq/j.t and 'P2 - qJI = the last eq u:ation in the form

(5.27)

This constant is called relaxation time. It can be seen from (5.26) that T is the time during which the capacitor charge decreases to 1Ie of its initial value. Differentiating (5.26) with respect to time, we obtain the law of variation of current:

1= -(it=/oe- tlt ,

.

we can

After separating the variables in this differential equation and integrating, we get q =., qoe- Itt, (5.26)

T

+

qJI qJ2 ~, . . of the circuit, includlllg where R is the total reslstalnce f source. Considering the internal resistance of t le e.m..

RI =

U.

=

COJisidering that I = - dqldt and U transform the above equation as follows:

dt

t 0 tl1e loop 1'f,R2, we get

'.

dq

.ccco

qlC, we can rewrite

•

dt"=

8-q/C R .

Separating the variables, we get

o

Rdq =dt. 8-Q/C

(5.28)

where 10 = qo/T is the current at the instant t = O. Figure 5.11 shows the dependence of the capacitor charge q on time t. The / vs. t dependence also has the same form. Charging of a Capacitor. Consider a circuit in which a capacitor G, a resistor R and a source of e.m.f. '0 are connected in series (Fig. 5.12). To begin with, the capacitor is not charged (key K is open). At the instant t == 0 the key is closed, and a current starts to flow through the circuit, thus charging the capacitor. The increasing charge on the capacitor plates will obstruct the passage of the cunent and gradually decrease it. The ClllTent in the circllit will now be assllmed positive if it flows toward the positive plate of the cnpacitor: I = dqldt. Applying Ohm's law for a non\luifonll sHbcircuit

Fig. 5.13

Integrating this equation under __ 0 we obtain the initial COli d 1't'1011 q -- 0 at t -- , RG In( 1 = - t,

;c )

_ (1 _e-t1t). . (5.29) q- qm · · t 'rh'\rge _ ",C is the limitillg value of the capHl~1 or ' ( Hero qm - _ RC . (for t -+ 00), HIIlI .~ 'tl' time according to the followlIlg The current vallCS WI 1. law: (5.30) 1= ~~ = loe- tlt ,

whence

=-

where 1 0 = 'f,1ll. f d I on t are shown in Fig. 5.13. The depemlences 0 q an

'i·

~

~ ;~:"""

._'

,;i_

';'i,

,

136

5. Direct Current

Problem'

Problems

conductors and C is the mutual capacitance of the conductors in the presence of the medium. Solution. Let us mentally supply the charges +q and -q to the conductors. Since the medium between them is poorly conduc.ting. the surfaces of the conductors are equipotential, and the neld configuration is the same as in the absence of the medium. Let us surround, for example, the positively charged conductol' by a closed surface S directly adjoining the conductor and calc:ulat.e R and C separately:

.• 5.~. Resistance of a conducting medium. A metallic s here of ~adlUs a IS surrounded by a concentric thin metallic shell of d' b fhe space bet,,:een the~ two ele~tr?des is filled with a homoa e~~~u~ poorly conductIng medIUm of reSIstIVity p. Find the resistanc: of th gap between the two electrodes. e

l

d ~olution. Let us ~nentally isolate a thin spherical layer between 11. rand r t dr. Lmes of current at all points of this layer are rpe~dlcular to It, and therefore such a layer can be treated as a cy\'fndrIca! conductor of length dr and a cross-sectional area 4 . Z U' formula (5.9), we can write nr. SIng ra

dr

dR=p -4nr Z

=..i!..4n:

(1.._1.) a b •

• 5.2. Two identical metallic balls of radius a are placed in a hOI.nogeneous boorly conducting medium with resistiVity p F'nd th ~hSlsJ~~~e of bt e medium b~tween the balls under tbe conditi~n tba~ e IS "nce etween them IS much larger than their size. . Solution. Let us mentally impart charges +q and _ to th b II Smce ~he ba}ls are at a large distance from one anothe/ elect~c fletd nhar t e sh" ace of each ball is practically determined only by the c ~rge 0 f t ~ n~arest sphere, and its charge can be considered to be uIlIformly dIstrIbuted over the surface. Surrounding the ositiv I ~hargedball. by a concentric sphere adjoining directly the Call'~ s~l­ face, we WrIte the expression for the current through this sphere; 1= 4lta 2j,

u

u

(J~EndS '

ee o ~ E" dS

q C=(f

Integrating this expression with respect to r between a and b, we get R

U

R=T-

137

U

where the integrals are taken over the given surface S. While calcu.luting R, we used Ohm's law in the form j = (JE, and C was calculated with the help of the Gauss theorem. Multiplying these expressions, we obtain ",' RC = eoe/a = eoEp. • 5.4. Conditions on the bounda~y of a conductor. A conductor with resistivity p borders OJ! a dielectric whose dielectric coustant is e. The electric induction at a certain point A at the surface of the conductor is equal to D, vectorD being directed away from the conductor and forming angle a with the normal to the surface. Find the surface charge density on th.e conductor and the current density near the point A. Solution. The surface charge density on the conductor is given by (J = Dn = D cos a.

:thiStreSfUtlht is vadl~d regardless of the magnitude of the dielectric con"an 0 erne mm .

The current density can be found with the help of Ohm's law: j = = E/p. It follows from the continuity equation (5.5) that the normal components of vector j are equal, and since in the dielectric ill = 0 (there is no current), in the conductor we also have in = O. HeDt:e, vector j in the conductor is tangent to its surface. The same applies to vector E inside the conductor. On the other hand, it follows from the theorem on circulation of vector E that its tangential components on. difIerent sides of tb;e interface are equal, and hence E = E t = J) SIn alee., where E", 15 the tangential component of vector E in the dielectric. Taking tJieBe arguments into account, we obtain E D sina j eoep

. . ., 5.3. Two conrl~ctors of arbitrary shape are placed into an in~~dtet~~od?~e~e~~s slIghtly cond.ucting medium with the resistivity p t.hi' w ec~n: cons~~nt e. FJ!Id the value of the product RC for Ii system. \It here R lli the rClillitunce of the medium between the

• 5.5. The gap between the plates of a parallel-plate capacitor is filled consecutively by two dielectric layers 1 and 2 with thicknesses 11 and 12 , dielectric constants CI and E2' an~ rcsi8tiviti~s PI and f,,' The capacitor is under a permanent voltage (;, the electrIC field bemg

where j is the current density. Using Ohm's law (/' formula E = q/4neoa2, we obtain

=

Elp) and the

1= qleop.

Let us now flIld the potential difference between the balls: U = lp + - lp_ ~ 2ql4nE oa. The sought resistance is given by R = UII = p/2na.

=p

t38

directed from layer 1 to 2 F' d h charges at the interface b~tw::n tthe sd'!rlfacte. delnsit y of extraneous . e Ie ec ric ayers. . S oI utJon. The required f h sur ace c arge density is given by a = D s n - ,D In- 8 02 8 E 2 - e 8 1 E l' (1) 0 In order to determine E and E h I since h = ii' it follows lhat E w~ sEall make use .of two conditions: = U. Solving the two e a ' I PI - 2 P2' and besides EIII + E 1 _ values into (t), we obtii~ tIOns, we find Eland E 2 • SUbstituting lh~;

7'

a=

U

+

Hence it follows that a = Of,or

_

, 8 l PI -

eo •

j21tr dr=

J

(Elp) 2nr dr.

The field intensity E is the same II' th!! given conductor, Le. is ind at pOI ts of the cross section of

d

:::.: :,t:;;e:.:~~~ :?:~f!e~l~. :!;~u~n:n f~h

C':;d~~~re~~ilih:t~~~

O{h:' and then applying to this con\th, exahmple, the conductor's axis vector E. our e t eorem on circulation of Thus, E can be taken out f th . gration we obtain 0 e lDtegral, and as a result of inteE = 2naI182. (2) The resistance of a u 't I h wHhthe help of the formul~1 R e~gt of t~e .c~nductor can be found equation by length lof the sectio;;oF/{ DlvdldlDg both sides of this R and voltage U, We find e con uctor, haVing resistance R u = Ell = 2na182. Ohm's law for no 'f rm

d

• 5.7. shown in Fig. 5.14 the emf' nunl suhcircuit. In the circuit Raland .R" and the capacita~~esc~ot~ Iffo?{ sourckes, the resistances n resistances of sources are ne l"blcapaci or are nown. The interplate 1 of the capacitor: g 19l Y small. Find the charge on Solution. In accordance ~itli ali ' I

RI = CPa - CPb

+ It,

-while lor the suhcircuit aCb we have

+ CPI -

CPI = CPb - CPa·

e2P2'

Solution. (1) In accordance w'th Oh ' '. l related to current density i wh·l . . m s dlaw , field lDtensity E is can write· , I e I IS rel ate to current I. Hence we

J

+

where the positive direction is chosen clockwise. On the other hand, for the nonuniform subcircuit aRb of the circuit we have

The joint solution of these equations gives

• 5.6. Nonhomogeneous c d t I cular cross section of area S . on DC or. A ong conductor of a cirpe dc !1ds only on the distance ~ f~~ethf a dterial w~lOse resistivity d e cpn uctor aXIS as p = a/r2 where a is a constant Th intensity E in the co~duct~rCon uctor came~ current I. Find (1) field of the conductor. and (2) the resistance of the unit length

1=

taining resistors Rand Ro, we can write (R R o) I = It - ~o,

~

B 2P2- eIPI PIll P2 12

139

Problem.

5. Direct Current

f m s aw or,a closed circuit con-

Fig. 5.14 The charge on plate 1 is determined by the formula ql = C (CPI - CPI)' Therefore, the final result is ...'

RC

ql= R+R

o

(~-t8'o)'

> 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
('k

Am

>

<

As

0,

0,

.1 W

<

O.

• 5.9. Transient proccSse A . . s~lUrce of e.m.f. tE, and a re~is~~r CircUIt co,!sists of a permanent nes. The internal resistance 'of th R and. c~paclto~ ~ connected in sem?ment t = 0, the capacitance' e source Is.negliglbly small. At the WIse) decreased by a factor of 11 °t,~hd r-hPacltor w~s abruptly (jumpfuncti~n ~f t·t e current III the circuit as a 1 Ime. Solution. We write Ohm's law 1 mhomogeneous part ltER2 f th fo~ t~e & .!.l2 (Fig. 5.15); 0 e CircUit

..

1

.

T~ R

RI =
Fig. 5.15 .

RI

.

=

t]q/C _

=

{!

=

1I -

(1)

dt.

Integration of this equation gives

ln~= -~ 1 RC 0

'

I

1

=. oe

-Tjl/RC

,

where J 0 is determined by condition (1) • I 11 dee d , we can wri te RIo = TJqo/C - e, where go = ,ffC is the char e f h . has changed. Therefore, g o t e capacitor hefore its capacitance 1 0 = (11 ~ .5.10. A charge qo was su C, which at the moment t =

lP

(1_0- 21 / RC ).

6. Magnetic Field in a Vacuum

Ji'

'" .

dl -r=T=R~

2

qlC' where C' -_ '

~Ve dd'Ierentillte this equation with r p . our case (q decreases) dq/dt = _l;es ect to tune, considering that

TJ ([t=-C I ,

Irom which it follows that first of all we must find the time dPpendence I (t). For this purpose, we shall use Ohm's law for the part JR2 of . t.he circuit (Fig. 5.16): RI = fill - lJl2 = 1I, or (2) RI = q/C. respect to Let. us differentiate (2) with t.ime: dE dt R~=_1 1 T-' RC' dt C' Integrating the last equation, we obtain It -IIRC Fig. 5.16 = RC' 1=100 , (3) ln 1o where I. is deterqrincd by condition (2) for q = qo, Le. 10 = qolRC. Substituting (3) into (1) and intewating over time, we get 2C

€].

(1)

U

Q=...!!!...

III

R dE

JllP dt,

1) ~/R.

lied t O . ,.' . h a cal,antor with capacitance the amount of heat liberated in J:eas s .unted hy resistor R. Hnd resIstor as a function ot time I.

6. t. Lorentz Force. Field B

Lorentz Force. Experiments show that the forc.e F acting on a point charge q generally depends not only on the position of this charge but also on its velocity v. Accordingly, the force F is decomposed into two components, viz. the electric component Fe (which does not depend on the motion of the charge) and the magnetic component F m (which depends on the charge veloc.ity). The direction and magn itude of the magnetic force -at any point of space depend on the velocity v of the charge, this force being always perpendicular to vector v. Besides, at any point, the magnetic force is perpend icnlar to the direction specifiCd at this point. Finally, the magnitude of this force is proportional to the velocity component which is perpendicular to thi~ direction.

143

6.1. Lorentz Force. F£eld B

142

6. Magnetic Field in a Vacuum

tltc!o'(' propcrtic' f . byAll introdll;;ill r the CO~IC~ magnetIc f~lI'e? can be described this field by v~ctOl" B wili~~ °II ;n~f:n.etl~ fteld. CI~al"acterizilJg at each pomt of sp'lce W(e el1ll\n~s the specIfic direction magnetic force ,in 'tl:e f~rme can WrIte the expression for Fm

= q [v

)( BI.

(6.1)

Tl.wn totalbyelectromagnetic force acting on charge q wIll bethe. gIven .

I F=qE+q Iv x RI· I

(6.2)

It is called the Lorentz for E· . ' . . it is valid for constant as c~~ll xpr~sslOn «(~.2)g IS un.i versa I: magnctic fIelds for any velo 't as orf vlarYIn electric and . . CI Y V 0 t lC charge . Th e actIOn of the Lorentz fo ciple be used for determinin tl~~e on a. charge can in prinof vectors E and B. Hence gth magmt.udes and directions force can be considered a~ e expre~s~on for the Lorentz magnetic fields (as was do t.he t~CfimtlOn of electric and It should be emphasizednet}~~t m~;::7iCoffi:~c~ric field). * on an electric charge at rest I }. oes not act essentially differs from el~ct;i 1 ~\~eS~Cl; m~gnetic field only on moving charges. c e. agnetIc fIeld acts Vector character'· . fIeld on aBmovin ~zes tl lC foree actlllg due to magnetic

an analog of vecro~ ~r~~, and l~e~ce in this respect it is to electric field. aracteflZIng the force acting due

A distinctive feature of m t' f . perpendicular to the veioci:;n~eI~ ore; that it is always fore, no work is done over the c or 0 t. Ie charge. Therech:zrge: TIllS means that the energy of a charged particle tic field alwa , . . movmg In a permanent magnemotion of the y;ar~~~:.ws unchanged h'respective of the

If

In the no (6.2), like

1 l' . .

a;~~e:tl~~~s~~~c::~[~:Si~~ii~~pe\~~~~r~~zcl~~~~:

Several methods measurin analysis, thevofall a b g fiIe ld B h ave been worked out. In the •fmal describeu by Eq. (G.2). • re ased on the phenomena that can be

of the (inertial) reference system. On the other hand, the magnetic component of the Lorontz force varies upon a transition from one frame of reference to another (because of v). Therefore, the electric component qE must also change. Hence it follows that the decomposition of the total force F (the Lorentz force) into the electric and magnetic components depends on the choice of the reference system. Without specifying the reference system, such a decomposition is meaningless. Magnetic Field of a Uniformly Moving Charge. Experiments shoW that magnetic field .is generated by moving C1harges (currents). As a result of the generalization of experimental results, the elementary law defining the field B of a point charge q moving at a constant nonrelativisticvelocity v was obtained. This law is written in the form* B=.fQ.. q[vx3 r ] 411:

r

\

(6.3)

,

\ wltere ~o is the magnetic constant, the coefficient If.:

~o/41t

= '10- 7 HIm,

and r is the radius vector from the point charge q to the point of observation. The tail of the radius vector r is B fixed in a given frame of reference, while its tip moves with velocity v (Fig. 6.1). Consequently, vector B in the given reference frame depends not only on the position of the point of observation but on time as well. Fig. 6.1 In accordance with formula (6.3), vector B is directed normally to the plane containing vectors v and r, the rotation around vector v in the direction of vector B forming • Formula (6.3) is also valid in the case when the charge moves with an acceleration, hut only at suffIciently small distances r from the charge (so small that the velocity v of the charge does not noticeably change during the time ric).

6. Magnetic Field in a

VOC1l1Jm

~------------

6.2. The Biot-Savart Law

the rig-ht-hallrlcll SystClll with vector v (Fig. li.1). It should be nQtHd that vector B is Hxial (pseutlovector). The quantity B is called maKfictic induction. Mag:nctic induction is measmetl in teslas (1'). The electric field of a point charge q moving at a nonrelativistic velo(~ity is described by the same law (1.2). Hence expression «U~) cnn be written in the form

B =, Eo!-to [v >< E]c= [v

X

Ellc\

First, we have to deal with beams of particles moving at velocities close to the velocity of light, for which this "correction" and the electric force become comparable (it should be noted that relation (6.5) is also valid for relativistic velocities). Second, during the motion, say, of electrons, along wires, their directional velocity amounts to several tens of a millimeter per second at normal densities, while the ratio (v/C)2 ~ 10-24 • It is indeed a negligible correction to the electric force! But as a matter of fact, in this case the magnetie force is practieally the only force since electric forces disappeared as a result of an almost ideal balance (much more perfect than 10- 24 ) of negative and positive charges in the wires. The participation of a vast number of charges in creating current compensates for the smallness of this term. In other words, the excess charges on the wires are negligibly small in"comparison with the total charge of carriers. For this reason, magnetic forces in this ease eonsiderably exceed the electric forcgs acting on excess charges of the wires.

(6.4)

where c is t.he electrollynamic constant (c -~, 1/V eo!-to), equal

L-1, 1

145

Z

Fig. 6.2

to the velocity of light in vacuum (this coincidence is not accidental). EKample. Comparison of forces of magnetic and electric interaction between moving charg~. Let two point charges q of a sufftciently large mass move in parallel to onc another with thc same nonrelativistic velocity v as is shown in Fig. 6.2. Find the ratio betwecn the magnetic Fm and electric Fe forces acting, for example, from charge 1 Oil

6.2. The Biot-Savart Law

cli8rge 2.

The Principle of Superposition. Experiments show that magnetic fields, as well as eleetric fields, obey the principle of superposition: the magnetic field created by several moving charges or currents is equal to the vector sum of the magnetic fields created by each charge or current separately:

According to (6.2), Fm = quB and Fe = qE where v is the velocity of charge 2 and Band E are the induction of the magnetic and the intensity of the electric fielns created by charge 1 at the point of location of charge 2. The ratio Fm/Fe -= vB/E. According to (6.4), in our case B = = IJEIc'l.. and hence (6.5)

(6.6)

Even at sufficiently high velocities, e.g. u = 300 km/s, this ratio is equal to 10-a, Le. the magnetic component of the force is a millionth fraction of the electric component and constitutes a negligible correction to the electric force.

The Biot-Savart Law. Let us consider the problem of determining the magnetic field created by a direct electric current. We shall solve this problem on the basis of law (6.3) determining the induction B of the fwld of it uniformly moving point charge. We substitute into (6')~) the charge pdF for q (where dV is the volume element and p is the volume charge density) and take into account that ill accord-

This example may give rise to the following question: Are snch forces wOI,th investigating? It turns ont that they are, and theI'e are two sound reasons hehin(l this.

, f.

10-0181

146

anre with (5.2) pv

=

j. Then formula (6.3) becomes

I dB ~ {;;-

Ii X,:I'1'

=

j

'B--~ I clIsada

.1

(6.7)

t!S dl = I dl,

where dl is an element of the length of the wire. Intl'Oducing vedor dt ill the direction of the current I, we can write this expression as follows: (6.8) j dV = I dt. , Vectors j dV and I dt are called volume and linear current elemen is resj)ecti vely. Replacing in formula (6.7) the volume current element by the linear one, we obtain

I dB=

* I[~~

4n.

r

'

b

Integrating this expression over all current elem~nts, which ~s equivalent to the integration over a between -n/2 and n/2, we lind -

B=~.Y:!4n

(IU1)

b

Example 2. Magnetic field at the axis of circular current. Figure 6.4 shows vector dB Ihlln the current element I dl located to the

B dBz -

dB

A

dl

r

xrl.

I

z

(6.9)

B=~~ I[dlxr] 411: 'j' r3

I

R

dB,B

(6.10) •

Generally, 'calculation of the ;.nagnetic induction of a current of ,111 arbitrary configuration by these formulas is complicated. However, the calculations can be considerably simplifIed if CllI'l'ent distribution has a certain symmetry. Let us consider several simple examples of drtel'min i Ilg' the ma~netic ind uctio 11 of CllI'ren t. ' EXaJpple f. Magnetic field or the line current. i.e. the c\Il"rent flowing along a thin straight wire of infinite length (Fig. 6.3). In accordance with (u.9), vectors dB from all current rlements have at a .poillt A the ,same direction, :,iz. are directed behind the plane of the hgure. There!orp, lhe summatlOlI of vectors dB can hp r(!placed by the summation of their magnitudes dB, where

b

0

A

I Fig. 6.4

Fig. 6.3

right. All curre~t deIlll'nts will form thl' cone of vectors dB, and it can be easily seen that the resultant \"I'dol' ,B at point A will be directed upwards along the Z-axis. This II1l'allS lhat in order to fmd th~ magnitude of vcdOl" H, it is sul'ricil'lIl 10 SUIH up the compollPnts 01 vectors dH along lhe Z-axis. Each such pl'ojl'c.liol1 has the form dR. = dB cos

f-l C~

4

flo -,-

'1:1

I dl

-2-

r

cos ~,

where we took into account that the angl~ between the element dl and the radius ve/'lor I' is equal to n12, and hrllce the sine is equal to IInity. lnterrrat ing this px pression over dl (whkh gives 2rrR) and taking into' accolI~t that {'os ~ ~- nl,. and r ~ (z~ : R2)1/2, we get Ilo

B=.Im

2nR21

(Z2-, R2)8;2 •

(ti.12)

Hence it follows that the rnagnitudp or v('etor B at the centre of HIP enrrpnt ring (z = 0) and at a distance z » R is given by

dB=c ~ I dl cos a

4n

-- I.n

I;t!

Formulas (6.7) and (6.9) express the Biot-Sauart law. In accordance with the principle of superposition the total field B is found as a result of integration of Eq. (6.7) or (6.9) over all current elements:

Bc=_,fA:.lL \' [jXrldV 8

'It is clear from the ligure that dl cos a = r da and r = b/cos a. Hence /

If the current I flows along a thin wire with the crosssectional area t!S, we have j dV

147

6.2. The Biot-Savart Law

6, Magnetic Field in a Vacuum

B z ,~'o

r2

10*

..l':!'... 2JlI 1 R m

B

z?/H

~~ ~ 4n z3

(' 13) 0 .•

148

6. Magnetic Field in a· Vacuum

6.3. Basic Laws 0/ Magnetic Field

6.3. Basic Laws of Magnetic Field

terminate. In other words, magnetic field has no sources. in contrast to electric fwld. . Theorem Oft Circulation of Vector B (for till' magnellc field of a direct. current in a vacllum). Circulation oj uector H around an arbitrary contour r is equal to the product oj ~~o by the algehraic sum oj the currel/ts enl'eZoped by the contour 1 :

Like electric field, magnetic field has two very important properties. These properties, which are also related to the flux and circulation of a vector field, express the basic laws of magnetic field. Before analysing these laws, we should consider tIle graphic representation of field B. Just as any vector field fi.eld B can be visually represented with the help of th~ h~es of vector B. They are drawn in a conventional way, VIZ. so that the tangent to these lines at any Doint coincides with the direction of vector B, and the ~density of the lin.es is proportional to the magnitude of vector B at' a given pOlDt. The geometrical pattern obtaiped in this way makes it possibl~ to easily judge about the configuration ofa given magnetic field and considerably simplifies the analysis of some situations. . Let us now consider the basic laws of magnetic field, I.e. the Gauss theorem and the theorem on circulation. The Gauss Theorem for Field B. The flux of B through any closed surface is equal to zero: . (6.14)

Iii,

Iii,'

This theorem is essentially a generalization of experience. It expresses in the form of a postulate the experimental result that the lines of vector B have neither beginning nor end .. Therefore, the number of lines of vector B, emerging from any volume bounded by a closed surface S, is always equal to the number of lines entering this volume. Hence follows an important corollary which will be repeatedly used below: the flux of B through a closed surface S bounded by a certain contour does not depend on the shape oj the surface S. This can be easily grasped with the help of the concept of the lines of vector B. Since· these lines are not discontinued anywhere, their number through the surface S bounded by a given contour (i.e. the flux of B) indeed must be independent of the shape of the surface S. Law (6.14) also expresses the fact that there are no magnetic charges in nature, on which the lines of vector B begin

(6.15 )

where I ~I,,, and I,. are algebraic 4I1alltitie~. The CUI'rent is assumed positive if its direction is eOllllecled with the direction of the circumvention of the contoui· through the right-hand serew rille. The current having the opposite directiOIl is considerc9 to be negat ive. This rule is illustrated. in Fig. 6.5: here currents I] and {a are positive sinctl their directions are 'connected with the direction of /1> 0 /2" 0 cont.Olir circumvention through Fig. 6.5 the right-hand screw rule, while . current I 2 is negative. The theorem on circulation (f).i5) can he proved Oil the basis of the Biot-Savart Jaw. In the general case of arbitrary currents, the proof is rather cumbersome and ~il1 not be considered here. We shall treat statement (h.i5) as a postulate verified experimentally. .' _ . Let us make one more remark. If CIlrrent I 1I1 (b.L)) IS distl'ibllted over the volume where contour r is located, it can l,e represented in the form I

= \

j dS.

(f).Hi)

In this expression, the integral is taken over an arbitrary surface S stretched on contour r. Current density j in the integrand corresponds to the point where the area element. dS is located, vector dS forming the right-handed system with the direetion of the contour circumventioll. Thus, in the general case Eq. (6.15) can be written a~

6.4. Theorem on Circulation of,Vector B

150

6. Magnetic Field in a Vacuum

. 1 t'on of vector B and then ~ee whcthc!' rem Oil Circu a I method is universal.

follows:

~Bdl=~IO f

jdS=f.to.\ indS.

(6.17)

TIIC fact that cil'c,dati ;,1 of vector B gene.r'llly differs from zero illdicates thalii! contr
6.4. Applications of the Theorem on Circulation of Vector B

Let us consider several practically i~po:tant examples illustrating' the effectiveness of the applIcatIOn of the theQ-

•

151 t II is

. fi Id' f·a strai~ht current. Direct current 1 Example t •. Ma~netlc Ie 0; ht ~'irp wi th a circular cross sedion flows along a~ mftmtely long. st~all 'nduction B outside and inside the of radius a. Fmd the magnetIc lie ( I wire. t of the probl~m t Ilat t he I'Illes 0 f It follows ~rom the s y mll1e r y f 'ircles with the centre at the vector B in thiS case have t he form 0 c

1/

/'

-- " .......

B

r

{ f,\

\

"

........

a

..-...._/ /

.fit

r

Fig. 6.7

Fig. 6.6

. d !Sf vector B must be the same at all points wire axis. The magm~u e" f the wire Therefore in accordance with at a distance r from.t e ax!s 0 . 'B for a ~ircular contour r l the theorem on nrBcu2latlO~ of /e~~:nce it follows that outside the (Fig. 6.6), we have . nr -- flO , wire (6.18) B = (flo/2n) Ilr (r> a).

pr~~I.emt (d'ith the help of the Biot-Savart law) turns out ~I bet:~:ei~~~iro~:th;t inside the From· the same symmetry cons". era I d' to the theorem' on \,:ire the. lines of vec~ornBf~~e :Is~i;~~l~~s. ~oC~t~l:nh (see Fig. 6"6~, circulatIOn of vcchto I = l ( la)2 is the current enveloped by the glB '2nr = flo l , were r r " d th .' ven contour. r Whence we I'mrl that illSI e e wIre (6.19) 2

It should be noted that a direct solution of this

B ~.~ (~lo/2n) lr/a

(r

<. a)

"d 't' B ( ~ is shown in Fig. 6.7. The dep~ndence r f tube the induction B outSI e I ~s If the wire has the s ape 0 a. . ' 'de the tube magnetic field IS determined by formula (6.1~1)' Wl~111e mS-\h the help of the theorem on absent. This can also be easl y s own WI circulation of vector ~. f' Id of a solenoid. Let current I flow al?n a Example 2: Magne Ie Ie h surface of a cylinder, Such cyhn ~r conductor hehcally wound o~ tell d a solenoid Suppose that a umt streamlined by the ~urrent !S ca e f h c~nductor. If the pitch length of the solenOid contams n turns 0 f'the solenoid can be appro xy of the helix is sufficientl smdall, each~r~hall also assume that the conimately replaced by a cl ose oop. ,

a

152

153

6. Magnetic Field in a Vacuum

6.4. Theorem on Circulation of Vector B

ductor cross-sectiollal "rea is so small that the CUITent in tIll' solenoid can be considered flowing over its surface. h 1xperiments and (,;lleulations show that the longer thl' solenoid t ~ ess the n!aWll'tic induction outside it. For an inlinitelv long sole~ nOId, ~agnetlc field ontside it is absent at all. . . * : .It IS clear from symmetry considerations that the lines of Vl'ctor B l~slde the solenoid are. directed along its axis, vector B forming the 1'Jght-handed system \nth the direction of the cnrrent in the soleIloid.

to infmity (nt a constant toroid cross section), we obtain in the limit the expression ((i.20) for the magnetic field of an infmitely long solenoid. . If the chosen circular contour passes outside the toroid, it does not envelop any currents, and hence B ·2J1r = 0 for such a contonr. This means that outside the toroid magnetic field is absent. In the above cOJl!'ic1erations it was assumed that the lines of current lie iUlIleridional plaues, i.l'. the planes passing "through the ·,axis 00' of the toroid. In'real toroids, the lines of current (turns) rlo not. lie' strictly in these planes, and hence therl' is a current component arounrl the 00' axis. This component creates an addiB tional field similar to the field of a circular current. Example 4. Magnetie field of a current-car· r~'ing plane. Let us consider an infmite conducting plane with unc1irectional current uniformly distribute!1 over its surface. Figure 6.10 shows t.he cross section of such a plane when the current flows behind the plane of the figure (this is marked BV crosses). Let us introduce the concept of lille~r clI!tTent density i directed along the lines of current. The malPlitude of this vector Fig. 6.10 is the current per unit length which plays the role of the "cToss-sectional <1l'ea". • l\lentally dividing the current-carrying plane into. thin ~urrent tilament::;, we can easily see that the resnltant field B wlll be thrected parallel to the plane, downwards to the right of the plan.e and up~ards to the left of it (Fif.{, 6.10) .. These directions can be easily estabhshed with the help of thl' right-hand screw rule. In orc1l'r to determine induetinn B of the field, we shall use the theorem on circulation of vector n. Knowing- the arrangement of the lines of vector B, we shall choose the contour in the form o! rectaJ;gle 1234 (Fig. 6.10). Then, in accordance with the theor.em on CIrculatIon, 2Bl ~, !Joil, where I is the lenl!th of the contour SIde parallel to the conducting plane. This g-ives 1 . (6.22) B =~ 2~tOI.

I , - ..... --, .......•.......... ~

:! :w ....lL -:

L

4_

.

I:

:oJ

• .......................... Fig. G.S

Fig. tUJ

'. Even the aho~'e consideratioIls abollt the magnetic fwld configuratIOn of the. sole,nOld indicate that we must choose a rectangular contour as shown In FIg. (i.H. Circulation of vector n around this contour is equal to BI, and the l'l!ntour envelops the current nlI. According t~1 the theorem or~ clrculatlOlI, BI "'" ~lonlI, whence it foHows that inside a long solenuld B = ~lonI,

(6.20)

i.e. the. Iield il.ls!d~ a long s()lenoi~1 is uniform (With the exception of ~he regIOns. ildJOIll!lIg the solenoid s endfaces, which is often ignored III caleul~tIOns). 11le product liT is called the Ilumber 0/ ampere-turns. For !/ ~ 20()(I turns/m and 1 = 2 A, the magnetic field inside the solenOId IS equal to 5 mT. Exnmple 3. l\fa~netie field of a torOid. A toroid is a wire wound arollnd a torus (hg. 6.9). . . It can he easily see,n from symmetry considerations that the Jines of v?ctor B mus~ be CIrcles whose centres lie on the axis 00' of the t?rold. Hence It IS c1t'ar that for the contour we must take one of SUcll CIrcles. a contour !iI'S in~ide I,h(~ toroid, it envelops the ('urrent N 1 wher!' !V IS th~ number of turns in tIle toroidal coil and 1 is the ~urrent III t~e wIre: Let the contour radius be r; then, according to the theorem on clrculatlOn B '2:rrr 1'01\'1, whence it follow!' that inside the toroid

.If

B

=

(/to/2n) N /h..

(6.21)

II compa.rison. ofyi.21) alltl (6.1H) shows that the lIIaglll'tie licit! illsidp til(' torOJ.d cOJflCldes with the magnetic lield fIT J of tl](' straight i'lll'rent flowmg along the 00' axis. Tellding N and radius H uf the toroid

This formula shows that the magnetic field is uniform on both sides of the plane. This result is also valid for a bounded current-carrying plane, but only for the points lying uear the plane and f~r from its ends.

General Consideralicms. The results obtained in the above examples could be found 'directly with the help of the Biot-Savart law. However, the theorem on circulation makes it possible to obtain "these results much more simply and quickly. However, the simplicity with which the field was calculated in these examples must not produce an erroneous

155

6.6. Ampere"s Force 154

6. Magnetic Field in

II

Vacuum

where

impression about the potentialities of the method based on the application of the theorem on circulation. Just as in the case of the Gauss theorem for electric field, the number of problems that can be easily solved by using the theorem on circulation of vector B is quite limited. It is sufficient to say that even for such a symmetric current configuration as the current ring, the theorem on circulation becomes helpless. D88pitean apparently high symmetry, the magnetic field configuration does not allow us, however, to find a simple contour required for calculations, and the problem has to be solved by other, much more cumbersome, methods. .

Oil

the right-hand side we have the projection of curl H onto

normal n. . . f Id B can be characterized by curl B lIence, ea~h POlllt of vedor I~ I' determined by the propertie.s of whose directlOll and. IIwglll~u~e1~1 lC. I'rection of curl B is detl'rllllllClI the field !tself at a given 1}010 tl: . ~~rface element S, for whi?h t.he by thtetl~Il('~c~z)n ~hf~l~md~ti~esothee J;laguitude of curl H, attainS Its quail I ) . , maximum value: .1 B .~ obtained in coordinate represen~In mathematiCs, cur k therfact is more importallt: It tion. However" for our purpos~sa:b~ considered as the cross pl'Oduct turns out that formally curl BB' 'tV X B. Wc shall he using the of the operator 'tV ~y vecto~ r 'n I~~~ce it allows us to represcllt the latter, morc c~nven~nt. ~htth~Ohelp of the determinant:

t'

cross product . X

~~x a~Zy iJ~~z

'tV\: B = \ a

Bx

6.5. Differential Forms of Basic laws of Magnetic Field

.

I

Bdl

IHn'-'--=(curl H)n,

8-0

S

(6.Y.)

Bz

By

.

r'

,

(6.23)

Le. the divergence of field B is equal to zero everywhere. As was mentioned above, this means that magnetic field has no sources (magnetic charges). Magnetic field is generated by electric currents and not by magnetic charges which do not exist in nature. Law (6.23) is of fundamental nature: it is valid not only for constant but for varying fields as well. Curl of field B. The important property of magnetic field expressed br the theorem on circulation of vector B motivates the representation o this theorem in the differential form which broadens its potentialities as a tool for investigations and calculations. For this purpose, we consider the ratio of circulation of vector B to the area S bounded by a contour. It turns out that this ratio tends to a certain limit as S -.. O. This limit depends on the contour orientation at a given point of space. The contour orientation is specified by vector n (the normal to the plane of the contour around which the circulation is calculated) the direction of n being connected with tlte direction of contour circumvention through the right-hand screw rule. The limit obtained as a result of this operation is a scalar quantity which behaves as the projection of a certain vector onto the direction of the normal n to the planp, of the contour around which the circulation is taken. This vector is caller! the curl of field B and is denoted as curl B. Thus, we can write

(Ii .25)

the unit vectors of Cartesian coordinates. ThiS where ~x' eZj.j~. th curl of not only field B, but of ,IllY other expressIOn I~ va I 01'.1' f f 11 E vector field a./rIso, iI!-IP:lrtthIC~:::;~r~lI1 ~~ ~irc~latioll of vector B. AccorltLet us now CoUSH 1'1 e I' tl f I : I.'q. (6 . 17) call bc representee ill Ie orn illg to. (u . 24) , c • ell'

Divergence of field B. The differential form of the Gauss theorem (6.14) for field B can be written as

\'

lim s-u or

(V

j

H dt

.

---c;- = f!oJr\' .

X Bl" = l.l.oill' Hence

\-'V-X-B-=-Il-oj-~ \

.

.

(fj.~6)

of the theorem on ~irculatlOn

.

~f~~~o~r~~i~t;i~~l:~~~de d~~~~J~F:~i~~~i~,c~~fl~~h::;:i~~d~ or viz. the current ensl y . t cJ'. ve 1 t Ii j E . al to zero IS equ • of V X B is equa ? rfio id circulation of vector In an electrostatIc e , and hence

\ 'V X E=O.

(fj.~7)

\

. I t zero everywhere is a potentIal A vector field whose curl IS equa?d l field Consequently, eleetroIield. Otherwise it is ~aned the,~?f:n~~ganet'c field is a solenoidal J1~ld. static field is a potential field,

II:

I

.

6.6. Am~re's Force , h 1 I' e carrier experiences the Ampere's Law.. Eac c~~egaction of this force is tr~llSaction of a magnetIc forc~. h which charges are lIlovmg. [nitted to a conductor t roug

156

6.6.

6. Magnetic Field in a Vacuum

~:s : resul~,

magnetic field acts with a certain force on a cur~ -carryIng conductor. Let us find this force th~~;~~i:~t~~t the volume density of charges ~hich are to .)'( W ~ cliIre~t (e.g. electrons in a metal) is equal con~~lctore m1tenta tY ' Isolate a volume element dV in the , . . con aInS the charcre ( t . to pdF. The tl P. f . «,., curren car:ner) equal f 'h , n , L orce actlllg on the volume element dV o t. e conductor can be' written by formula (6.1) as follows: dF = p [u X Bl dV. Since i = pu, we have >'

IdF

= [j X B)

dv·1

(6.28)

If. the current flows along a thin cordmg to (6.8), j dV = I dl, and conductor, then, ac-

I

dF = I fdl X BI,

I

(6.29)

where dJ is. the vector coinciding in direction with the cUITdent and characterizing an element of length of the thO con uctor. , ' I n ,.

F~rmtdas

(6.28)

a~d

(6.29) express Ampere's law. 1nte-

~~af!ng th)ese expreSSIOns over the curren. t elements (vol ume

meal', we can find the rna t' f tain vo] ume ~f d gne I.C o~'ee acting on a cerTh e f a . can uctor or on Its lmear part. called A(),rce~. ~ctfmg on currents in a magnetic field are npel e s orces. Example. The force of 'nt t· b ". Find the Arrwere's force with wh.ah l?~ .etw~en parallel currents. ?urrents /1 and I. interact in IC LW? JllfillJ~ely long- wires with IS b. Calculate the force per ~n~tcluumt'hlf rtheh distance between them eng 0 t e system. . Each current element of I is in th . .. I.e. aeeording to (6.19), the 2 field B e~agn~t1C field of ~he current 1 I, t.ween the element of current / . d I - (flI/4:r~ 2f1/b. fhe angle befrom fonnula (6.29) that the io~~e v:~.tor B1 IS 1(/2. Hence, it follows ductor with current I I'S F _ laB> mg per unIt length of the con2 u -- 2 It or

Fu=~ 2/ 1 12 4rt b

(6.30)

Naturally, the same expression . b. b ' . per unit leDgth of the d car~ e 0 tallied lor the force actin/.( con uctor With current fl'

Amp~re's

Force

157

Finally, it can he easily seen that currents of the same direction are attracted, while the opposite currents repell each other. Here we speak only about a magnetic force. It should be recalled, however, that besides magnetic forces there are electric forces, i.e. the forces due to excess charges on the surface of conductors. Consequently, if we speak aho~t the total force of interaction between the wires, it can either be attractive or repulsive, depending on the ratio between the magnetic and electric components of the total force (see Problem 6.7). . The Force Acting on a Current Loop. The resultant Ampere's force acting on a current loop in a magnetic field is dermed, in accordance with (6.29), by

F = I ~ rdl X B],

(H.31)

where the integration is performed along the given loop with current [. If the m~netic field is uniform, vector Bean be taken out of the integral, and the problem is reduced to the calculation of the integral ~~ dl. This integral is a dosed chain of elementary vectorsdl and hence is equal to zero. Consequently, F = 0, i.e. the resultant Ampere's force in a uniform magnetic fIeld is equal to zero. If, however, the magnetic lield is nonuniform, the resultant force (6.31) generally differs from zero and ill each concrete case is determined with the help of (6.31). The most interesting case for further analysis is that of a plane current loop of sufficiently small size. Such a current loop is called elementary. The behaviour of an elementary current loop can be conveniently described with the help of magnetic llwrnent Pin' By delin ition, magnetic moment Pm is given by Pm = ISn,

(6.:32) where [ is tJw cUlTent, S is the areil hounded hy lhe loop and n is the normal to the loop whose direction is eonnected with the direction of the current in .tho !f;op through the rig-lit-hand screw rule (Fig. 1).11). In terms of magnetic lipid, the clemenU;' \' loop is eompletely charaeteT'iz,~d by its magnl.'lic moment Pm'

iSS

159

6. Magnetic Field in a Vacuum

6.7. Torque Acting on a Current Loop

1\n j~volved calculation with the help of formula (6.31), taklIlg 1I1to accoullt the smaJl size of the loop, y ields th~ foHowing expression for the force acting Oil all elementary cnrrent loop in a nOlluniform magnetic field: '

(0.3:1) in terms of the projections onto this direction. This gives

IF= Pm"* '

(6.33)

where Pm is the magnitude of the magnetic moment of the loop, and oBion is the derivative of vector B along the

10 0

b

Fig. 6.11

F~~Pm

10 G).----~ Pm

F~ 10 €lfolo'---~O Pill

1

F=O Fig. 6.12

direction of normal n or along the direction of vector p . This formula is similar to expression (1.39) for the for~e acting on an electric dipole in an electric field. It follows from forml.da (6.33) that, like in the case of electric dipole, (1) in a uniform: magnetic field F = 0 since oBian = 0; (2) the direction of vector F g-enerally does not coincide with vector B or with vector Pm; vector F.coincides in direction only with the elementary increment of vector B taken in the direction of vector Pm at the locus of the loop. This is illustrated by Fig. 6.12, where three arrangements of the loop in the magnetic field of straight current To are shown. The vector of the resultant force F actillg on the loop in each case is abo shown in the flgllre (it is useflll to verify independently that it is really so). If we are interested in the projection of force ..~ onto a certain dit'ectioll X, it is sufficient to write expression

iJB x

F x= Pm an-

(6.34)

where iJlJxiiJn is the derivative of 'the corresponding projection of vector B again with respect to normal n (or with respect to Pm) to the loop. Example. An elementary current loop having a magnetic moment Pm is arranged perpendicularly to the symmetry .axis of a nonuniform magnetic field, vector Pm being directed along vector B. Let us choose the positive direction of the X-axis as is shown in Fig. 6.13. Since the increment of projection B x will be Fig. 6.13 negative in the direction of vector Pm. F < O. Hence .yector F is directed to tlfe left, i.e. to .fli.e side where B is greater. If we rotate the current loop (and vector Pm) through 90° so that the loop centre -coincides with the symmetry axil\- of field B, in this position F x = 0, aIlji vector F is directed perpendicularly to the X-axis and to the same 'side as vector Pm·

6.7. Torque Acting on a Current Loop Let us consider a plane loop with a current I in a uniform magnetic field B. It was shown above (see p. 157) th?t the resultant force (5.31) acting on the current loop In a uniform magnetic field is equal to zero. It is known from mechanics that if the resultant of the forces acting on any system is equal to zero, the resultant moment of these for?es does not depend on the position of point 0 relativ~ to wInch the tot'ques are determined. Therefore, we can speak about the resultant moment of Ampere's forces in our case. ny deflllitinll, the resultant moment of Ampere's forces is gin~n by

M = (~ !r:< dFl,

(6.35)

where dF is defined by formula (5.29). If we perform calculation by formula (6.35) (it is cumbersome and hardly interesting, hence we omit it here), it turns out that the

16,)

6. Magnetic Field in a Vacuum

moment of fOl'ces for an arbitrary shape of the loop can be represented in the form '

I

M = {Pm >: 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
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 <~-

161

6.8. Work Done upon Displacement of Current Loop

i ... IIfil u

=

l'

';'d<:.....

~t

where the integnt! is LuLen OVer the slHIael' S s\.rr'tehed over th,; current loop. This intf'g'l'dl does nM depend 011 the choice' of til(> surfacl: S and depends only on the loop over which it is stretched.

= Id<1J,

(6,37)

F @

B,n where d
162

6. Magnettc Field In

II

Problem,

Vacuum

(6.37), where d is the increment of the magnetic flux , through the entire circuit. In order to find the work done by Ampere's forces upon the total displacement of the current loop from the initial position 1 to the final position 2, it is sufficient to integrate expression (6.37):

work 6A

I) 'I, I

=

F dx

=

IBZ dx

=

IB dS,

(6.38)

where dS is the increment of the area bounded by the contour. In order to determine the sign of magnetic flux , we shall agree to take t~e normal n to the surface bounded by t~e c~ntour always In such a way that it forms with the dl~ectlOn of the c~rrent in the circuit a right-handed system ~Ig. 6.15): In thIS case, the current I will always be a positIV~ ~uantIty. On .the other hand, the flux may be either posItIve or negatIve. In our case, however, both and dCf> = 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(<1>2 - <1>1)' (6.40)

",

where <1>1 and <1>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 <0. It should be noted that work (6.40) is accomplished not at the expense of the energy of the external magnetic field (which does not change) but at the expense of the e.m.f. source maintaining the cu~re~t in the loop. (This question will be considered in greater detaIl III Chap. 9.) Problems • 6.1. Direct calculation of induction B. Current I flows along a thfn conductor bent as is shown in Fig. 6.16. Find the magnetic induction B at point O. The required data are presented in the figure. Solution. The required quantity B = B _ B~, where B - is the

+

1;

,~t'

I ·~·.,

" ,..•.,' "".:

.....•....••.

, Y;;

ft ..

164

8. Magnetic Pield in a Vacuum

m a(l'netic ~eld created by the rectilin a r by Its curvilinear part. According t th . p~rt of the loop and B , 0 e IOt- avart law (see Examplef on p. 146), we have

B

!-tol cos a da 4na cos a o

Substituting (1) and (3) into (2) and integrating the result over r between a and b, we obtain IN In.!: B= !-to 2 (b-a)

!-tol

165

Problems

•

2rui"' tg a"

a

(2) The magnetic moment of a turn of radius r is Pml = Irrr 2 •

B~= ~ I (2n- 2a o) a _ !-tol 4n a2 2na (n-ao).

and of all the turns, Pm =

~

Pml dN, here dN is defined by formula

(3). The integration gives

As a result, we obtain

__ nIN bS - as _ nUl ( 2 b _ a -3- 3 a

+

B = (n - ao tg ao) !-toI /2na. It is interesting to show that f or ao - 0, we arrive at the familiar eXI,ression (6.13).

Pm -

• 6.3. Current I flows in a long straight conductor having th~ shape of a groove with a cross section in the form of a thin half-ring

Fig. 6.19

Fig. 6.18

Fig. 6.16

Fig. 6.17

I . • 6.2. A thin isolated wire fo a large number N of closely packed ~s \ [ ane spIral consisting of rent I is flowing. The radii of th . urns I rough which a direct curand b (Fig. 6.17). Find (1) the r::.aIn er.na. and ~xternal loops are a of the spiral (point 0) and (2) th gnebc .InductIOn B at the centre the given current.' e magnetIc moment of the spiral for Solution. (1) According to (6 13) h radius r to the magnetic I'nduct: ~ t e colntribution of one turn of IOn IS equa to (1) and of all the turns, B 1 = f.to Il2r, B=JBldN,

where dN is the number of turns in the interval (r, r+ drl,

dN=~dr. h-n

(2)

+ ab + b2) .

of radius R (Fig. 6.18). The current is directed from the reader behind the plane of the drawing. Find the magnetic induction B on the axis O. Solution. First of all, let us determine the direction of vector B at point O. For this purpose, we mentally' divide the entire conductor into elementary filaments with currents dI. Then it is clear that the sum of any two symmetric filaments gives vector dB directeli to the . right (Fig. 6.19). Consequently, vector B will also be directed to the right. Therefore, for calculating the field B at point 0 it is suffIcient to find the sum of the projections of elementary vectors dB from each current filament onto the direction of vector B: B=.\dBsincp

In accordance with (6.11), we have dB = /lo III /2rrR, where dI

","-C

(1)

(2)

(IIn) dq; (see Fig. 6.19). Substituting (2) into (1), we get

166

6. Magnetic Field in a Vacuum

..• 6.4. Theorem on circulati lIOn.. ' nsid.e a homogeneous Ion ,on .of 8 ~nd pr.!ncipl(. of sUpt"rposi_ jhere IS a .clrcular cylindrical c; s.~ralg~t wlre?f ~lrcular cross section uctor aXIs ~nd displaced relati:~ ~t se aXI~ IS parallel to the con~ ~ent of deusl ty j flows along th 0.1 Y.a dIstance I. A direct curtnsldp the cavity. I.' wire. Ftnd magnetic induction B

i b

S.olution. In accordance with h' . . required quantity can be t e prmclple of superposition the represented as follows: ' B = 80 - B' (1 ) where 8 0 is the magnetic induction of th~ conductor without cavity,

Problems

• 6.5. Principle of superposition. Current I flows through a long solenoid whose cross-sectional area is S and the number of turns n. Find th~ magnetic flux through the endface of this solenoid. Solution. Let the nux of 8 through the solenoid endface be <1>. If we place another similar solenoid in contact with the endface of our

(a) Fig. 6.22

(b)

Fig. 6.23

+

solenoid, the flux through the contacting endfaees will be lI> 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

>

<

a),

a),

Where we used the theore . I . ' side the solenoid' 2nrB ~ on2cIr~u atlOn while calculating B outTh . a - f!b ltal a W .. us, h~~ing represented the current' h . poslt!,on of transverse" and "Ion 't d' i~ t I.' solenoId as the superthe conclusion that onl th I gl. U I.na components, we arrive at e;Xist~ inside such a sole~oidea:dn~I~ldlilial component of the field B SIde It (as in the case of a strai~ht y I.' transverse component outBeside 'f' d current). d . S, I we ecrease the and 'dth . " eDSlty unchanged, 1-+0 as h 0 hI Imamtammg the current ~.nldYal~?e field inside the solenoidre:Uai~~ I.h =thconstl' In. this case, I I.' ,I.e. I.' so enold becomes ~ ..6.7. Interaction of parallel neghgtble resistance are shunted- at ~hr~ntB'd Tbwo lo~g wires with eIr en s y reSIstor R and at the other 'Cnds are connected to a sou~ce of constant voltage. The ra.dlU~ of the cross section of each WIre IS smaller than the di~tance be~een t~eir axes by a factor of TJ. - 20. Fmd the value of the reSIstance !i at which the resultant fo:ce of I~teraction between the WIres vamshes. Fig.

:

. ' Solution. There are excess surtIVe of whether or not the . face .charges on each wire (irrespec~ence, in.addition to the mC::~~~~If flowkng through them) (Fig. 6.24). t I.' e.lee-trlC force Fe. Suppose that orce m, we must take into account an exces~ charge A corresponds to a unIt length of the wi~e Th length of the wire by the' other en. the elebctrlc force -exerted per unit Gauss theorem: WIre can I.' found with the help of the

E:.-_

Fe=AE=A ~ 2"-2 4m,0 l - - 4m l ' where I is the distance betwe h 0 force acting per unit length :Pt~ee a~es of tbhe wires. The magnetic WIre can e found with the help

169

of the theorem on cirulation of vector ~: Fm = (f!0/4n) 212ll, where I is the current in the wire. . It should be noted that the two forces, electric and magnetic, are directed oppositely. The electric force is resp'onsible for the. attraction between the wires, while the magnetic force causes their repulsion. Let us find the ratio of these forces: FmlF e = eOf!0/2/A 2. (1) There is a certain relation between the quantities / Problem 2.8): '),.=C1U= Ine o U,

nl]

where U = RI. Hence it follows from relation (2) that /1'),. = In TJineoR. Substituting (3) into (1), we obtain Fm f!o In 2 l] ----2 ,,' Fe 80 n R2'

and t.. (see (2)

(3)

(4)

The resultant force of interaction vanishes when this ratio is equal to unity. This is possible w~en R = Ro, where Ro= .. /f!o V eo

Inl] =360Q. n

If R < RTJ then F m > 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<:ation of the given current element, but created by all other current elements excluding the given one. In order to find B', we proceed as for calculatin~ electric force (Sec. 2.3). Let B, be the magnetic induction

172

7. Magnetic Field in a Substance

7.1. MagneUzation of a Substance. Vector J

of the field created by the surface 1 ' to its surface (see Fig 6 29 wh e ~f~nt Itself at a point very close flowing from ~s). Acc~rding' to (~~~2) ~. a~u(T/ed th?t the current is . i. l 2) ~Ol. Further, USIng the theorem considerations we can easH OUCIrcu atwn of vector B and symmetry field outside the cYlinder n!ars~et that the .magnetic induction of the I s surface IS equal to . . . B = ~oI/21tR, (3) whIle InsIde the cylinder the field' b The latter fact indicates that th fis ld s~nt. rent elements at points 1 and 2 e e B from the remaining curFig. 6.29~ must be the same and ;:~fsf~l~h }O 11he ~ylinder. s!lrfac.e (see and outsIde the cylinder surface: e 0 OWIng condItIOns mside B'

=

Bt

and

=

B

Hence it follows that .

.

B'

=

B'

+ B t = 2B'.

B/2.

(4

SubstItutIng this result into (2) b' ) for the required pressure: ' we 0 tam the following expression dF

p=- = dS

2B' iB' = - - B' ~o

-

Taking into account (3), we finally get p =

B2 2~

. 0

~oI2/81t2R2.

It is clear from formula (2) th t h l' pression. a t e cy mder experiences lateral com-

7. Magnetic Field in a Substance 7.1. Magnetization of a Substance Magnetization Vector J •

Field in Magnetics If a t ' into a magnetic field form~~r ~m sUb.stanc~ is introduced the field will change. This is I y. c~r~ents m conductors, substance is a magnetic . eX~t~le y theJact that each a magnetic moment) unde~·eih I, IS. magnetIzed (acquires A magnetized substance creat e .;ctlOn of mag~Ietic field. which forms, together with t~ I s .own magnetIc field B', by conduction currents, the re=uft:~n:a~~l:eld B o created B

=

Bo

+ B'.

H~re B~ an~ B stand for the fields

physIcal mfimtely small volume.

(7.1)

averaged over a

173

The field B', like the field Eo of conduction currents, has no sources (magnetic charges). Ilence, for the resultant field B in the presence of a magnetic, the Gaus.~ theorem is applicable:

I fBdS~o·1

(7.2)

This means that the field lines of B remain continuous everywhere in the presence of 11 substance. Mechanism of Magnetization. At present, it is estab. lished that molecules of many substances have intrinsic magnetic moments due to the motion of intrinsic charges in them. Each magnetic moment corresponds to a circular current creating a magnetic field in the surrounding space. In the absence of external magnetic field, the magnetic moments of molecules are oriented at random, and hence the resultant' magnetic field. due to these moments is equal to zero, as well as the total magnetic moment of the snbstance. This also appHes to the substances that have no magnetic moments in· the absence of an external field. If a substance is placed into an external magnetic field, under the action of this field the magnetic moments of the molecules acquire a predominant orientation, and the substance is magnetized, viz. its resultant magnetic moment becomes other than zero. In this case, the magnetic moments of individual molecules no longer c.ompensate each other, and as a result the field B' appears. Magnetization of a substance' whose molecules have no magnetic moments in the absence of external fIeld proceeds differently. When such materials are introduced into an external field elementary circular currents are induced in the molecules, and the entire substanc.e acquires a magnetic moment, which also results in the generation of the field B'. Most of materials are weakly magnetized when introduced inlo a' magnetic. field. Only ferromagnetic materials such as iron, nickel, cobalt and their many alloys have clearly pronounced magnetic. properties. Magnetization. The degree of magnetization of a magnetic is characterized by the magnetic moment of a unit

114

1.1. M41,.etnatlon

7. Magnetic Field in a $ubdance

volume. This quantity is called magnetization and denoted by J. By de~nition, .

(7.3)

= n

(Pm),

(7.4)

where n is molecular concentration and (Pm) is the average mllgnetic moment of a molecule. This formula shows that vector J is collinear precisely to the average vector (Pm)' Hence for further analysis it is sufficient to know the behaviour of vector (Pm) and assume that all the molecules within the volume L\ V have the same magnetic moment (Pm)' This will considerably simplify the understanding of questions associated with the phenomenon of magnetization. For example, an increase in magnetization J of a material implies the corresponding increase in vector (Pm): if J = 0, (Pal) = 0 as well. If vector J is the same at each point of a substance, it is said that the substance is uniformly magnetized. Magnetization Currents 1'. As was mentioned above, magnetization of a substance is caused by the preferential orientation of magnetic moments of individual molecules in one cl irection. The same can be said about the elementary circular cmrents associated with each molecule and called "molecular currents. It will he shown that such a behaviour of molecular cllrrents leads to the appearance of macroscopic currents J' called magnetization currents. Ordinary currents flowing in the conductors are associated with the motion of charge carriers in a substance and are called conduction currents (I). In order to understand how magnetization currents appear, let us first imagine II cylinder made of a homogeneous magnet ic whose magnetizatiou J is uniform and directed along the axis. Molecular cuneuts in the magnetized mag-netic lIre oriented as shown in Fig. 7.1. Molecular CIIITcnts of

115

J

. . he oint:- of their contact have opadjacent mol~cules at t p tach othel'. The Oil Iy uncom. posite directiOns and comp~nsa_ etriOSe emerging 011 the lateral pensated molecular curren ale currents form macroscopic ItesJ' circulating over the lateral surface of the. cyl.inrler. IUTjace magnetization curren.

;1

where L\ V is a physical infmitely small volume surrounding a given point and Pm is the magnetic moment of an individual molecule. The summation is performed over all molecules in volume L\ V. By anillogy with what was done for polarization P (see (3.3)1, magnetization can be represented in the form J

01 4 Sub,tance. Vector

Y4

___. . l' ~

-Fit. 7.1

I'

'I x Fig. 7.2

rent]' induces the same macsurface of the cyl.inder-lhe c~~at of all molecular currents roscopic magnetic fiel as taken together.. er case' a magnetized magnetic Let us now conSider anot~ t . tile molecular currents Let for msance, .' k : F' ~ 2 where the Ime thiC 'ness is nonhomogeneous. be arranged as sho~n Ill. Ig~f ~oieclllar clll'rents. It follows corresponds to the mtenslty J' directed behind the plane from the figure th.at vector. ~a nitude with the coord iof the figure and Illcrea~st Irhe ;olecular cnrrents are not nate x. It can be seen t a h eneous magnetic, aIllI as compensated inside a I?-0n omog rna ~etization current l' a result, th~ macrosc.OPIC vO:~~ve d1rection of the Y-axi~; appears, which flows III t~ Pb t the linear l' and surface ~ Accordingly, we can spea a ou I I'n Aim and j' in A/m~. ., ., being measurec h t current denSItIeS, t. • M netic. It can be stated t ~ Calculation of Field B 10 a f ~g a magnetized magnetIC the contribution to field. B t~O~ would be created by the is equal to the contribution ~" a vacuum. In other same distribution o~ curre~tsd' tl':~ution of magnetization words, having established t e _t~S the help of the Biotcurrents 1', we can find, WI I'ng to them and thell ' h field B' corresponc I Savart law, t e fi Id B by formula (7.1). caIc.ulate the resultant e

t76

7. Magnettc Fteld in a Subl1tance

7.2. Circulation of Vector J

U Iifortunately, . the distribut' 0 not only on the confi uratio 1 n of currents /' depends netic but on the field as lY ~nd ~~e properties of a magcase the problem of findin;~: or t IS re~son, in the general directly. I t remains for us to ~n a magnetic cannot be solved ~o the solution of this problem and find a~othe.r approach IS the establ ishment of . ' he first step III Hus direction ..' an Important reI at . b . IOn etween magn etIzatlOH current /' a I viz. its circulation. nt a certam property of vector J,

J

i

7.2. Circulation of Vector J

It tUl'1lS out that for a statio . magnetization J around bJl?IY case the circulation of . an ar Itrary contour r is equal

II I

I I

I

&6W

dl

r

r

Fig. 7.3

F' _ 19. i.4

to the algebraic sum of . '. by the contour r: magnetIzatIOn currents I' enveloped

r

ItJdJ~I'·1

(7.5)

where l' j' dS d' .. - J an mtegration is performed Ov arhltrary surface stretched on th er an In order to prove this theo e contour r. algebraic sum of molecular cu ,rem, we shall calculate the Let us stretch an arbl"raI'y nenfts enSvelo ped by contour r. '. sur ace 0 tl (F'Ig. 7'.3). It can be seen from tJ f . n Ie contour r currents intersect the surface S Ie /gure that, some molecular and for the second time in 'the t~ ,ce (~H1ce .In o~le direction such currents make no crnt 'b .pposlte directIOn). Hence IIe t'IzatlOn . ) rI1I lItlOn to the,::;nant reo It current throllgl magHowe. th . .d Ie surface S \er, e molecular currents that are', wound around

r

I

t77

the contour r intersect the sUl'face S only once. Such molecular currents create a macroscopic magnetization current ]' piercing the surface S. Let each molecular current be equal to 1 m and the area embraced by it be Sm' Then, as is shown in Fig. 7.4, the element dt of the contour r is wound by those molecular currents whose centres get into the oblique cylinder with the volume dll c= Sill cos a dl, where a is the angle between the contour element dl and the direction of vector J at the point under consideration. All these molecular currents intersect the surface S once, and their contribution to the magnetization current is dI' = I mn dV, where n is the molecular concentration. Substituting into this formula the expression for dV, we obtain dI' = l rn S mn cos a dl = J cos a dl = J dl, where we took into account that I mS 111 = Pm is the magnetic moment of 1m individual molecular current, and I mS mn is the magnetic moment of a unit volume of the material. Integrating this expression over the entire contour r, we obtain (7.5). The theorem is proved. It remains for us to note that if a magnetic is nonhomogeneous, magnetization current I' generally pierces the entire surface (see Fig. 7.3) and not only the region near its boundary, adjoining the coutour 1'. This is why we can use the expression I' c= j' as, where integration is performed over the entire surface S, bounded by the contour r. In the above proof, we managed as if to "drive" the entire cmrent I' to the boundary of the surface S. The only goal of this method is to simplify the calculation of the current I'.

I

Differential form of Eq.(7.5):

IvxJ=r, I

t

(7.6)

Le. the curl of magnetization J is equal to the magnetization current density at the same point of space.

A Remark about the Field of J. The properties of the field of vector J, expressed by Eqs. (7.5) and (7.6), do not imply at all that the field J itself is determined only by the cmrents I'. The field of J (which is hounded only by the t 2-0 181

Ii

7.8. Vector H ~!!

.

7._ Magnetic Field in a Substance

u

S}?ltial magnet,'c) depends on all currents, ,." l'jlregionthfilled by . . ',Z,]l.' 1 on ,0 mag~letlZatlOn currents l' and the conductl?ll currents I. However, ill some cases wIth a definite symmetry the situation is such as if the field of vector J were determined only by currents 1'. .Example. Fi.nd the surface magnetization c"!ll"ent per umt length of a cylinder made ?f ~ ho~ogeneous magnetic, if its magnetIzatIOn IS J, vector J being directed everywhere along the cylinder axis. Let us .apply Eq. (7.5) to the contour ch D.sen as I,S shown in Fig. 7.5. It can be eaSIly see? that the circulation of vector J Fig', 7.5 around thl.S contour is equal to the product (l. In thIS case, the magnetization current " ' _, ., IS the surface ~urrent. If we denote its linear J.rnslty ly ! ,the,contour under consideration embraces the ma netlzatlOll current i l. It follows from the equality It = i'l that g if

=

J.

(7.7)

lated to the circulation of the magnetization:

~ J dl =1'.

7.3. Vector H

on Circulation of Vector H (for magnetic fields

dlre~t currents). In magnetics placed into an external ~agnetI~ fiel~l magnetization currents are induced. Hence

of

cl~c~latlOn .0;'

vector B will now be de.termined not only b) conductIOn currents but by magnetIzation currents as well:

~Bdl=/lo(I+I'),

(7.8)

where I and l' are the conduction .and magnetization CUI'embraced by a given contour r. . SInce the .determination of currents l' in the general case IS. a c0m.-phcated. problem, formula (7.8) becomes inapplIcable In practlcal respect. It turns out, however that w~ can find a. certain auxiliary vector whose circuiation wIll be determIned only by the conduction currents embracfJd by the contour r. Indeed, as we know, the current l' 'is

ren~s

(7.9)

Assuming that the cirenlation of vectors Band J is taken vel' the same contour r, we express l' in Eq. (7.8) throngh ormula (7.9). Then

~

(:u -J) dl=I.

(7.10)

Let us denote the integrand of this expression by H. Thus, we have found an auxiliary vector H,

\H~;'-J'\

(7.11)

whose circulation around an arbitrary contour r is equal , to the algebraiO'tsum of the eonduction currents I embraced by this contour: (7.12)

It should be noted, by the way that vectors i' 8:111 J are mutually perpend iClllar: i' 1. J. '

T~eorem

179

This formula expresses the theor{!m on circulation of vector H: the circulation of vector H around an arbitrary closed contour is equal to the algebraic sum of the conduction c'urrcnts embraced by this contour. The sign rule for curr~nts is the same as ill the case of drculation of yector B (see p. 149). It should be noted that vector H is a combination of two quite different quantities, B//l o and J. Hence, vector H is indeed an auxiliary vector which does not haye any deep physical meaning. * However, the important property of vector H expressed in the theorem on its circulation justifIOs the introduction of this vector: in many cases, it considerably simplifies the inyestigation of the field in magnetics. Relations (7.11) and (7.12) are val id for any magnetics, induding anisotropic ones.

* The quantity II is often called magnetic field intensity, hut we shall not be using tlJis term to emphasize again its auxiliary nature. 12*

('

180

7. Magnetic Field in a

Sllhstaw'(,

7.3. Vector B

It follows from formula (7 12) I . vector H has. .the cli' menslOUS . wI, lhe IIl:\
I'

I

I

' , rem on cIrculatwn or \'eCtor H:

I

V x H=j, (7.13) i.e. the curl of vector H . I -, current at the same pointlsofeqtuha tO tthe density of the conduction e su b s ance. Relation Between Vectors J d that magnetization J d d an H. It was shown above . d · epen' bs on the lllag ne I'IC 111' point of the uction B a t a gIven vect or J'IS related not withsuB stance . , Howev"C!, customarily ourselves to the analvsi" of flit WIth H. \\e slwll coniine the depeudence bet~~enU J . on YHt~lOS~ magnetics fOl' which cl1l d IS hllecu', viz. J = XH

where . h . ' (7.14) . X IS t e magnetic susce tibil't I ' , , ' . quantIty typical of eacl p . l y. t IS a dllnenSlOnless magnetIc (tl t . J' f·01'I ows from the fact tl I' t . .' Ia X IS hlmension1ess · . 0 f H and J aI'e tl la , accol'lllllg' to (7 SIOns . "11) t I LC ( f'1men· ,e S<:.me ) Un 1II~e electric sllsceptibilit' " . magnetIc susceptibility X x ~\ Illch IS always positive Accordingly the mag t~nay _ e .eltI~er positive hI' negative' "d . ., ne ICS ,.,atlsf\"l f 1, ~ . • d IVI ed mtD paramagnetics ( fig . orIllU ,1 (i .14) are In paramagnetics, J tt H X ~ .0) a.JlII d~amagndit's (z < 0). It should be noted' that' iI~ ~~e. I.n (hamp'{!If'hl: J H H. there exist ferromagnetic~ for' 'I ~f~tlOr to tll(',' !!lag-netics, ~laS a rather complicated' fol'!~ ,w i ~I~.' t le ~epeHde',cD J (H) and besides-, It has . a hysteresi~" J' ,". J d epen d' sons tnot It 1111ear, I . netIc. (Ferromarrnetic" will b I .. e pre IIstory of it magin Sec. 7.6.) ,., ~ e (eSenbell in greatel' dptail

h

n

Relation Between Band H F

expression (7 11) . • or magnetics olw,Xilw U·l1), . acqllJres the forlll (1 I.,) H ence , Z II h B/~o.

=

B

where ~l is lhe !Jermeahilt·t.t/ .

~ =

p~loH, ()f

(7.1:) )

I

lie medilllll.

1 .:- Z.

(7.1lj)

For paramagneties

~t

>

1, while for diamagnetics fL

18t

<

1.

It should be noted that in both cases fL differs from unity only slightly, Le. the magnetic properties of these magnetics are manifested very weakly. A Remark about the Field of H. Let us consider a question which often involves misunderstanding: which currents determine the fIeld of veetor H? Generally, the fwld of H (just as the field of B) depends on all currents, both conduction and magnetization. This can be seen even from (7.i5). However, in some cases the field H is determined only by conduction enrrents. Vector H is very helpful for these cases in particular. At the same time, this makes grounds for the erroneous conclusion that the field of H always lkpends on conduction currents and for incorrect interpretations of the theorem on circulation of vector H and Eq. (7.13). This theorem expresses only a certain property of tite lIeltl,of H without deHning the ~field itself.
Example. A system consists of a long straight wire with current I and an arbitrary piece of a papmagnetic (Fig. 7.6). Let us lind the changes in the lieilis of Vl'ctors B and If ami in the circulation of vector II arour.d a certain fixed contour r upon the removal of the magnetic. The field B at each point of space is determined by the conduction current I as well as by the magnetization currents in the paramagnetic. Since in our case, in accordance with (7.15), If = B/l!l!o' this also applies to the Held of vector II, i.e. it also depends on the conduction Fig. 7.6 current I and the magnetization currents. The removal of t.he piece of paramagnetic will lead to a change in the lield B, ami hence in the lield ll. The circulation of vector B around the contour r will also change, since the surface st.retched on the contour r will no longer be pierced by the magnetization currents. Only the conduction current will remain. However, the circulation of v(!ctor II around the contour r will remain unchanged in spite of the variation of the tield II itself.

r

When = 0 Inside a Magnetic? We shall show now that magonetization ellrren'ls inside a magnetic are absent if (1) tlte magnetic is homogeneolls nnd (2) there nrc no condllction ClllTellt.S in it (j - 0). I II this case. fot' any shape of a 1l111gnetic mill allY conliglll'ation of the magnetic Held, we call be

7,4. Boundary Conditions for 8 and B 182

7. Magnetic Field in a Substance

sure that volume magnetizat' ual and only surface magnetizat:~~ currents are e9 to zero, In order to prove this we . currents remam. c,II1atioll of vector J ar~ d shall use the theorem on cirpletely lying inside the ~~gn:~itr~itrary contour r comgeneous magnetic we can : ~ the H case of a homoill Eq. (7.5) out ~f the i~teSgurbaSltItu~/ng .X for J, take 'X. anu WrIte

in Fig. 7.7. Then the flux of B in the outward direction (we ignore the flux through the lateral surface) can be written in the form

+-

B 2n /).8 BIn' /).8 = O. Taking the two projections of vector P onto the normal n to the interface, we obtain BIn' = -13]11' and after cancelling /).8, the previous equation becomes (7.20)

l' =X~ Hdl. In accordance with (7.12.) h ' .. . to the algebraic SUIll of th~ ~oe ~~m?mlIlg mtegral is equal by the contour r. Hence f . n ~,ctlOn currents I embraced have 01 a lOmogeneous magnetic, we

l' = 'X/ '1'1 " ,1" . (7.17) . us Ie atlOllslup lJetweell tl ,. ' ' any contour inside the llHlgll~~·Clll~ClltsI. and I is valid for ~~nall ~ontour, when l' :-+ ~C:, III p.artlcular, for a. very Ih01l l' d8 ~ . dS' ln d8 and I -+dI - ] d8 " , r: ,,- ~ Xln ,and aftet' cancellinO' dS n .. L~ __ XllI.IJus eqnation holds f). t : : ' .we ohtam small contour, Le. for an' d" (.t any orwntatIOn of the lIcllce the vectors i" au/' t~rect~oJIl, of the normal n to it. tile same equation: J temse \ es are related through

dI'

j'

.

= xi.

'I'll if J'liS it0 [olio P Ews that'III a IlOtllogcucous maguetie , -(. '.D.

(7.18) J"

ce_' 0

7.4. Boundarv Conditions for Band H We slwll analyse tli> C d" the interface between \ onhltlOns for vectors nand H at conditions will be obtai w~ .omoge!IOOIlS magnetics. These with tlte Iw.lp of the C' n~~ 'Illist as III the case of uielectrics cnlatio/l Wo' l'ol'"lll 11 "ltllf!:iS t leOl'ern ctlt.d the theorem on or vocto .. B alit I H these theorems have llw. form , . lil . ' I:>

cir~

1,BdS'O,

I

~Hdl~l.

(7.19)

Condition for V('dol' B I·,f'.t liS" I III [I g'1ne .. located dt tlte il t >'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<:e nl·t·u., . .. 10vectors being. 1'0 1ate d'In the same way as the corresponding currents, VIZ. (7.24)

°

B'

=--0

XBo.

. ()f t.lle resultant field B Theu the inductlOll . =(1 X) B o 01'

+

n

,--~

B o -i' B'

fJB o.

=

(i.25)

I ti 'e space is filled with a homoThis means that. when. t Ie en SI . times. In other words, the geneous magnotlc, B Increase .tJ the magnetic iielrl B upon 't -lows the increase In . . quantI y ft S. I. . d 1 y the f18ld with a magnetlc. fIlling the entll'o spaI· ce OCCt·\lP~e (7 )25) by tJ-f.\. we obtain If we divide bot 1 par s o . 0' f. ' ) ' \7.",6) H -"l - H0 . . fIe ld II. tUfllS• out to be (in the case :lllder 50nsHleratlOn, the same as 1II a vacllum). I valid when a homogeneous Form. \lIas (7.24)-(~.26) lare abso d d by thp s~rfaces formed t' fill' the entlre /..'0 ume oun e I magne lC. l S . (I field of the conduction current. .n o by the lmes of B t le .. . d tio·n B inside the magnetic this case too the magnetic 1Il uc . 1 . . t' larger than B o· 13' IS tJ- IIDes. .I 1. hove the maanetie IIll uctlOn In the cases ~orrsl( e:~~ .at · I ~lIrrents '\s connected with of the .jiel~l ofJ ~a~:~e :;~~~~iC through the simple relation magnetlzatlOlI 0 ., (7.27) B = 1l0J. v . .

')

. b en<:ilv obtained from the formula This c:xpresslOll can 0 L~. J l' t H' -- X and -- B'-~- B' if we take into account t Ja _B 0 B--- 0 ; ., _._ J/. H B ~-~ tJ-tJ-o , where H -~. . I . > in other cases the As has <~.ll'oadY.1 been,~Il~ntlOpY~cl;t'~~O\~;ld forrilult;s (7.24)situatioll IS lilliC I mouo c o m , . (7.27) become ,ilwppl~cablei' II" ro))';ider two c;imple examConcluding iilO :'('('.11011, ec , . . ple,. Example 1: Field H

.

I 'd A solenoid having nl amp,'rea so \~~ltOl: ; ho'lllogene(Jus magnetic of the HIler! "

In

turn~ per unl t length is

~~_ 7.6. 188

permeability the magTletic.J..l

7. Magnetic Field in a Subs/ane,·

>

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
Experiments show that a ferromagnetic i~ he~ted upon reverse magnetization. It can be shown that til thiS case the amount of heat Qu liberated per unit volume of the ferro mal?netic is numerically equal to the "area" Sn of the hysteresIs loop: ",' (7.29)

Curie Point. As the temperature increases, the ability ?f ferromagnetics to get magnetized becomes weaker, and 111 particular, saturation magnetization becomes lower. At.1l certain temperature called the Curie point, ferromagnetIc properties of the material disappear. At temperatures above the Curie point, a ferromagnetic . becomes a paramagnetic. . On the Theory of Ferromagnetism. The phYSical nat1ll'~ of ferromagnetism could be explained only with the help of quantum mechanics. The so-called exchange forces may appear in crystals under certain condi~ions. These forces orient parallel to each bther the magnetic moments of electrons. As a result, regions (1-10 !lm in size) of spontaneolls magnetization, which are called domains, appear il.l c~'ystals. Within the limits of each domain, a ferromagnetIc IS magnetized to saturation and has a certain magnetic lllomcn t. The directions of these moments are different for rlifferent domains. Hence, in the absence of external fIeld, the resultant magnetic moment of the sample is equal to zero,. and the s<'lmple as a whole is macroscopically TlonmagnetlZed. When an external magnetic field is switche(l on, tllO

II

If12

7, Magnetic Field in a Suhstance

Problems

dO/llailiS ol'ipliled a/olll{ II . f' f I ,'. Ie Ie t I{I'OW al tit t I O/lldIlIS. ')/'I('lllpd £lg-aillsl Ihe fipld (' ~XPCllse f!f g'l'Owlh IS ,·p\'P!'sih/(. 1,1 ' I ' : 111 weak i1e]tls, thIs . s rOllger' the/ds tI ' 1 l'l'ol'iPlilalioll or III 'lg'lIl'l,'(' I" , Ie SI/IIII lalleOIlS ' l I l O J l I P I I S wilhill II I' I I la,ps plaep This pl' '. . IC I'll 1['(' f omaill '. oepss IS IITp veJ'si hie I' I 1 IIj',slel'esis alld I'('sidll't1 'll't" I' . . ., W lie I {'xp aills , , ',.,lIl· lzallOlI.

radial direction, We shall use the theorem on circulation of magneti~­ ation vector J, taking as a ,con,tonr a smal.l rec.tangl? w~~se plane IS perpendicular to the magnetIzatIOn current III thiS regIOn. Ihe ~rr~nge­ ment of this contour is shown in Fig. 7. i8, where the crosses llldIc~te the direction of the surface magnetization current. From the equalIty Jl = i'l we get i' = J. , Then we can write J = XH, where H can be rletermmed from the circulation of vector II around the circle of radius r with the centre

J>roblcllls • 7.1. Contlitions at the interface I (Fig. 7,10) on a III (' • II the Vicinity of point A i1gone Ic-vacuum inferfilcl', thl' magnl'tic inductjon I

Fig, 7.18

Fig. 7.16

Fig. 7.i9

Fig'. 7.17 on the conductor axis: 2nrH = I (it is clear from symmetry considerations that the' lines of II lIlust be circles lying in the planes perpendicular to the conductor with the current l). ThiR gives

in the vacuum is B vector B f . to the ,interfan'. Th'e p('rmeab)iIi~~JI~fg'ti1I,1 angle a~ w,ith the normal mag-netIc illrfuction B ill tI " . ,Il nla/:rJletlc IS /I' Find Ull' . If' m"g'nl'tlc lJl till' vidnity or ' Sullittun. The requirerl quantity , flomt A. 11= V-B~ +/Jl "

t'

Considerin/:r conditions (7.20) and (7.22) ,'tt.

the inll'rfacf', we lind Bn = B o cos ao, IJT=-'llflolfT~llItllf{ --'til '- B . . • 0' t 0" - II 0 sIn a o, ~here H 0-, IS the tangential component o' Substituting" these expression~ . t (1) j veefo~ U o in the vacuum. ~ lJl 0 ,We obtam B~B l·/ c 2 a o T' ft" . sin 2 a , oOS

Solution. The lines of H are clearly also circles. Unlike vector B, vector H undergoes a discontinuity at the vacuum-magnetic interface. We denote by Hand H o magnetic fIelds in the magnetic and vacuum respectively. Then, in accordance with the theorem on circultaion of vector H around the contour in the form of a circle of radius r with the centre on the conductor axis, we have

o

. • 7.2. Surfacemagnelizationf'urrenl " , JIlg conductor with current 1 " ' . A lOng' thIn currenl-carr.·_ vacu ' " 1~ al'ran<~ed perpI'n I' I 'j J . UJn-IIl,lgn('tH' mtprfare IF'" ., 1- ~ 1'1 .1 JCU 111 .\ to a plaoe netIc IS II. Find the lllleal' eu'r J,.,./ 'f' 11. \I' jlr'rnleabiU I" 01 the malY' , . n'll (ell-If\, i' ()' th f' " t lOll current 011 thiE in(.prJ'ac' 's .- ,,' ,', -c. .' (' sur ace magtl:ctizarOllduetor. l'
nrH

I

,";olt:tiu/I Let n" Hrst ('()/ISIIII'I' the

netlZatwlI current It ,

j. 'j ,_ . ' .. (Colt

f

COl j'" . .. I 1.,~ratIOn

rOIll 1- 19, 7,!, lIwt

Besides, B

o[ the surfac~ HlCigcurrent has tltt'

=

+ nrHo =

I.

(1)

B o at the interface, or ~H

,.

t1li:s

(~- i) J/2nr•

• 7.3. Circulation of vector H, A long thin conductor with current I lies in the plane separating the space fIlled with a nonconducting magnetic of permeability ~ from a vacuum. Find the magnetic induction B in the entire space as a function of the distance r from the conductor. Bear in mind that the lines of B are circles with centres on the conductor axis.

(t)

1; ,

=

= lI o,

(2)

Solving Eqs. (1) and (2) together, we obtain

.J

II =

-j'

"

.

"',

~

.

}:

:!tlh .

,,.

13-01RI

I

(i-j-ft)nr'

H=

I t

Ifo

1I =

[!~oI

(i+t.t)nr·

The configurations of ttl(: lields n anti II in this case are shown in l·'ig. 7.1!J. We advise the reader to convince himself that for ~t = 1, we arrive aL the well-known formulas for B and II in a vacuum. • 7.4. Circulation of vectors Hand J. L\ 'direct current I flows along a long homogeneous cylindrical wire with a circular cross section of radius R. The wire is made of a paramagnetic with the susceptibility X. Find (1) fIeld B as a function of the distance r from the axis of the wire and (2) magnetization current density j' inside the wire. Solution. (1) From circulation of vector H around the circle of radius r with the centre at the wire axis, it follows that r

<

R,

r> R,

211rH

=

. I tI d ·t· width. Hence where I is the contour hClg It an r I S dJ '2lLH o j;'''-~ --ar=--~ r.

• .' . ,clor ' f is direeted agai~st t!le norm.a1 The minus sIgn wdlcates t~hat d J tern with the (IIrecllOn of Clfvector n which forms the flght- an sys

'h.

-

ex: r), (H ex: l/r).

I (rIR)2

(H

2nrH = I

I

Since dI' follows:

~~,

j' =

I

I'

.!...r +!:!... . dr

Let us now consider that J = XH = (XI12nR2) r. This gives j'

=

xI/nR2.

It can be easily seen that this current has the same direction as the conduction current (unlike the surface magnetization current which has the opposite direction).

• 7.5. A long solenoid is filled wi th a nonhomogeneous isotropic paramagnetic whose susceptibility depends only on the distance r from the s()lenoid axis as X = ar2 , where a is a constant. The magnetic induction on the solenoid axis is B o• Find (1) magnetization J and (2) magnetization current density j' as functions of r. Solution. (1) J = XH. In our case, H is independent of r (this directly follows from circulation of vector II around the contour shown in the left part of Fig. 7.21). Hence H = Ho on the solenoid's axis, and we obtain J = ar 2Ho = ar2B o//!o. (2) From the theorem on circulation of magnetization J aroun d an inflllitc1y narrow contour shown in Fig. 7.21 on the right, it follows that Jl -

(J

+ dJ) I =

j;'l dr,

..--,

I--- ~

r

ro~, dr

-~ Fig. 7.21

Fig. 7.20

j'2nr dr, the above equation can bl' transforml'd as

II

..,::::J

",.-

Figure 7.20 shows the plots of the dependences H (r) and B (r). (2) Let us use the theorem on circulation of magnetization J arouh'l the circle of radius r (see Fig. 7.20): 211rJ = r, where I' is the magnetization current embraced by this contour. We find the differential of this expression (going over from r to r + dr): 2n d (rJ) = dI'. I'

195

Problems

7. Magnetic Field in a Substance

194

• h. ds j' is directed toward" us cumvenLion of the c~ntour. In ot.~rth:ofi u're, i.e. volumr JIlagnetizain the region of the rJ~hht con~our~ the le!h-hand system. tion currents lorm Wit vec or J h s th hape of a ring with a narrow i' 7.6. A (Jermanent magn~tdl: dia~~ter of the ring is d. The ,!?ap gap between the poles. :rh~ ~j'd . f the field in the gap is n. F 111<1 width is b, the magnetic wHuctHin/inside the magnet, ignoring the th magnitude" of vectors an< h di~sipation of 'the field at the edges of t e gap. . ; ." on circulation of vector II around th.~ Solutwn. USIll~ the the~re(r 7 22) and considering that there IS dashed eirele of dwmeter 19':t no conduction current, we can Wfl e (nd - b) l/-r; + bB//!o = 0, ., , I I onto the direclion of cont?ur where 1I-r; is the pI:oJe~honko f veet~hat it coincides with the (liredlOn circumvention (which IS ta en so • of vector n in the gap). Hence bB bB (1 ) Il-r;= - tJ-o(rtd-b) ;::::: - tJ-o nd . . .' h h direction of vector II inside the The minus sign mdlcates ~ at t .he \. 't'on of vector B at the same . t·. I' OpPOSI te to t eb ( IO rcc Il l __ 0 as we 11 . magnetic ma erld IS 'be found by formula (7.11), point. It should be noted t~wLt'as J The magnitude of magnetJzalon

7:10

13*

_.Hili

.~

1f' lid Ie . ,_.._'_'2.:'

. FlCld ill

II

taliiug into account (1): II/I'll

J

-'1'---':"'1'-,/"-n-d- ~

II -,I-II

. .The relationship. IlI'tween Vt'r1ors B/IlO, II and J at Inside the magnet IS shown in Fig. 7.23.' any point

Solulion. (1) Let, for the sake of defmiteriess, vector B on the axis be oirecteo upwards and coincide with the X-axis. Then, in accordance with (6.34), Fx 0= I'm aB/ax, where we took into account that the IDagnetic mompnt Pm is directed as vector B (for paramagnetics) ant! h('nc(' replaced an by ax. Next, since I'm 0= IV =-XllV and eIB/ax = _2aBoxe- ax ', we have Fx = -

B/I-Jo

Axe-2ax2 ,

Fig. 7.23

. • 7.7. i\ Winding contailliw' "or t u I . flls' ' , . In the form of a torlls with the m~1 'II,e r "dOUI\( on an 11'011 core slot of wid I h b (Firr 7 2?) I' .. t . (t , (lallle er : A narrow transversI' .... . w S CII 11\ Ie core \Vh'l t I . . I Ilrough Ihe winding, Ihl' magnptic illdu.t: ,I' ~hurrlen .. IS pas...o;ed the permeability of iroll under the ( IOJI 111 e sol IS B. Find siJ)ation of tIl!' field at the edrres o l'setI CO~II(httIOIIS, neglecting the disb ICso·. Sululio/!. In accordance with tI 't1 . arollnd the circle of diaml'ter I (It ·FI~orelll.?1l CIrculation of v('Ctor n . (see 'Ig. 7.22), we have (rrd - !J) HlblIo~' NI, whl're JI and /10 are the magnitud f II' . . spectivPly. Blsides, the nbsence of /i~l r .. In tl~on and In the slot n.', that I I Isst!'n 1011 at the edges lll('ans

It

(1)

where A = 2aB:! XV/'IllOHaving cal;ulated the derivative dF,,/dx and equating it to zero, we obtain the following equation for determining x m : 1 - 4ax 2 = 0, whence X m = 1/2 y~. (2) (2) Substituting

Fig. 7.22

197

Problems

SII!Jsfanrr

(2)

into

(1), we

find ___ ,toF III X- BltI

..

/,c

Fig. 7.24

V a'

where we took into account that f1 ~ 1 for paramagnetics.

• 7.9. A long thin cylim{rical rod made of a paramagnetic with the magnetic susceptibility X and cross-sectional area S is arranged

x

dx

I

B B o• lJsillg th",.:" two I'qll
It

Jl

dB

J!oN l---bB ,. • 7.k. The foree adin« on a g t· TI . I· Ig. 7.24 is used for 1Il('~'l1ri~" (w' I 'j3 c Ie" It' deVice shown in which a small par:JIl1a~~eticgsph~~ I t I{ 1C1p 01 a b:~lanee) a force with pole of magnet .1/. The Ill~g' n r. ~ OJ v~ I1me ~. IS attracted to the ' e Ie IllI uctlOn on the axis of the rv I I I Sloe (l'pends 011 Ihe height as B ~ B ,-"x~ . ,., 1',1 e stants. Find (1) tlip h,.j ht ~!, whue En aud a arc conso tha I thp allr;lcliVt' fcrrce at ,,:llJch the sphere TIlust be placed eeptibijily of lhe !"ll'alll,w;".t7c l\\~~~mum a.nd (2) the m~g-lletic SIlSIS F , , . , . Ie lIlaXllIllI1ll forcC' ot atlr:lctioll

t

l:"

m

Fig. 7.25 along the axis of a cllI'!'l'ntearrying coil. One end of the rod is at the cenlre of the (,oil, whl'!'l' IhI' luagnetic fIeld is B, while the olh~r end lies in the regioll when' IhI' lIlagnetic fwld is practically absent. Fint! the force with which the coil acts 011 the rod. Solutiol/. Let us IIll'ntally isolate all element of the rod of length (Fi£:,_ 7.25). The force acting o!,! this elemellt is given by

d;T

x d"~d . ',. Pm aB an .

~lIflPoSC that vector B on the axis of the roil is directed to the right (see the figure), toward::; positive valu!'s of x. Then B x = B, an = ax,

I ....

198

8. Relative Natllre of El

and since dpm

=

J S dx _

ec

t. riC

and Magnetic Fields

8.1. Electromagnetic Field. Charge Invariance

dFx=XHSdx~_ XS B

ax

-~

Having integrated h t is equation, we obtain

dB.

()

Fx

Th

=

- XS /l/lo

\' B dHC'

-

-

XSB2 2/l,J'

. B 0 e !llmus sign indicates th rod IS attracted to the cu::e:~tor ~ is dir~cted to the left, i. th arrymg coIl ,.c. e . • 7.10. A small s h magnetic susceptih T P ere of volume V, made of carrying coil from thity ~, was displaced along th par~magnetic with region wher th e pomt where the m ' . e aXIS of a C1lITentagainst the emag~~i~gfetic field is practf:~i~cab~~~~t~~ dis B to the S I . orees accomplished in th' . In the work o utlOn. Let us d' h IS case. . the elementar k Irect t eX-axis alon th placement of ill(7~rph donbe agaiI!st the ma~net~cafis o~ the coil. Then orces Upon the disere y dx IS equal to 6A= -F dx-.

iJB x --Pm-d x

x

Where F . h Th . x IS! e projection of the

.

an

'

(1)

leS~;;~:eSlih~fn~~~:~~s ~hat thr~~~kt;~ ~~~: ~~~~~s~~t?

tll e X-axis :a ues of x. Then B = Bon dt e axis is directed tow IS orce... vn = -ax' h x . an an = a ( t h ' ard s posItIve

i~ ~h~tf~r~~'~otn~1:~I;:tfh:/:'~/:Z~~is~I~tI1?~~do~Xth; dir~ct~~~ X

,we rewrIte Eq. (1)

6A= ~XHV!!!!... d _ XV a x - - H dB ntegratmg this equation bet x /l/lo . . ween Band 0 we hI . (J , 0 ,am

I.

A=_~o rr

19!1

XHS dx, we get

-

j' B

B2V BdB=-J--

2/l/lo .

8. Relative Nature of Electric and Magnetic Fields 8. f. Electromagnetic Field Ch . So far . arge JAvallance se ' we have considered ele . parately, without establishi . ctnc and magnetic field~ ng any clear reIat·IOn 1>etween

them. 'I'll is could be done only because the two fields were static. In other cases, however, it is impo~sible. It will be shown that electric and magnetic fields must nlways be considered together as a single total electromagnetic flCld. In othel' words, it turns out that electric and mag-netie fields are in a certain sense the components of a single physical object which we call the electromagnetic field. The division of the electromagnetic field into electric and magnetic fields is of relative nature since it depends to a very large extent on the reference system in which the phenomena are considered. The field wldch is constant ill one reference frame in the general case is found to vary in another reference frame. Let us consider some examples. A charge moves in an inertial system K at a constant velocity v. In this system, we shall observe both the electric and magnetic fields of this ch.u·ge, the two flelds varying with time. If, however, we go over to an iner~ial system K' moving together with the charge, the charge will be'1it rest in this' system, and we shall observe only the electric field. . Suppose two identical chatges move in the system K towards each other at the same velocity'v. In this system, we shall observe both electric and magnetic fields, both these fwlds varying. In this case it is im:rossible to find such a system K' where only one of the lields woul be observed. In the system K, there exists a permanent nonuniform magnetic field (e.g., the field of a fixed permanent magnet). Then in the system K' moving relative to the system K we ~hall ob~crve a varying magnetic field and, as will be shown below, an ele<;tric fleld as well.

Thus, it is clear that in different reference frames different relations are observed between electric and magnetic fields. Before considering the main points of this chapter, viz. the laws of transformation of fields upon a transition from one reference system to another, we must answer the question which is important for further discussion: how do an electric charge q itself and the Gauss theorem for vector E behave upon such transitions? Charge Invariance. At present, there is exhaustive evidence that the total charge of an isolated system does not change with the change in the motion of c.harge carriers. This maybe proved by the neutrality of a gas consisting of hydrogen molecules. Electrons in these molecules move at much higher velocities than protons. Therefore, if the

---~\~

200

8. Relative Nature of Electric and Magnetic Fields

charge depended on the velocity, the charges of the electrons and protons would not compensate each other and the gas would be charged. However, no charge (to 20 decimal places!) has been observed in hydrogen gas. Another proof can be obtained while observing the heating of a piece of a substance. Since the electron mass is considembly smaller than the mass of nuclei, the velocity of electrons upon heating must increase more than that of nuclei. And if the charge depended on the velocity, the substance would become charged upon heating, Nothing ·of this kind '.vas ever observed. 'Further, if the electron charge depended on its velocity, then the total charge of a substance would change as a result of chemical reactions, since the average velocities of electrons in a substance depend on its chemical composition. Calculations show that even a weak dependance of the charge on the velocity would result in extremely strong electric fields even in the simplest chemical reactions, But nothing of this kind was observed in this case either. Finally, the design and operation of all modem particle accelerators are based on the assumption that a particle's charge is invariable when its velocity is changed. Thus, we arrive at the conclusion that the charge of any particle is a relativistic invariant quantity and does not depend on the"particle's velocity or on the choice of the reference system. Invariance of the Gauss Theorem for Field E. It tlll'llS out, as follows from the generalization of experimental results, that the Gauss theorem ~ E dS = qle o is valid not only for fIXed charges but for moving charges also. I Tl the latter case, the s1ll'face integral must be taken for a definite instant of tirnl' in a given reference system. Besides, since different inerti,ll reference systems ilre physically eq uivalent (ill accordaJICo with 1110' relati vi ty principle), we can state that the (;aIl8s theorelll is valid for all inertial systern~ of reference.

. ,;t blished in the special theory of this transformatIon are e,. a}' ted way For this reason, of relativity in a rather comp IC:es ondi~g derivations and we shall not present h,ere ~het~~r co~tent of these laws, the will pay mam at~entI~n 0 th m and the application of corollaries followm~ rom e : e~ific problems, these laws for solvlJl~ s~;n §uppose we have two inertial Formulation of the ro em., K nd the system K' moving . systems of reference: the system a 1 't v We know the . th fi st system at a ve OCI Y 0' relatIve to e r, ld E d B at a certain point in space magn~tud~s of th~ fie s K a~hat are the magnitudes of the and tIme m th~ systfm ' o'nt in space and time in the .fields E' and B at t Ie same P a~e oint in space and time system K'? (Recall tl~t t~e s d trme in the two reference is a point whose COOl' lila es ::e Lorentz transformations~) systems arc related through to this question is As was mentioned above,. t~Ie answer} . 1 't follows that . h th O'y of relatIvItv from w llC I I I given III t e e I . f fi'elds are expressed by tIe the laws of transformatwn 0 formulas _ _"'-"---.----:;::----;;-;----:;:;:-----\ Ell = Ell, BII - BI" \ B - [voxEl/c z ' E.L+[voXBl B' ~.L_= E .L = V -pz' .L = V1-~z . 1

\Vltile going over from olle referellce system to anotller, fields E and B are transformed in a certain .way. The laws

(8.1)

I

I i I "k the IOn'fitudinal and transHere the s~Jllb()ls II ~ll( J - ~nal com ol~ents of the elec.tric verse (relative to the vectol/ v o~ d I?s t,lle velocity of light and mag-ne t ·IC c.ne I (\s , Rp ---- V 0',c ' an C I in a vacuum (c 2 = 1/eof!o)'

" th formulas have the form Written in terms of prOJectIOns, ese •

\

E~=EX' " E'I- voE z E'l"" '

7' .,"

EI.-c.

.1

J

y 1-(IZ

B~=B;.,:+(!ORZIC2 ,

'

E Z ',l-l1 o lJ y

V1-~I'

where it is assumed that t h e X - all

8.2. Laws of Transformation for Fields E a'nd B

201

8.2. Laws of Transformation for Fields E and B

1'>' Ju~ B' z

1!1_~Z •

llz- v OE y /c

=

\ \

(8,2)

2

V1-~2

d X'-axes of coordinates are directed t - xvo/c 2

t/~·

V 1-(00/C)2

202

R. Relative Nature of EI('ctric and Magnetic Ficlds

8.2. Laws of Transformation for Fields E and

along the vector va, the Y'-axis is parallel to th y_,' I h " to the Z-axis. . e aXIS aUt t e Z-axIs

E' It

f~II~~s. from

Eqs. (8.1) alld (8.2) that each of the vectors IS expressed both ill terms of E alld 8. This i~ dll eVIdence of Il common nature of "],, t " 1 . rIe-I I Is. 'I' aI · "fiellls
(I Il I.

It sh?llld be emphasized that the properties of electromagnetic fwld expressed by the laws of its trllllsformation a~e lo~al; ~he values of E' ,alld 8' at a certain point ill ~pace d tire HI he system K are uniqnely determined only by . Ietlva IICS 0 E and 8 at the same point in space and time III Ie system K. The following- features of the laws of field t.. ' f ' . IclllS OJlllatlOn . al'e also . very Important. 1. Unlike the transverse components of E· I 8 I' I ch'lJlge 0 . t " , .tli( ,w IIC I ( U~.II a .ran~ItlOn hom Olle referollce system to Rn°t]thor, theu. longtludmal components do not change and l'emaill Ie same HI all systems. .2. Vectors E and B in different systems are COJiliccted WIth each other through highly symmetric relations TI'. can be. seen most clearly ill the form of the Jaws of 'trm;~~ formatlOll ,f~r tlte pro~ections.of the fields (see (8.2)1. 3. In. o.I del' to obtalll the fOJ'llltlla~ . " ()f tll(", In v,er<;c t 1"1JI st' ormatIOn (from K' to K), it is sufficient to I'eplac~ ill 'fo~'/Ill/las (8.1). and (8.2) all pl'imed quantities by llnprimod ones (and VIce versa) as well as t.he sian of " S . I C b "0' peCta ase of Field Transformation (u '~c) If tl _ system K' moves relative to tl1 svst"ln K O a"t' . 1 'tIC J' " ( , a ve OCI y L o «c, the rad lCals III tire denominators of fonn uhs (8 1') call be replaced by IInity, alld we obtain (, .

:t

i

0

,

'

.

Eir=E", E~

=

E.l + (vo X B},

"

C

Bir = 8",

B~=B.l-(voxEl/c2.

(8.3)

Hence it follows that

IE' ~ E+ (va X Bl,

B' = B- [vo X E}/c 2 •

I J

(8.4)

n

203

It should be noted that the first formula in (8.1) can he directly obtained in a very simple wa.y. Let a charge q in the system K have a velocity v o at a certain instant t. The Lorentz force actin~ on this charge is F = qE Iq [v o X 81. We go over to the inertial system K', movilll{ relative to the system K at the salUe velocity as the charge q at the instant t, viz. the velocity v o' At this instant, the charge q is at rest in the system K', and the force actiJl~ on the immobile ehurge is purely electric: F' ..co gE'. If Vo «c (as in the case under eonsiderlltioll), this {orec is illvariant (F' = F), whellce we ol)taill the Ihst fOl'lIlula ill (8.4). . However, the transformation fOl'llIulas fol' magnetic Held canbe obtained only with the help of the theory of relativity as a result of rather cllmbersome calculations. Let us consider a simple example of the application of formulas (8,1).

+

Example. A.Jal'ge metallic plate moVl'S at a constant nonrelativistic velocity v in a'unifol'In magnetic Held 1J (Fig. 8.1). Fillll tlIP surface density of charges iIHluced on the surfaces of the plate as a resultof its motion. • Let us go over to the reference system fixed to the plate. In accordance with the first of formulas (8.4), in this, system of reference a constant uni\' form electric field

Bt

E' =

Iv

X

Bl

will be observed. It is :directl'd towards liS. Under the action of this external Fig. 8.1 field, the charges will be displaced so that positive charges will appear on the front surface of the plate and negative charges, on the lJinrl surface. The surface density a of these charges will be such that the Held created by these charges inside the plate completely con!pe.nsat~s ~he external field E', since in equilibrium the resultant electrIc held IIIslde the plate must be equal to zeru. Taking into acconnt relation (1.11), we ubtain ' a

=

foE'

=

fovR.

It should be noted that while solving this problem, we coulll follow another line of reasoning, viz. from Ihe point of vil'W of the reference frame in whieh the plate J1loves at the velocity v. III this system there will be an electric jidd inside the plnte. It appears as a result of the action of the magnetic cOlllponent of the Lorentz force causing the

204

8.2. Laws of Transformation for Fields E and

fi. Rell/tive Nature of Electric and Magnetic Fields

displacement of all the electrons in the plate behind the plane of Fi;t. 8.1. As a result, the front surface of the plate will be char~ed positively and the hind surface, negatively, and the electric field appearing- in the plate will be such that the electric force qE compensates the magnetic component of the Lorentz force, q [v X Bl, whence E = - [ y X BI. This field is connected with the surface charge density through the same formula lJ = BovB. , Both approaches to the· solution of the given problem are equally justified. .

Relativistic Nature of Magnetism. Formulas (8.1) and (8.2) for the transformation of fields lead to a remarkable conclusion: "the appearance of the magnetic field is a purely relativistic effect following from the existence in nature of the limiting velocity c equal to the velocity of light in a vacuum. If this velocity (and hence the velocity of propagation of interactions) were infinite, no' magnetism would exist at all. Indeed, let us consider a free electric charge. Only electric field exists in the system K where this charge is at rest. Consequently, according to (8.1), no magnetic field would appear in any other system K' if c tended to infinity. This field appears only due to the fact that c is finite, i.e. in the long run due to the relativistic effect. The relativistic nature of magnetism is a universal physical fact, and its origin is associated with the absence of magnetic charges. Unlike most of relativistic effects, magnetism can be easily observed in many cases, for example, the magnetic field of a current-carrying conductor. The reason behind such favourable circumstances is that the magnetic field may be created by a very large number of moving electric charges under the conditions of the almost complete vanishing of the electric field due to practically ideal balance of the numbers of electrons and protons in cond Hctors. I TJ this case the magnetic interaction is predominant. The almost complete compensation of electric charges made it possible for physicists to study relativistic effects (Le. magnetism) and discover correct laws. For this reaSOJl, the laws' of electromagnetism, unlike Newton's laws, have not
n

205

. the fields E aud B lUust he to an.other refcrence, sys~em,. '11 the eoneept of the tl 'u .. J:< 01' Iustance, eve . . treaLe(! Wi i C ie. . " char re IJy a muving magnetICforce exerted. Ull a mov lU", . ~. The force is deterfield tioes not have any preCise m~a.llln~. and B at the point mined by the vah~es of the q¥:ntIt~e:esult of motion of the wllere the charge ISElo~~t;\heira~;lue~ at this point change, sources of th~ nellis a, well Otherwise the motion of the the force Will ch,mge as e ~laO'nitude of the lorce. . ,;\' . sources does not a~ect th ob"'lem about the force actlllg Pr 1 B at the point of location ~t Thus, while solvmg the E ,;, :, an( 't' ,\> on a cIlarge, we must know . 1 . 't v All these quantI les of the charge as ~~~l a~ 1~~I:~,~e~~i~l ;eference frame we are

i'·

'~.~'.;, ~~I~:~e~~e~t~~I.l re a lve 0 The expression "nlOY -

K'I

.j.•.: ,._: ,:

K

B

I

ing field" (if it is lised) ... t -v-o.......... should be int~rpreted .as " •q I a convenient way of the ,I • t'OIl of. ~.\ verlJal ucscnp I L _ varying neh! unllor C('ltain conditions, anll nothFig. 8.2 ing more. . The following silllple tthat upon a transition example will illllstraf.t.c the fa~n~ther the nel(l must be from one refcrcnc,c [ame to t1'oatell with care.

\

.. t between the poles of a magExample A charge(l partlde IS at res to the system K' that moves net lixed in 'the system K. Let us gOt?~~~ic velocity Yo relative to the to the right (Fig. 8.2) at a t~n[~l~harged particle moves i~ the n~ar system K. (t) Can we stat~~ ,~a(2) I~ind the force acting on thiS partlc e netic tielll in the system . . the system K'. . elic field. It should be no tell , m (t) Yes, the particle ~lUves In the rn~~'\ield and not relative to ~he however, that it moves m. the magne • tlwt a particle moves relatIve magnetic field. It is meaIllngful to rS~iher objects. How:~ver,:~e cannot to the reference system, a magnlti'~e to the magnetic held. lhe ~at~:r state that a particle I.novles ~e :: I All this refrrs not 'lIlly to magnl' IC statement has no physlca ml'unmg. but to electric tields as we}l. must take into aceount that t~l (2) In order to lind the °lrc ! weted towards us (Fig. 8.2), WI electric lield E' =- l vD.'>; B ~ ~recstem the charge will move to t ~Po in the svst"1ll I\. . III thIS sy . ' 'n occur in crossL',ld<'CtrH' appear I '. , lI,l this mottOll W I ' th·.t the left at the ve O~·,!l.i ·_~o~ ' \ ' I of llefmiteness, we suppose • and magnetic ileitis. l' 01 t Ie sa
l

206

8. Relative Nature of Electric and lUagnetic Fields

~harge of the particle is q

IS

>

8.9. Field Transformation Laws. Corollaries

O. Then the Lorentz force in the system K'

F' =qE' +q [-v o X B]=q [v o X B]-q[vo X B]=O

which, h.~~v~,:er, dire~tly fo,nows from the invariance of the force upon nonrelatr ,Jst1C transformatIOns from one reference frame to the other.

8.3. Corollaries of the Laws of Field Transformation

. Some Simple Corollaries. Several simple and useful relatIOns follow in some cases from transformation formulas

(8.1). '. If the ~lectric field E alone is present in the system K

(the magnetIc field B = 0), the following relation exists hetween the fields E' and B' in the system K': B' = -[Yo X E'lIc 2 •

(8.5)

°

= 0, then E~= El.lV 1 - ~2 and B' = c=. -[Yo X El/c 2 V 1 - ~~ c-= -[v o X 'E'1/c 2 , wh;re w~ t~ok mto account that in the cross product we can write elthe~ E .01' El. (the ,same applies to the primed quantities).

Indeed, if B

B~

I

I

Consldermg that B = Bj, + B~ = B~, we arrive at formula (8.5). 2•. If the magnetic fwld B alone is present in the system K (wIllIe tIte electric field E = 0), then for the system K' we have E'

=

[Yo X B'l,

(8.6)

In ieed, if E = 0, then B~ = Bl.IV 1 - ~2 and Ell = 0 Ef. = [Yo, x. BlIV 1 - V Suhstituting Bl. for B and the~ rs.6~~r BJ. m the last cross product, we arJ'ive at formula Form~Ilas (8.5) and (8.6) lead to the following important conclusIOn: If only one at the fields (E or B) is present in the system K, the electric and magnetic fields in the system K' are mutually p,erpendicular (E' -L B'). It should be noted t~a.t the ~p~oslte statement is true only under certain addItiOnal lImitations imposed on the magnitudes of the vectors E and B. Finally, since Eqs. (8.5) and (8.6) contain only the quantities referring to the same system of reference, these equa-

207

tions can be easily applied to the fields varying in space and time. An illustrative example is the field of a uniformly moving point charge. Field of Freely Moving Relativistic Charge. The formulas for field transformation are interesting above all by that they express remarkable properties of electromagnetic field. Besides, they are important from a purely practical point of view since they allow us to solve some problems in a simpler way. For example, the problem of determining the field of a uniformly movv ing point charge can be solved. ,as a result of transformation of a purely COUlomb field which is observed in the reference system fixed to the charge. Calculations sh9w (see Problem 8.10) that the lines 61 the field E created by a freely moving charge q have the form Fig. 8.3 shown in Fig. 8.3, where'v is the velocity of the charge. This pattern corresponds to the instant "picture" of the electric field configuration. At an arbitrary point P of the reference system, vector E is directed along the radius vector r drawn from the point of location of the charge q to point P. The magnitude of vector E is determined by the formula q

1

1_~2

E "''' -4:rt-e-o -r-2 7":(1:-_--;;-~2;;-s-;i.:..n;;-2n{j-";;;)3'";;;/2'

(8.7)

where ~ = vIc, it is the angle between the radius vector r and the vector v corresponding to the velocity of the charge. The electric field is "flattened" in the direction of motion of the charge (see Fig. 8.3), the degree of flattening being the larger the closer the velocity vof the diarge to the velocity c of light. It should be also borne in mind that the field shoWJl in the figure "moves" together with the charge. That is why the field E in the reference frame relative to which the charge moves varies with time. Knowing the field E, we can find the field B in the same reference system: 1,

110

B = ~ [v< El = 4it

q[vxr]

,.a

1-~2

(1- ~2 sin2 {j-)3/2 •

(8.8)

209

Problems 208

8. Relative Nature of Electric and Magnetic Fields

This formula is a corollary of relation (8.5) iu which we have replaced the primed quantities by unprimed ones and v by -v. When v ~ c (~ «1), expressions (8.7) and (8.8) are transformed to (1.2) and (6.3) respectively.

. . 'f E - 0 l' B = 0 in some reference system, opposite IS Ba~s? true. Ither s-yste~ (This conclusion has alreadY heen then E'..l III any 0 .

drA~ifinS:llY 8~~;~ t h at genera II Y

and E2-c2B2=inv.1

(8.9)

The invariance of these quantities (relative to the Lorentz transformations) is a consequence of field transformation formulas (8.1) or (8.2). This question is considered in greater detail in Problem 8.9. The application of these invariants allows us in some cases to find the solution quickly and simply and make the corresponding conclusions and predictions. We shall list the most important· conclusions. L The invariance of E'B immediately implies that if E ..1 B in some reference system, Le. E· B = 0, in all other inertial systems E' ..1 B' as well. 2. It follows from the invariance of E2 - c 2B2 that if E = cB (Le. E2 - c2B2 = 0), in any other inertial system E' = cB' as well. 3. If in a certain reference system the angle betwl'cn vectors E and B is acute (ohtuse) (this means that E· B is greater (Ies:;) than zero), in any other reference system the ang-Ie between vectors E' and H' will he acute (ohtus(') as well. 4. If in a certain reference system F > cB (or E < cB). (which means that, E2 - r 2B2 is greater (or l!!ss) than zero), in any other reference system E' > cB' (or E' <; clJ 0) as well. 5. If hoth invariants are equal to zero, then E 1. R ana E = cB in all rf'ference systems. It will he shown later that this is precisely ohserved in electromagnetic waves. 6. If the invariant I>B alorw is equ.. 1 to zero, we can fmd a reference system in which either E' = 0 or B' = O. The sign vI ihe other invariant determines which of these vectors is equal to zero. The

c

B

1

K'I

K

jE.B=inv.

".

~

r~o I

8.4. ElectYomagnetic Field Invariants

Since vectors E and B characterizing electromagnetic field depend on the system' of reference (at the same point in space and time), a natural qu~stion arises concerning the invariants of electromagnetic field, i.e. the quantitative characteristics independent of the reference system. It can be shown that there exist two such invariants which are the combinations of vectors E and B, viz.

more remark: it should be borne in.mind nilis E and B depend both on coordlllates fi

--vo

1

r

I

I

I .f(11c"

----

----

q

q

v

~.

Fj~

Fig. 8.01

8.5

. .... t f (8 \J) refers . d' d on time COllsequently, eac Il lIlvallHl 0 . ' . 'n s ace and time of the field, the COOl' 1~::~eas:::i~~~~~he ~oint in different systems of .reference being connected through the Lorentz transformations.

an

Problems . I f f Id transformation. A nonrelativistic • 8.1. Specla case 0 n I~ant velocity v. Using the transformapoint charge q mo;e at a co field B of this charge at a point whose . tion formulas,. fin t e ma~ne rc. defined hy the radius vector r. h position relatIve to tee arge IS t the s stem of reference K' fixed to the Solution'h~et us go ov r ~he coJlomh fIeld with the intensity charge. In t IS system, on Y

t

l.

1

1

q

41tEo

r

E ' = - - , r, is resent where we took into account that the radius :~~tof~o~~~h~ in ~he sy~tem K' (nonrelativistic case). Leltt~s ntowth:\ystem K' at stem K that moves re a Ive 0 h ' t th system K 0 e sy. we shall use the formula for t I'

}.~i/~lf~~~(~;), ir~~v~?~(rl fi:~~l:~:r~~~m~~JI~:~g~i:Sth;I~~I~~Ai~~ by unprimed quantIties an vice '. d ' v and hence b -Vo (Fig. 8.4). In the case under conSI ~ratJOn, vo ~ 'K' B' = 0 l = B' + [v X E'l/c Considering that the system ~

2.

III

and that c2 = 1/Eoflo, we fInd B=

110

4Jt

q [v X rJ

r3

•

We have obtained formula (6.3) which was earlier postulated as · t'JOn of experimental data. a result of the genera 1lza 14-0181

C'

210

8. Relative Nature 0/ Electric and Magnetic Fields

Problem•

•. 8..2.. A large plate made of homogeneous dielectric with the permlttlvltY.F 1I.lOves at a cOllstant nonrelativistic velocity v in a uniform m~gneLJc. held H as shown in Fig. 8.5. Find the polarization P of the 'dlelectnc and the surface density cr' of hound charges.

induction B' of the llIagnetic lield at the same distance frolll the beam in the system K' that moves relative to the system K at the velocity Vo in the direction of the proton beam. Solution. This problem can be solved in the most simple way with the help of formulas (8.1). But first we must find the induction B in the syst.em K at the same distance from the beam where the intensity E is given. Using the theorem on circulation of vector B and the Gauss theorem for vector E, we find R = !JIlI/2rrr, F A/2rtfor,

SolUtion,. In the refereIl;ce system fixe~ to the plate, in addition to the ~agne~lc field, there .wIlI he an elec,trlc field which we shall denote bY Eo· In ,lccord:lllce with formulas (8.4) of field transformation we h ave ' E~ .= [v X B). The polarization of the diP!ectrie is gin'lI by

P= xeoE

,

e-1

=. e o - - [v X HI e

where r is the distance from the beam, I = A.V is the current, alI,1 'A is the charge per unit length of the heam. It follows from these formulas that BIA' = fO!-tOJ/I.. = v/c 2 ,

'

where we to~k into account that according to (3.29), E' surface denSIty of hound charges is .

=

Eolt:. The

, e-1 Icr I =P=lI o ---vB.

where c 2 = lleo!-to. Substituting the expression for B fmlll this formula into the last of transformation formulas (8.1), we obtain

e

At the front surface of the plate (Fig. 8.5) cr' face cr' < o. "

>

2ft

0 while on the opposit

e

B' ..

,,'

=

1~'lv-vol

c2

V1- (v o/C)2

• 8.3. S,~ppose we have an uncharged long straight wire with a current I. ~ Illd the char~e per uni ~ length of this wire in the reference system moving translatlOnally WIth a nonrelativistic velocity v along the conductor in the direction of the current I. 0

If in this case Vo < u, the Jines of B' form the right-handed system with the vector vo. If va > v, they form the left-handed system (since the current ]' in the system K' in this case will flow in the opposite direction).

Soluti?n . . In aC,cordance with the transformation formulas (B.4), the electnc held E = [va X B) will appear in the moving reference system, or

• 8.5. A relativistic charged particle moves in the space occupied by. uniform mutually perpendicular electric and magnetic Helds E and B. The particle moves rectilinearly in the direction perpendicular to the vectors E and B. Find E' and B' in the reference system moving translationally with the particle. Solution. It follows from the description of motion of the particle that its velocity must satisfy the condition vB = E. (1)

E~ = -1'0!-toII2nr

(1)

where r is the distance from the wire axis. The expression for B in this formula was obtained with the help of the theorem on circulation. On the other hand, according to the Gauss theorem (in the moving system) we have E; = 'A'/2m: or, where 1..' is t~le charge per ulli t length of the wire. A companson of (1) and (2) gives

(2)

'A' = -vOIlc 2 , 2

,where c = Ih 0 !:l0' The o,rigin of this charge is associated with the ~orentz contractIOn expeflenced by the "chains" of positive and negative charges (they have different velocities!). . • 8.~. Th?re is a narrow beam of protons mOVing with a relativistiC velOCity v II.! the ~ystem of reference K. At a certain distance from the beam, the IlltenSlty of the electric field is equal to E. Find the

In accordance with the transformation formulas (B.l), we have

E'= E+[vxBI =0 Vl-~2 ' since in the case under consideration the Lorentz force (and hence the quantHy E [v X B) is equal to zero. According to the same transformation formulas, the magnetic field is given hy 2 H' = B-Iv X EI/e •

+

l"

1_~2

The arrangement of vectors is shown in l<'ig. 8.6, from which it follows that l v X EJ tt B. Consequently, considering that according to (1) 14*

213

Problems

212

8, Relative Nature

0/ Electric and Magnetic Fields

v = EIB, we can write B' =

B-E'IBe'

Y1-1i' or in the vector form B'=BV1-(ElcB)'. We the reader to ver'f . expressions satisfy bothadvise field invariants. I y th a t th e 0 b tamed ~ ~.~. The lIIotion of a charge ill crossed field ].' d relatIvIstIc particle with a specific charge qlrn movessI'n' aanregIOn ~. Anlonw Iere y

E

x B

q

Z

Fig. 8.6

Fig. 8.7

u~iform, mutually perpendicular fields E. and B have been created (FIg..8.7). At the moment t = 0 the partlcle was at point 0 d 't velocIty was equal to zero. Find the law of motion of the pant' II s x (t) and y (t). , a r IC e,

Solutton. The particle mov~s under the action of the Lorentz f ~t can b.e easily seen tha.t the particle always remains in the plane ~~. tS motIOn can b~ descn~ed most easily in a certain system K' h . onI y the magnetIc field IS present. Let us find this refe n ,were It follows from transformations .re ce sfystem. . (84) ' that E' -_ 0 mare erence .h sys t em -th a t mov~s WIt a velocIty Vo satisfying the reI a t' E= -!vo XB!. It .IS more convenient to choose the s stem ~~ h ve~ocIty Vo. IS ~hrected towards the positive value/on the ;a~~e (FIg. 8.7), smce m such a system the particle will move perpendicula 1s r y to vector B' and its motion will be the simplest v 2h;iB·i~h t~e l~~~er:: K' that moves to the. right at the velocity a~ce with (8e4)e d F-=- 0 8an7d onlyhthe field B' IS observed. In accord. an Ig. ., we ave B' = B - [yO X E]/c 2 = B (1 - v3Ic2).

Since for a nonrelat~vistic par~icle vo ~ c, we can assume that B' = B , In th~ syst~m K , the partlcle will move only in the rna net' fi I . Its ve~ocIty ~em!!, perpendicular to this field. The eqUatio~ of I~ :- d, 0 IOn for thIS partIcle m the system K' will have the form rnv~/R = qvoB. (1)

This equation is written for the instant t = 0, when the particle moved in the system K' as is shown in Fig. 8.8. Since the Lorentz force F is always perpendicular to thE: velocity of the particle, vo = = const, and it follows from (1) that in the system K' the particle will move in a circle of radius R = mvo/qB. , Thus, the particle moves uniformly with the velocity vo in a circle in the system K' which, in turn, moves uniformly to the right with the same velocity vo = EIB. This motion can be compared with the motion of the point q at the rim of a wheel (Fig. 8.9) rolling with the angular velocity w = voiR = qBlm. Figure 8.9 readily shows that the coordinates of the particle q at the instant t are given by x=vot-R sinwI=a (wI-sin WI), y=R-Rcoswt=a (1-cos wt), where a = mElqB2 and w = qBlm. • 8.7. There is only a uniform electric field E in an inertial sys tern K. Find the magnitudes and directions of vectors E' and B' in the system K' II14Yving relative to the system K at a constant relativistic velocity vo at an angle a to vector E. Solution. According to transformation formulas (8.1) and taking into account that B = 0 in the system K, we obtain E'n=Ecosa, E~=Esinan/i-lit, Ii=vol c . Hence the magnitude of vector E' is given by R'-~VE~~ I-Ei-El/(1-Bt cos' a)/(1-1i')

and the angle a' between vectors E' and vo can be determined from the formula tan a' = E~I Ell = tan al -lit.

Vi

The magnitude and direction of vector B' can be found in way: B"=O,

11

similar

B~=-[voXEJ/(c'V1-1i'), B'=B~.

This means that B' .l vo, and B' = voE sin al(c' V i-Ii')· , • 8.8. Uniform electric and magnetic fields E and B of the same direction exist in a system of 'reference K. Find the magnitudes of vectors E' and B' and the angle between these vectors in the system K' moving at a constant relativistic velocity Vo in the direction perpendicular to vectors E and B. Solution. In accordance with formulas (8.1), in the syste lll

Fe

21~

215

8. Relative Nature 0/ Electric and Magnetic Fields

Problem.

the tWo vectors E' and D' will be also perpendicular to the vector v 0 (Fig. 8.10). The magnitudes of the vectors E' and B' can be found by . the form1l\as

where we used the fact that [YO X BJ.Hvo X E.L] = v~J. E.L X X cos a = v3B.L ·E.L (Fig. 8.11). The remaining two scalar products in (2) are equal to zero since the vectors are mutually perpendicular. 1'hus, the right-hand side of (1) becomes Erl·B{, +EJ. .BJ. =EII·B" +EJ. .B.L =E·B

+

E' =" ;- EI (VOB)I V i-(volc)I'

B'

= .. ;- BI+ (voElcl)S

V 1-(volc)' The angle between the vectors E' and B' can be found from its tangent through the formula Ian a' = tan (as+a B)= (tan as -Han a B)/(1- tan ai: tan ail). Since tan

ai: =

voBIE and tan

Q.E.D. • 8.tO. Field E of a uniformly moving charge. A point charge q moves uniformly and rectilinearly with a relativistic veloaity v.

aB =

voElelB (Fig. 8.10), we obtain tan '"-= Vo (BI + Ellc l ) a (1-~1) EB

This formula shows that as vo

-+

e (~ -+ 1), the angle a'

--+

n12. The

+y' /

"

_I-, I

l~

lli--_

Vo

0

X'

Fig. X.X

Fil{. 8.!!

opposite is also tru!': if we know E and 8 in one reference system, and the angle between these vectors is less than 90°, there exist reference systems where the two vectors E' and B' are parallel to one anotheL .

•

tl:9. Invaria~t E: B.. Using transformation formulas (8.1)

show thaI the quantIty E· n IS all invariant. Solution. In the system K', this product will be E' . B' = (E'II-tE'.lHBII+BJ.)=EII·BII+EJ..BJ.' , , " , ,

=

(1)

(EJ. + [Yo X BJ. I)· (BJ. - [Yo X EJ. I/e l ) 1-~1

(2)

Considerf'hg that the vectors E J. and BJ. are perpendicular to the vector Yo, we transform the numerator of (2) to the form E.l.B.l-(v o!e) 2E .l.Bl. =E.l·B.l (1_~2),

Find the intensity E of the fle'ld created by this charge at a point whose radius vector relative to the charge is equal to r and forms an angle ~ with the vector v. Solution. Suppose that the charge moves in the positive direction of the X-axis in the system of reference K. Let us go over to the system K' at whose origin this charge is at rest (the axes X' and X of these systems coincide while the 'Y'- and Y-axes are parallel). In 'the system K' the field E' of the charge has a simpler form: E'

(3)

=

1 4n8o

q

---,:;r r

,

,

and in the X'Y' plane we have

E~ = 4~80

Let us write the last term with the help of formulas (8.1): EJ. .BJ.

Fig. 8.11

Fig. 8.10

i)D

;'3 x',

E~ 4~80 ='=

: ' y'.

(1)

Let us now perform the reverse transition to the initial system K, which moves relative to the system K' with the velocity -v. At the m
--

216

9.1. Faraday's Law. Lenz's Law

8. Relative Nature of Electric and Magnetic Field.

to (8.2), we obtain E:x:= E~,

= vElc

Ey=E~1 Vl-~I.

Substituting into these expressions (1) where x' and y' are replaced by the corresponding expressions from formulas (2), we get E%=_l_ L 4neo r'a

E __ l 1/- 4neo

x

Vl-~2'

q_ II ria J/I-1\2'

zS+yl 1 _ pl.

a ( 1- ~2 sin 2 {t ) 3/2 r 1- ~I , in accordance with (2), the electric field intensity will be given by

=

R- _1_..!L. 1-~2 • -- 4np. o r 2 (1-1\2 sin2 ,'})3/2 • 8.11. Interaction between two moving charges. The relativistic particles having the same charge q raove parallel to each other at the same velocity v as is shown in Fig. 8.12. The separation between the particles is [. Using expression (8.7), find the force of interaction between the particles. Solution: In this case, the angle between the vector v of one of the particles and the direction towarrls the other particle is {t = 900. Hence, in accordance with formula (8.7), the electric component of the Lorentz force is given by

1

q2

II Vi-I\I

(i)

while the magnetic component of the Lorentz force is It Fm=qvB= _0_ 4n

[I

ql v 2 Vi-~I

,

e2

= 11 follo· It should be noted that I 2 the ratio F IF. = Follov l = (v e) ,

. ~ .e (6 5) It canI be seen that as v ---+ F' relatiVIstIc h just as in tenon f case h force. . F ten( s to e' b the magnetic componentfo. ~ eaction (~pulsive force) is given y The resultant force 0 III er

.

Fig. 8.12

Fe=qE===- 4nE o

,

_ _ 1_!fl_

l=Fe-·F n1 -

It should be noted that E:x:IEII= xly, Le. vector E is directed radially along the vector r. The situation is such as if the effect of lag were absent altogether. But this is true only when v = const. If, however, the c~arge moves with an acceleration, the field E is no longer radial. K I t remains for us to find the magnitude of vector E: ... v

- V~ - _1_...!L / II EEx, E y - 4neo r 'S J q ••- -...~ v Since x 2 + y2 = r 2 and L _

2

(2)

where we took into account that in the case under consideration B is related to E through formula (8.5) from which it follows that B =

217

4ne

o

9. Electromagnetic

12

C,

yl_1\2. .

Induction

d ' Law of Electromagnetic Induction. 9.1. Fara ay S Lenz's Law . "t R8 established that the relation In the previous chapter,~ w . 'electromagnetic field, viz. between the "components 0 f ~n · and tu a f7 netic fielrls, IS e Iec t nc ..-.• '": bv the C mainly determmed . choice of the reference sy~tem. In . ds., the two components otI leI' wor I in . of electromRgnetic fie dare d. We shall show here t e rlate re". '11 ]osel' that there eXIsts a stl r, I .conneetlon . hetween the. ,E- IalH ' B-fields which is exhlllltC( I.n f7 netlc , henomena 0 f e I ectroma ,.., induction.. I 1831 Fig. 9.1 Faraday's DIscovery. n I . tl Ic, . phenoFRraday mal e.onc. of ,,' the ,. '11most ('1('('lrod~'lIamlcs-, h t fl1l((lam'~lIlal ,1Iscov( IICSt, I . duction. It eonsists m t a menon of electromaKne lC {:~ induced current) appe~rs 1I1 an electric Cl.lrrent (cRlled t hange in the maKnetlc flux ' l0 op 11pon a c loop . d t tnR ,tJ· a closed conl.lc (Le. the nux of B) enclose~ ce~:-~e\lrre~t indicates that the Tile appearance of the ml U . . tile loop as a result ' . f orce.r·'nappears induced electromotwe It ism important i here t t l~ of a change of the mRgne~~ h~~'the ('lInnge in the magl\etl~ '{. rloes not at Rl! rleprnd . . I ()lllv by the rate . o! i ' . I l ' dctel'llll1l(,(. Oil " {f-. was realIzc( alH IS n'rl '1 CII'III f7 e ill the SigH I ' . . . .I.e. by d/dt. ueSl es" . < " its -,vanatlOll, ,

;y

"'ll

218

9. Electromagnetic Induction

9,1. Faraday', lAw. Len.', Law

of the del'ivative d/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 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 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 dldt 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 through the surface stretched over such a complex loop can be represented as

eb

(n.2)

This quantity is called the total magnetic flux, or the magnetic-flux linkage. In this case, the e.m.f. induced in the loop is defined, in accordance with (9.1), by the formula 'L'.

C,

=_

)\T

l'

d
dt

.

(9.3)

'" If these two directions were connected through the left-hand screw rule, Eq. (9.1) would not contain the minus si~n.

220

9. Electromagnetic Induction

9.2. Origin of Electromagnetic Induction Let us now consider the h . I appearance of induced e f p rICa reaso~s behind· the induction (9.1) from whaim~e' a~n dtry to derIve the law of two cases. . rea y know. Let us consider A Loop Moves in a Per t M . . first take a l ' man.en agnetIc Field. Let us oop wIth a movable Jumpe~ o~ l?ngth 1 (Fig. 9.3) .. Su~pose that It IS III a uniform magE=[V"B] netIc field perpendicular to the plane r-----€I-__ of the loop and directed behind :\ the plane of the figure. We start ( : ) movmg the jumper to the right with v velo.city v. The charge carriers in tfe Jumper, viz. the electrons, will Fig. 9.3 a so ~tart to move with the same . v~loClty. As a result, each electron force F = -e [v X BJ w~ll be acted upon by a magnetic electrons will start mov·dlrected along the jumper. The which is equivalent to ~f! d?wnwards al~:mg the jumper, This is just the ind d ctrlC current dIrected upwards. over the surfaces ot~~e cur~nt. The ~harges redistributed field which will ind con ucto~s wIll create an electric the loop. uce a current III the remaining parts of The magnetic force F I h The field correspondingP t~y\t .e r~~e of an extraneous force. It should be noted that h' 1 IS . = F/(-e) = [v X Bl. with the help of field t t .IS expre~sIOn can also be Obtained Circulation of vector r~~sformatIOn formulas (8.4). by definition, the magnitud:r~~~~ a. cldosed contour gives, case under consideratI'on h e III uced e.m.f. In the ,we ave

~i = -vBl, . (9.4) where the minus sign is tak . sign rule: the normal n t~n tf: accordance with the adopted loop was chosen so that ·t· ~. surface stretched over the Fig. 9.3 (towards the fi:ld ISB) Irect~d behind the plane of the right-hand screw rule ,~~ he!lce, according to vention is the clockwise l the .pOSltIV~ dlrecf.ion of circumIn this case, the extraneo~sefit~f;'E~s~s ds~own in th~ figure. IS . Irected agaIllst thf

9.2. Origin 01 Elet..romagnetic Induction

221

positive direction of loop circumvention, and hence ~l is a negative quantity. The product vl in (9.4) is the increment of the area bounded by the loop per unit time (dS/dt). Hence vBl = B dS/dt = = det>/dt, where d<1> is the increment of the magnetic flux through the area of the loop (in our case, d<1> > 0). Thus, ~l

=

-d<1>/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
. Betatron. The vortex electric field' I ' 1D ~e induction electron accel~ratol'uavs: Ol~~ a hrilli8.~t aPillication CODSlSts of a toroidal evacuated ~h " .z. B.atron. TblS aecelerato" 0lf an (Fig 94) ThamDer. at:rang"d between the e ectromagnet wmding cr~at~s ; e .varlstlon of the current in ihn a vortex electric field. Ttl~ latt':r y\ng ,magnetic field which iIlduce; taneously retains them on an eq UJ a'~?be ~r.. 1.e3. the eillctmna and aimuI.. rlUm cIrcular Drbit of a certain

electroma~et

v8:

pol<;~

<\lu E dl =,

all) t' - - --\. (\) [v

at .

Bj dl.

(9.9)

J

The expression on the right-hand side of this equation is the total deri vative deDI dt. Here the first term is d Ul' to the time variation of the magnetic field, while the second

224

9 "

- - - - - , - - , - - - '. :~Irctromagnetic Induction

9.2. Origin of Electromagnetic Induction

is due. to tile mution of the 100 T1 " term IS explained in "Teater de~"1 ~e 0pl'lglll of the seconu Possible Difficulties'''' s< . al III roblern 9.2. t' J • ometllnes we cc a IOn w wn tile law of ('Iectro . . Ime across a situ(9.1) tlll'f1~ Ollt to In' if!' I~agnetlc IIldllction in the form difficlllti('~ associated' wi;;P~1 Ica~le. (mainly Lecallse of the I Ie, c JOIce of the contoul' itself). <

'I

'I

~s' L! ~/"':\"'}_--J.rFig. 9.5 Fig. 9.6

In such cases it is necessar the Lorentz force qE t~ resort to the Lasic laws, viz. q X BI and the law v E _ 1= -aBI8t. These 1aws express tl J' v /,aw of electromagnetic ind 1- l~ P lyslcal meaning of the Let us consider two . utc 1O~1 In all cases. . Ins ructlve examples.

+

fv

E~ample 1. A conduct' t.. . a regIOn in which a m Illg ~ rIp IS dIsplaced at a velocit

'

~~~~J~~::fj~~~~ lh:~;ebI~ n;1ddo~t;J ~fr~f~e~~::!'t::St:1~oJ~~

~h~~~:~e~~~c~:n! ct;retVi~;o~~.r ~iIli ctoh~n~~\~~~~m~~~~ ~~:U~:: At first sight th . . is difficult t h e questIOn seems not so sim I . he "closed" ? cthoos e ~he contour itself: it is n ~ e'l smce In th!s case it d' III e strIp and how tho . 0 c ear where It should f urmg. the motion of the stri I IS part of the contour will hehave tthJe Lorentz S III Ie movin . h ace upwards h' h . In =hlui~b~o~~~~r thi~c~t,. 'd7re~~et~lo~~~~i~e an electric curren1

s~:i;'~IlIhb~o:rI.eSpslimmdediatelrc1:~rh~h:~v~:;~;lecrterSoonr~ ~o

1\

~~~~?~;~:y~~g,:~~~{;:i~r!N1I{I~f:Z~!1~i~fi;r:!:;~(:~~~j s

t e case of the conducting r tr~spects, the situation is the same S J'Ip. Exam Ie 2 ., P . lhe dotted circle in F' Ig. 9.6. shows the region in Which

III

225

a permanent magnetic fIeld B is localized (it is directed perpendicularly to the plane of the figure). This region is encircled by a fixed metallic ring R. By moving the sliding contacts to the other side of the ring (see Fig. 9.6), we introduce the magnetic flux l1> into the closed contour containing a galvanometer G (i-initial position, 2-final position). Will the galvanometer show a current pulse? Applying formally the law (9.1), we should conclude that there will be an induced current. But it is not so. There is no current since in this case both oB/iJt and the Lorentz force are equal to zero: the magnetic field B is constant and the closed loop moves in the region where there is no magnetic field. Thus, we have none of the physical reasons underlying the law of electromagnetic induction.

On an Apparent Paradox. We know that the force acting on an electric charge in a magnetic field is perpendicular to its velocity and hence accomplishes no work. Meanwhile, during the motion of a current-carrying conductor (moving charges!) Ampere's forces undoubtedly accomplish some work (electric motor!). What is the matter? This seeming contradiction disappears if we t,ake into account that the*,IDotion of the current-carrying conductor in the magnetic neld is inevitably accompanied by electromagnetic induction. And since the e.m.f. induced in the conductor performs work ort the charges, the total work of the forces of the magnetic field (the work done by Ampere's forces and by the induced e.m.f.) is equal to zero. Indeed, upon an elementary displacement of a current loop in the magnetic field, Ampere's forces perform the work (see Sec. 6.8) (9.10) .. ' . 6A A = 1 d
The work of Ampere's forces is done not at the expense of the energy of the external magnetic field but at the expense of the power supply which maintains the current in the loop. In this case, the source performs additional work against the induced e.m.f., equal" to 6Aad 0= -'til dt = I dW, which turns out to be the same as the work 6A A of Ampere's forces. The work 6A which is done during the displacement of the loop against the impeding Ampere's forces (which appear due to the current induced in accordance with Lenz's law) is transformed into the work of the induced e.m.f.: 6A = -6A A = 6.,1 i. (9.13) From the point of view of energy, this is the essence of the operation' of all induction generators. 9.3. Self-indudion

d' If the loop is rigid and there are no ferromagnet~cs ~e i;:~~ighbourhood, the inductance is a constant quantlty independent of the current /. . h The unit of inductance is henry (H). In acco~da~ce w:~ (9 14) the inductance of 1 H corresponds to t e oop e m~gnetic flux through which is equal to 1 Wb at the current of 1 A which means that 1 H = 1 Wb/A.

E ' I Find the inductance of a solenoid, neglecting edge effec~t xamp e. f th olenoid be V, the number of turns p~r um Let the voluIDde hO e s t"c permeability of it substance inside the length n, an t e magne I " solenoid fJ.. d' t (9 t4) L = $/1. Consequently, the problem IS .re'.'. f th t tal magnetic flux $ assummg Accor lllg 0 duced to the dete~ml!1atlOn. or thee cu~rent I the magnetic held in the that t~e ~urre~ I IS ;re~il~~agnetic flux through a turn of, th~ sole~ solen?ld IS B - fJ.~o . IS whole the total magnetic flux piercIng II noid IS $1 = BS - 11I1oll , I turns is N'" l B(' = r IlIlon2Vl