Introduction to Electrodynamics (solutions) - ch08

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Chapter 8
Conservation Laws
Problem 8.1
Example7.13.
E- A 1
}
-
2
--8
11"1008
B ~ Pol!. s=~(E x B) =~ 1 .
211"8 cp 411"210082 Zj
b b
I I >.I I I >.Ip = S. da= S211"8d8=_2 -d8=_2 In(b/a).11"100 8 11"100
a a
b b
ButV= /E.dl=-2A l ~d8=~ln(b/a), solp=IV.111"100 8 211"100
a a
Problem 7.58.
E =~z
}
100 1 a1
S =-(E x B) =-Y;
~ J.Lol~ J.Lo €oW
B =J.LoKx =- xw
p=/S.da=SWh=alh, butV= /E'dl=~h, solp=IV.1100 100
Problem 8.2
a~ Q ~t ~(a) E =- z; a =---:i";Q(t) =It => E(t) = ~ z.~ ~ 1I"~a
BE 2 111"82 J.Lo18 A
B 211"8=J.LO€O-;:;-1I"8 =J.Lo€o~ =>B(8, t) = ~ cpout 1I"€oa 211"a
( ) [ ( )
2
( )
2
]
1 2 1 2 1 It 1 J.Lo18
(b) Uem=2" €oE + J.LoB =2" 100 1I"€oa2 + J.Lo 211"a2 =
S - ~(E x B) - ~(~ )(J.LOI8)(-8) - - 12t 88- J.Lo - J.Lo 1I"€oa2 211"a2 - 211"2€oa4.
J.Lo12
211"2a4[(ct)2 + (8/2)2] .
146
147
QUem =/l0[2 2c2t=~j -V. S = [2t V. (8S) =~ =8Uem. ./
Qt 27r2a4 7r2€oa4 27r2€oa4 7r2€oa2 8t
I /l [2 {b /l W[2 [ 82 184 ] I
b
(C) Uem= UemW27r8dS=27rW2:2a4Jo [(ct)2+(8/2)2]sds= :a4 (ct)2"2+ 44 0
w[2b2
[
b2
] I [2t=IJL027ra4(ct)2+ 16 . Overasurfaceat radiusb: .Rn=- S. da=27r2€oa4[bs.(27rbws)]=
dUem /low[2b2 2 [2wtb2-
d = 4 2c t =~ =.Rn../ (Setb=a for total.)t 27ra 7r€oa
Problem8.3
[2wtb2
7r€oa4.
F =fT' da- /lo€o~I Sdr.
Thefieldsareconstant,sothesecondtermis zero.The forceis clearlyin thez direction,soweneed
++ 1
( 1 2 )(T' da)z = Tzxdax+Tzyday+Tzzdaz="/l0 BzBxdax+BzByday+ BzBz daz-"2B daz
= :0 [Bz(B'da)-~B2daz].
NowB = ~/loaRc.vz(inside)and B =4/lon;(2cosOr+sinO8)(outside),wherem = ~37rR3(awR). (From3 7rr
Eq.5.68,Frob. 5.36,and Eq. 5.86.) We wanta surfacethat enclosesthe entireupperhemisphere-saya
hemisphericalcap just outside r =R plus the equatorial circular disk.
Hemisphere:
/lom [ (A ) . n ((}
A
) ] /lom [ 2 n . 2 n] /lom ( 2 )Bz = 47rR32cosO r z+ smu z = 47rR32cos u - sm u =47rR33eos 0 - 1 .
da = R2sin0dOdcPr; B. da=::;(2cosO)R2sin0dOdcP;daz= R2sin0dOdcPeosOJ
(/lom)
2
(/lom)
2
B2 = 47rR3 (4COS20+sin2B)= 47rR3 (3COS20+1).
H 1
(/lom)
2
[
1
](T' da)z = /lo 47rR3 (3COS20- 1)2eosOR2sinOdOdcP- "2(3cos20+ 1)R2sinOcosOdOdcP
(
aWR
)
2
[
1.
]= /lo 3" "2R2smOcosOdOdcP (12cos20- 4 - 3eos20-1)
/lo (
aWR2
)2 2 .= 2 ~ (geos0 - 5)smOeosOdOdcP.
2 1r/2 2
1
1r/2
/lo awR2 . awR2 9 5
(Fhemi)z = 2 (~) 27rI (9cos30- 5eosO)smOdO=/l07r (~) [-4eos40+"2eos20]
0 0
= /l07r(aw3R2r (0+ ~- ~) =_/l~7r(a~R2r .
148 CHAPTER 8. CONSERVATIONLAWS
Disk:
2 ~
Bz = -Po(JRI.JJ; da=rdrd</J<p= -rdrd</Jz;3
2
(
2
)
2
B.da = -"3po(JRl.JJrdrd</J; B2= "3po(JRI.JJ ; daz=-rdrd</J.
1
(
2
)
2
[
1
]
1
(
2
)
2
ei.da)z = PO "3Po(JRi.JJ -rdrd</J+ 2rdrd</J = - 2po "3PO(JRi.JJrdrd</J.
(
(JI.JJR
)
2
I
R
(
(JI.JJR2
)
2
(Fdisk)z = -2po 3 211" rdr =-211"po ~ .
0
Total:
F =-1I"Po((J1.JJ3R2)2 (2 + ~) z =1-1I"PO((J1.JJ2R2)2ZI(agreeswith Prob.5.42).
Problem 8.4
H
(a) (T .da)z=Tzx dax +Tzyday+Tzzdaz.
But for thexy planedax= day= 0, anddaz=
-r drd</J(I'llcalculatetheforceontheuppercharge).
z
+q
(T' da)z=EO(EzEz - ~E2) (-r drd</J). x
Now E =_
41 2; casal',andcosO= ~,soEz =
( ;100)
~ r2 '"
0, E2= _2 3' Therefore11"100 (r2 +a2)
y
Fz =
2 Ef~ 00
1
( q )
.
1
r3 dr q2 1
1
udu . 2-Eo - 211" =-- lettm u =r
2 211"100 (r2+ a2)3 411"1002 (u +a2)3 ( g - )0 0
[ ] 1
00
~q2 1 1 a2 q2 1 1 a2 q2 1411"1002-(u+a2) + 2(u+a2)3 0 =411"1002[0+a2- 2a4]=411"Eo(2a)2,1(=
(b) In this caseE =-~2; sinOz,andsinO=~,so411"100'" '"
( )
2
( )
2
qa 1 H EO qa r dr d</J
E2 =E; = _2 3' andhence(T' da)z= --2 _2 3' Therefore1I"EO (r2 +a2) 1I"EO (r2 +a2)
( )
2
1
00
d 2 2
[
1 1
]
00 2 2
[
1
] I
2
Fz =- EO ~ 211" r r =- q a. - - =_!L!! 0+- = - L ~ 1.(
2 211"EO 0 (r2 +a2)3 411"EO 4 (r2 +a2)2 0 411"100 4a4 411"EO(2a)2'
149
Problem8.5
(a)Ex =Ey =0,Ez=-a/fa.Therefore
T - T - - foE2- -~.xx - yy - - ,
2 2fO
Txy= Txz= Tyz= ... = OJ T (E
2 1 2
)
fO a2
zz= fO - -E =-E2 =-
z 2 2 2fO.
2
(
-1 0 0
)
H a
T =- 0 -1 0 .
2fO 0 0 +1
(b)F =IT. da (8 = 0,sinceB = O)jintegrateoverthexy plane:da=-dxdyz (negativebecause
outwardwithrespectto a surfaceenclosingtheupperplate).Therefore
Fz =!Tn daz=- ;f: A, andtheforceperunitareais f =~=1-~:z.1
(c) -Tn =I a2/2fo I is themomentumin thez directioncrossinga surfaceperpendicularto z, perunit
area,perunit time(Eq. 8.31).
(d)Therecoilforceis themomentumdeliveredperunit time,sotheforceperunit areaonthetopplateis
~~ (sameas (b)).
Problem8.6
(a)Pem=faCEx B) =foEBy; Pem=I foEBAdy.1
(b)1= 100 Fdt =100 1(1x B)dt =100 lBd(z x x)dt=(Bdy) 100 (- ~~)dt
=-(Bdy)[Q(oo)- Q(O)]=BQdy. But the-originalfieldwasE =a/fa =Q/foA,soQ=foEA,andhence
1=I foEBAdy; I as expected,themomentumoriginallystoredin the fields(a) is deliveredas a kick to the
capacitor.
(c)IE. dl =- ~~=- ~~ld (fora length1 in they direction).-tEed) + lE(O) = -ld~~ =>
E(d)- E(O)=d~. F = -aAE(d)y +aAE(O)y = -aA[E(d)- E(O)]Y = -aAd~ y. 1=100F dt=
-(aAdy) rooddB dt=-(aAdy)[B(oo)- B(O)]=aAdBy.ButE =~, soI =I foEBAdy, I asbefore.h t ~
Problem8.7
B =/.Lonl2(fora <r <Rj outsidethesolenoidB =0). The forceon a segmentdr of spokeis
dF=l'dl x B =l'/.Lonldr(ix 2)=-l'/.Lonldr<f,.
Thetorqueonthespokeis
R
N =!r x dF =l'/.Lonl!rdr(-i x /jJ)=I'JLonl~ (R2- a2)(-2).
a
150 CHAPTER 8. CONSERVATION LAWS
Thereforetheangularmomentumofthecylindersis L =/N dt=-~J.tonI(R2- a2)Z/ I'dt. But J I'dt=Q,
so
I L =-~J.tonIQ(R2- a2)ZI (inagreementwithEq.8.35).
Problem 8.8
(a)
{
0, (r <R)
} {
~J.toMZ, (r<R)
E- . B-
- 4:€O~i', (r>R) , - ~~~[2cosOi'+sinOO],(r>R)
4 3 J.to Qm A A A
(wherem ="371"RM); p =€o(Ex B) =(471")2-;:s(i' x 0)sinO,and(i' x 0) = <jJ,so
J.to mQ. A A
l = r x p = (471")2-;::4smO(rx <jJ).
But (i' x (/J)=-0,andonlythez componentwillsurviveintegration,so(since(O)z= -sinO):
} (Ex.6.1)
J.tomQ A/ sin20( 2 . ) /
21f
/1f . 3 4 /00 1 ( 1)1 00 1L = (471")2Z -;:4 r smOdrdOd<jJ. d<jJ=271"j sm OdO="3; r2 dr = -;: R =Ii,"
0 0 R
J.tomQ A ) (
4
)(
1
) 12 2 A IL =(471")2Z (271" "3 R = gJ.toMQR z.
z
(b) Apply Faraday'slawto theringshown:
f E' dl =E(271"rsinO)=- dd<P=-7I"(rsinO)2(~J.todM ). t 3 dt
J.todM . A
=>IE =-3&(rsmO) <jJ.
The forceon a patchof surface(da) is dF =O"Eda=_J.t~0"~~(r sin 0)da(/J (0"=471"~2)'
J.toO"dM ( 2 A A AThe torqueon thepatchis dN =r x dF =-3& r sinO) da (i' x <jJ).But (i' x <jJ)= -0, andwewant
onlythez component(Oz=- sinO):
N =_J.t~0" d:: z/r2 sin2 0 (r2 sinO dOd<jJ).
1f 21f
Here r = Rj /sin3OdO=~j / d<jJ = 271",soN = _J.t~0"d:: zR4 (~) (271")=1-2~oQR2d::z.
0 0
0
L =/Ndt =- 2~oQR2z/dM =I i!-MQR2ZI (sameas (an.M
151
(c) Let the chargeon thesphereat timet be q(t); thechargedensityis a = q(Rt)2. The chargebelow47r
("southof') theringin thefigureis
11"
qs =a (27rR2)/sine'dB'=~ (- cosB')I;=~(1+cosO).
0
Sothetotalcurrentcrossingthering (flowing"north")is I (t) = - ~ddq(1+cosB),andhence2 t
I A 1 dq(1+cosB) A .
K(t)= . B(-0) =- R
-
d . B o. The forceon a patchof areada ISdF =(K x B) da.27rR sm 47r t sm
[
2 Po t7rR3M A
]
1 PoM A
Bave= 3poMi+ 47r R3 (2cosBf+sinBO) 2"=~[2i+2cosBf+sinBO];
K x B =~ dqPoM (1+cosB)[2(0x i) +2cosB(0 x f)].47rR dt 6 sinB --..-.-
-4>
dN = RfXdF=POM (dq)(l+cosB)2[ fx(Oxi)247r dt sinB ~
O(f . i) - i(f . 0)
poM (dq) 2 A A PoM R2 (dq)
A
= 127r dt (1+cosB)R [cosB0 + cosB0]dBd4>= 67r dt (1+ cosB)cosBdBd4>O.
- cosB(f x (fJ)]R2sin BdBd4>'---v---'
-9
211"
Thex and y componentsintegrateto zero; (O)z =- sinB,so (using/d4> =27r):
0
PoMR2 (dq) /11" . poMR2 (dq)Nz = - 67r dt (27r) (l+cosB)cosBsmBdB=- 3 dt0
- _poMR2 (dq)(~)=- 2poMR2dq. IN =- 2poMR2dq i,l3 dt 3 9 dt 9 dt
(Si~2 B - CO~3B)[
Therefore
0
L =/N dt=- 2~0MR2 i /dq=I ~MR2Qi I (sameas (a)).
Q
(1usedtheaveragefieldat thediscontinuity-whichisthecorrectthingto do-but in this caseyou'dgetthe
sameanswerusingeithertheinsidefieldor theoutsidefield.)
Problem8.9
dif> 2 I 1
(
2
)
dIs
(a)E =-di; if>= 7raB; B = ponIs; E = IrR. 80 Ir = -Ii po7ran yt'
f dif> 2 dIs 1 dIs A paIr b2 A(b) E. dl =--d =>E(27ra)=-PO7ran-d =>E =--2Poan-d cPoB =_2 3/2z (Eq. 5.38).t t t (b2+z2)
8=~(E x B) =~(_poan dIs)(
poIr b2
)
((fJX i) = -~poIr dIs ab2n f.
Po Po 2 dt 2 (b2+z2)3/2 4 dt (b2+ Z2)3/2
152 CHAPTER 8. CONSERVATION LAWS
Power:
= =
! ! 1 dIs ! 1P = S.da= (S)(27ra)dz=--7rpoa2b2nIn-d 3/2dx2 t (b2+Z2)-= -=
. . Z 100 1
(
1
)
2
The mtegralls .
I
=
b2
- -
b? =b2'b2yz2 +b2 -= ~
(' 2 dls) 2= - 7rpoandI Ir = (RIr)Ir = Ir R. Qed
Problem 8.10
Accordingto Eqs.3.104,4.14,5.87,and6.16,thefieldsare
{
1
} {
2--
3 P, (r <R), -110M£0 31""' ,
E= B=
1 1 ~~ pom ~~
--[3(p.r)r-p], (r>R), --[3(m.r)r-m],
47r£0r3 47rr3
(r <R),
}(r >R),
whereP = (4/3)7rR3P,andm = (4/3)7rR3M.NowP = £0f(E x B) dT,andtherearetwocontributions,one
frominsidethesphereandonefromoutside.
Inside:
!( 1 ) (2 ) 2 ! 2 4 3 8 3Pin = £0 - 3£0P X 3fLoM dT =-gfLo(P X M) dT =-gfLO(P X M)37rR =27fLO7rR(M X F).
Outside:
Pout= £0-4
1
4
Po! 16{[3(p.f) f - p] X [3(m.f) f - ill]} dT.7r£07r r
Nowf X(p x m) =p(f.m)-m(f.p), sof x [fx (pxill)]= (f.m)(fxp)- (f.p)(fxill), whereasusingtheBAC-
CAB ruledirectlygivesf x [f x (px ill)] =f[f.(p x ill)] - (pxm)(f.f). So{[3(p.f) f - p] x [3(m.f) f - mn =
-3(p.f)(fxm)+3(m.f)(fxp)+(pxm) =3{f[f. (p x ill)] - (p x m)}+(pxm) =-2(pxm)+3f[f.(pxm)].
Pout= 1~;2!r16{-2(p x m) +3f[f . (p x ill)]} r2sin 0dr dOdl/>.
To evaluatethe integral,set the z axisalong(p x ill); thenf. (p x m) =Ip x mlcosO.Meanwhile,f =
sin0cosI/>x +sin0sinI/>y +cos0z. But sinI/>andcosI/>integratetozero,sothex andy terms drop out, leaving
Pout = 1~;2 (100 r14dr){-2(p x m) !sin0dOdl/>+31px mlz!COS2 0sin0dOdl/>}
Po
(
1
)1
=
[
47r
]
Po
= - -- -2(p x m)47r+3(px m)- =--(p x m)
167r2 3r3 R 3 127rR3
= -~ (~7rR3p)x (~7rR3M)=4poR3(M x F).127rR3 3 3 27
Ptot = (287+ 2~)POR3(Mx P) =I ~POR3(Mx p).1
153
Problem8.11
(a)FromEq. 5.68andFrob. 5.36,
{
r <R: E =0,B =~J.Loo"Rc.vZ, with 0"= 4
e
R2;
3 ~ 4r >R: E = -
41 ~r, B =4J.LOn; (2cosBr+sinBO), with m = -3~0"c.vR4.~fOr ~r
Theenergystoredin theelectricfieldis (Ex. 2.8):
1 e2
WE=8~€0R'
Theenergydensity of the internal magneticfield is:
1 2 1 (
2 e
)2 J.Loc.v2e2 J.Loc.v2e24 3 J.Loe2c.v2RuB =2J.LoB =2J.Lo 3"J.LoRc.v 4~R2 = 72~2R2'so WE;"= 72~2R23"~R = 54~ .
Theenergydensityin theexternalmagneticfieldis:
1 J.L5 m2 2 . 2 e2c.v2R4 J.Lo 1 2
UB =2J.Lo16~2-;:6(4cos B+sm B) =18(16~2)r6 (3cos B+ 1), so
WBout=
00 ~ 2~
J.Loe2c.v2R4! 1 2 ! 2 . ! J.Loe2c.v2R4 ( 1 ) J.Loe2c.v2R(18)(16)~2 r6r dr (3cos B+ 1)smBdB d<jJ=(18)(16)~23R3 (4)(2~)= 1O8~ .
ROO
J.Loe2c.v2R J.Loe2c.v2R 1 e2 J.Loe2c.v2R
WB = WB;" +Wbout= 1O8~(2+1)= 36~ ; W =WE+WB= 8~foR + 36~ .
(b)SameasFrob. 8.8(a),with Q -t e andm -t ~ec.vR2:
I
L = J.Loe2c.vRZ.
3 18~
(c)J-loe2c.vR=~:::}c.vR=9~n= (9)(~)(1.05x 10-34) =19.23X 1010m/s.!18~ 2 J.Loe2 (4~x 10-7)(1.60x 10-19)2
~ e2
[
1 ~(
c.vR
)
2
]
= 2.
[
1 ~(
c.vR
)
2
]
= ~(
9.23X 1010
)2 =210 X 104.SITEOR + 9 e me , + 9 e 1+ 9 3 x 108 . ,
- (2.01x 104)(1.6x 10-19)2 -I -11 .1 - 9.23x 10-10-I 21 IR- 8~(8.85x 10-12)(9.11x 10-31)(3 X 108)2- 2.95x 10 m, c.v- 2.95X 10-11- 3.13x 10 rad/s.
Since(;JR,thespeedof a pointontheequator,is 300timesthespeedof light,this "classical"modelis clearly
unrealistic.
Problem8.12 z
E-~~'
- 4~€0r3'
B=J.Loqm~=J-loqm (r-dz)
4~ r,3 4~ (r2+d2- 2rdcosB)3/2'
qm,
xqe
154 CHAPTER 8. CONSERVATION LAWS
Momentumdensity(Eq. 8.33):
- (E B) - ./loqeqm (-d)(r x z)p - fO X -
( )2 3/2.411" r3 (r2 +d,2 - 2rd cos0)
Angularmomentumdensity(Eq. 8.34):
/loqeqmd r x (r x z)
(
~
) ~ 2 ~ 2 ~ 2 ~
l=(rxp)=- (4)2 3/2' Butrx rxz =r(r.z)-r z=r cosOr-r z.11" r3 (r2 +d,2- 2rdcosO)
The x andy componentswill integrateto zero;using(f)z =cosO,wehave:
L - - /loqeqmdZ/ r2 (COS20- 1) 3 2r2 sin 0dr dOdc/>.- (411")2 r3 (r2 + d,2- 2rdcosO) /
1 00 )= /loqeqmdZ(211")// r (1 - U2 3 2 dudr.(411")2 (r2 +d2 - 2rdu) /-1 0
Let u ==cosO:
Do ther integralfirst:
00
/ r dr (ru - d) 1 00 u d u + 1 1(r2+ d,2- 2rdu)3/2=d(l - u2)v'r2+d,2- 2rdu0 =d (1- u2) +d(1- u2)d =d(1- U2) =d(l - u)'0
Then
1 1 1
L =/loqeqmd~.!./ (1- U2)d =/loqeqmA /(1 + )d =/lOqeqm ~ ( +U2)I =811" Z d (1 - u) u 811" Z U U 811" Z u 2 -1-1 -1
Problem 8.13
(a) The rotating shell at radius b producesa solenoidalmagneticfield:
B =/loK z, whereK = O"bWbb,andO"b= - 2~brSoB = - /l~~~Qz (a < 8 < b).
The shellat a alsoproducesa magneticfield (/lowaQ/211"1)z, in theregion8 < a, sothetotalfieldinsidethe
innershellis
B =~:~(wa -Wb) z, (8 <a).
Meanwhile,theelectricfieldis
E 1 A ~ Q ~ ( b)=_2 - 5 =-2 1 5, a <8 < .1I"fO8 1I"fO8
( Q )( /lOWbQ ) ~ ~ /lOWbQ2 ~ /lOWbQ2 ~P =fo(Ex B) =1':0211"1':018\-2;1 (s x z) = 411"2[28</J; l =r x p = 411"2128(r x </J).
Now r x (fJ=(8S+ zz) x (fJ=8 Z- Z S,andthes termintegratestozero,so
L = /lOWbQ2A /d =/lOWbQ2 (b2 - 2)1A =I /lOWbQ2(b2 - a2) A411"212z T 411"2[211" a z 411"l z.
155
(b)Theextraelectricfieldinducedby thechangingmagneticfielddueto therotatingshellsis givenby
E 2
d<I>
E 1 d<I>;.. d . h . b1!'S= - -
d
=> =- _
2
-
d '/",an m t eregIOna <s <t 1!'St
/loQ
(
2 J..LOQWb( 2 2)
/loQ
( 2 2)
1 /loQ (2 dwa 2 dWb)
~
<P=- Wa-Wb)1!'a --1!' S -a =- waa -Wbs ; E(s)=--- a --s - cp,
21!'l 21!'l 2l 21!'s 2l dt dt
Inparticular.
E(a) =_/loQa (dwa dwb)
~
41!'l dt - dt cP, andE(b) = _/loQ (a2dWa - 2dWb )
~
41!'lb dt b dt cpo
Thetorqueon a shellis N =r x qE =qsEz, so
Nb =
( /loQa)(dwa dwb) ~ ["XJ /loQ2a2 ~Qa - 41!'l dt - dt z; La =Jo Nadt = - 41!'l (Wa- Wb)z.
Qb( /loQ)(2dwa b2dwb)
~ L 1
00
d /loQ2(
2 b2 )
~
- -- a - - - z; b = Nb t =- a Wa- Wb z.
~b & & 0 ~
L L /loQ2 (
2 b2 2 2 )
~ J..LoQ2Wb
(
2 2
)
~
a + b = 41!'l a Wa- Wb- a Wa+ a Wb Z = 41!'l b - a z.
Na =
Ltot =
Thusthereductionin thefinalmechanicalangularmomentum(b) is equalto theresidualangularmomentum
inthefields(a), .(
Problem8.14
~ q Il.
B=/lonlz, (s<R); E=- 4 3' whereIl.= (x-a,y,z).1!'€o1-
p =€o(Ex B) =€o(J..LonI) (-4q ) 13(Il.X z) =/lOq~I[yx - (x - a)y].1!'€o 1- 41!'1-
Linear Momentum.
J
J..LoqnI
J
yx- (x - a)y ~.' ..
p = pdT = 4;- [(x - a)2+y2+ z2J3/2dxdydz.The x termISoddm y; It Integratesto zero.
/loqnI ~
J
(x - a) ,= - - y
[( )2 2 2]3/2dxdydz. Do thez Integralfirst:41!' x - a +y + z
z
1
00 2
[(x - a)2 + y2h/(x - a)2 + y2+ Z2 -00 - [(x - a)2 + y2]'
/loqnI ~
J
(x - a) d d S . h I d.= -- y
[( )2 2]
X y. Wltc to poar coor mates:
21!' x - a + y
X=8COSt/J, y=8sint/J, dxdy =>8d8dt/J; [(x-a)2+y2]=82+a2-28acost/J,
/loqnI ~
J
(8 COSt/J - a) d d,!.= --y 8 8 'f'
21!' (82 +a2 - 28acast/J)
Now {27r cost/Jdt/J - 21!'(1- A ). {27r dt/J - 21!'Jo (A+Bcost/J)- B VA2_B2' Jo (A+Bcost/J)- VA2_B2'
HereA2 - B2 =(82 +a2)2 - 482a2 =84 +282a2 +a4 - 482a2 =(82 - a2)2; viA2 - B2 =a2 - 82.
/loqnI ~
J [1 (
a2+82
)
2a2
]
d /loqnI ~ 1
R
d J..LoqnIR2~= -y - + 8 8 =-y 8 8 = y,
2a a2- 82 (a2- 82) a 0 2a
156 CHAPTER 8. CONSERVATION LAWS
AngularMomentum.
l = r x p =J-Loq~Ir x [yx - (x - a)y]=J-Loq~I{z(x- a)x +zyY- [x(x- a)+y2]z}.4~ 4~
The x andy termsareoddin z, andintegrateto zero,so
J-LoqnIA ! X2 +y2- xa ..L = - - z [( )2 2 2]3/2dxdydz. The z mtegrallsthesameasbefore.411" x - a +y +z
= - J-LoqnIZ! X2 +y2- xa dxdy=- J-LoqnIz! 8 - a cos <p 82d8 d<p211" [(x - a)2 +y2] 211" (82+a2- 28acos<p)
![ 82 ( a2+82)] lR 82 - 82= -J-Loqnlz 2 2+ 1- 2 2 8d8=-J-Loqnlz 2. 28d8=lzero.1a -8 a -8 0 a -8
Problem 8.15
(a) If we'reonly interestedin theworkdoneonfreechargesandcurrents,Eq. 8.6becomes
dW r aD aD
dt =lv(E. J I) dT. But J I =V x H - at (Eq.7.55),so E. J I =E. (V x H) - E. at. Fromproductrule#6,V. (E x H) =H(V x E) - E. (V x H), whileV x E =- ~~,so
aB aB aDE. (V x H) =-H. - - V. (E x H). ThereforeE. J I =-H. - - E. - - V. (E x H) andhenceat at at '
dW
1 (
aD aB
) i- =- E .- +H . - dT- (E x H) .da.dt v at at s
This is Poynting'stheoremfor thefieldsin matter.EvidentlythePoyntingvector,representingthepowerper
unit areatransportedby thefields,is S = E x H, andtherateofchangeof theelectromagneticenergydensity
is auem=E . aD +H . aB.at at at
For linearmedia,D = €E andH = .!.B, with €andJ-Lconstant(in time);then
J-L
auem aE 1 aB 1 a 1 a 1a
- =€E. - +-B. - = -€-(E. E) +--(B. B) = -- (E. D+ B. H)at at J-L at 2 at 2J-Lat 2at '
soUem= ~(E.D +B .H). qed
(b)If we'reonlyinterestedin theforceonfreechargesandcurrents,Eq.8.15becomesf =PIE +J I xB.
00
(
00
)But PI =V. D, andJI =V x H - at'sof =E(V. D) + (V x H) x B - at x B. Now
a aD
(
aB
)
aB aD a
at(Dx B) =atx B +D x at ' andat=-V x E, soatx B =at(D x B) +D x (V x E), and
hencef =E(V .D) - D x (V x E) - B x (V x H) - %t(D x B). As before,wecanwith impunityaddthe
termH(V .B), so
f =([E(V .D) - D x (V x E)]+[H(V .B) - B x (V x H)]}- %t(D x B). I
The termin curly bracketscanbe writtenas thedivergenceof a stresstensor(asin Eq. 8.21),andthelast j
termis (minus)therateof changeof themomentumdensity,p = D x B.