文档介绍：Chapter 8
Conservation Laws
Problem
Example .
E- - --8A 1
211"1008
B ~ Pol!. s =~ (E x B) =~ 1 .
211" 8 cp } 411"210082 Zj
b b
>.I I >.I
p = S. da= S211"8d8= _ -d8= _ In(b/a).
I I 211"100I 8 211"100
a a
b b
A
ButV= =- ~d8=~ln(b/a), solp=
/ 211"100l 8 211"100
a a
Problem .
E = ~z
100 1 a1
S = -(E x B) = -Y;
~ ~ €oW
B = = -w x }
p= =SWh=alh, butV= E'dl=~h, solp=
/ 100 / 100
Problem
a~ Q t ~
(a) E = - z; a = ---:i"; Q(t) = It => E(t) = ~ z.
~ ~ ~1I"~a
BE 2 111"82 A
B 211"8= €O-;:;-1I"8ut = €o~1I"€oa => B(8, t) = 211"a~ cpo
2 2
1 2 1 2 1 It 1
(b) Uem= 2"(€oE + ) = 2" [100(1I"€oa2) + (211"a2) ] = 211"2a4 [(ct)2 + (8/2)2] .
S - ~(E x B) - ~ ~ (-8) - - 12t 88
- - (1I"€oa2)(211"a2) - 211"2€oa4 .
146
147
QUem = /l0[2 2c2t = ~j -V. S = [2t V. (8 S) = = 8Uem. ./
Qt 27r2a4 7r2€oa4 27r2€oa4 ~7r2€oa2 8t
b
/l [2 {b /l W[2 82 184
(C) Uem = IUemW27r8dS= 27rW2:2a4Jo [(ct)2+ (8/2)2]sds= :a4 [ (ct)2"2 + 44 ] I 0
w[2b2 b2 [2t [2wtb2
= IJL027ra4 [ (ct)2 + 16 ] . Over a surface at radius b: .Rn = - IS. da = 27r2€oa4[bs. (27rbws)] = 7r€oa4 .
-dUem /low[2b2 2 [2wtb2
dt = 27ra4 2c t = ~7r€oa = .Rn../ (Set b = a for total.)
Problem
F =fT' da - /lo€o ~ I S dr.
The fields are constant, so the second term is zero. The force is clearly in the z direction, so we need
++ 1 1 2
(T' da)z = Tzx dax + Tzy day + Tzz daz ="/l0 (BzBx dax + BzBy day + BzBz daz -"2B daz)
= :0 [Bz(B'da)-~B2daz].
NowB = ~/ (inside) and B 7rR3(awR). (From
3 = 4/lon;7rr (2cosOr + sinO8) (outside), where m = ~3
Eq. , Frob. , and Eq. .) We want a surface that encloses the entire upper hemisphere-say a
hemispherical cap just outside r = R plus the equatorial circular disk.
Hemisphere:
A
/lom A . n /lom 2 n . 2 n /lom 2
( ) ((})
Bz = 47rR3 [2cosO r z + smu z] = 47rR3 [2cos u - sm u] = 47rR3 (3eo