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Introduction to Electrodynamics - ch08 - Chapter 8...

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Chapter 8 Conservation Laws Problem 8.1 Example 7.13. E- A 1 } - 2 --8 11"1008 B ~ Pol!. s =~ (E x B) =~ 1 . 211" 8 cp 411"2100 82 Zj b b I I >.I I I >.I p = S. da = S211"8d8 = _ 2 -d8= _ 2 In(b/a). 11"100 8 11"100 a a b b ButV= / E.dl=- 2 A l ~d8=~ln(b/a), solp=IV.1 11"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.LoK x = - x w p= / S.da=SWh=alh, butV= / E'dl=~h, solp=IV.1 100 100 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) = ~ cpo ut 1I"€oa 211"a ( ) [ ( ) 2 ( ) 2 ] 1 2 1 2 1 It 1 J.Lo18 (b) Uem = 2" €oE + J.Lo B = 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
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147 QUem = /l0[2 2c2t = ~j -V. S = [2t V. (8 S) = ~ = 8Uem. ./ Qt 27r2a4 7r2€oa4 27r2€oa4 7r2€oa2 8t I /l [2 {b /l W[2 [ 82 184 ] I b (C) Uem = UemW27r8dS = 27rW2:2a4 Jo [(ct)2 + (8/2)2]sds = :a4 (ct)2"2 + 44 0 w[2b2 [ b2 ] I [2t = IJL027ra4 (ct)2 + 16 . Over a surface at radius b: .Rn = - S. da = 27r2€oa4 [bs. (27rbws)] = dUem /low[2b2 2 [2wtb2 - d = 4 2c t = ~ = .Rn. ./ (Set b = a for total.) t 27ra 7r€oa Problem 8.3 [2wtb2 7r€oa4 . 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 = ~/loaRc.vz (inside) and B = 4 /lon; (2cosOr + sinO8) (outside), where m = ~ 3 7rR3(awR). (From 3 7rr Eq. 5.68, Frob. 5.36, and Eq. 5.86.) We want a surface that encloses the entire upper hemisphere-say a hemispherical cap just outside r = R plus the equatorial circular disk. Hemisphere: /lom [ ( A ) . n ((} A ) ] /lom [ 2 n . 2 n ] /lom ( 2 ) Bz = 47rR3 2cosO r z + smu z = 47rR3 2cos u - sm u = 47rR3 3eos 0 - 1 . da = R2 sin 0 dOdcP r; B. da = ::; (2 cos O)R2 sin 0 dOdcP;daz = R2 sin 0 dOdcP eos OJ ( /lom ) 2 ( /lom ) 2 B2 = 47rR3 (4COS20 + sin2B) = 47rR3 (3COS20+ 1). H 1 ( /lom ) 2 [ 1 ] (T' da)z = /lo 47rR3 (3COS20 - 1) 2eosOR2 sinO dOdcP - "2(3cos2 0 + 1) R2 sinO cosO dOdcP ( aWR ) 2 [ 1. ] = /lo 3" "2R2 smOcosOdOdcP (12cos20 - 4 - 3eos2 0 -1) /lo ( aWR2 ) 2 2 . = 2 ~ (geos 0 - 5) smOeosOdOdcP. 2 1r/2 2 1 1r/2 /lo awR2 . awR2 9 5 (Fhemi)z = 2 (~) 27r I (9cos3 0 - 5eosO) smOdO = /l07r (~) [-4 eos4 0 + "2eos2 0] 0 0 = /l07r (aw3R2r (0 + ~ - ~) = _/l~7r (a~R2r .
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148 CHAPTER 8. CONSERVATION LAWS 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) 2 Z I (agrees with Prob. 5.42). Problem 8.4 H (a) (T . da)z = Tzx dax + Tzy day + Tzz daz. But for the x y plane dax = day = 0, and daz = -r dr d</J (I'll calculate the forceon the upper charge). z +q (T' da)z = EO ( EzEz - ~E2) (-r dr d</J). x Now E = _ 4 1 2; casal', and cosO = ~, so Ez = ( ;100 ) ~ r2 '" 0, E2 = _ 2 3' Therefore 11"100 (r2 + a2) y Fz = 2 Ef~ 00 1 ( q ) . 1 r3 dr q2 1 1 u du . 2 -Eo - 211" = -- lettm u = r 2 211"100 (r2 + a2)3 411"100 2 (u + a2)3 ( g - ) 0 0 [ ] 1 00 ~ q2 1 1 a2 q2 1 1 a2 q2 1 411"1002 -(u+a2) + 2(u+a2)3 0 = 411"1002 [0+ a2 - 2a4] = 411"Eo(2a)2,1( = (b) In this case E = -~2; sinOz,and sinO= ~,so 411"100 '" '" ( ) 2 ( ) 2 qa 1 H EO qa r dr d</J E2 = E; = _ 2 3' and hence (T' da)z = -- 2 _ 2 3' Therefore 1I"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
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