Introduction to Electrodynamics - ch08 - Chapter 8...

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Unformatted text preview: 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 a 1 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.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. (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 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- 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 + 1) R2 sinO cosO dOdcP ( aWR ) 2 [ 1. ] = /lo 3" "2R2 smOcosOdOdcP (12cos20- 4- 3eos2-1) /lo ( aWR2 ) 2 2 . = 2 ~ (geos- 5) smOeosOdOdcP. 2 1r/2 2 1 1r/2 /lo awR2 . awR2 9 5 (Fhemi)z = 2 (~) 27r I (9cos3- 5eosO) smOdO = /l07r (~) [-4 eos4 + "2eos2 0] = /l07r (aw3R2r (0 + ~ - ~) = _/l~7r (a~R2r . 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....
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This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.

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Introduction to Electrodynamics - ch08 - Chapter 8...

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