Introduction to Electrodynamics - ch09

# Introduction to Electrodynamics - ch09 - Chapter 9...

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Unformatted text preview: Chapter 9 Electromagnet ic Waves Problem 9.1 a:: =-2Ab(z- vt)e-b(z-vt)2; ~~1 =-2Ab [e-b(Z-vt)2- 2b(z- vt)2e-b(Z-vt)2] ; 8ft = 2Abv ( z- vt ) e-b(z-vt)2, 82ft = 24bv [-ve-b(Z-vt)2 + 2bv ( z- vt ) 2e-b(Z-vt)2 ] = V2 82ft .; m '~ . M' M ~h 2. 8z = Abcos[b(z- vt)]; 8z2 =-Ab sm[b(z- vt)]; 8 h 82 h 2 2 . 282 h 8t =-Abvcos[b(z- vt)]; 8t2 =-Ab v sm[b(z:- vt)] = v 8z2"; 813-2Ab(z- vt) 82h-2Ab BAb2(z- vt)2-- . -- + . 8z- [b(z- vt)2 + 1]2' 8z2- [b(z- vt)2 + 1]2 [b(z- vt)2 + 1]3' 813 = 2Abv(z-vt); 82h =-2Abv2 + BAb2v2(Z-vt)2 =v282h..; 8t [b(z- vt)2 + 1]2 8t2 [b(z- vt)2 + 1]2 [b(z- vt)2 + 1]3 8z2 8i4 =-2Ab2ze-b(bz2+vt). 8214 =-2Ab2 [ e-b(bZ2+vt)- 2b2z2e-b(bZ2+vt) ] . 8z I 8Z2 ' 8i4-- Ab-b(bz2+vt). 8214 - Ab 2 2-b(bz2+vt)-J. 2 82 i4 8t- ve , 8t2- v e-r v 8Z2' 8 is 3 82 is 2 . 3 8 is 3 3 2 . . 3 8z = Abcos(bz) cos(bvt) ; 8z2 =-Ab sm(bz) cos(bvt) ; 8t =-3Ab v t sm(bz) sm(bvt) ; 8;~5 =-6Ab3v3t sin(bz) sin(bvt)3- 9Ab6v6t4 sin(bz) coS(bvt)3 =I V2 ~~5 . Problem 9.2 M ~i 8z = Akcos(kz)cos(kvt); 8z2 =-Ak2 sin(kz) cos(kvt); ~~ =-Akvsin(kz) sin(kvt); ~:; =-Ak2v2 sin(kz) cos(kvt) = v2 ~:{. ./ Usethe trig identity sin a cos 13 = ~[sin(a + t3) + sin(a- 13)] to write i = I ~ {sin[k(z + vt)] + sin[k(z - vt)]}, 157 158 CHAPTER 9. ELECTROMAGNETIC WAVES which is of the form 9.6, with 9 = (Aj2) sin[k(z- vt)] and h = (Aj2) sin[k(z + vt)]. Problem 9.3 (A3)2 = (A3eiO3) (A3e-iO3) = (A1eiOl + A2eio2) (A1e-iol + A2e-io2) = (Ad2 + (A2)2 + A1A2 (eiOle-io2 + e-iOleio2) = (Ad2 + (A2)2 + A1A22cos(61- 62); A3 = I v(Ad2 + (A2)2 + 2A1A2 cos(61 - 62).1 A3eio3 = A3 (cos 63 + i sin 63) = Al (cos 61 + i sin 61) + A2 (cos 62 + i sin 62) ... A3 sin 63 Al sin 61 + A2 sin 62- (AI COS61 + A2 cos (h) + z(AI sm 61 + A2 sm 62). tan 63 = A 6 = ; 3COS 3 A1cos61 +A2coS62-1 ( AI sin 61 + A2 sin 82 ) tan . Al COS81 + A2 cos 82 83- Problem 9.4 82 j 1 82 j The wave equation (Eq. 9.2) says 8 2 = -:1 8 2. Look for solutions of the form j(z, t) = Z(z)T(t). Plug z v t . d?Z 1 d2T 1 d2Z 1 d2T this in: T dZ2 = V2 Z dt2 . Divide by ZT : Z dZ2 = v2T dt2 . The left side depends only on z, and the righ { t Si~~ ;nly on t, 2 S0 both must be constan k t. Call th k e con } stant - k2.- =-k Z =} Z ( z ) = Ae1 Z + Be-1 Z dz2 , ~t:; = _(kV)2T =} T(t) = Ceikvt + De-ikvt. (Note that k must be real, else Z and T blow up; with no loss of generality we can assume k is positive.) j(z, t) = (AeikZ + Be-ikz) (Ceikvt + De-ikvt) = A1ei(kz+kvt) + A2ei(kz-kvt) + A3ei(-kz+kvt) + A4ei(-kz-kvt). The general linear combination of separable solutions is therefore j(z,t) = 100 [AI (k)ei(kz+U.lt) +A2(k)ei(kz-U.lt) +A3(k)ei(-kz+U.lt) +A4(k)ei(-kz-U.lt)] dk, where W == kv. But we can combine the third term with the first, by allowing k to run negative (w = Iklv remains positive); likewise the second and the fourth: j(z,t) = i: [Al(k)ei(kzwtJ +A2(k)ei(kZ-U.ltJ] dk....
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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 - ch09 - Chapter 9...

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