3_10 - PHY 5346 HW Set 6 Solutions – Kimel 2. 3.10 This...

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Unformatted text preview: PHY 5346 HW Set 6 Solutions – Kimel 2. 3.10 This problem is described by a) From the class notes  , z ,   n A n sin   B n cos  I  n L sin nz L where 1 A n  2 L I  nb L 1 B n  2 L I  nb L Þ0 Þ0 L L 2 V , z sin na sin  ddz, L V , z sin na cos  ddz, L 2 0  0 0 Þ0 Þ0 2 1 B n  1  L I  n b L Noting Þ0 Þ0 L V , z sin na ddz, L Þ Þ  2  2  2  2 sin d 3 2  2 Þ 3 2  2 sin d 0 we conclude A n  0. Similarly, noting cos d Þ cos d  4 1m , 2m  1 m  0, 1, 2, . . . . where I’ve recognized that  must be odd, ie,   2m  1. Also L 2  Þ 0 sin nL z dz  2l  1  , l  0, 1, 2, . . . . . where again I’ve recognized that n must be odd, ie, n  2l  1. Thus 16 1 m V B n  2 nb  I 2m1 L 2l  1 2m  1 b) Now z  L/2, L  b, L  . Then from the class notes I 2m1 Also 2l  1  L sin  1 2m  2   b 2l  1  2L 1 l m1 2l  1  L so  , z ,   l ,m 16 1 lm V  2l  1 2m  1 2 Ý 2m1 cos 2m  1  Using tan 1x  l0 1 x 2l1 1l  1 Ý 1 l   tan 4 so   , z ,   4V  m 1 l0 1l 2l  1 cos 2m  1  1m 2m  1  b 2m1 Remembering from problem 2.13 that 1m 2m  1  b 2m1 m cos 2m  1   1 tan 2 1 2  b cos  2 b2 1 we find  , z,   2V tan  which is the answer for problem 2.13. 1 2  b cos  2 b2 1 ...
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