Unformatted text preview: ½ ï¿½ ï¿½ 0 ï¿½ a ï¿½ J ï¿½ k r ï¿½ rdr ï¿½ ï¿½ï¿½ ï¿½ ï¿½ ï¿½ s,z ï¿½k 0 0 4ï¿½ sin ï¿½ ï¿½ k s , z ï¿½ k ï¿½ h / 2 ï¿½
ï¿½
ï¿½ ï¿½k s,ï¿½ 0 4ï¿½ sin ï¿½ï¿½ k s , z ï¿½ k ï¿½ h / 2 ï¿½ ks , ï¿½ a
ï¿½
ï¿½ ï¿½ k ï¿½ k s2,ï¿½ ï¿½uJ ï¿½ u ï¿½ ï¿½ ï¿½ ï¿½ï¿½ ï¿½ ï¿½ ï¿½
k s ,ï¿½ a 1 ï¿½ ï¿½ k ï¿½ k s2,ï¿½ ï¿½k s,z J 0 ï¿½ u ï¿½ udu 0 4ï¿½ a sin ï¿½ ï¿½ k s , z ï¿½ k ï¿½ h / 2 ï¿½
ï¿½
ï¿½ 0 0 s,z ï¿½ k ï¿½ ks,ï¿½ J1 ï¿½ ks,ï¿½ a ï¿½ (1.9) Note that in spherical coordinates k s , z ï¿½ k cos ï¿½ s , k s , ï¿½ ï¿½ k sin ï¿½ s . As a result, in spherical
coordinates we have the result given by ï¿½ï¿½ ï¿½ ï¿½ 0 ï¿½ 4ï¿½ a sin ï¿½ï¿½ cos ï¿½ s ï¿½ 1ï¿½ kh / 2 ï¿½
ï¿½
ï¿½
k 2 ï¿½ cos ï¿½ s ï¿½ 1ï¿½ sin ï¿½ s J 1 ï¿½ ka sin ï¿½ s ï¿½ (1.10) There is no dependence on ï¿½s here but remember that we have to include the vector
Ë†
Ë†
Ë†
Ë†Ë†Ë†
Ë†
Ë†
(1.11)
e ï¿½ e ï¿½ k k ï¿½ x ï¿½ k k ï¿½ x ï¿½ sin ï¿½ cos ï¿½ k
i ï¿½ i s ï¿½ s s ,x s s ss The final result is ï¿½ ï¿½ Ë†Ë†
Qï¿½ ks , ki ï¿½ E0
ï¿½ E0 4ï¿½ a ï¿½ï¿½ ï¿½ ï¿½ 0 ï¿½ sin ï¿½ ï¿½ cos ï¿½ s ï¿½ 1ï¿½ kh / 2 ï¿½
ï¿½
ï¿½
k 2 ï¿½ cos ï¿½ s ï¿½ 1ï¿½ sin ï¿½ s 4ï¿½ a ï¿½ï¿½ ï¿½ ï¿½ 0 ï¿½ sin ï¿½ ï¿½ cos ï¿½ s ï¿½ 1ï¿½ kh / 2 ï¿½
ï¿½
ï¿½
k 2 ï¿½ cos ï¿½ s ï¿½ 1ï¿½ sin ï¿½ s Ë†
Ë†
J 1 ï¿½ ka sin ï¿½ s ï¿½ ï¿½ x ï¿½ sin ï¿½ s cos ï¿½s ks ï¿½
ï¿½
ï¿½
(1.12) Ë†
J 1 ï¿½ ka sin ï¿½ s ï¿½ ï¿½cos ï¿½ s cos ï¿½s Î¸s ï¿½ sin ï¿½s Ë† ï¿½
sï¿½
ï¿½ W here we have used Ë†
ï¿½ cosï¿½ s cos ï¿½s ,cos ï¿½ s sin ï...
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This note was uploaded on 02/26/2013 for the course EE 25227 taught by Professor Akbabi during the Spring '13 term at Sharif University of Technology.
 Spring '13
 akbabi

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