# FinalExam_EM_A_Sep_2007_Answers - ΠE∆IA A 26/09/2007 Θ...

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ΠEΔIA A2 6 / 09 / 2007 Θ ±μα 1 ( α ) ² E = ˆ i x - 1 2 ± 0 ± σ 1 + σ 2 + 2 0 π ² , -∞ <x< - a - ( d/ 2), - 1 2 ± 0 ± - σ 1 + σ 2 + 2 0 π ² , - a - ( d/ 2) ( d/ 2), + 1 2 ± 0 ± σ 1 - σ 2 + 2 0 π sin( πx d ) ² , - ( d/ 2) +( d/ 2), + 1 2 ± 0 ± σ 1 - σ 2 + 2 0 π ² , +( d/ 2) <x<b +( d/ 2), + 1 2 ± 0 ± σ 1 + σ 2 + 2 0 π ² ,b d/ 2) + . ( β ) σ 1 + σ 2 + 2 0 π =0 ( γ ) W e /S = d 8 ± 0 [( σ 1 - σ 2 ) 2 + 4 d 2 ρ 2 0 π 2 ] ( δ ) ² f = ˆ i x 2 0 π σ 1 - σ 2 2 ± 0 ( ± ) ² f = ˆ i x σ 2 2 ± 0 ± σ 1 + 2 0 π ² Θ ±μα 2 ( α ) ² K 1 = I/ 2 2 πr T ˆ i r T , I ( z )= I 2 exp( - α | z | ) , ² J ( r T ,z ˆ i r T α I 2 1 2 T exp( - α | z | ) .

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( β ) ± H = ˆ i φ - I 2 πr T , for z> 0, - I/ 2 2 T exp( - α | z | ) , for z< 0. ( γ ) ± f = - ˆ i z 3 μ 0 I 2 32 π 2 r 2 T ( δ ) W m = μ 0 I 2 32 πα ± 1 - exp( - 2 α² ) ² ln( b a ) exp( - 2 αd ) Θ ³μα 3 ( α ) ± E = ˆ i y E 0 exp( - jk 1 z )+ E r exp(+ 1 z ) , for 0, E 1 exp( - 2 z E 2 exp(+ 2 z ) , for 0 <z<d . ( β ) ± H = ˆ i x - E 0 Z 1 exp( - 1 z E r Z 1 exp(+ 1 z ) , for 0, - E 1 Z 2 exp( - 2 z E 2 Z 2 exp(+ 2 z ) , for 0 . ( γ ) E 0 + E r = E 1 + E 2
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## This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

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FinalExam_EM_A_Sep_2007_Answers - ΠE∆IA A 26/09/2007 Θ...

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