FinalExam_EM_A_Jul_2005_Answers - 0.4 Normalized Electric...

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-5 0 5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 Normalized Distance y/a Normalized Electric Field, E/( λ /2 πε 0 a) a = L/2 A Σ KH Σ H 1 ( α ) ± E = ˆ i y λ 2 π² 0 1 y ± 1 - L/ 2 ² y 2 +( L/ 2) 2 ³ ( β ) ± E = ˆ i z ± - λ 2 0 ³ z ( L/ 2) 2 - z 2 1
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-1 -0.5 0 0.5 1 -10 -8 -6 -4 -2 0 2 4 6 8 10 Normalized Distance z/a Normalized Electric Field, E/( λ /2 πε 0 a) a = L/2 A Σ KH Σ H 2 ( α ) ± H = ˆ i φ - I 2 πr T z>R , - I 2 T 0 <z<R , κai r T > R 2 - z 2 , 00 , κai r T < R 2 - z 2 , 0 z< 0 . ( β ) ± K s = ˆ i θ I 2 πR sin θ (1) ± K p = ˆ i r T I 2 T (2) ( γ ) ± f s = - ˆ i z I 2 μ 0 8 π 2 r 2 T (3) ± f p = - ˆ i r I 2 μ 0 8 π 2 R 2 sin 2 θ (4) A Σ Σ H 3 ( α ) H x = - β ωμ 0 exp( - jβz ) E c exp[ - γ c ( x - h/ 2)] h/ 2 <x< , E f cos( k f x ) - h/ 2 <x<h/ 2 , E c exp[ γ c ( x + h/ 2)] -∞ - h/ 2 . 2
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H z = 1 jωμ 0 exp( - jβz ) γ c E c exp[ - γ c ( x - h/ 2)] h/ 2 <x< , k f E f sin( k f x ) - h/ 2 <x<h/ 2 , - γ c E c exp[ γ c ( x + h/ 2)] -∞ - h/ 2 . ( β ) E c = E f cos( k f h/ 2) (5) γ c E c = k f E f sin( k f h/ 2) (6) tan[ ± k 2 0 n 2 f - β 2 h 2 ]= ± β 2 - k 2 ) n 2 c ± k 2 0 n 2 f - β 2 (7) ( γ ) ± E = ˆ i y E c exp[ - γ c ( x - h/
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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_Jul_2005_Answers - 0.4 Normalized Electric...

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