# FinalExam_EM_A_Sep_2005_Answers - E = 8a3ρ/3ε d2 0 0 0.8...

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Unformatted text preview: E = 8a3ρ /3ε d2 0 0 0.8 0.7 Normalized Electric Field, E/E0 0.6 0.5 0.4 0.3 0.2 0.1 (a2−d2/4)1/2 −(a2−d2/4)1/2 0 −4 AΣKHΣH −3 −2 −1 0 1 Normalized Distance, z/(d/2) 1 (α) E = ˆx i ρd 30 (β) E = ˆx i a3 ρ d 2 + (d/2)2]3/2 3 0 [z 1 2 3 4 AΣKHΣH 2 (α) √ k = ω µ0 [sin θˆx + cos θˆz ] i i (β) Er = ˆy (−E0) exp[−jk(x sin θ − z cos θ)] i √ k = ω µ0 (γ) E0 cos θ cos(kz cos θ) cos(ωt − kx sin θ) η E0 Hz (x, z, t) = +2 sin θ sin(kz cos θ) sin(ωt − kx sin θ) η Hx (x, z, t) = −2 (1) (2) (3) η= µ0 / (δ) K = ˆy 2 i E0 cos θ exp[−jkx sin θ] η (ε) 2 f (x, t) = ˆz 2E0 cos2 θ cos2(ωt − kx sin θ) i f(x) = ˆz E0 cos2 θ i 2 AΣKHΣH 3 (α) H = ˆz σω0 a i (β) Wm = 1 2 µ0 σ 2a4 ω0 πL 2 (γ) E = ˆφ[−µ0 σaΛ i rT 2 (δ) We = 1 2 2 6 2 0 µ0 σ a Λ πL 16 2 ...
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FinalExam_EM_A_Sep_2005_Answers - E = 8a3ρ/3ε d2 0 0 0.8...

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