sp08hw4sol1 - ECE329 Spring 2008 Homework 4 Problem 1 x a)B...

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Unformatted text preview: ECE329 Spring 2008 Homework 4 Problem 1 x a)B = −β E0 cos(ωt − βz )ˆ ω ∂ b) · D = 0 E0 ∂y cos(ωt − βz ) = 0 c) ∂Dy ∂Hx = ∂t ∂z ∂Dy β 2 E0 = sin(ωt − βz ) ∂t ωµ0 Ey = β 2 E0 cos(ωt − βz ) ω 2 0 µ0 E = y E0 cos(ωt − βz ) ˆ d)c = 3 · 108 m/s 2π 2π e)Ey = E0 cos( 300 z ) and Ey = E0 cos( π − 300 z ) 4 f)The sign of β determines the direction of propagation of the wave. 2π g)Ey = E0 cos( π + 300 z ) 4 h)−z ˆ 2π i)β = 300 /m and λ = 300m Problem 2 a) For E = Ex (z, t)ˆ, Faraday's law simplies to ∂Ex = − ∂By . x ∂z ∂t b) For H = Hy (z, t)ˆ, Ampere's law simplies to ∂Hy = − ∂Dx . y ∂z ∂t c)Put the pieces together. ∂ 2 Ex ∂ 2 Hy = −µ0 = µ0 ∂z 2 ∂t∂z 0 d) Ex = f (t + 1 ∂ 2 Ex ∂ 2 Ex =2 ∂t2 c ∂t2 z ) vp ∂Ex 1 ∂f (t + =− ∂z vp ∂z z vp ) 2 ∂ 2 Ex 1 ∂ f (t + =2 ∂z 2 vp ∂z 2 z vp ) 1 ∂ 2 Ex ∂ 2 Ex =2 2 ∂z c ∂t2 e)Plots... Problem 3 a)For E = Ex (z, t)ˆ,Faraday's law simplies to x ∂Ex ∂z =− ∂By ∂t . For this problem, ∂Ex ∂By A z B z =− =− f t− + g t+ ∂z ∂t c c c c A z B z f t− − g t+ c c c c z z A B f t− g t+ Hy = − cµ0 c cµ0 c By = Hy = b) Ohms Ω A c) H = −0 f t − η z c + B η0 g t+ z c A z B z f t− − g t+ η0 c η0 c x ˆ 1 Problem 4 a) ω = π · 1015 rad/s and f = 5 · 1014 Hz b) β = π · 107 /m and λ = 600nm 3 c)−y ˆ d) E = −η0 cos(π · 1015 t + π · 107 y )ˆ z 3 e) optical 2 ...
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This note was uploaded on 04/14/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

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sp08hw4sol1 - ECE329 Spring 2008 Homework 4 Problem 1 x a)B...

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