Fall2010 hw7sol - ECE 329 Fall 2010 Homework 7 Solution Due...

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ˆ P = - ˆ x v = 1 . 5 10 - 1 = 15 ( m/s ) = 1 μ 0 v 2 = 3536 ( F/m ) H = 1 η ˆ P × E = - ˆ z q μ 0 sin 2 π · 10 8 t + 10 - 1 1 . 5 x ( A/m ) ˆ P = - ˆ z v = 2 10 - 8 = 2 × 10 8 ( m/s ) = 1 μ 0 v 2 = 1 . 22 × 10 - 5 ( F/m ) = 2 . 25 0 E = η ˆ H × ˆ P = ˆ y q μ 0 Δ 10 3 t + 10 - 8 2 z ( A/m ) ˆ P = ˆ z v = ω β = π 10 8 π/ 2 = 2 × 10 8 ( m/s ) = 1 μ 0 v 2 = 1 . 22 × 10 - 5 ( F/m ) = 2 . 25 0 E = η ˆ H × ˆ P = q μ 0 ( ˆ x cos ( π · 10 8 t - π 2 z ) + ˆ y sin ( π · 10 8 t - π 2 z )) ( A/m )
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s so ( t τ ) The corresponding electric field is E ( x, t ) = η o 2 J so t x c ˆ y = η o t x c rect t x c τ ˆ y V m for x 0 and the magnetic field is H ( x, t ) = ± 1 2 J so t x c ˆ z = ± t x c rect t x c τ ˆ z A m for x 0 , where η o = 120 π Ω is the intrinsic impedance in free-space. a) Plotting E y ( x, t ) and H z ( x, t ) for x = ± 1200 m . Time [ μ s] E y ( x, t ) [ μ V m ] 0 2 4 6 8 10 - η o 0 η o x = 1200 m Time [ μ s] H z ( x, t ) [ μ A m ] 0 2 4 6 8 10 - 1 0 1 x = 1200 m Time [ μ s] E y ( x, t ) [ μ V m ] 0 2 4 6 8 10 - η o 0 η o x = - 1200 m Time [ μ s] H z ( x, t ) [ μ A m ] 0 2 4 6 8 10 - 1 0 1 x = - 1200 m b) Plotting E y ( x, t ) and H z ( x, t ) for t = 5 μ s . 3 ECE-329 fall 2010 2. On the x = 0 plane, we have a sheet current J ( t ) = - J ( t ) y ˆ = - 2 t rect y ˆ A / m , where τ = 2 μ s .
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x -axis [km] E y ( x, t ) [ μ V m ] - 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 2 - η o 0 η o t = 5 μ s x -axis [km] H z ( x, t ) [ μ A m ] - 2 - 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5 2 - 1 0 1
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