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HW 1 solutions (Ch. 24,25)

# HW 1 solutions (Ch. 24,25) - Physics 6C HW#1 solution 24.7...

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Physics 6C HW#1 solution April 10, 2008 24.7 a) = c f (50 . 0 m ) = 3 . 00 × 10 8 m/s f = 6 . 00 × 10 6 Hz b Remember for E& M wave, E field and B field satisfy the following equation. E B = c 22 . 0 B max = 3 . 00 × 10 8 Thus, -→ B = - 73 . 3 ˆ k nT c) We need to find k and ω , k = 2 π λ = 2 π 50 . 0 = 0.126 m - 1 ω = 2 πf = 2 π × 6 . 00 × 10 6 = 3 . 77 × 10 7 rad/s Thus the B equation is, -→ B = ---→ B max cos( kx - ωt ) = - 73 . 3 cos(0 . 126 x - 3 . 77 × 10 7 t ) ˆ k nT 24.8 a) B = E c = 100 V/m 3 × 10 8 m/s = 3 . 33 × 10 - 7 T b) λ = 2 π k = 2 π 1 . 00 × 10 7 m - 1 = 6 . 28 × 10 - 7 m c) f = c λ = 3 . 00 × 10 8 m/s 6 . 28 × 10 - 7 m = 4 . 77 × 10 14 Hz 1

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24.9 We need to prove that E = E max cos( kx - ωt ) and B = B max cos( kx - ωt ) satisfy the wave equation. For E field, ∂E ∂x = - kE max sin( kx - ωt ) 2 E ∂x 2 = - k 2 E max cos( kx - ωt ) ∂E ∂t = ωE max sin( kx - ωt ) 2 E ∂t 2 = - ( - ω ) 2 E max cos( kx - ωt ) Plug the derivatives into the wave equation. 2 E ∂x 2 = μ 0 0 2 E ∂t 2 - k 2 E max cos( kx - ωt ) = - μ 0 0 ( - ω ) 2 E max cos( kx - ωt ) - k 2 E max cos( kx - ωt ) = - μ 0 0 ( - ω ) 2 E max cos( kx - ωt ) Recall k 2 ω 2 = 1 c 2 = μ 0 0 . Thus, k 2 = ω 2 μ 0 0 .
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