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936_PartUniversity Physics Solution

# 936_PartUniversity Physics Solution - Electromagnetic Waves...

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Electromagnetic Waves 32-9 32.37. I DENTIFY and S ET U P : Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem. E XECUTE : Eq.(32.12): y z E B x t = − Taking t of both sides of this equation gives 2 2 2 y z E B x t t = − ∂ ∂ . Eq.(32.14) says 0 0 . y z E B x t μ = P Taking x of both sides of this equation gives 2 2 2 2 0 0 2 2 0 0 1 , so y y z z E E B B x t x t x x μ μ = = − ∂ ∂ ∂ ∂ P P . But 2 2 y y E E x t t x = ∂ ∂ ∂ ∂ (The order in which the partial derivatives are taken doesn’t change the result.) So 2 2 2 2 0 0 1 z z B B t x μ = − P and 2 2 0 0 2 2 , z z B B x t μ = P as was to be shown. E VALUATE : Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that both fields propagate with the same speed 0 0 1/ . v μ = P 32.38. I DENTIFY : The average energy density in the electric field is 2 1 ,av 0 av 2 ( ) E u E = P and the average energy density in the magnetic field is 2 ,av av 0 1 ( ) 2 B u B μ = . S ET U P : ( ) 2 1 2 av cos ( ) kx t ω = . E XECUTE : max ( , ) cos( ) y E x t E kx t ω = . 2 2 2 1 1 0 0 max 2 2 cos ( ) E y u E E kx t ω = = P P and 2 1 ,av 0 max 4 E u E = P . max ( , ) cos( ) z B x t B kx t ω = , so 2 2 2 max 0 0 1 1 cos ( ) 2 2 B z u B B kx t ω μ μ = = and 2 ,av max 0 1 4 B u B μ = . max max E cB = , so 2 2 1 ,av 0 max 4 E u c B = P . 0 0 1 c μ = P , so 2 ,av max 0 1 2 E u B μ = , which equals ,av B u . E VALUATE : Our result allows us to write 2 1 av ,av 0 max 2 2 E u u E = = P and 2 av ,av max 0 1 2 2 B u u B μ = = . 32.39. I DENTIFY : The intensity of an electromagnetic wave depends on the amplitude of the electric and magnetic fields. Such a wave exerts a force because it carries energy. S ET U P : The intensity of the wave is 2 1 av 0 max 2 / I P A cE = = P , and the force is av F P A = where av / P I c = . E XECUTE : (a) I = P av / A = (25,000 W)/[4 π (575 m) 2 ] = 0.00602 W/m 2 (b) 2 1 0 max 2 I cE = P , so max 0 2 I E c = P = 2 12 2 2 8 2(0.00602 W/m ) (8.85 10 C /N m )(3.00 10 m/s) × × = 2.13 N/C. B max = E max / c = (2.13 N/C)/(3.00 × 10 8 m/s) = 7.10 × 9 10 T (c) F =P av A = ( I / c ) A = (0.00602 W/m 2 )(0.150 m)(0.400 m)/(3.00 × 10 8 m/s) = 1.20 × 12 10 N E VALUATE : The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth worrying about! 32.40. I DENTIFY : c f λ = . max max E cB = . 2 1 0 max 2 I cE = P . For a totally absorbing surface the radiation pressure is . I c S ET U P : The wave speed in air is 8 3.00 10 m/s c = × . E XECUTE : (a) 8 9 2 3.00 10 m/s 7.81 10 Hz 3.84 10 m c f λ × = = = × × (b) 9 max max 8 1.35 V/m 4.50 10 T 3.00 10 m/s E B c = = = × × (c) 2 12 2 2 8 2 3 2 1 1 0 max 2 2 (8.854 10 C /N m )(3.00 10 m/s)(1.35 V/m) 2.42 10 W/m I cE = = × × = × P (d) 3 2 2 12 8 (2.42 10 W/m )(0.240 m ) (pressure) 1.94 10 N 3.00 10 m/s IA F A c × = = = = × × E VALUATE : The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the wavelength of the light.

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32-10 Chapter 32 32.41. (a) I DENTIFY and S ET U P : Calculate I and then use Eq.(32.29) to calculate max E and Eq.(32.18) to calculate max . B E XECUTE : The intensity is power per unit area: 3 2 3 2 3.20 10 W 652 W/m .
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936_PartUniversity Physics Solution - Electromagnetic Waves...

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