936_PartUniversity Physics Solution

# 936_PartUniversity Physics Solution - Electromagnetic Waves...

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Electromagnetic Waves 32-9 32.37. IDENTIFY and SET UP: Take partial derivatives of Eqs.(32.12) and (32.14), as specified in the problem. EXECUTE: 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 tt . Eq.(32.14) says 00 . y z E B x t μ −= P Taking x of both sides of this equation gives 22 1 , so yy z z EE B B x tx x = P P . But y y E E x x = (The order in which the partial derivatives are taken doesn’t change the result.) So 1 z z B B P and , z z B B x t = P as was to be shown. EVALUATE: Both fields, electric and magnetic, satisfy the wave equation, Eq.(32.10). We have also shown that both fields propagate with the same speed 1/ . v = P 32.38. IDENTIFY: The average energy density in the electric field is 2 1 ,av 0 av 2 () E uE = P and the average energy density in the magnetic field is 2 ,av av 0 1 2 B uB = . SET UP: ( ) 2 1 2 av cos ( ) kx t ω . EXECUTE: max (, ) c o s ( ) y E xt E k x t . 2 11 m a x cos ( ) Ey uEE k x t == PP and 2 1 ,av 0 max 4 E = P . max ) c o s ( ) z Bxt B k x t , so 2 max cos ( ) Bz uBBk x t μμ and 2 ,av max 0 1 4 B = . max max E cB = , so 1 ,av 0 max 4 E uc B = P . 1 c = P , so 2 ,av max 0 1 2 E = , which equals ,av B u . EVALUATE: Our result allows us to write 2 1 av ,av 0 max 2 2 E uu E P and 2 av ,av max 0 1 2 2 B B . 32.39. IDENTIFY: 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. SET UP: The intensity of the wave is 2 1 av 0 max 2 / IPA c E P , and the force is av FP A = where av / PI c = . EXECUTE: (a) I = P av / A = (25,000 W)/[4 π (575 m) 2 ] = 0.00602 W/m 2 (b) 2 1 0m a x 2 Ic E = 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 EVALUATE: The fields are very weak compared to ordinary laboratory fields, and the force is hardly worth worrying about! 32.40. IDENTIFY: cf λ = . max max E cB = . 2 1 a x 2 E = P . For a totally absorbing surface the radiation pressure is . I c SET UP: The wave speed in air is 8 3.00 10 m/s c . EXECUTE: (a) 8 9 2 7.81 10 Hz 3.84 10 m c f × = × × (b) 9 max max 8 1.35 V/m 4.50 10 T E B c = × × (c) 21 2 2 2 8 2 3 2 a x (8.854 10 C /N m )(3.00 10 m/s)(1.35 V/m) 2.42 10 W/m E × ⋅× = × P (d) 32 2 12 8 (2.42 10 W/m )(0.240 m ) (pressure) 1.94 10 N IA FA c × = = × × EVALUATE: 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) IDENTIFY and SET UP: Calculate I and then use Eq.(32.29) to calculate max E and Eq.(32.18) to calculate max . B EXECUTE: The intensity is power per unit area: 3 2 32 3.20 10 W 652 W/m . (1.25 10 m) P I A π × == = × 2 max max 0 0 , so 2 2 E IE c I c μ 78 2 max 2(4 10 T m/A)(2.998 10 m/s)(652 W/m ) 701 V/m E × = 6 max max 8 701 V/m 2.34 10 T 2.998 10 m/s E B c = × × EVALUATE: The magnetic field amplitude is quite small.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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936_PartUniversity Physics Solution - Electromagnetic Waves...

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