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Unformatted text preview: Physics 6C HW#1 solution April 10, 2008 24.7 a) f = c f (50 . 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 . B max = 3 . 00 10 8 Thus, B = 73 . 3 k nT c) We need to find k and , k = 2 = 2 50 . = 0.126 m 1 = 2 f = 2 6 . 00 10 6 = 3 . 77 10 7 rad/s Thus the B equation is, B = B max cos( kx t ) = 73 . 3cos(0 . 126 x 3 . 77 10 7 t ) k nT 24.8 a) B = E c = 100 V/m 3 10 8 m/s = 3 . 33 10 7 T b) = 2 k = 2 1 . 00 10 7 m 1 = 6 . 28 10 7 m c) f = c = 3 . 00 10 8 m/s 6 . 28 10 7 m = 4 . 77 10 14 Hz 1 24.9 We need to prove that E = E max cos( kx t ) and B = B max cos( kx t ) satisfy the wave equation. For E field, E x = kE max sin( kx t ) 2 E x 2 = k 2 E max cos( kx t ) E t = E max sin( kx t ) 2 E t 2 = ( ) 2 E max cos( kx t ) Plug the derivatives into the wave equation.Plug the derivatives into the wave equation....
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This note was uploaded on 07/02/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.
 Spring '01
 staff
 Physics, Thermodynamics, Light

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