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Unformatted text preview: Rodriguez, Brandon – Homework 20 – Due: Oct 31 2006, 11:00 am – Inst: Andrew J Rader 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA TIONS TO AT LEAST THREE SIGNIFI CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 2) 4 points A microwave transmitter emits electromag netic waves of a single wavelength. The maxi mum electric field 1 . 38 km from the transmit ter is 5 . 77 V / m. The speed of light is 2 . 99792 × 10 8 m / s and the permeability of free space is 4 π × 10 7 N / A 2 . Assuming that the transmitter is a point source and neglecting waves reflected from the Earth, calculate the maximum magnetic field at this distance. Correct answer: 1 . 92466 × 10 8 T. Explanation: Let : E = 5 . 77 V / m and c = 2 . 99792 × 10 8 m / s . The maximum magnetic field is B = E c = 5 . 77 V / m 2 . 99792 × 10 8 m / s = 1 . 92466 × 10 8 T . 002 (part 2 of 2) 3 points Calculate the total power emitted by the transmitter. Correct answer: 1 . 05745 × 10 6 W. Explanation: Let : μ = 4 π × 10 7 N / A 2 and r = 1380 m . I = E max B max 2 μ = (5 . 77 V / m)(1 . 92466 × 10 8 T) 2(4 π × 10 7 N / A 2 ) = 0 . 0441866 W / m 2 , thus the total power emitted is P = I A = 4 π r 2 I = 4 π r 2 µ E max B max 2 μ ¶ = 4 π (1380 m) 2 (0 . 0441866 W / m 2 ) = 1 . 05745 × 10 6 W . keywords: 003 (part 1 of 3) 4 points A plane electromagnetic sinusoidal wave of frequency 51 MHz travels in free space....
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 Spring '08
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 Work, Light, Correct Answer, Andrew J Rader

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