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Unformatted text preview: Brannon (blb675) HW #11 Shih (58250) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A point source of electromagnetic radiation has an average power output of 843 W. The speed of light is 2 . 99792 10 8 m / s and the permeability of free space is 4 10 7 T N / A. Calculate the maximum value of the electric field at a point 7 . 6 m from the source. Correct answer: 29 . 5819 V / m. Explanation: Let : = 4 10 7 T N / A , c = 2 . 99792 10 8 m / s , P ave = 843 W , and r = 7 . 6 m . Recall that the wave intensity, I , a distance r from the point source is I = P av 4 r 2 , where P av is the average power output of the source and 4 r 2 is the area of a sphere of radius r centered on the source. Because the intensity of an electromagnetic wave is also given by the equation I = E 2 max 2 c , we have P av 4 r 2 = E 2 max 2 c . E max = c P av 2 r 2 = (4 10 7 T N / A) (2 . 99792 10 8 m / s) (843 W) 2 (7 . 6 m) 2 = 29 . 5819 V / m . 002 (part 2 of 2) 10.0 points Determine the maximum value of the mag netic field at a point 7 . 6 m from the source. Correct answer: 9 . 86745 10 8 T. Explanation: The maximum values of the electric and the magnetic fields are related by E max B max = c B max = E max c = 29 . 5819 V / m 2 . 99792 10 8 m / s = 9 . 86745 10 8 T . 003 (part 1 of 3) 10.0 points A current of 16 A flows into a capacitor having plates with areas of 0 . 5 m 2 . The permittivity of free space is 8 . 85 10 12 C 2 / N m 2 ....
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This note was uploaded on 05/03/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Radiation

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