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Unformatted text preview: nguyen (jmn727) oldhomework 33 Turner (59070) 1 This printout should have 12 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 0 . 3071 A current is charging a capacitor that has circular plates, 11 . 14 cm in radius. The plate separation is 3 . 49 mm. The permitivity or free space is 8 . 85 10 12 and the permeability of free space is 1 . 25664 10 6 T m / A. What is the time rate of increase of electric field between the plates? Correct answer: 8 . 90054 10 11 V / m s. Explanation: Let : I = 0 . 3071 A , r = 11 . 14 cm = 0 . 1114 m , A = r 2 = 0 . 038987 m 2 , and = 8 . 85 10 12 . Let E be the flux of the electric field, defined as E = integraldisplay vector E d vector A . Thus d E dt = d dt ( E A ) = I Time rate of increase of electric field between the plates is P = d E dt = I A = . 3071 A (8 . 85 10 12 )(0 . 038987 m 2 ) = 8 . 90054 10 11 V / m s . 002 (part 2 of 2) 10.0 points What is the magnetic field between the plates . 0595 m from the center? Correct answer: 2 . 9448 10 7 T. Explanation: Let : r = 0 . 0595 m and = 1 . 25664 10 6 T m / A . Since contintegraldisplay vector B dvectors = d dt E Then magnetic field, B ,at the distance r from the center between the plates satisfies 2 r B = d dt parenleftbigg Q A r 2 parenrightbigg Hence B = I r 2 A = (0 . 3071 A)(0 . 0595 m) 2(0 . 038987 m 2 ) (1 . 25664 10 6 T m / A) = 2 . 9448 10 7 T . 003 10.0 points A sinusoidal voltage is applied directly across a 19 . 23 F capacitor. The frequency of the source is 5 . 2 kHz, and the voltage amplitude is 41 . 4 V. Find the maximum value of the displace ment current in the capacitor. Correct answer: 26 . 0113 A. Explanation: Let : C = 19 . 23 F = 1 . 923 10 5 F , f = 5 . 2 kHz = 5200 Hz , and V max = 41 . 4 V ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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