hw15b - aljabr (faa335) Hw15 Ross (89251) 1 This print-out...

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Unformatted text preview: aljabr (faa335) Hw15 Ross (89251) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 62 . 4 turns circular coil with radius 4 . 12 cm and resistance 6 . 57 is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = a 1 t + a 2 t 2 , where a 1 = 0 . 0197 T / s, a 2 = 0 . 0884 T / s 2 are constants, time t is in seconds and field B is in Tesla. Find the magnitude of the induced emf in the coil at t = 8 . 23 s. Correct answer: 0 . 49074 V. Explanation: Let : n = 62 . 4 turns , r = 4 . 12 cm = 0 . 0412 m , R = 6 . 57 , a 1 = 0 . 0197 T / s , a 2 = 0 . 0884 T / s 2 , and t = 8 . 23 s . The area of the circular coil is A = r 2 = (0 . 0412 m) 2 = 0 . 00533266 m 2 , so from Faradays law, E = n d B dt = n dAB dt = nA dB dt = nA ( a 1 + 2 a 2 t ) = (62 . 4 turns) (0 . 00533266 m 2 ) [0 . 0197 T / s + 2 (0 . 0884 T / s 2 ) (8 . 23 s)] = . 49074 V , which has a magnitude of . 49074 V . 002 10.0 points A car with a 0 . 567 m long radio antenna travels at 91 . 6 km / h in a place where the Earths magnetic field is 3 . 9 10 5 T. What is the maximum possible induced emf in the antenna as it moves through the Earths magnetic field? Correct answer: 0 . 000562653 V. Explanation: Let : = 0 . 567 m , v = 91 . 6 km / h = 25 . 4444 m / s , and B = 3 . 9 10 5 T . If the antenna moves perpendicular to the Earths field, the induced emf between the ends of the antenna becomes the maximum: E = B v = (3 . 9 10 5 T) (0 . 567 m) (25 . 4444 m / s) = . 000562653 V . 003 10.0 points A metal bar is sliding across the surface of a metal ring, as shown in the figure. There is a uniform magnetic field directed into the paper. 50 cm / s 1 1 c m . 4 T . 4 T What is the magnitude of the induced emf , E , in the rod when it is a distance 6 . 5 cm from the center of the ring? Correct answer: 0 . 0354965 V. Explanation: aljabr (faa335) Hw15 Ross (89251) 2 v R B B Let : R = 11 cm , B = 0 . 4 T , and v = 50 cm / s ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue University-West Lafayette.

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hw15b - aljabr (faa335) Hw15 Ross (89251) 1 This print-out...

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