hw 20 - yang (ey942) ohw20 turner (56705) 1 This print-out...

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Unformatted text preview: yang (ey942) ohw20 turner (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A toroid having a rectangular cross section ( a = 1 . 85 cm by b = 4 . 11 cm) and inner radius 1 . 69 cm consists of N = 450 turns of wire that carries a current I = I sin t , with I = 73 . 3 A and a frequency f = 67 . 1 Hz. A loop that consists of N = 12 turns of wire links the toroid, as in the figure. b a N R N l Determine the maximum E induced in the loop by the changing current I . Correct answer: 0 . 761396 V. Explanation: Basic Concept: Faradays Law E =- d B dt . Magnetic field in a toroid B = N I 2 r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = N I 2 r . So, the flux through the loop of wire is B 1 = integraldisplay B dA = N I 2 sin( t ) integraldisplay b + R R a dr r = N I 2 a sin( t ) ln parenleftbigg b + R R parenrightbigg . Applying Faradays law, the induced emf can be calculated as follows E =- N d B 1 dt =- N N I 2 a ln parenleftbigg b + R R parenrightbigg cos( t ) =-E cos( t ) where = 2 f was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( t ). E =- N d B 1 dt =- N N I 2 a ln bracketleftbigg b + R R bracketrightbigg =- (12 turns) (450 turns) (73 . 3 A) (67 . 1 Hz) (1 . 85 cm) ln bracketleftbigg (4 . 11 cm) + (1 . 69 cm) (1 . 69 cm) bracketrightbigg =- . 761396 V |E| = 0 . 761396 V . 002 10.0 points The two-loop wire circuit is 71 . 619 cm wide and 47 . 746 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0 . 001 T / s) t and its direction is into the page. Assume The resistance per length of the wire is 0 . 145 / m. B B P Q 23 . 873 cm 47 . 746 cm 47 . 746 cm yang (ey942) ohw20 turner (56705) 2 When the magnetic field is 0 . 8 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 274 . 402 A. Explanation: Let : = 0 . 47746 m , A l = 1 2 2 = 0 . 227968 m 2 / s , A r = 2 = 0 . 455936 m 2 / s , = 0 . 145 / m , and d B dt = d dt t = = 0 . 001 T / s . Basic Concept: Faradays Law is E =- d B dt , where B = A B . Ohms Law is V = I R . Solution: The instantaneous value of the magnetic field ( B = 0 . 8 T) is not germane to this problem. B B P Q 2 I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 47 . 746 cm, the resistance is R = = (0 . 145 / m) (0 . 47746 m) = 0 . 0692317 ....
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hw 20 - yang (ey942) ohw20 turner (56705) 1 This print-out...

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