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Unformatted text preview: Reilly (mar3978) – ohw20 – turner – (56725) 1 This printout should have 15 questions. Multiplechoice 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 . 27 cm by b = 2 . 76 cm) and inner radius 2 . 38 cm consists of N = 280 turns of wire that carries a current I = I sin ω t , with I = 31 . 1 A and a frequency f = 82 . 7 Hz. A loop that consists of N ℓ = 22 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 . 194681 V. Explanation: Basic Concept: Faraday’s Law E = d Φ B dt . Magnetic field in a toroid B = μ N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ N I 2 π sin( ω t ) integraldisplay b + R R a dr r = μ N I 2 π a sin( ω t ) ln parenleftbigg b + R R parenrightbigg . Applying Faraday’s law, the induced emf can be calculated as follows E = N ℓ d Φ B 1 dt = N ℓ μ N I 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) =E cos( ω t ) where ω = 2 πf was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( ω t ). E = N ℓ d Φ B 1 dt = N ℓ μ N I ω 2 π a ln bracketleftbigg b + R R bracketrightbigg = (22 turns) μ (280 turns) × (31 . 1 A) (82 . 7 Hz) (1 . 27 cm) × ln bracketleftbigg (2 . 76 cm) + (2 . 38 cm) (2 . 38 cm) bracketrightbigg = . 194681 V E = 0 . 194681 V . 002 10.0 points The twoloop wire circuit is 111 . 591 cm wide and 74 . 3942 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 out of the page. Assume The resistance per length of the wire is 0 . 105 Ω / m. B B P Q 37 . 1971 cm 74 . 3942 cm 74 . 3942 cm Reilly (mar3978) – ohw20 – turner – (56725) 2 When the magnetic field is 0 . 5 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 590 . 43 μ A. Explanation: Let : ℓ = 0 . 743942 m , A l = 1 2 ℓ 2 = 0 . 55345 m 2 / s , A r = ℓ 2 = 1 . 1069 m 2 / s , δ = 0 . 105 Ω / m , and d B dt = d dt α t = α = 0 . 001 T / s . Basic Concept: Faraday’s Law is E = d Φ B dt , where Φ B = A B . Ohm’s Law is V = I R . Solution: The instantaneous value of the magnetic field ( B = 0 . 5 T) is not germane to this problem. B B P Q ℓ 2 ℓ ℓ I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 74 . 3942 cm, the resistance is R = δ ℓ = (0 . 105 Ω / m) (0 . 743942m) = 0 . 0781139 Ω ....
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This note was uploaded on 12/01/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas at Austin.
 Spring '10
 TURNER,J

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