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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 ω t , with I = 73 . 3 A and a frequency f = 67 . 1 Hz. A loop that consists of N ℓ = 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: 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 =- (12 turns) μ (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 Ω / 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 μ A. Explanation: Let : ℓ = 0 . 47746 m , A l = 1 2 ℓ 2 = 0 . 227968 m 2 / s , A r = ℓ 2 = 0 . 455936 m 2 / s , δ = 0 . 145 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 . 8 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 47 . 746 cm, the resistance is R = δ ℓ = (0 . 145 Ω / m) (0 . 47746 m) = 0 . 0692317 Ω ....
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- Spring '10
- Magnetic Field, bar magnet