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Unformatted text preview: Kapoor (mk9499) – homework24 – Turner – (60230) 1 This printout should have 12 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 adr 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 A long straight wire carries a current 40 A. A rectangular loop with two sides parallel to the straight wire has sides 2 . 5 cm and 15 cm, with its near side a distance 1 cm from the straight wire, as shown in the figure. The permeability of free space is 4 π × 10 7 T · m / A. 1cm 2 . 5 cm 15cm 40A Kapoor (mk9499) – homework24 – Turner – (60230) 2 Find the magnetic flux through the rectan gular loop. Correct answer: 1 . 50332 × 10 6 Wb. Explanation: Let : I = 40 A , a = 2 . 5 cm = 0 . 025 m , b = 15 cm = 0 . 15 m , d = 1 cm = 0 . 01 m , and μ 4 π = 1 × 10 7 N / A 2 . d a b x dx I The magnetic flux through the strip of area dA is d Φ = B dA = μ 2 π I x bdx = μ 4 π 2 bI dx x , so the total magnetic flux through the rectan gular loop is Φ total = integraldisplay d + a d d Φ = μ 4 π (2 bI ) integraldisplay d + a d dx x = μ 4 π (2 bI ) ln d + a d = (1 × 10 7 N / A 2 ) 2 (0 . 15 m) (40 A) × ln parenleftbigg . 01 m + 0 . 025 m . 01 m parenrightbigg = 1 . 50332 × 10 6 Wb . 003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 17 kg in the figure below is pulled horizon tally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 610 g. The uniform magnetic field has a magnitude of 940 mT, and the distance between the rails is 13 cm.and the distance between the rails is 13 cm....
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 Fall '08
 Turner
 Physics, Work, Magnetic Field, Kapoor

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