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# oldhw 24 - nguyen(jmn727 oldhomework 24 Turner(59070 This...

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nguyen (jmn727) – oldhomework 24 – Turner – (59070) 1 This print-out should have 12 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 . 22 cm by b = 3 . 08 cm) and inner radius 5 . 75 cm consists of N = 230 turns of wire that carries a current I = I 0 sin ω t , with I 0 = 69 . 4 A and a frequency f = 92 . 4 Hz. A loop that consists of N = 13 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 . 126091 V. Explanation: Basic Concept: Faraday’s Law E = - d Φ B dt . Magnetic field in a toroid B = μ 0 N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ 0 N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ 0 N I 0 2 π sin( ω t ) integraldisplay b + R R a dr r = μ 0 N I 0 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 d Φ B 1 dt = - N μ 0 N I 0 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) = -E 0 cos( ω t ) where ω = 2 πf was used. The maximum magnitude of the induced emf , E 0 , is the coefficient in front of cos( ω t ). E 0 = - N d Φ B 1 dt = - N μ 0 N I 0 ω 2 π a ln bracketleftbigg b + R R bracketrightbigg = - (13 turns) μ 0 (230 turns) × (69 . 4 A) (92 . 4 Hz) (1 . 22 cm) × ln bracketleftbigg (3 . 08 cm) + (5 . 75 cm) (5 . 75 cm) bracketrightbigg = - 0 . 126091 V |E| = 0 . 126091 V . 002 10.0 points A long straight wire carries a current 28 A. A rectangular loop with two sides parallel to the straight wire has sides 5 cm and 15 cm, with its near side a distance 3 cm from the straight wire, as shown in the figure. 3 cm 5 cm 15 cm 28 A Find the magnetic flux through the rectan- gular loop.

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nguyen (jmn727) – oldhomework 24 – Turner – (59070) 2 The permeability of free space is 4 π × 10 - 7 T · m / A. Correct answer: 8 . 23897 × 10 - 7 Wb. Explanation: Let : I = 28 A , a = 5 cm = 0 . 05 m , b = 15 cm = 0 . 15 m , d = 3 cm = 0 . 03 m , and μ 0 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 0 2 π I x b dx = μ 0 4 π 2 b I dx x , so the total magnetic flux through the rectan- gular loop is Φ total = integraldisplay d + a d d Φ = μ 0 4 π (2 b I ) integraldisplay d + a d dx x = μ 0 4 π (2 b I ) ln d + a d = (1 × 10 - 7 N / A 2 ) 2 (0 . 15 m) (28 A) × ln parenleftbigg 0 . 03 m + 0 . 05 m 0 . 03 m parenrightbigg = 8 . 23897 × 10 - 7 Wb . 003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 99 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 760 g. The uniform magnetic field has a magnitude of 200 mT, and the distance between the rails is 35 cm. The rails are connected at one end by a load resistor of 51 mΩ.
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