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# HW24_pdf - mcconnell(kam2342 – homework24 – Turner...

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mcconnell (kam2342) – homework24 – Turner – (60230) 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 = 2 . 44 cm by b = 3 . 87 cm) and inner radius 3 . 21 cm consists of N = 630 turns of wire that carries a current I = I 0 sin ω t , with I 0 = 43 A and a frequency f = 42 . 1 Hz. A loop that consists of N = 16 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 . 442577 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 = - (16 turns) μ 0 (630 turns) × (43 A) (42 . 1 Hz) (2 . 44 cm) × ln bracketleftbigg (3 . 87 cm) + (3 . 21 cm) (3 . 21 cm) bracketrightbigg = - 0 . 442577 V |E| = 0 . 442577 V . 002 10.0 points A long straight wire carries a current 36 A. A rectangular loop with two sides parallel to the straight wire has sides 5 cm and 10 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. 1 cm 5 cm 10 cm 36 A

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mcconnell (kam2342) – homework24 – Turner – (60230) 2 Find the magnetic flux through the rectan- gular loop. Correct answer: 1 . 29007 × 10 - 6 Wb. Explanation: Let : I = 36 A , a = 5 cm = 0 . 05 m , b = 10 cm = 0 . 1 m , d = 1 cm = 0 . 01 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 . 1 m) (36 A) × ln parenleftbigg 0 . 01 m + 0 . 05 m 0 . 01 m parenrightbigg = 1 . 29007 × 10 - 6 Wb . 003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 95 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 110 g. The uniform magnetic field has a magnitude of 630 mT, and the distance between the rails is 45 cm. The rails are connected at one end by a load resistor of 91 mΩ.
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HW24_pdf - mcconnell(kam2342 – homework24 – Turner...

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