24 - Kapoor (mk9499) homework24 Turner (60230) 1 This...

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Unformatted text preview: Kapoor (mk9499) 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 = 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 t , with I = 31 . 1 A and a frequency f = 82 . 7 Hz. A loop that consists of N = 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: Faradays Law E =- d B dt . Magnetic field in a toroid B = N I 2 r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = N I 2 r . So, the flux through the loop of wire is B 1 = integraldisplay B dA = N I 2 sin( t ) integraldisplay b + R R adr r = N I 2 a sin( t ) ln parenleftbigg b + R R parenrightbigg . Applying Faradays law, the induced emf can be calculated as follows E =- N d B 1 dt =- N N I 2 a ln parenleftbigg b + R R parenrightbigg cos( t ) =-E cos( t ) where = 2 f was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( t ). E =- N d B 1 dt =- N N I 2 a ln bracketleftbigg b + R R bracketrightbigg =- (22 turns) (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 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 4 = 1 10- 7 N / A 2 . d a b x dx I The magnetic flux through the strip of area dA is d = B dA = 2 I x bdx = 4 2 bI dx x , so the total magnetic flux through the rectan- gular loop is total = integraldisplay d + a d d = 4 (2 bI ) integraldisplay d + a d dx x = 4 (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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24 - Kapoor (mk9499) homework24 Turner (60230) 1 This...

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