OHW 24 - terry (ect328) oldhomework 24 Turner (59130) 1...

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Unformatted text preview: terry (ect328) oldhomework 24 Turner (59130) 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 = 0 . 96 cm by b = 4 . 56 cm) and inner radius 2 . 3 cm consists of N = 650 turns of wire that carries a current I = I sin t , with I = 41 . 7 A and a frequency f = 32 . 6 Hz. A loop that consists of N = 15 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 . 174735 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 =- (15 turns) (650 turns) (41 . 7 A) (32 . 6 Hz) (0 . 96 cm) ln bracketleftbigg (4 . 56 cm) + (2 . 3 cm) (2 . 3 cm) bracketrightbigg =- . 174735 V |E| = 0 . 174735 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 5 cm and 12 . 5 cm, with its near side a distance 3 cm from the straight wire, as shown in the figure. 3cm 5 cm 12 . 5cm 40A Find the magnetic flux through the rectan- gular loop. terry (ect328) oldhomework 24 Turner (59130) 2 The permeability of free space is 4 10- 7 T m / A. Correct answer: 9 . 80829 10- 7 Wb. Explanation: Let : I = 40 A , a = 5 cm = 0 . 05 m , b = 12 . 5 cm = 0 . 125 m , d = 3 cm = 0 . 03 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 . 125 m) (40 A) ln parenleftbigg . 03 m + 0 . 05 m . 03 m parenrightbigg = 9 . 80829 10- 7 Wb . 003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 84 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 270 g. The uniform magnetic field has a magnitude of 720 mT,...
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OHW 24 - terry (ect328) oldhomework 24 Turner (59130) 1...

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