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Unformatted text preview: mcconnell (kam2342) homework27 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 small aircore solenoid has a length of 20 cm and a radius of 0 . 43 cm. The permeability of free space is 1 . 25664 10 6 N / A 2 . If the inductance is to be 0 . 91 mH, how many turns per centimeter (whole turns plus fractional turns) are required? Correct answer: 78 . 951 turns / cm. Explanation: Let : = 20 cm = 0 . 2 m , = 1 . 25664 10 6 N / A 2 , r = 0 . 43 cm = 0 . 0043 m , and L = 0 . 91 mH = 0 . 2 m . Let n be the number of turns per centimeter. The area is A = r 2 = (0 . 0043 m) 2 = 5 . 8088 10 5 m 2 . The inductance of the aircore solenoid is L = n 2 A n 2 = L A = 1 1 . 25664 10 6 N / A 2 . 00091 H (5 . 8088 10 5 m 2 )(0 . 2 m) = 6 . 23325 10 7 (turns / m) 2 Therefore, n = radicalBig 6 . 23325 10 7 (turns / m) 2 = 7895 . 1 turns / m = 78 . 951 turns / cm . 002 10.0 points An emf of 29 . 4 mV is induced in a 430turn coil when the current is changing at a rate of 16 . 3 A / s. What is the magnetic flux through each turn of the coil at an instant when the current is 2 . 49 A? Correct answer: 1 . 04446 10 5 T m 2 . Explanation: Let : E = 29 . 4 mV = 0 . 0294 V , N = 430 , I = 2 . 49 A , and I t = 16 . 3 A / s . From Faradays Law E = N t = L I t E = L I t L = E I t = . 0294 V 16 . 3 A / s = 0 . 00180368 H At the instant where the current is 2 . 49 A , the flux through each turn is = L I N = (0 . 00180368 H) (2 . 49 A) 430 = 1 . 04446 10 5 T m 2 . keywords: 003 (part 1 of 3) 10.0 points One application of an RL circuit is the gen eration of timevarying highvoltage from a lowvoltage source, as shown in the figure....
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 Spring '10
 Turner

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