27 - Kapoor (mk9499) homework27 Turner (60230) 1 This...

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Unformatted text preview: Kapoor (mk9499) homework27 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 small air-core solenoid has a length of 1 cm and a radius of 0 . 17 cm. The permeability of free space is 1 . 25664 10- 6 N / A 2 . If the inductance is to be 0 . 47 mH, how many turns per centimeter (whole turns plus fractional turns) are required? Correct answer: 641 . 83 turns / cm. Explanation: Let : = 1 cm = 0 . 01 m , = 1 . 25664 10- 6 N / A 2 , r = 0 . 17 cm = 0 . 0017 m , and L = 0 . 47 mH = 0 . 01 m . Let n be the number of turns per centimeter. The area is A = r 2 = (0 . 0017 m) 2 = 9 . 0792 10- 6 m 2 . The inductance of the air-core solenoid is L = n 2 A n 2 = L A = 1 1 . 25664 10- 6 N / A 2 . 00047 H (9 . 0792 10- 6 m 2 )(0 . 01 m) = 4 . 11946 10 9 (turns / m) 2 Therefore, n = radicalBig 4 . 11946 10 9 (turns / m) 2 = 64183 turns / m = 641 . 83 turns / cm . 002 10.0 points An emf of 57 mV is induced in a 494-turn coil when the current is changing at a rate of 11 . 5 A / s. What is the magnetic flux through each turn of the coil at an instant when the current is 2 . 94 A? Correct answer: 2 . 94983 10- 5 T m 2 . Explanation: Let : E = 57 mV = 0 . 057 V , N = 494 , I = 2 . 94 A , and I t = 11 . 5 A / s . From Faradays Law E =- N t =- L I t |E| = L I t L = |E| I t = . 057 V 11 . 5 A / s = 0 . 00495652 H At the instant where the current is 2 . 94 A , the flux through each turn is = L I N = (0 . 00495652 H) (2 . 94 A) 494 = 2 . 94983 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 time-varying high-voltage from a low-voltage source, as shown in the figure....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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27 - Kapoor (mk9499) homework27 Turner (60230) 1 This...

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