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Unformatted text preview: ragsdale (zdr82) HW9 ditmire (58335) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A coil has an inductance of 4 . 5 mH, and the current through it changes from 0 . 22 A to . 7 A in 0 . 32 s. Find the magnitude of the average induced emf in the coil during this period. Correct answer: 6 . 75 mV. Explanation: E = L I t = (0 . 0045 H) parenleftbigg . 7 A . 22 A . 32 s parenrightbigg = 0 . 00675 V = 6 . 75 mV . 002 10.0 points A small aircore solenoid has a length of 2 cm and a radius of 0 . 32 cm. The permeability of free space is 1 . 25664 10 6 N / A 2 . If the inductance is to be 0 . 85 mH, how many turns per centimeter (whole turns plus fractional turns) are required? Correct answer: 324 . 239 turns / cm. Explanation: Let : = 2 cm = 0 . 02 m , = 1 . 25664 10 6 N / A 2 , r = 0 . 32 cm = 0 . 0032 m , and L = 0 . 85 mH = 0 . 02 m . Let n be the number of turns per centimeter. The area is A = r 2 = (0 . 0032 m) 2 = 3 . 21699 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 . 00085 H (3 . 21699 10 5 m 2 )(0 . 02 m) = 1 . 05131 10 9 (turns / m) 2 Therefore, n = radicalBig 1 . 05131 10 9 (turns / m) 2 = 32423 . 9 turns / m = 324 . 239 turns / cm . 003 10.0 points A long solenoid has inside a coil of fine wire coaxial with it. I Outside solenoid has n turns per meter Inside coil has N total turns r R What is the mutual inductance between the solenoid and the inner coil? 1. M = 2 R n N 2. M = 2 R 2 2 n N 3. M = R 2 n N correct 4. M = r 2 n 5. M = R 2 n 6. M = 2 r 2 2 n N 7. M = 2 r n N 8. M = r 2 n N 9. M = 2 R n ragsdale (zdr82) HW9 ditmire (58335) 2 10. M = 2 r n Explanation: The magnetic field of a solenoid is B = n I . The magnetic flux is B = B A = ( n I )( R 2 ) , so the emf is E = N d B dt = R 2 n N d I dt . (Because we are interested in the emf in the inner coil, the area to use is the area of the inner coils rather than the solenoid area.) The emf created through a mutual induc tance M is E =M d I dt , so we have M = R 2 n N . keywords: 004 (part 1 of 5) 10.0 points A circuit is set up as shown in the figure. L R 1 R 2 E S I 1 I 2 I The switch is closed at t = 0. The current I through the inductor takes the form I = E R x parenleftBig 1 e t/ x parenrightBig where R x and x are to be determined. Find I immediately after the circuit is closed. 1. I = E R 1 2. I = E R 1 + R 2 3. I = E R 2 4. I = 0 correct Explanation: Before the circuit is closed, no current is flowing. When we have just closed the circuit we are at t = 0 + , a mathematical nota tion meaning a very short time after t = 0....
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.
 Spring '10
 SHIH
 Current, Inductance

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