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Unformatted text preview: lim (kl9356) – Homework 17 – Weathers – (22202) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Assume that the length of the solenoid is much larger than the solenoid’s radius and that the core of the solenoid is air. Calculate the inductance of a uniformly wound solenoid having 230 turns if the length of the solenoid is 37 cm and its crosssectional area is 4 cm 2 . Correct answer: 0 . 0718661 mH. Explanation: Let : N = 230 , μ = 1 . 25664 × 10 − 6 N / A 2 , ℓ = 37 cm = 0 . 37 m , and A = 4 cm 2 = 0 . 0004 m 2 . In this case, we can take the interior magnetic field to be uniform and given by the equation B = μ n I = μ N ℓ I , where n is the number of turns per unit length, N ℓ . The magnetic flux through each turn is Φ B = B A = μ N A ℓ I , where A is the crosssectional area of the solenoid. Using this expression and the equa tion for the inductance, L = N Φ B I , we find that L = μ N 2 A ℓ = (1 . 25664 × 10 − 6 N / A 2 ) × (230) 2 (0 . 0004 m 2 ) . 37 m = 7 . 18661 × 10 − 5 H = . 0718661 mH . 002 (part 2 of 2) 10.0 points Calculate the selfinduced emf in the solenoid described in the first part if the current through it is decreasing at the rate of 49 A / s. Correct answer: 3 . 52144 mV. Explanation: Using the equation E = − N d Φ B d t , and given that d I d t = − 49 A / s , we get E = − L d I d t = − (7 . 18661 × 10 − 5 H ) ( − 49 A / s) = 0 . 00352144 V = 3 . 52144 mV . 003 10.0 points An automobile starter motor draws a current of 1 . 7 A from a 17 . 2 V battery when operating at normal speed. A broken pulley locks the ar mature in position, and the current increases to 12 . 3 A. What was the back emf of the motor when operating normally? Correct answer: 14 . 8228 V. Explanation: Let : I = 12 . 3 A , I ′ = 1 . 7 A , and E = 17 . 2 V . When not rotating, E = I R , and from this, R = E I = 17 . 2 V 12 . 3 A = 1 . 39837 Ω , When rotating, E −E back = I ′ R , or E back = E − I ′ R = 17 . 2 V − (1 . 7 A) (1 . 39837 Ω) = 14 . 8228 V . lim (kl9356) – Homework 17 – Weathers – (22202) 2 004 10.0 points In an RL series circuit, an inductor of 4 . 26 H and a resistor of 6 . 45 Ω are connected to a 28 . 4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Correct answer: 41 . 2949 J. Explanation: Let : L = 4 . 26 H , R = 6 . 45 Ω , and E = 28 . 4 V . The current in an RL circuit is I = E R parenleftBig 1 − e − Rt/L parenrightBig ....
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Magnetism, Work

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