This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: montes (dam2772) Homework5 shih (56310) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A bar magnet is held above the center of a conducting wire loop in a horizontal plane with the south end of the magnet toward the loop. The magnet is dropped. S N counter- clockwise clockwise bar magnet Find the direction of the current, as viewed from above, around the wire loop while the magnet is falling toward the loop. 1. Unable to determine the direction. 2. clockwise correct 3. counter-clockwise Explanation: To oppose the field due to the south end of the magnet as it approaches the wire loop, the magnetic field should be directed downward along the axis of the loop, so the current must be clockwise when viewed from above the loop. 002 (part 2 of 2) 10.0 points Find the direction of the current around the loop after the magnet has passed through the wire loop and moves away from it. 1. counter-clockwise correct 2. Unable to determine the direction. 3. clockwise Explanation: To enhance the field due to the north end of the magnet as it recedes from the wire loop, the magnetic field should be directed upward along the axis of the loop, so the current must be counter-clockwise when viewed from above the loop. 003 (part 1 of 2) 10.0 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. 167 mH 5 . 51 8 . 2 V S b a If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches 164 mA? Correct answer: 3 . 53878 ms. Explanation: Let : R = 5 . 51 , L = 167 mH = 0 . 167 H , and E = 8 . 2 V . L R E S b a The time constant of an RL circuit is = L R = . 167 H 5 . 51 = 0 . 0303085 s . The final current reached in the circuit is I = E R = 8 . 2 V 5 . 51 = 1 . 4882 A . montes (dam2772) Homework5 shih (56310) 2 The switch is in position a in an RL circuit connected to a battery at t = 0 when I = 0. Then the current vs time is I = I parenleftBig 1 e t / parenrightBig . Solving the above expression for t , when I = I 1 gives t 1 = ln parenleftbigg 1 I 1 I parenrightbigg = (0 . 0303085 s) ln parenleftbigg 1 . 164 A 1 . 4882 A parenrightbigg = 3 . 53878 ms . 004 (part 2 of 2) 10.0 points What is the maximum current in the inductor a long time after the switch is in position a ? Correct answer: 1 . 4882 A. Explanation: After a long time compared to , we have a d.c. circuit with a battery supplying an emf E , which is equal to the voltage drop I R across the resistor. Thus I = E R = 8 . 2 V 5 . 51 = 1 . 4882 A ....
View Full Document