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Unformatted text preview: Flores, Michael Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Assume: The induced emf for the closed loop octagonal CXDY C is E . A solenoid (with magnetic field B ) pro duces a steadily increasing uniform magnetic flux through its circular cross section. A octagonal circuit surrounds the solenoid as shown in the figure. The wires connecting in the circuit are ideal, having no resistance. The circuit consists of two identical light bulbs (labeled X and Y ) in series. A wire connects points C and D . The ratio of the solenoids area A L left of the wire CD and the solenoids area A R right of the wire CD is A L A R = 4 . B B B B Y X D C i 3 i 1 i 2 A L A R The equations for the (right) loop CXDC and the (left) loop CDY C are respectively given by 1. 4 E 5 + i 1 R = 0 and E 5 i 2 R = 0 . 2. E 5 + i 1 R = 0 and 4 E 5 i 2 R = 0 . 3. E 5 i 1 R = 0 and 4 E 5 + i 2 R = 0 . 4. 4 E 5 + i 1 R = 0 and E 5 + i 2 R = 0 . 5. E 5 + i 1 R = 0 and 4 E 5 + i 2 R = 0 . cor rect 6. E 5 i 1 R = 0 and 4 E 5 i 2 R = 0 . 7. 4 E 5 i 1 R = 0 and E 5 i 2 R = 0 . 8. 4 E 5 i 1 R = 0 and E 5 + i 2 R = 0 . Explanation: By definition, the areas of the left and right loops are related by A = A L + A R . Since A L A R = 4, we can solve for A L and A R in terms of A . A L = 4 A 5 A R = A 5 . Then we can compute the magnitude of the induced emf around the right and left loops. E R = A R dB dt = A 5 dB dt = 1 5 E E L = A L dB dt = 4 A 5 dB dt = 4 5 E . The induced emf and the changing mag netic flux are related by E = d dt = A dB dt . Since the magnetic flux is increasing, the in duced emf is in the clockwise direction and the direction of the current is counterclockwise, as shown in the figure. From Kirchoffs laws, the loop equations for the right and left loops respectively are right loop : 1 5 E + i 1 R = 0 (1) left loop : 4 5 E + i 2 R = 0 (2) 002 (part 1 of 1) 10 points Given: Assume the bar and rails have neg ligible resistance and friction. In the arrangement shown in the figure, the resistor is 9 and a 6 T magnetic field is directed into the paper. The separation Flores, Michael Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban 2 between the rails is 6 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 7 m / s . m 1g 7 m / s 9 6 T 6 T I 6m Calculate the applied force required to move the bar to the right at a constant speed of 7 m / s. Correct answer: 1008 N....
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 Spring '04
 Panab

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