exam3sol

# exam3sol - Flores Michael – Exam 3 – Due 10:00 pm –...

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Unformatted text preview: Flores, Michael – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban 1 This print-out should have 16 questions. Multiple-choice 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 solenoid’s area A L left of the wire CD and the solenoid’s 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 Φ 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 counter-clockwise, as shown in the figure. From Kirchoff’s 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 Ω 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 9Ω 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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exam3sol - Flores Michael – Exam 3 – Due 10:00 pm –...

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