# quiz3 - Daniels Matthew – Quiz 3 – Due 9:00 pm – Inst...

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Unformatted text preview: Daniels, Matthew – Quiz 3 – Due: Nov 16 2006, 9:00 pm – Inst: Ditmire 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A wire 2.6 m long experiences a magnetic force of 0.30 N due to a perpendicular uniform magnetic field. If the wire carries a current of 10.0 A, what is the magnitude of the magnetic field? Correct answer: 0 . 0115385 T. Explanation: Let : ‘ = 2 . 6 m , F m = 0 . 30 N , and I = 10 . 0 A . The magnetic force is F m = I ‘ B B = F m I ‘ = . 3 N (10 A)(2 . 6 m) = . 0115385 T keywords: 002 (part 1 of 2) 10 points Consider the circuit shown. L R R E S P What is the instantaneous current at point P immediately after the switch is closed? 1. I P (0) = 8 E R 2. I P (0) = E L 2 R 3. I P (0) = E 2 R 4. I P (0) = E R L 5. I P (0) = 2 E R 6. I P (0) = 3 E R 7. I P (0) = 4 E R 8. I P (0) = 0 correct 9. I P (0) = E R 10. I P (0) = 16 E R Explanation: The current in L has to change gradually, so immediately after the switch is closed, there is no current going through point P . 003 (part 2 of 2) 10 points After the switch has been closed for a long time, it is opened at time t = 0. Which of the following graphs best repre- sents the subsequent current I at point P as a function of time t ? 1. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] 2. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] Daniels, Matthew – Quiz 3 – Due: Nov 16 2006, 9:00 pm – Inst: Ditmire 2 3. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] 4. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] 5. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] 6. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] 7. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] cor- rect 8. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] Explanation: At time t = 0, the current at P is not zero because the current going through L cannot be changed abruptly. Also from Kirchhoff’s rule, we have L d I dt + I R + I R = 0 , which will give the expression for the current I as I = I e- 2 R t/L . The correct graph is 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Time, t [s] Current, I [A] keywords: 004 (part 1 of 1) 10 points A static uniform magnetic field is directed into the page. A charged particle moves in the plane of the page following a counter- clockwise spiral of decreasing radius as shown. B B Neglect the effect due to gravity. What is a reasonable explanation? 1. The charge is positive and with a constant speed. 2. None of these 3. The charge is neutral and with a constant speed....
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quiz3 - Daniels Matthew – Quiz 3 – Due 9:00 pm – Inst...

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