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# Mid3 - Answers - Arb Kellen – Midterm 3 – Due Nov 9...

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Unformatted text preview: Arb, Kellen – Midterm 3 – Due: Nov 9 2004, 10:00 pm – Inst: Turner 1 This print-out should have 25 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 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) = 16 E R 2. I P (0) = 3 E R 3. I P (0) = E R 4. I P (0) = E RL 5. I P (0) = E L 2 R 6. I P (0) = E 2 R 7. I P (0) = 2 E R 8. I P (0) = 4 E R 9. I P (0) = 8 E R 10. I P (0) = 0 correct Explanation: The current in L has to change gradually. So immediately after the switch is closed, there is no current going through point P . 002 (part 2 of 2) 10 points When the switch has been closed for a long time, what is the energy stored in the induc- tor? 1. U L = L E 2 4 R 2 2. U L = L E 16 R 3. U L = L E 2 R 4. U L = L E 2 2 R 2 correct 5. U L = E 2 R 2 4 L 6. U L = L E 32 R 7. U L = LR 2 2 E 2 8. U L = L E 8 R 9. U L = L E 3 R 10. U L = L E 4 R Explanation: After the switch has been closed for a long time, the current in L does not change any more. So there is no voltage increase or de- crease across L . Therefore, the current going through L is I = E R , which gives the energy stored in L as U L = 1 2 LI 2 = L E 2 2 R 2 . 003 (part 1 of 3) 10 points The figure below shows a straight cylinderical coaxial cable of radii a , b , and c in which equal, uniformly distributed, but antiparallel currents i exist in the two conductors. Arb, Kellen – Midterm 3 – Due: Nov 9 2004, 10:00 pm – Inst: Turner 2 O i out i in ⊗ F E D C r 1 r 2 r 3 r 4 c b a Which expression gives the magnitude B ( r 1 ) at F of the magnetic field in the re- gion r 1 < c ? 1. B ( r 1 ) = μ i ( a 2- r 2 ) 2 π r 1 ( a 2- b 2 ) 2. B ( r 1 ) = μ i 2 π r 1 3. B ( r 1 ) = μ ir 1 2 π a 2 4. B ( r 1 ) = μ ir 1 2 π b 2 5. B ( r 1 ) = μ i π r 1 6. B ( r 1 ) = μ i ( r 2 1- b 2 ) 2 π r 1 ( a 2- b 2 ) 7. B ( r 1 ) = μ i ( a 2- b 2 ) 2 π r 1 ( r 2- b 2 ) 8. B ( r 1 ) = 0 9. B ( r 1 ) = μ ir 1 2 π c 2 correct 10. B ( r 1 ) = μ i ( a 2 + r 2 1- 2 b 2 ) 2 π r 1 ( a 2- b 2 ) Explanation: Ampere’s Law states that the line inte- gral 6 B. 6 dl around any closed path equals μ I , where I is the total steady current pass- ing through any surface bounded by the closed path. Considering the symmetry of this problem, we choose a circular path, then Ampere’s Law is simplified to: B (2 π r 1 ) = μ i in , where r 1 is the radius of the circle and i in is the current enclosed. For Part 1, r 1 < c , B = μ I in 2 π r 1 = μ i π r 2 1 π c 2 2 π r 1 = μ i r 2 1 c 2 2 π r 1 = μ ir 1 2 π c 2 ....
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Mid3 - Answers - Arb Kellen – Midterm 3 – Due Nov 9...

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