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Mid3 - Answers - Arb Kellen Midterm 3 Due Nov 9 2004 10:00...

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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 R L 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 = L R 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 L I 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.
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Arb, Kellen – Midterm 3 – Due: Nov 9 2004, 10:00 pm – Inst: Turner 2 O i out fl 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 0 i ( a 2 - r 2 ) 2 π r 1 ( a 2 - b 2 ) 2. B ( r 1 ) = μ 0 i 2 π r 1 3. B ( r 1 ) = μ 0 i r 1 2 π a 2 4. B ( r 1 ) = μ 0 i r 1 2 π b 2 5. B ( r 1 ) = μ 0 i π r 1 6. B ( r 1 ) = μ 0 i ( r 2 1 - b 2 ) 2 π r 1 ( a 2 - b 2 ) 7. B ( r 1 ) = μ 0 i ( a 2 - b 2 ) 2 π r 1 ( r 2 - b 2 ) 8. B ( r 1 ) = 0 9. B ( r 1 ) = μ 0 i r 1 2 π c 2 correct 10. B ( r 1 ) = μ 0 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 I ~ B. ~ dl around any closed path equals μ 0 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 0 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 0 I in 2 π r 1 = μ 0 i π r 2 1 π c 2 2 π r 1 = μ 0 i r 2 1 c 2 2 π r 1 = μ 0 i r 1 2 π c 2 . 004 (part 2 of 3) 10 points Which expression gives the magnitude B ( r 2 ) E of the magnetic field in the region c < r 2 < b ? 1. B ( r 2 ) = μ 0 i ( a 2 - b 2 ) 2 π r 2 ( r 2 2 - b 2 ) 2. B ( r 2 ) = μ 0 i r 2 2 π c 2 3. B ( r 2 ) = μ 0 i π r 2 4. B ( r 2 ) = μ 0 i r 2 2 π a 2 5. B ( r 2 ) = μ 0 i ( a 2 - r 2 2 ) 2 π r 2 ( a 2 - b 2 ) 6. B ( r 2 ) = μ 0 i r 2 2 π b 2 7. B ( r 2 ) = μ 0 i ( a 2 + r 2 2 - 2 b 2 ) 2 π r 2 ( a 2 - b 2 ) 8. B ( r 2 ) = μ 0 i 2 π r 2 correct 9. B ( r 2 ) = μ 0 i ( r 2 2 - b 2 ) 2 π r 2 ( a 2 - b 2 ) 10. B ( r 2 ) = 0 Explanation:
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Arb, Kellen – Midterm 3 – Due: Nov 9 2004, 10:00 pm – Inst: Turner 3 For Part 2, c < r 2 < b , B = μ 0 I in 2 π r 2 = μ 0 ( i ) 2 π r 2 = μ 0 i 2 π r 2 .
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