{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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:
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern