# hw14b - aljabr(faa335 – Hw14 – Ross –(89251 1 This...

This preview shows pages 1–2. Sign up to view the full content.

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.

Unformatted text preview: aljabr (faa335) – Hw14 – Ross – (89251) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points The figure below shows a straight cylindrical coaxial cable of radii a , b , and c in which equal, uniformly distributed, but antiparallel currents i exist in the two conductors. 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 of the magnetic field in the region r 1 < c (at F )? 1. B ( r 1 ) = μ ir 1 2 π c 2 correct 2. B ( r 1 ) = μ ir 1 2 π b 2 3. B ( r 1 ) = μ i π r 1 4. B ( r 1 ) = μ i ( r 2 1 − b 2 ) 2 π r 1 ( a 2 − b 2 ) 5. B ( r 1 ) = μ i ( a 2 + r 2 1 − 2 b 2 ) 2 π r 1 ( a 2 − b 2 ) 6. B ( r 1 ) = 0 7. B ( r 1 ) = μ i 2 π r 1 8. B ( r 1 ) = μ i ( a 2 − r 2 1 ) 2 π r 1 ( a 2 − b 2 ) 9. B ( r 1 ) = μ i ( a 2 − b 2 ) 2 π r 1 ( r 2 1 − b 2 ) 10. B ( r 1 ) = μ ir 1 2 π a 2 Explanation: Ampere’s Law states that the line inte- gral contintegraldisplay vector B · d vector ℓ 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, so Ampere’s Law simplifies to B (2 π r 1 ) = μ I en , where r 1 is the radius of the circle and I en is the current enclosed. For r 1 < c , B = μ I en 2 π r 1 = μ parenleftbigg i π r 2 1 π c 2 parenrightbigg 2 π r 1 = μ i parenleftbigg r 2 1 c 2 parenrightbigg 2 π r 1 = μ ir 1 2 π c 2 . 002 (part 2 of 4) 10.0 points Which expression gives the magnitude of the magnetic field in the region c < r 2 < b (at E )? 1. B ( r 2 ) = μ i 2 π r 2 correct 2. B ( r 2 ) = μ i ( a 2 − r 2 2 ) 2 π r 2 ( a 2 − b 2 ) 3. B ( r 2 ) = μ i π r 2 4. B ( r 2 ) = μ ir 2 2 π a 2 5. B ( r 2 ) = μ i ( a 2 − b 2 ) 2 π r 2 ( r 2 2 − b 2 ) 6. B ( r 2 ) = μ i ( r 2 2 − b 2 ) 2 π r 2 ( a 2 − b 2 ) 7. B ( r 2 ) = μ i ( a 2 + r 2 2 − 2 b 2 ) 2 π r 2 ( a 2 − b 2 ) aljabr (faa335) – Hw14 – Ross – (89251)...
View Full Document

## This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.

### Page1 / 5

hw14b - aljabr(faa335 – Hw14 – Ross –(89251 1 This...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online