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Unformatted text preview: Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih 1 This printout should have 27 questions. Multiplechoice 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 The figure below shows a 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 B ( r 3 ) at D of the magnetic field in the re gion b < r 3 < a ? 1. B ( r 3 ) = μ i ( a 2 r 2 3 ) 2 π r 3 ( a 2 b 2 ) correct 2. B ( r 3 ) = 0 3. B ( r 3 ) = μ ir 3 2 π b 2 4. B ( r 3 ) = μ i ( a 2 + r 2 3 2 b 2 ) 2 π r 3 ( a 2 b 2 ) 5. B ( r 3 ) = μ ir 3 2 π c 2 6. B ( r 3 ) = μ i ( r 2 3 b 2 ) 2 π r 3 ( a 2 b 2 ) 7. B ( r 3 ) = μ ir 3 2 π a 2 8. B ( r 3 ) = μ i π r 3 9. B ( r 3 ) = μ i ( a 2 b 2 ) 2 π r 3 ( r 2 3 b 2 ) 10. B ( r 3 ) = μ i 2 π r 3 Explanation: Ampere’s Law states that the line inte gral 6 B · d 6 5 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 is simplified to B (2 π r 3 ) = μ i in , where r 3 is the radius of the circle and i in is the current enclosed. Since, when b < r 3 < a , for the cylinder, A in A cylinder = π ( r 2 3 b 2 ) π ( a 2 b 2 ) , we have B = μ I in 2 π r 3 = μ i i π ( r 2 3 b 2 ) π ( a 2 b 2 ) 2 π r 3 = μ i a 2 r 2 3 a 2 b 2 2 π r 3 = μ i ( a 2 r 2 3 ) 2 π r 3 ( a 2 b 2 ) . 002 (part 1 of 1) 10 points Calculate the resonance frequency of a se ries RLC circuit for which the capacitance is 99 μ F , the resistance is 33 kΩ , and the induc tance is 56 mH . Correct answer: 67 . 5941 Hz. Explanation: Let : R = 33 kΩ = 33000 Ω , L = 56 mH = 0 . 056 H , and C = 99 μ F = 9 . 9 × 10 5 F . The resonance frequency is the frequency at which the current becomes maximum, or the Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih 2 impedance becomes minimum. This occurs when X L = X C ω L = 1 ω C . From this condition, the resonance frequency is given by f = 1 2 π √ LC = 1 2 π (0 . 056 H) (9 . 9 × 10 5 F) = 67 . 5941 Hz . 003 (part 1 of 1) 10 points Consider an electromagnetic wave pattern as shown in the figure below. E B The wave is 1. traveling left to right. correct 2. traveling right to left. 3. a standing wave and is stationary. Explanation: The 6 E vector and 6 B vector are not at the same point on the velocity axis. Pick an instant in time, where the E and B fields are at the same point on the velocity axis....
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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