EXAM 3 - Platt David Quiz 3 Due 10:00 pm Inst Ken Shih This...

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Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih 1 This print-out should have 27 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 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 0 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 0 i r 3 2 π b 2 4. B ( r 3 ) = μ 0 i ( a 2 + r 2 3 - 2 b 2 ) 2 π r 3 ( a 2 - b 2 ) 5. B ( r 3 ) = μ 0 i r 3 2 π c 2 6. B ( r 3 ) = μ 0 i ( r 2 3 - b 2 ) 2 π r 3 ( a 2 - b 2 ) 7. B ( r 3 ) = μ 0 i r 3 2 π a 2 8. B ( r 3 ) = μ 0 i π r 3 9. B ( r 3 ) = μ 0 i ( a 2 - b 2 ) 2 π r 3 ( r 2 3 - b 2 ) 10. B ( r 3 ) = μ 0 i 2 π r 3 Explanation: Ampere’s Law states that the line inte- gral B · d 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, so Ampere’s Law is simplified to B (2 π r 3 ) = μ 0 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 0 I in 2 π r 3 = μ 0 i - i π ( r 2 3 - b 2 ) π ( a 2 - b 2 ) 2 π r 3 = μ 0 i a 2 - r 2 3 a 2 - b 2 2 π r 3 = μ 0 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 k Ω , and the induc- tance is 56 mH . Correct answer: 67 . 5941 Hz. Explanation: Let : R = 33 k Ω = 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
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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 L C = 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 E vector and 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. z v x y E B For instance, let us choose the point where the E vector is along the x axis, as shown in the above figures. At this same instant, the B vector is along the negative y axis (at a point with a phase di ff erence of 360 from the place on the velocity ( z ) axis where the E vector is drawn).
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