EXAM 3 - Platt, David Quiz 3 Due: Nov 15 2005, 10:00 pm...

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Unformatted text preview: 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 ) = i ( a 2- r 2 3 ) 2 r 3 ( a 2- b 2 ) correct 2. B ( r 3 ) = 0 3. B ( r 3 ) = ir 3 2 b 2 4. B ( r 3 ) = i ( a 2 + r 2 3- 2 b 2 ) 2 r 3 ( a 2- b 2 ) 5. B ( r 3 ) = ir 3 2 c 2 6. B ( r 3 ) = i ( r 2 3- b 2 ) 2 r 3 ( a 2- b 2 ) 7. B ( r 3 ) = ir 3 2 a 2 8. B ( r 3 ) = i r 3 9. B ( r 3 ) = i ( a 2- b 2 ) 2 r 3 ( r 2 3- b 2 ) 10. B ( r 3 ) = i 2 r 3 Explanation: Amperes Law states that the line inte- gral 6 B d 6 5 around any closed path equals 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 Amperes Law is simplified to B (2 r 3 ) = 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 = ( r 2 3- b 2 ) ( a 2- b 2 ) , we have B = I in 2 r 3 = i- i ( r 2 3- b 2 ) ( a 2- b 2 ) 2 r 3 = i a 2- r 2 3 a 2- b 2 2 r 3 = i ( a 2- r 2 3 ) 2 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 F , the resistance is 33 k , and the induc- tance is 56 mH . Correct answer: 67 . 5941 Hz. Explanation: Let : R = 33 k = 33000 , L = 56 mH = 0 . 056 H , and C = 99 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 L = 1 C . From this condition, the resonance frequency is given by f = 1 2 LC = 1 2 (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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EXAM 3 - Platt, David Quiz 3 Due: Nov 15 2005, 10:00 pm...

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