This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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 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 I ~ B d ~ 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 p (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....
View
Full
Document
This note was uploaded on 07/26/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

Click to edit the document details