Physics 2 - gilvin (jg47854) - 19. Amperes law - meyers -...

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gilvin (jg47854) - 19. Ampere’s law - meyers - (21235) This print-out should have 13 questions. 1 Multiple-choice questions may continue on the next column or page - find all choices before answering. This homework is due Tuesday, april19, at midnight Tucson time. 001 10.0 points A lightening bolt may carry a current of 4200 A for a short period of time. The permeability of free space is 1.25664 × 10 −6 T · m/A. What is the resulting magnetic field 68 m from the bolt? 4. B = 0 , r > R; B = 3 µ 0 I 2 π r , r < R 5. B = 0 , r > R; B = µ 0 I 2 π r , r < R 6. B = 0 , r > R; B = 5 µ 0 I 2 π r , r < R Explanation: Applying Ampère’s law for r > R, B · dl = B (2 π r) = µ 0 I C µ 0 I Suppose that the bolt extends far above and below the point of observation. Correct answer: 12.3529 µT. Explanation: Given : L = 68 m and I = 4200 A . Treat the lighting bolt as a long, straight conductor. Then the magnetic field is B = µ 0 I ( π L (4200 A) B r>R = 2 π r . applying Ampère’s law for r < R, B · dl = B (2 π r) = 0 C B r<R = 0 . 003 (part 1 of 3) 10.0 points A wire of radius 1 cm carries a current of 90 A that is uniformly distributed over its cross-sectional area. The permeability of free space is 4 π × 10 −7 T · m/A. Find the magnetic field at a distance of = 1.25664 × 10 6 T · m/A) 2π(68 m) 0.1 cm from the center of the wire. ( × 1.0 × 10 6 µT 1 T ) Correct answer: 0.00018 T. = 12.3529 µT . 002 10.0 points A long, straight, thin-walled cylindrical shell of radius R carries a current I . Find the magnetic field inside the cylinder and outside the cylinder. 1. B = 5 µ 0 I 2 π r , r > R; B = 0 , r < R Explanation: Let : I = 90 A , R = 1 cm = 0.01 m , r = 0.1 cm = 0.001 m , and µ 0 = 4 π × 10 −7 T · m/A . Using Ampère’s law, we have B 2. B = µ 0 I 2 π r , r > R; B = 0 , r < R correct 3. B = 3 µ 0 I 2 π r , r > R; B = 0 , r < R · dl = µ 0 I C . C Because r < R, I C = r 2 R 2 I , so
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2 ? R = 2 4? 0 I 2 ? r ' = 2 4? 0 I gilvin (jg47854) - 19. Ampere’s law - meyers - (21235) 2 Function: B(r) 10 9 B (2 π r) = µ 0 r 2 I R 2 B = µ 0 r 2 π R 2 I = (4 π × 10 7 T · m/A) (0.001 m) 2 π (0.01 m) 2 × (90 A) = 0.00018 T . 004 (part 2 of 3) 10.0 points Find the magnetic field at the surface of the wire. Correct answer: 0.0018 T.
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Physics 2 - gilvin (jg47854) - 19. Amperes law - meyers -...

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