19. Ampere's law-solutions

19 Ampere's - gilvin(jg47854 – 19 Ampere’s law – meyers –(21235 1 This print-out should have 13 questions Multiple-choice questions may

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Unformatted text preview: gilvin (jg47854) – 19. Ampere’s law – meyers – (21235) 1 This print-out should have 13 questions. 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? Suppose that the bolt extends far above and below the point of observation. Correct answer: 12 . 3529 μ T. 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 I 2 π L = ( 1 . 25664 × 10 − 6 T · m / A ) (4200 A) 2 π (68 m) × parenleftbigg 1 . × 10 6 μ T 1 T parenrightbigg = 12 . 3529 μ T . 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 I 2 π r , r > R ; B = 0 , r < R 2. B = μ I 2 π r , r > R ; B = 0 , r < R correct 3. B = 3 μ I 2 π r , r > R ; B = 0 , r < R 4. B = 0 , r > R ; B = 3 μ I 2 π r , r < R 5. B = 0 , r > R ; B = μ I 2 π r , r < R 6. B = 0 , r > R ; B = 5 μ I 2 π r , r < R Explanation: Applying Amp` ere’s law for r > R , contintegraldisplay C vector B · dl = B (2 π r ) = μ I B r>R = μ I 2 π r . applying Amp` ere’s law for r < R , contintegraldisplay C vector B · dl = B (2 π r ) = 0 B r<R = . 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 cm from the center of the wire. Correct answer: 0 . 00018 T. Explanation: Let : I = 90 A , R = 1 cm = 0 . 01 m , r = 0 . 1 cm = 0 . 001 m , and μ = 4 π × 10 − 7 T · m / A . Using Amp` ere’s law, we have contintegraldisplay C vector B · d vector l = μ I C . Because r < R , I C = r 2 R 2 I , so gilvin (jg47854) – 19. Ampere’s law – meyers – (21235) 2 B (2 π r ) = μ r 2 R 2 I B = μ r 2 π R 2 I = (4 π × 10 − 7 T · m / A) (0 . 001 m) 2 π (0 . 01 m) 2 × (90 A) = . 00018 T ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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19 Ampere's - gilvin(jg47854 – 19 Ampere’s law – meyers –(21235 1 This print-out should have 13 questions Multiple-choice questions may

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