quiz7solutions - 2 . 4. From Ampere’s law we have the...

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Quiz 7 Solution Guide Hi Everyone, Here are the answers for quiz 7 along with brief explanations of the problems. The same quiz was used for both classes. If you have any questions, please email me ( [email protected] ). Jason Quiz 7 Version A Version B Version C Version D 1 A D B D 2 B D D C 3 B A A B 4 D B B B 5 D A B A 6 D B D D
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Solutions 1. F=qvB. Since the particle is moving perpendicular to the magnetic field, the sin term is 1. F=8C * 6m/s * 100T = 4800 N.
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2. F=qvBsin(27) To determine the direction of the field, use the right hand rule, k j i ! ! ! = " , so the force is in the positive z direction. 3. Remember that qvB=mv 2 /r. Solve this expression for v, the solve for kinetic energy remembering that it is (1/2)mv
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Unformatted text preview: 2 . 4. From Ampere’s law we have the result that B= µ I/(2 π r). Solve this expression for the current. Remember that 1G = 10-4 T. 5. Use Ampere’s Law. The current enclosed is 1A out of the page. If you us the right hand rule you find that along the y axis, outside of the loop the field is pointing in the negative x direction. The answer is -2.5*10-6 T. Note that the problem said to pick the closest answer. The closest answer is -6*10-6 T. 6. Remember that F = ILB (where B is the field perpendicular to the direction of the current flow). In this case the perpendicular component of the B field is simply the y component. L = 22N/(7T*2A) = 1.57m....
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This note was uploaded on 05/22/2008 for the course PHYS 2b taught by Professor Schuller during the Spring '08 term at UCSD.

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quiz7solutions - 2 . 4. From Ampere’s law we have the...

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