Test+3--Fundamentals+of+Physics+II--key - Test 3 Magnetism...

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Test 3 Magnetism key Principles of physics II R.D. Knight 1. B 2. B 3. C 4. A 5. B 6. E 7. C 8. E 9. D 10. B 11. E 12. A 13. D 14. D 15. D 16. E 17. C 18. C 19. A 20. (a) Eq. 30-8 leads to (1.2T)(0.10 m)(5.0 m/s) 0.60 V . BLv ε = = = (b) By Lenz’s law, the induced emf is clockwise. In the rod itself, we would say the emf is directed up the page. (c) By Ohm’s law, the induced current is i = 0.60 V/0.40 = 1.5 A. (d) The direction is clockwise. (e) P = i 2 R = 0.90 W. (f) The force on the rod associated with the uniform magnetic field is directed rightward and has magnitude F iLB = = = ( . )( . . 15 010 018 A m)(1.2 T) N . To keep the rod moving at constant velocity, therefore, a leftward force (due to some external agent) having that same magnitude must be continuously supplied to the rod. (g) Using Eq. 7-48, we find the power associated with the force being exerted by the external agent: P = Fv = (0.18 N)(5.0 m/s) = 0.90 W, which is the same as our result from part (e). 21. As the problem states near the end, some idealizations are being made here to keep the
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This note was uploaded on 06/07/2009 for the course PHYSICS 2212 taught by Professor Geist during the Fall '09 term at Georgia Perimeter.

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Test+3--Fundamentals+of+Physics+II--key - Test 3 Magnetism...

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