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

# 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 calculation straightforward (but are slightly unrealistic). For circular motion (with speed v which represents the magnitude of the component of the velocity perpendicular to the magnetic

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