sp01em_sol - d B F E & M Proficiency Exam,...

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Unformatted text preview: d B F E & M Proficiency Exam, Spring 2001 Useful constants: e = 1 . 60 10- 19 C hc = 1240 eV nm c = 3 . 00 10 8 m s m e = . 511 MeV c 2 k = 1 4 = 8 . 99 10 9 Nm 2 C 2 = 8 . 85 10- 12 C 2 Nm 2 1. The two rails of a superconducting track are separated by a distance d . A conductor can slide along the track. The conductor, initially at rest, is pulled to the right by a constant force F . The friction force between the conductor and the track is directly proportional to its velocity with a proportionality constant . The portion of the conductor between the rails has a resistance of R . The entire setup is in a uniform magnetic field ~ B as shown in the figure. The field ~ B points into the page. (a) What is the direction of the induced current in the conductor? We can answer to this questions as follows. When the conductor moves to right the magnetic flux through the area enclosed by the conductor and the superconducting track increases. According to the Lenzs law a current will be induced in the closed circuit to oppose the increase of the magnetic flux. Therefore the induced magnetic field points out of paper in the closed loop. The current that produces this magnetic flux must go counterclockwise. The other way to reach this conclusion is to use the Lorentzs law. According to this law when a charged particle moves in a magnetic field a force will act on it which is given by ~ F L = q ~v ~ B. (1) F B x y z 1 2 With the directions as in the figure above we get...
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This note was uploaded on 03/11/2012 for the course PHY 3900 taught by Professor Staff during the Fall '1 term at FSU.

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sp01em_sol - d B F E & M Proficiency Exam,...

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