problem29_58 - 29.58 a) According to Example 29.6 the...

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Unformatted text preview: 29.58: a) According to Example 29.6 the induced emf is ε = BLv = (8 × 10 −5 T) (0.004 m) (300 m s ) = 96µ V ≈ 0.1 mV. Note that L is the size of the bar measured in a direction that is perpendicular to both the magnetic field and the velocity of the bar. Since a positive charge moving to the east would be deflected upward, the top of the bullet will be at a higher potential. b) For a bullet that travels south, the induced emf is zero. c) In the direction parallel to the velocity the induced emf is zero. ...
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This homework help was uploaded on 02/03/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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