P31_029

# P31_029 - . 10 m)(1 . 2 T) = 0 . 18 N . To keep the rod...

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29. (a) By Lenz’s law, the induced emf is clockwise. In the rod itself, we would say the emf is directed up the page. Eq. 31-10 leads to E = BLv =(1 . 2 T)(0 . 10 m)(5 . 0m / s) = 0 . 60 V . (b) By Ohm’s law, the (clockwise) induced current is i =0 . 60 V / 0 . 40 Ω = 1 . 5A . (c) Eq. 27-22 leads to P = i 2 R =0 . 90 W. (d) From Eq. 29-2, we ±nd that the force on the rod associated with the uniform magnetic ±eld is directed rightward and has magnitude F = iLB =(1 . 5 A)(0
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Unformatted text preview: . 10 m)(1 . 2 T) = 0 . 18 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. (e) Using Eq. 7-48, we nd 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 (c)....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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