# hw16 - Physics 21 Fall 2004 Solution to HW-16 29 P 65 A...

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Unformatted text preview: Physics 21 Fall, 2004 Solution to HW-16 29 P 65 A thin metal rod of length L rotates with angular velocity ω about an axis through one end. The rotation axis is perpendicular to the rod and is parallel to a uniform magnetic field B . Determine the emf developed between the ends of the rods. x y z B L r ω A charge on the rotating rod experiences a v × B force. We calculate this force using the instantaneous velocity of the segment of the rod a distance r from the axis of rotation. Assume the rod rotates in the xy plane and is at the position shown, aligned along the x axis. The velocity of the rod at the point r is- ωr ˆ k , and B = B ˆ j , so the force on a charge q on the rod would be F = q v × B = q (- ωr ˆ k ) × B ˆ j = qωrB ˆ i The work W to move the charge q from L to 0 is W = q Z L ωrB dR = 1 2 qωBL 2 , and the potential difference ∆ V is the work divided by the charge: ∆ V = W/q = 1 2 ωBL 2 ....
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## This note was uploaded on 08/06/2008 for the course PHYS 21 taught by Professor Hickman during the Fall '07 term at Lehigh University .

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