This preview shows page 1. Sign up to view the full content.
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 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 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- r k , and B = B j , so the force on a charge q on the rod would be F = q v B = q (- r k ) B j = qrB i The work W to move the charge q from L to 0 is W = q Z L rB dR = 1 2 qBL 2 , and the potential difference V is the work divided by the charge: V = W/q = 1 2 BL 2 ....
View Full Document
- Fall '07