(b) Yes, an excess amount of charge is built up at the ends of the rod when key is open. When key is closed, the excess charge is maintained by the continuous flow of current. P is the positive end and Q is the negative end. The magnitude of the induced emf is given by ε = B v I (a) Here, speed with which the rod is moved, v = 12 cm s –1 (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? (f) How much power is dissipated as heat in the closed circuit? What is the source of this power ? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s –1) when K is closed? How much power is required when K is open? (d) What is the retarding force on the rod when K is closed? (c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. (b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? Give the polarity and magnitude of the induced emf. (a) Suppose K is open and the rod is moved with a speed of 12 cm s –1 in the direction shown. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. A galvanometer G connects the rails through a switch K. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet.
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