Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s^–1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if 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. Explain.
(d) What is the retarding force on the rod when K is closed?
(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?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power ?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

(a) Here, speed with which the rod is moved, v = 12 cm s^–1
The magnitude of the induced emf is given by ε = B v I
       = 0.50 x 0.12 x 0.15 volt
       = 9 x 10^–3volt
       = 9 mV
P is the positive end and Q is the negative end. 

(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.

(c) There is no net force as the magnetic force F_m = – e (v + B) is cancelled by the electric force [F_e = eE] which is set up due to the excess charge of opposite signs at the ends of the rod. 

(d) Induced current, $\mathrm{I} = \frac{\mathrm{e}}{\mathrm{R}} = \frac{9 \times {10}^{-3}}{9 \times {10}^{-3}} = 1\mathrm{A}$ 

Retarding force on the rod F = BIl
                                        = 0.5 x 1 x 0.15
                                        = 7.5 x 10 ^–2 N

(e) Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm/s is given by, 
          P = F.v
          P = 75 x 10^–3 x 12 x 10^–2 W
             = 9.0 x 10^–3 W 

(f) Power dissipated as heat = I² R
                                        = 1² (9 x 10^–3)
                                        = 9 x 10^–3 W 
Source of this power is an external agent which keeps rod in motion, against magnetic retarding force.

(g) When the permanent magnet is rotated in a vertical position, the field becomes parallel to rails. The motion of rod will not cut across the lines of magnetic field and hence no e.m.f. is induced.