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 mutually perpendicular directions. A galvonometer G connects the rails through a switch K. Length of rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = `9.0mOmega`. Assume the field to be uniform.

a. Suppose K is open and the rod is moved with a speed of `12cms^(-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 eventhough 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 `(=12cms^(-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?

Explanation

a. `varepsilon=Blv=0.5xx15xx10^(-2)xx12xx10^(-2)=9.0mV`
P is the positive end and Q the negative end. [The conductor moves towards this end)
b. Yes. On closing K, current flows but the excess charge is maintained as long as the change continues.
c. Magnetic force on the charge is cancelled by the electric force set up due to the excess charge of opposite signs at the ends of the rod.
d. `F=BIl=75xx10^(-3)N`
e. Power = `E.v=9xx10^(-3)W`. No power is expended when K is open (no current)
f. `I^(2)R=9.0xx10^(-3)W`. Source of power is supplied by the external agent.
g. Zero. No change of flux.