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 m`Omega`. 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 ?
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 ?

Explanation

(a)Here `vecB bot vecv bot vecl` and so induced emf is ,
`epsilon=Bvl`
=(0.5)(0.12)(0.15)
`therefore epsilon=9xx10^(-3)` V
Here along with the motion of rod, free electrons acquire velocity `vecv` towards right. Magnetic field `vecB` is vertically downward. Now, magnetic force exerted on the free electrons in the rod is `vecF_m=e(vecBxxvecv)` which is in the direction from P to Q. Hence end P of rod will become positive and end Q of rod will become negative because electrons displace from P to Q.
(b) When K is open, induced charge gets accumulated more and more across the two ends P and Q of moving rod, till equilibrium is reached. Here, there is no induced current.
When K is closed, excess charge at Pand Q is steadily maintained by the continuous flow of induced current.
(c) After the induced emf reaches its maximum value `epsilon` = Bvl, electric force `F_e = eE` (from Q to P) and magnetic force `F_m` = Bev (from P to Q) become equal in magnitude and opposite in direction. Hence, there is no net force on the electrons in the rod. Here `F_m` =Bev is constant in magnitude but `F_e = eE` goes on increasing because of increase in E (Because of increase in p.d. across the rod). Finally when `F_e = F_m`, net force on the electron becomes zero.
(d) Here retarding force (also known as Lenz.s force) is given by Ampere.s law,
`vecF=I(veclxxvecB)`
Direction of above force is opposite to `vecv` (velocity of rod)
`therefore F=IlBsin90^@ " " (because vecl bot vecB)`
`=(epsilon/R)lB`
`=(9xx10^(-3))/(9xx10^(-3))xx0.15xx0.5`
`therefore` F=0.075 N
(e) For making velocity of rod constant mechanical power required is equal to electrical power (assuming no resistance provided by the rails on the rod)
`therefore P_m=P_e`
`=epsilon^2/R`
`=(9xx10^(-3))^2/(9xx10^(-3))`
`=9xx10^(-3)` W
If K is open and if rails offer no resistance then no power is required to keep the rod moving with constant velocity.
(f) In the absence of friction, power dissipated in the form of heat will be equal to electrical power produced (= mechanical power spent). Hence, it will be
`P=I^2R=(epsilon/R)^2 R=epsilon^2/R=9xx10^(-3)` W
Source of above power is the external agent which does mechanical work on the rod.
(g)If `vecB || vecv` , no emf gets induced across the rod (because of no magnetic force acting on the electrons in the rod).