Prove that mechanical power which needed to move the rod in uniform magnetic field is converted into electrical power.

Explanation

Let r be the resistance of movable arm PQ of the rectangular conductor shown in figure. We assume that the remaining arms QR, RS and SP have negligible resistances compared to r. Thus, the overall resistance of the rectangular loop is r and this does not change as PQ is moved. The current I in the loop is,
`I=epsilon/r=(Bvl)/r`...(1)

On account of the presence of the magnetic field, there will be a force on the arm PQ. This force `vecF = Ivecl xxvecB` is acting in the direction opposite to the velocity of the rod. The magnitude of this force is,
`F=IlB=(B^2l^2v)/r`...(2)
This force is due to drift velocity of charge and Lorentz force.
Power delivered by external force : Now to push the arm PQ with a constant speed u, we need to apply same force in the direction of velocity. So, the power required to do this is,
P=Fv
`=(B^2l^2v^2)/r`...(3)
Electrical power produced : Now electric power produced can be found by,
`P_e`=voltage x current
`therefore P_e= epsilonl=(Bvl)xx((Bvl)/R)=(B^2v^2l^2)/R`...(4)
which is identical to equation (3),
Thus, mechanical energy which was needed to move the arm PQ is converted into electrical energy (the induced emf) and then to thermal energy.