Two parallel wires PQ and ST placed a distance w apart are connected by a resistor R as show in figure and placed in a magnetic field B which is perpendicular to the plane containing the wires. A rod CD connects the two wires. The power spent to slide the rod CD with a velocity v along the wires is (neglect the resistance of the wires and the rod)

When wire CD is made to slide on wires PQ and ST, the flux linked with the circuit changes with time and hence an emf is induced in the circuit, which is given by:

$\left|\mathrm{e}\right|=\frac{\mathrm{d\varphi }}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{BA}\right)=\mathrm{B}\frac{\mathrm{dA}}{\mathrm{dt}}$

If wire CD moves a distance dx is time dt, then dA = wdx (here w = CD) and 

$\left|\mathrm{e}\right|=\mathrm{B}\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{wdx}\right)=\mathrm{Bw}\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{Bwv}$

The induced current is: $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{\mathrm{Bwv}}{\mathrm{R}}$

This current is caused by the motion of wire CD. From Lenz's law, the current I opposes the motion of wire CD. Therefore, work has to be done to slide the wire CD. Now, the magnetic force on wire CD (of length w) is

 $\mathrm{F}=\mathrm{BIw}=\mathrm{B}\left(\frac{\mathrm{Bwv}}{\mathrm{R}}\right)\mathrm{w}=\frac{{\mathrm{B}}^2{\mathrm{w}}^2\mathrm{v}}{\mathrm{R}}$   ……..(1)

Work done is sliding wire CD through a small distance dx in time dt is

  $\mathrm{dW}=\mathrm{Fdx}$

Therefore, the work done per second is

$\mathrm{P}=\frac{\mathrm{dW}}{\mathrm{dt}}=\mathrm{F}\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{Fv}$

Using (1), we get

$\mathrm{P}=\frac{{\mathrm{B}}^2{\mathrm{w}}^2{\mathrm{v}}^2}{\mathrm{R}}$