A straight wire PQ of length 50 cm and resistance 0.8 $\Omega$  slides on parallel metal rails CD and EF with a velocity of 4 cms^-1 in a uniform magnetic field of 2 T directed into the page as shown in Fig. Two resistances 3 $\Omega$  and 2 $\Omega$  are connected as shown in the figure. The external force to be applied to PQ to keep it moving at a constant velocity of 4 cms^-1 is

Motional emf induced between the ends of PQ is:

$\mathrm{e}=\mathrm{Blv}=2 \mathrm{x} 0.5 \mathrm{x} (4 \mathrm{x} {10}^{-2}) = 4 \mathrm{x} {10}^{-2} \mathrm{V}$

From Fleming's right hand rule, the direction of the induced current in PQ is from Q to P. The circuit can be assumed to have a cell of emf e = 4 x 10^-2 V and internal resistance r = 0.8$\Omega$ connected as shown in Fig. (a).

Resistances 3$\Omega$ and 2$\Omega$ are in parallel. Their combined resistance is

$R=\frac{3\times 2}{3+2}=1.2 \mathrm{\Omega }$

The circuit can be redrawn as shown in Fig. (b). Current in the circuit is

$\mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}+\mathrm{r}}=\frac{4\times {10}^{-2}}{1.2\times 0.8}=2\times {10}^{-2} \mathrm{A}$

Therefore, force acting on wire PQ is:

$\mathrm{F}=\mathrm{BIL}=2\times \left(2\times {10}^{-2}\right)\times 0.5=2\times {10}^{-2}\mathrm{N}$