A rectangular loop with a sliding connector of length $l=1.0m$ is situated in a uniform magnetic field $\text{B}=\text{2T}$   perpendicular to the plane of loop. Resistance of connector is $r=2\Omega$. Two resistances of $6\Omega$ and $3\Omega$ are connected as shown in figure. The external force (in N) required to keep the connector moving with a constant velocity $u=2m/s$ is

Motional emf $e=Bvl$

$e=\left(2\right)\left(2\right)\left(1\right)=4V$

This acts as a cell of emf  $\text{E }=\text{ 4V}$and internal resistance $r=2\Omega$. 

The simple circuit can be drawn as follows

$\therefore$ Current through the connector

 $i=\frac{4}{2+2}=1A$

Magnetic force on connector $F_m=ilB$

$=\left(1\right)\left(1\right)\left(2\right)=2N$ (towards left)

Therefore, to keep the connector moving with a constant velocity a force of 2N will have to be applied towards right.