Question

Figure shows a rectangular loop conducting $PQRS$ in which the arm $PQ$ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm $PQ$ is moved with the velocity $v$ towards the arm $RS$. Assuming that the arms $QR,RS$ and $SP$ have negligible resistance and the moving arm $PQ$ has the resistance $r$, obtain the expression for
The force

Answer 1

The force required to keep the arm $PQ$ in
Constant motion
$F=BIl=B\left(\frac{Blv}{r}\right)l=\frac{B^2{l}^{2}v}{r}$