Question

A rectangular loop with a sliding connector of length l = 10 m is situated in a uniform magnetic field B = 2T perpendicular to the plane of loop. Resistance of connector is $r=2\Omega$. Two resistance of $6\Omega$ and $3\Omega$ are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v= 2 m/s is

Answer 1

Motional emf e = Bvl
e = (2) (2) (1) = 4 V
This sets as a cell of emf E = 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 $F_m=ilB$
=(1)(1)(2)=2N (towards left)
Therefore, to keep the connector moving with a constant velocity a force of 2 N will have to be applied towards right.