Question

A square metallic wire loop of side $0.1m$ and resistance of $1\Omega$ is moved with a constant velocity in a magnetic field of $2Wb$-$m^{-2}$ as shown in the figure. The magnetic field is perpendicular to the plane of the loop which is connected to a network of resistances. If the steady state current in the loop is $1mA$, then the velocity of loop is (in $cm/sec$)

Answer 1

Equivalent resistance of the given Wheatstone bridge circuit (balanced) is $3\Omega$
So total resistance in circuit is $R=3+1=4\Omega .$ The emf induced in the loop is $e=Bvl$.
So induced current $i=\frac{e}{R}=\frac{Bvl}{R}$
$\Rightarrow {10}^{-3}=\frac{2\times v\times \left(10\times {10}^{-2}\right)}{4}\Rightarrow v=2cm/sec$