$A$ conducting rod moves towards the right with a constant velocity $v$ in a uniform transverse magnetic field. Determine the nature of the graphs between the force applied by the external agent versus velocity and the power supplied by the external agent versus velocity.

$A$ square loop $PQRS$ having $10$ turns, area $3.6 \times 10^{-3} \, m^2$ and resistance $100 \, \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $B=0.5 \, T$ as shown. Work done in pulling the loop out of the field in $1.0 \, s$ is . . . . . $\times 10^{-6} \, J$.

$A$ thin wire of length $2 \ m$ is perpendicular to the $xy$-plane. It moves with a velocity $v = (2\hat{i} + 3\hat{j} + \hat{k}) \ m/s$ in a magnetic field $B = (\hat{i} + 2\hat{j}) \ Wb/m^2$. What is the induced potential difference (emf) across the ends of the wire?

$A$ current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \, T$. If the radius of the loop starts shrinking at a uniform rate of $2 \, mm/s$, then the emf induced in the loop at the instant, when its radius is $20 \, cm$ will be (in $\pi \, \mu V$)

$A$ wire of length $1 \, m$ is moving at a speed of $2 \, ms^{-1}$ perpendicular to its length and a homogeneous magnetic field of $0.5 \, T$. The ends of the wire are joined to a circuit of resistance $6 \, \Omega$. The rate at which work is being done to keep the wire moving at constant speed is:

Suppose a long rectangular loop of width $w$ is moving along the $x$-direction with its left arm in a magnetic field perpendicular to the plane of the loop (see figure). The resistance of the loop is zero and it has an inductance $L$. At time $t=0$,its left arm passes the origin,$O$. If for $t \geq 0$,the current in the loop is $I$ and the distance of its left arm from the origin is $x$,then $I$ versus $x$ graph will be