As shown in the figure,a metal rod makes contact and completes the circuit. The circuit is perpendicular to the magnetic field with $B = 0.15 \, T$. If the resistance is $3 \, \Omega$,the force needed to move the rod as indicated with a constant speed of $2 \, m/s$ is:

The radius of a circular loop placed in a perpendicular uniform magnetic field is increasing at a constant rate of $r_0 \ m s^{-1}$. If at any instant the radius of the loop is $r$,then the emf induced in the loop at that instant will be:

One conducting $U$ tube can slide inside another as shown in the figure,maintaining electrical contacts between the tubes. The magnetic field $B$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $v$,then the emf induced in the circuit in terms of $B, l$ and $v$,where $l$ is the width of each tube,will be

$A$ conductor wire $ABCDE$ with each arm $10 \ cm$ in length is placed in a magnetic field of $\frac{1}{\sqrt{2}} \ T$,perpendicular to its plane. When the conductor is pulled towards the right with a constant velocity of $10 \ cm/s$,the induced emf between points $A$ and $E$ is . . . . . . $mV$.

$A$ copper disc of radius $0.1 \ m$ is rotated about its centre with $10$ revolutions per second in a uniform magnetic field of $0.1 \ T$ with its plane perpendicular to the field. The e.m.f. induced across the radius of the disc is:

$A$ metallic rod of $1\; m$ length is rotated with a frequency of $50\; rev/s$,with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $1\; m$,about an axis passing through the centre and perpendicular to the plane of the ring (Figure). $A$ constant and uniform magnetic field of $1\; T$ parallel to the axis is present everywhere. What is the $emf$ between the centre and the metallic ring?