$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ square loop of area $25 \ cm^2$ has a resistance of $10 \ \Omega$. The loop is placed in a uniform magnetic field of magnitude $40 \ T$. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1 \ s$ will be:

$A$ rod of length $80 \ cm$ rotates about its midpoint with a frequency of $10 \ rev/s$. The potential difference (in volts) between the two ends of the rod due to a magnetic field $B = 0.5 \ T$ directed perpendicular to the rod is:

$A$ square loop of side $20\, {cm}$ and resistance $1\, \Omega$ is moved towards the right with a constant speed ${v}_{0}$. The right arm of the loop is in a uniform magnetic field of $5\, {T}$. The field is perpendicular to the plane of the loop and is directed into it. The loop is connected to a network of resistors,each of value $4\, \Omega$. What should be the value of ${v}_{0}$ so that a steady current of $2\, {mA}$ flows in the loop?

$A$ square loop of side $a$ and resistance $R$ is moved in a region of uniform magnetic field $B$ (the loop remaining completely inside the field) with a velocity $v$ through a distance $x$. The work done is:

$A$ metallic disc of radius $0.3 \ m$ is rotating with a constant angular speed of $60 \ rad \ s^{-1}$ in a plane perpendicular to a uniform magnetic field of $5 \times 10^{-2} \ T$. The emf induced between a point on the rim and the centre of the disc is: (in $V$)