$A$ metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction $B$ acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is:

$A$ rod of length $20 \text{ cm}$ is rotating with an angular speed of $100 \text{ rps}$ in a magnetic field of strength $0.5 \text{ T}$ about one of its ends. What is the potential difference between the two ends of the rod? $(V)$

$A$ rod of length $1.0 \,m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \,T$ with a frequency of $12 \,rev/s$. The induced emf across the ends of the rod is (in $\,V$)

$A$ cycle wheel of radius $0.5 \; m$ is rotated with a constant angular velocity of $10 \; rad/s$ in a region of magnetic field of $0.1 \; T$ which is perpendicular to the plane of the wheel. The $EMF$ generated between its centre and the rim is.....$V$

$A$ rectangular loop $PQMN$ with a movable arm $PQ$ of length $12 \, cm$ and resistance $2 \, \Omega$ is placed in a uniform magnetic field of $0.1 \, T$ acting perpendicular to the plane of the loop as shown in the figure. The resistance of the arms $MN$, $NP$, and $MQ$ are negligible. The current induced in the loop when arm $PQ$ is moved with a velocity of $20 \, ms^{-1}$ is (in $ \, A$)

$A$ 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?