$A$ closed loop is moving with constant velocity in a uniform magnetic field as shown. The potential difference and current in element $AB$ will be:

When a $J$-shaped conducting rod is rotating in its own plane with constant angular velocity $\omega$,about one of its ends $P$,in a uniform magnetic field $\vec B$ directed normally into the plane of the paper,then the magnitude of the emf induced across it will be:

The direction of current induced in a wire moving in a magnetic field is found using

$A$ metal conductor of length $1 \ m$ rotates vertically about one of its ends at an angular velocity of $5 \ rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4} \ T$,then the e.m.f. developed between the two ends of the conductor is:

$A$ conducting rod of length $l$ is falling with a velocity $v$ perpendicular to a uniform horizontal magnetic field $B$. The potential difference between its two ends will be

$A$ long,rectangular conducting loop of width $l$,mass $m$,and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $v$ so that it may continue to fall freely. The velocity $v$ is ($g=$ acceleration due to gravity).