$A$ 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

$A$ conductor is moving in a magnetic field $B$ and the induced current is $I$. If the magnetic field is doubled,the induced current will

$A$ wire of length $1 \, m$ is moving at a speed of $2 \, m/s$ perpendicular to a homogeneous magnetic field of $0.5 \, T$. The ends of the wire are joined to a resistance of $6 \, \Omega$. The rate at which work is being done to keep the wire moving at that speed is:

$A$ conducting rod of length $\ell$ is moving with velocity $v$ in a uniform magnetic field $B$ directed into the plane of the paper,as shown in the diagram. Which end of the rod is at a lower potential?

$A$ square loop of area $25\,cm^2$ has a resistance of $10\,\Omega$. The loop is placed in a uniform magnetic field of magnitude $40.0\,T$. 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 $1.0\,s$ will be $..........\times 10^{-3}\,J$.

$A$ 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: