$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \, \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are:

$A$ conducting rod of length $2l$ is rotating with constant angular speed $\omega$ about its perpendicular bisector. $A$ uniform magnetic field $\vec{B}$ exists parallel to the axis of rotation. The $e.m.f.$ induced between two ends of the rod is

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

$A$ 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:

$A$ 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?

$A$ 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