If current $I$ is flowing in a closed circuit with total resistance $R$,the rate of production of heat energy in the loop as we pull it with a constant speed $V$ is ($L=$ length of conductor,$B=$ magnetic field).

$A$ rectangular wire loop of sides $8 \text{ cm}$ and $3 \text{ cm}$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \text{ T}$ directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is $2 \text{ cm s}^{-1}$ in a direction normal to the shorter side of the loop, will be:

$A$ circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is

$A$ rod of mass $m$ and resistance $R$ slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\theta$ with respect to the horizontal as shown in the figure. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field $\vec{B}$ along the vertical direction. If the rod is initially at rest,find the velocity of the rod as a function of time.

$A$ $1\,m$ long metal rod $XY$ completes the circuit as shown in the figure. The plane of the circuit is perpendicular to the magnetic field of flux density $0.15\,T$. If the resistance of the circuit is $5\,\Omega$,the force needed to move the rod in the direction indicated with a constant speed of $4\,m/s$ will be $................\,10^{-3}\,N$.

$A$ horizontal wire is free to slide on the vertical rails of a conducting frame as shown in the figure. The wire has a mass $m$ and length $l$,and the resistance of the circuit is $R$. If a uniform magnetic field $B$ is directed perpendicular to the frame,find the terminal speed of the wire as it falls under the force of gravity.