$A$ rod closing the circuit shown in the figure moves along a $U$-shaped wire at a constant speed $v$ under the action of a force $F$. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate $F$ if the rate of heat generation in the circuit is $Q$.

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

$A$ 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?

$A$ rod of length $80 \ cm$ rotates about its midpoint with a frequency of $10 \ rev/s$. The potential difference (in volts) between the two ends of the rod due to a magnetic field $B = 0.5 \ T$ directed perpendicular to the rod is:

$A$ 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).

Consider the situation shown in the figure. The wire $PQ$ has a negligible resistance and is made to slide on the three rails with a constant speed of $5 \, cm/s$. Find the current in the $10 \, \Omega$ resistor when the switch $S$ is connected to the middle rail. (Given: $B = 1.0 \, T$) (in $mA$)