$A$ copper rod of length $l$ is rotated about one end perpendicular to the magnetic field $B$ with constant angular velocity $\omega$. The induced e.m.f. between the two ends is

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 $mA$)

The figure shows a straight wire placed between the pole pieces of a magnet. An induced emf will be developed across the ends of the wire when it is moved towards:

$A$ rod of length $l$ is rotated with angular velocity $\omega$ about one of its ends,in a region perpendicular to a uniform magnetic field of induction $B$. The induced e.m.f. in the rod is:

Consider a metal ball of radius $r$ moving at a constant velocity $v$ in a uniform magnetic field of induction $\vec{B}$. Assuming that the direction of velocity forms an angle $\alpha$ with the direction of $\vec{B}$,the maximum potential difference between points on the ball is

$A$ square conducting frame of side $a$ and a long straight wire carrying current $I$ are placed as shown in the figure. If the frame is moved to the right with a constant velocity $V$,then the induced $emf$ in the frame is proportional to: