$A$ conducting rod $PQ$ of length $l = 5 \ m$ is oriented as shown in the figure. It is moving with a velocity $\vec{V} = (2 \ m/s) \hat{i}$ without any rotation in a uniform magnetic field $\vec{B} = (3 \hat{j} + 4 \hat{k}) \ T$. The induced $Emf$ in the rod is...........$V$.

$A$ 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 $Q$.

$A$ wire loop is rotated in a magnetic field. The frequency of change of direction of the induced $e.m.f.$ is

An aeroplane,with its wings spread $10 \, m$,is flying at a speed of $180 \, km/h$ in a horizontal direction. The total intensity of the Earth's magnetic field at that location is $2.5 \times 10^{-4} \, Wb/m^2$ and the angle of dip is $60^{\circ}$. The $EMF$ induced between the tips of the plane's wings will be ...... $mV$.

$A$ conducting rod $AB$ moves parallel to the $X-$ axis in a uniform magnetic field,which is directed perpendicular to the plane of motion (outwards). The end $A$ of the rod gets:

$A$ simple pendulum made of a mass of $10 \ g$ and a metallic wire of length $10 \ cm$ is suspended vertically in a uniform magnetic field of $2 \ T$. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with the vertical, then the maximum induced $EMF$ between the point of suspension and the point of oscillation is . . . . . . $mV$. (Take $g = 10 \ m/s^2$)