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

$A$ thin wire of length $2\,m$ is placed perpendicular to the $x-y$ plane. It is moved with velocity $\overrightarrow v = (2\hat i + 3\hat j + \hat k)\,m/s$ through a region of magnetic induction $\overrightarrow B = (\hat i + 2\hat j)\,Wb/m^2$. The potential difference induced between the ends of the wire is......$V$.

$A$ rectangular coil of single turn,having area $A$,rotates in a uniform magnetic field $B$ with an angular velocity $\omega$ about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the field,then the average induced $e.m.f.$ when it has rotated through $90^{\circ}$ is

$A$ vertical rod of length $l$ is moved with constant velocity $v$ towards the east. If the vertical component of the Earth's magnetic field is $B$ and the angle of dip is $\theta$,then the induced $emf$ in the rod is:

$A$ square frame of metallic wire is moving in a uniform magnetic field $(\vec{B})$ acting perpendicular to the paper inward as shown. $LP$ and $QN$ are also metallic wires. Find the potential difference between $L$ and $N$.

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed $v$ in a uniform magnetic field $B$ directed into the plane of the paper (See figure). If charge densities $\sigma_1$ and $\sigma_2$ are induced on the left and right surfaces,respectively,of the sheet,then (ignore fringe effects):