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

Derive the equation for the induced $emf$ in a rod of length $l$ sliding with velocity $v$ on a $U$-shaped frame placed perpendicular to a uniform magnetic field $B$.

$A$ conducting circular loop is placed in a uniform magnetic field of $0.04\, T$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at a rate of $2\, mm/s$. The induced $emf$ in the loop when the radius is $2\, cm$ is:

$A$ $1\,m$ 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 $................\,10^{-3}\,N$.

$A$ conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $R_{2}$ respectively is:

$A$ metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction $B$ acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is: