$A$ square loop of side $15 \ cm$ is being moved towards the right at a constant speed of $2 \ cm/s$ as shown in the figure. The front edge enters the $50 \ cm$ wide magnetic field region at $t=0$. The value of the induced emf in the loop at $t=10 \ s$ will be:

$A$ $10 \, m$ wire kept in the east-west direction is falling with a velocity of $5 \, m/s$ perpendicular to the magnetic field of $0.3 \times 10^{-4} \, Wb/m^2$. The induced e.m.f. across the terminals of the wire will be:

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then:

In the figure shown, the conductor $PQ$ of length $l$ is moved from $x = 0$ to $x = b$ and then up to $x = 2b$ with a constant velocity $\vec{v}$. $A$ uniform magnetic field $\vec{B}$ is perpendicular to the plane of the paper and extends from $x = 0$ to $x = b$, and it is zero for $x > b$. The magnitude of the emf induced in the conductor is:

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 $B$.

The horizontal component of the earth's magnetic field at a place is $3 \times 10^{-4} \, T$ and the angle of dip is $53^\circ$. $A$ metal rod of length $0.25 \, m$ is placed in the north-south position and is moved at a constant speed of $10 \, cm/s$ towards the east. The emf induced in the rod will be ...... $\mu V$. (Given $\tan 53^\circ = 4/3$)