$A$ metallic rod of length $1 \,m$ held along the east-west direction is allowed to fall down freely. Given the horizontal component of the Earth's magnetic field $B_H = 3 \times 10^{-5} \,T$. The emf induced in the rod at an instant $t = 2 \,s$ after it is released is (Take $g = 10 \,m/s^2$):

$A$ conducting rod of mass $m$ and length $l$ is free to move without friction on two parallel long conducting rails,as shown in the figure. There is a resistance $R$ across the rails. In the entire space,there is a uniform magnetic field $B$ normal to the plane of the rod and rails. The rod is given an impulsive velocity $v_0$. What happens to the initial kinetic energy $\frac{1}{2} m v_0^2$?

$A$ $2 \; m$ wire is moving with a velocity of $1 \; m/s$ perpendicular to a magnetic field of $0.5 \; Wb/m^2$. The induced e.m.f. in it will be $... \; V$.

The magnitude of the earth's magnetic field at a place is $B_0$ and the angle of dip is $\delta$. $A$ horizontal conductor of length $l$ lying along the magnetic north-south moves eastwards with a velocity $v$. The emf induced across the conductor is

$A$ conducting circular loop is placed in a uniform magnetic field of $0.4\,T$ with its plane perpendicular to the field. The radius of the loop starts expanding at a constant rate of $1\,mm/s$. The magnitude of the induced emf in the loop at an instant when the radius of the loop is $2\,cm$ will be $...........\,\mu V$.

Two $U$-shaped tubes are moving inside each other as shown in the figure. If the length of each tube is $l$ and they move with velocity $v$ in a magnetic field $B$,what is the induced $emf$ in the circuit?