$A$ 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 $g = 10 \,m/s^2$):

$A$ metallic conductor of length $1 \; m$ rotates in a vertical plane parallel to the east-west direction about one of its ends with an angular velocity of $5 \; rad/s$. If the horizontal component of the Earth's magnetic field is $0.2 \times 10^{-4} \; T$,then the emf induced between the two ends of the conductor is .............

$A$ conducting wire is dropped along the east-west direction. Then,

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

$A$ 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 $\frac{1}{2} m v_0^2$?

$A$ long metal bar of $30\,cm$ length is aligned along a north-south line and moves eastward at a speed of $10\,ms^{-1}$. $A$ uniform magnetic field of $4.0\,T$ points vertically downwards. If the south end of the bar has a potential of $0\,V$,the induced potential at the north end of the bar is.....$V$.