$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 2\, m/s$ 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

$A$ rod closing the circuit shown in the figure moves along a $U$-shaped wire at a constant speed $v$ under the action of a force $F$. The circuit is in a uniform magnetic field perpendicular to the plane. Calculate $F$ if the rate of heat generation in the circuit is $Q$.

$A$ straight line conductor of length $0.4 \ m$ is moved with a speed of $7.0 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.8 \ Wb \ m^{-2}$. The induced e.m.f. across the conductor is (in $V$)

$A$ rod of length $20 \text{ cm}$ is rotating with an angular speed of $100 \text{ rps}$ in a magnetic field of strength $0.5 \text{ T}$ about one of its ends. What is the potential difference between the two ends of the rod? $(V)$

$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

$A$ circular coil of radius $30 \ cm$,having $1$ turn and a resistance of ${\pi ^2} \ \Omega$,is placed in a magnetic field of $10^{-2} \ T$. The coil rotates about its diameter with a speed of $200 \ rpm$ in a direction perpendicular to the magnetic field. The value of the induced $AC$ current in the coil will be . . . . . . $mA$.