$A$ 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:

$A$ $1.0 \; m$ long metallic rod is rotated with an angular frequency of $400 \; rad \; s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. $A$ constant and uniform magnetic field of $0.5 \; T$ parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (in $; V$)

$A$ $20 \ m$ long uniform copper wire held horizontally is allowed to fall under gravity $(g = 10 \ m/s^2)$ through a uniform horizontal magnetic field of $0.5 \ Gauss$ perpendicular to the length of the wire. The induced $EMF$ across the wire after it travels a vertical distance of $200 \ m$ is . . . . . . $mV$.

$A$ conducting rod,with a resistor of resistance $R$,is pulled with constant speed $v$ on a smooth conducting rail as shown in the figure. $A$ constant magnetic field $B$ is directed into the page. If the speed of the bar is doubled,by what factor does the rate of heat dissipation across the resistance $R$ change?

$A$ rectangular loop of wire is placed in the $XY$-plane with its side of length $3 \,cm$ parallel to the $X$-axis and the side of length $4 \,cm$ parallel to the $Y$-axis. It is moving in the positive $X$-direction with the speed $10 \,cm/s$. $A$ magnetic field exists in the space with its direction parallel to the $Z$-axis. The field decreases by $2 \times 10^{-3} \,T/cm$ along the positive $X$-axis and increases in time by $2 \times 10^{-2} \,T/s$. The induced emf in the wire is

$A$ horizontal telegraph wire of length $30 \ m$ spread east to west falls freely from a height of $20 \ m$. If the resistance of the wire is $40 \ \Omega$ and the horizontal component of the earth's magnetic field at the place is $2 \times 10^{-5} \ T$,then the induced current when the wire reaches the ground is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)