$A$ metallic cube of side $15\,cm$ is moving along the $y$-axis at a uniform velocity of $2\,m/s$ in a region of uniform magnetic field of magnitude $0.5\,T$ directed along the $z$-axis. In equilibrium,the potential difference between the faces of higher and lower potential developed because of the motion through the field will be $..........mV$.

$A$ simple pendulum made of a mass of $10 \ g$ and a metallic wire of length $10 \ cm$ is suspended vertically in a uniform magnetic field of $2 \ T$. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^{\circ}$ with the vertical,then the maximum induced $EMF$ between the point of suspension and the point of oscillation is . . . . . . $mV$. (Take $g = 10 \ m/s^2$)

$A$ conducting rod is moving towards the right with a velocity '$V$' in a uniform magnetic field '$B$'. If the direction of the induced current '$i$' is as shown in the figure,then the direction of '$B$' is:

An electric potential difference will be induced between the ends of the conductor shown in the diagram,when the conductor moves in the direction

$A$ 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? $(V)$

$A$ wire having mass $m$ and length $l$ can freely slide on a pair of parallel smooth horizontal rails placed in a vertical magnetic field $B$. The rails are connected by a capacitor of capacitance $C$. The electric resistance of the rails and the wire is zero. If a constant force $F$ acts on the wire as shown in the figure,then the acceleration of the wire is given by: