$A$ wheel with radial metal spokes $1 \ m$ in length is rotated in a magnetic field of $0.5 \times 10^{-4} \ T$ normal to the plane of the wheel. If the induced emf between the rim and axle is $\pi / 3000 \ V$,then the rotational speed of the wheel in revolutions per minute is

$A$ rod of length $80 \ cm$ rotates about its midpoint with a frequency of $10 \ rev/s$. The potential difference (in volts) between the two ends of the rod due to a magnetic field $B = 0.5 \ T$ directed perpendicular to the rod is:

Consider the situation shown in the figure. The wire $AB$ is sliding on the fixed rails with a constant velocity. If the wire $AB$ is replaced by a semicircular wire,the magnitude of the induced current will

$A$ bicycle wheel of radius $R$ has $n$ spokes. It is rotating at the rate of $F$ r.p.m. perpendicular to the horizontal component of earth's magnetic field $\vec{B}$. The e.m.f. induced between the rim and the centre of the wheel is

Two conducting rings $P$ and $Q$ of radii $r$ and $2r$ rotate uniformly in opposite directions with centre of mass velocities $2v$ and $v$ respectively on a conducting surface $S$. There is a uniform magnetic field of magnitude $B$ perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

As shown in the figure,a metal rod makes contact with a partial circuit and completes the circuit. The circuit area is perpendicular to a magnetic field with $B = 0.15\, T$. If the resistance of the total circuit is $3\,\Omega$,the force needed to move the rod as indicated with a constant speed of $2\, ms^{-1}$ will be equal to: