When a $J$-shaped conducting rod is rotating in its own plane with constant angular velocity $\omega$,about one of its ends $P$,in a uniform magnetic field $\vec B$ directed normally into the plane of the paper,then the magnitude of the emf induced across it will be:

There is a uniform magnetic field $B$ normal to the $xy$ plane. $A$ conductor $ABC$ has length $AB = l_1$,parallel to the $x$-axis,and length $BC = l_2$,parallel to the $y$-axis. $ABC$ moves in the $xy$ plane with velocity $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The potential difference between $A$ and $C$ is proportional to

Two long parallel horizontal rails,a distance $l$ apart,each having a resistance $\lambda$ per unit length,are joined at one end by a resistance $R$. $A$ perfectly conducting rod $MN$ of mass $m$ is free to slide along the rails without friction. There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. $A$ variable force $F$ is applied to the rod $MN$ such that,as the rod moves,a constant current $i$ flows through the circuit. The applied force $F$ as a function of distance $x$ of the rod from $R$ is:

$A$ metal rod of length $1\,m$ is rotated about one of its ends in a plane at right angles to a magnetic field of induction $2.5 \times 10^{-3}\,Wb/m^2$. If it makes $1800\,rpm$,calculate the induced e.m.f. between its ends in $V$.

$A$ copper disc of radius $0.1 \, m$ is rotated about its centre with $10 \, rev/s$ in a uniform magnetic field of $0.1 \, T$ with its plane perpendicular to the field. The emf induced across the radius of the disc is ........... $V$.

$A$ 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