$A$ metallic rod of $1\; m$ length is rotated with a frequency of $50\; rev/s$,with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $1\; m$,about an axis passing through the centre and perpendicular to the plane of the ring (Figure). $A$ constant and uniform magnetic field of $1\; T$ parallel to the axis is present everywhere. What is the $emf$ between the centre and the metallic ring?

(157 V) Method $I$
As the rod is rotated,free electrons in the rod move towards the outer end due to the Lorentz force and get distributed over the ring. Thus,the resulting separation of charges produces an $emf$ across the ends of the rod. At a certain value of $emf$,there is no more flow of electrons and a steady state is reached. The magnitude of the $emf$ generated across a length $dr$ of the rod as it moves at right angles to the magnetic field is given by $d\varepsilon = B v dr$.
Hence,$\varepsilon = \int d\varepsilon = \int_{0}^{R} B v dr = \int_{0}^{R} B \omega r dr = \frac{B \omega R^{2}}{2}$.
Using $v = \omega r$,we get $\varepsilon = \frac{1}{2} \times 1.0 \times (2 \pi \times 50) \times (1^{2}) = 157\; V$.
Method $II$
To calculate the $emf$,we can imagine a closed loop $OPQ$ in which point $O$ and $P$ are connected with a resistor and $OQ$ is the rotating rod. The potential difference across the resistor is then equal to the induced $emf$ and equals $B \times$ (rate of change of area of loop). If $\theta$ is the angle between the rod and the radius of the circle at $P$ at time $t$,the area of the sector $OPQ$ is given by $\pi R^{2} \times \frac{\theta}{2 \pi} = \frac{1}{2} R^{2} \theta$,where $R$ is the radius of the circle. Hence,the induced $emf$ is $\varepsilon = B \times \frac{d}{dt} [\frac{1}{2} R^{2} \theta] = \frac{1}{2} B R^{2} \frac{d\theta}{dt} = \frac{B \omega R^{2}}{2} = 157\; V$.