A metallic rod of length L is rotated with angular frequency ω with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

All points on the rod are moving perpendicular to the magnetic field. Hence, all elementary small elements of the rod induce a small potential difference and the net potential difference in the rod is the integration of the potential differences along the rod.
Motional emf in a conductor moving perpendicular to the field is given by:
ε=Bvl
The potential difference across a small element of rod at a distance l from the center,
dε=Bv(dl)
But v=ωl
⇒dε=Blωdl
Hence total emf produced across the rod,
ε=∫₀ᴸB(lω)dl=1/2BωL²