Question

A metallic rod of length L is rotated with angular frequency of co 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.​

Answer 1

Answer:

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