Question

A metallic rod of length ${}^{\prime }{l}^{\prime }$ is rotated with a frequency $v$ 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 uniform magnetic field $B$ parallel to the axis is present everywhere. Using Lorentz force, explain how e.m.f. is induced between the centre and the metallic ring and hence obtain the expression for it.

Answer 1

With the rotation of the rod,free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring.This separation of charges produces an emf across the ends of the rod.After a certain emf there is no more flow of electrons and a steady state is reached.

Let a metallic rod of length $l$,rotating with angular velocity in an uniform magnetic field $B$,the plane of rotation being perpendicular to he magnetic field. Let an external length,$dx$ at a distance $x$ from the center have a linear velocity $v$.

Area swept by $dx=vdxdt$

$dE=B\frac{dA}{dt}=Bvdx$ But $v=x\omega$

$dE=Bx\omega dx$

$E={\int }_0^{l}Bx\omega dx=B\omega \frac{l^2}{2}$