Question

A metallic rod of ${}^{\prime }{L}^{\prime }$ length is rotated with angular frequency of ${}^{\prime }{\omega }^{\prime }$ 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

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:

$\epsilon =Bvl$

The potential difference across a small element of rod at a distance $l$ from the center, $d\epsilon =Bv\left(dl\right)$

But $v=\omega l⟹d\epsilon =Bl\omega dl$

Hence total emf produced across the rod,

$\epsilon ={\int }_0^{L}B\left(l\omega \right)dl$

$=\frac{1}{2}B\omega L^2$