Question

A metallic rod of length 'I' 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 r, 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 emf is induced between the centre and the metallic ring and hence obtained the expression for it.

Answer 1

Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction .
Suppose the direction of the magnetic field is along +z direction.
Then, using Lorentz law, we get the following

$\overrightarrow{F}=-e(\overrightarrow{v}\times \overrightarrow{B})$

$F=-e(v\hat{j}\times B\hat{k})$
F¯=−e(v¯×B¯)$\overrightarrow{F}=-evB\hat{i}$
Thus, the direction of force on the electrons is along -x axis.
So, the electrons will move towards the centre i.e the fixed end of the rod. This movement of electron will effect in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring.
Let $\theta$ be the angle between the rod inside the circle $=\frac{1}{2}\pi r^2\theta$=12πr2θ
Induced emf

$=B\times \frac{d}{dt}\left(\frac{1}{2}\pi r^2\theta \right)=\frac{1}{2}\pi r^{2}B\frac{d\theta }{dt}=\frac{1}{2}\pi r^{2}B\omega$