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 circualr metallic ring of a 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 every where. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it.

Explanation

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 `vecF=-e(upsilonhatjxxBhatk) rArr vecF=-eupsilonBhati`
Thus, the direction of force on the electrons is along `x`-axis.
Thus, the electrons will move towards the center i.e., the fixed end of the rod.This movement of electron will result in current and hence it will produce `emf` in the rod between the fixed end and the point touching the ring Let `theta` be the angle between the rod and radius of the circle at any time `t`.
Then, area swept by the rod inside the circle
=`1/2pir^(2)theta`
Induced `emf`
=`Bxxd/(dt)(1/2pir^(2)theta)=1/2pir^(2)B(d theta)/(dt)=1/2pir^(2)omega =1/2pir^(2)B(2piupsilon)=pi^(2)r^(2)Bupsilon`.
Note:There will be an induced `emf` between the two ends of the rod also.