A metallic rod of length ‘l’ 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 obtain the expression for it. 

As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.

Expression for Induced emf in a Rotating Rod

Consider a metallic rod OA of length l, which is rotating with angular velocity  in a uniform magnetic field B, the plane of rotation being perpendicular to the magnetic field. A rod may be supposed to be formed of a large number of small elements. Consider a small element of length dx at a distance x from centre. 

If v is the linear velocity of this element, then area swept by the element per second = v.dx 

Emf induced across the rod , 
But, v = 



Therefore,

Emf induced across the rod is given by, 

 

Power dissipated, when circuit is closed is, P = .