A copper disc of radius `1m` is rotated about its natural axis with an angular velocity `2 "rad"//"sec"` in a uniform magnetic field `5` telsa with its plane perpendicular to the field. Find the emf induced between the centre of the disc and its rim.

Explanation

`e=1/2 Bomega r^2 , e=1/2 xx5xx2xx1xx1=5` volt
As current flows in the conductor PQ from Q t o p of the conductor. So, an equal and opposite force F has to be applied on the conductor to move the conductor with a constant velocity v.
Thus `F=F_m=(B^2l^2v)/R`
The rate at which work is done by the applied force to move the rod is
`R_("applied")=Fv=(B^2l^2v^2)/R`
The rate at which energy is dissipated in the circuit is ,
`P_("dissipated")=i^2R=((Bvl)/R)^2R=(B^2l^2v^2)/R`
This is just equal to the rate at which work is done by the applied force.