A flat insulating disc of radius 'a' carries an excess charge on its surface is of surface charge density `sigma C//m^(2)`. Consider disc to rotate around the axis passing through its centre and perpendicular to its plane with angular speed `omega` rad/s. If magnetic field `vecB` is directed perpendicular to the rotation axis, then find the torque acting on the disc.

Explanation

Suppose the disc is placed in xy-plane and is rotated about the z-axis. Consider an annular ring of radius r and of thickness dr, the charge on this ring
`dq=sigma(2pirdr)`
As the ring rotates with angular velocity `omega`, so the current
`i=(dq)/(dt)=(sigma(2pidr))/((2pi)/(omega))=sigmaomegardr`
The torque on the current loop
`vectau=ivecAxxvecB`.

Hence the torque on this annular ring `dvectau=i(dvecAxxvecB)`
`=sigmaomegardr(pir^(2)Bsin90^(@))=pisigmaomegar^(3)Bdr`
and `tau=pisigmaomegaBint_(0)^(a)r^(3dr)=(pisigmaomegaBa^(4))/4`