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 perpendicular to its plane with angular speed `omega rad//s`/ If magnetic field `vec(B)` is directed perpendicular to the roation 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 rand of thickness dr, the charge on this ring.
` dq = sigma (2pi dr)`
As the ring rotates with angular velocity `omega` , so the current `i = (dq)/(dt) = (sigma (2pi r dr))/((2pi)/(omega) ) = sigma omega r dr`

The torque on the current loop `vec tau = i vecA xx vecB`
Hence the torque on this annular ring `d vectau = i (dvecA xx vecB) =sigma omega r dr (pi r^2 B sin 90^@)`
` = pi sigma omega r^3 Bdr " and " tau = pi sigma omega B int^(a) r^3 dr = (pi sigma omega Ba^4)/(4)`