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 `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(2pirdr))/((2pi)/(omega))=sigmaomegardr`

Torque on the current loop `vec(tau)=vec(iA)xxvec(B)`
Hence the torque on this annular ring
`dvec(tau)=i(dvec(A)xxvec(B))=sigma omegardr(pir^(2)B sin 90^(@))`
`=pi sigmaomegar^(3)dr` and `tau=pi sigmaomegaBunderset(0)overset(a)int r^(3)dr=(pisigmaomegarBa^(4))/(4)`