A thin non conducting horizontal disc of mass `m` having total charge `q` distributed uniformly over its surface, can rotate freely about its own axis. Initially when the disc is stationery a magnetic field `B` directed perpendicular to the plane is switched on at `t=0`. Find the angular velocity `omega` acquired by disc as a function of time, if `B=kt`, where `t` is time.

Explanation

Consider a ring of radius `r`, width `dr` and charge on ring
`dq=(2pirdr)-(Q)/(piR^(2))`
Induced emf in ring
`epsilon=ointvecE.dvecl= -(-d)/(dt)intvecB.dvecS`
Let `E` is electric field in circular loop.
Then `E.2pi= -pir^(2)(dB)/(dt)= -pir^(2)k`
`:. E= -(r.k)/(2)`
Force on the ring `=dF=dqE= -(dq)(rk)/(2)`
Torque `tau=int_(0)^(R )rdF=int_(0)^(R ) r((2piQdr)/(piT^(2)))(rk)/(2)=int_(0)^(R )(kQr^(3))/(R^(2))dr=(-kQR^(2))/(4)=Ialpha`
`:. alpha=-(kQR^(2))/(4(mR^(2)//2))=(-kQ)/(2m)`
`:. omega=-(kQ)/(2m)t`
`:. |omega|=(kQ)/(2m)t`