A magnetic field B is confined to a region `r le` a and points out of the paper (the z-axis), r = 0 being the centre of the cicular region. A charged ring (charge = Q) of radius b,bgt a and mass m lie in the x-y plane with its centre at origin. The ring is free to rotate and is at rest. The magnetic field is brougth to zero in time `Delta t`. Find the angular velocity `omega or the ring after the field vanishes.

Explanation

Since, the magnetic field is brought to zero in time `Deltat`, the magnetic flux linked with the ring also reduces from maximum to zero. Thus, in turn, induces an emf in ring by the phenomenon of EMI. The induces emf causes the electric field E generation around the ring.
The induced emf = electric field E x (`2pib)` (Because V=`E xx d)`............(i)
By Faraday's law of EMI
The induced emf = rate of change of magnetic flux = rate of change of magnetic field x area
`=(Bpia^(2))/(Deltat)`.............(ii)
From eqs. (i) and (ii), we have
`2pibE="emf"=(Bpia^2)/(Deltat)`
Since, the charged ring experienced a electric force = QE
This force try to rotate the coil, and the torque is given by
Torque = b x Force
`=QEb= Q[(Bpia^(2))/(2pibDeltat)]b`
`= Q(Ba^(2))/(2Deltat)`
If `DeltaL` is the change in angualr momentum
`DeltaL = "Torque" xx Deltat= Q(Ba^(2))/(2)`
Since, initial angular momentum `= 0`
Now,since Torque x `Deltat`= Change in angular momentum
Final angular momentum `=mb^(2)omega= (QBa^(2))/(2)`
`omega= (QBa^(2))/(2mb^(2))`
On rearranging the terms, we have the required expression of angular speed.