A line charge `lamda` per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has a light non-conducting spokes and is free to rotate without friction about its axis. A uniform magnetic field extends over a circular region within the rim.

It is given by `vecB=-B_0 K (r le a , a lt R)`
What is the angular velocity of the wheel alfter the field is suddenly switched off?

Explanation

Let E be the electric field generated at points situated at radius a, so
Induced emf, `e =- (dphi_B)/(dt) = ointE.dI`
or `ointE.dI = (dphi_B)/(dt)`
For constant E

`Eointdl=d/(dt)(pia^2B)`
`implies E(2pia)=pia^2(dB)/(dt)implies E= a/2(dB)/(dt)`
On rotation, charge on outer part will produce same effect on inner part situated at radius a.
So, force acting, F=charge x electric field
`=[(2pia)lamda]xx(a/2(dB)/(dt))`
By Newton.s law, `=(2pia)lamdaxxa/2(dB)/(dt)= M d/(dt)(Romega)`
`implies (2pia^2lamda)/(2) dB = MRdomega`
When switched is OFF, magnetic field changes from `B_o` to zero and angular velocity changes from 0 to `omega`
So, on integrating, `implies " "(2pia^2lamda)/2dB=MRdomega`
`pia^2lamda(-B_0)=MRomega`
`omega = - (pia^2lamdaB_0)/(MR)`