A uniformly conducting wire is bent to form a ring of mass `'m'` and radius `'r'`, and the ring is placed on a rough horizontal surface with its plane horizontal. There exists a uniform and constant horizontal magnetic field of induction `B`. Now a charge `q` is passed through the ring in a very small time interval `Deltat`. As a result the ring ultimately just becomes vertical. Calculate the value of `g`(acceleration due to gravity) in terms of other given quantities. Assume that friction is sufficient to prevent slipping and ignore any loss in energy.

Explanation

Angular momentum, `((3)/(2)mr^(2))omega=int taudt=int(dq)/(dt)pir^(2)Bdt=qpir^(3)B`
For the ring to become vertical,
`(1)/(2)((3)/(2)mr^(2))omega^(2)=mgr`
i.e. `g=(pi^(2))/(3)((qB)/(m))^(2)r`