A conducting ring of radius r is placed perpendicular inside a time varying magnetic field as shown in figure. The magnetic field changes with time according to `B = B_0 + alphat` where `B_0` and `alpha` are positive constants. Find the electric field on the circumference of the ring.

Explanation

The magnetic field linked with the ring at time t,
`B=B_0 + alpha t`
`phi=B(pir^2)=(B_0 + alphat)pir^2` …(1)
From Faraday.s law, emf produced in a ring.
`epsilon =-(dphi)/(dt)`
`=-d/(dt)[(B_0 + alphat)pir^2]`
`therefore epsilon =-alpha pir^2`...(2)
Now by definition, emf is the work done by the electric field for one complete revolution of a unit positive charge on the circumference of the ring. If `vecE` is the electric field intensity the work done is,
`=int vecE. dvecl` since `vecE` and `d vecl` are in the same direction.
`int vecE . dvecl =E int dl =E(2pir)`...(3)
Comparing equations (2) and (3) ,
`E(2pir) = alphapir^2` (neglecting negative sign)
`therefore E=(alphar)/2`