A long solenoid of cross-sectional radius `a` has a thin insulates wiere ring tightly put on its winding, one half of the ring has the resistance `eta` times that of the other half. The magneticv induction produced by the solenoid varies with the time as `B = bt`, where `b` is a constant. Find the magnitude of the electric field strength in the ring.

Explanation

The changing magnetic field will induce an emf in the ring, which is obviously equal, in the two parts by symmetry (the emf induced by electromagnetic induction does not depend on resistance). The current, that will flow due to this, will be different in the two parts. This will cause an acceleration of charge leading to the setting up of an electric firld `E` which has opposite sign in the two parts. Thus,
`(xi)/(2) - pi a E = r I` and , `(xi)/(2) + pi a E = eta r I`,
where `xi` is the total induced e.m.f. Form this,
`xi = (eta + 1) rI`,
and `E = (1)/(2pi a) (eta - 1) rI = (1)/(2pi a) (eta - 1)/(eta + 1) xi`
But by Faraday's law, `xi = pi a^(2) b`
so, `E = (1)/(2) ab (eta - 1)/(eta + 1)`