Question

A supercondutig round ring of radius `a` and inductance `L` was located in a unifrom magnetic fied of induction `B`. The ring plane was parallel to the vector `B`, and the current in the ring was equal to zero. Them the ring was turned through `90^(@)` so that its plane became perpendicular to the feild. FInd:
(a) the current induced in the ring after the turn,
(b) the work perfromed during the turn.

Answer 1

In a superconductor there is no resistance, Hence,
`L (dI)/(dt) + (d Phi)/(dt)`
So intergating `I = (Delta Phi)/(L) = (pi a^(2) B)/(L)`
because `Delta Phi = Phi_(f) - Phi_(i), Phi_(f) = pi a^(2) B, Phi_(i) = 0`
Also, the work done is `A = int xi I dt = int (d Phi)/(dt) = (1)/(2) L I^(2) = (1)/(2) (pi^(2) a^(4) B^(2))/(L)`