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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. |
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Answer» 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)` |
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