Question

A superconducting round ring of radius $$a$$ and inductance $$L$$ was located in a uniform magnetic field of induction $$B$$. The ring plane was parallel to the vector $$B$$, and the current in the ring was equal to zero. Then the ring was turned through $$90^{\circ}$$ so that its plane became perpendicular to the field. Find:
(a) the current induced in the ring after the turn;
(b) the work performed during the turn.

Answer 1

In a superconductor there is no resistance, Hence,
$$L \dfrac {dI}{dt} = + \dfrac {d\Phi}{dt}$$
So integrating, $$I = \dfrac {\triangle \Phi}{L} = \dfrac {\pi a^{2} B}{L}$$
because, $$\triangle \Phi = \Phi_{f} - \Phi_{i}, \Phi_{f} = \pi a^{2}B, \Phi_{i} = 0$$
Also, the work done is, $$A = \int \xi I\ dt = \int I\ dt \dfrac {d\Phi}{dt} = \dfrac {1}{2} L\ I^{2} = \dfrac {1}{2} \dfrac {\pi^{2} a^{4} B^{2}}{L}$$.