For a superconducting loop, the electrical resistance is zero ($R=0$). According to Faraday’s law, the induced EMF is $\mathcal{E} = – \frac{d\Phi}{dt}$. Since $R=0$, the voltage drop $IR$ is zero, implying the net EMF around the loop is zero.

Therefore, $\frac{d\Phi_{total}}{dt} = 0$, meaning the total magnetic flux through the loop must remain constant.

Initial State:
The loop carries a current $I_i$. The magnetic moment is given by $m = I_i A$. Therefore, the initial current is $I_i = \frac{m}{A}$.
The external field is zero. The flux is solely due to self-inductance:
$$ \Phi_i = L I_i = L \left(\frac{m}{A}\right) $$

Final State:
An external magnetic field $B$ is switched on. The current changes to $I_f$.
The total flux is now the sum of the self-flux and the external flux:
$$ \Phi_f = L I_f + BA $$

Equating initial and final flux ($\Phi_i = \Phi_f$):

$$ L \left(\frac{m}{A}\right) = L I_f + BA $$

$$ L I_f = \frac{Lm}{A} – BA $$

$$ I_f = \frac{m}{A} – \frac{BA}{L} $$