Separating two superconducting rings

Since the ring is superconducting, the emf in the ring equal to IR is always zero and unchanged. Thus the total flux through it is a constant. When the two currents are very close to each other, the flux through one ring can be considered to be caused by one ring conducting the two currents ,i.e $2I_{o}$ . When one ring is taken apart very far, the magnetic flux is caused only by its own current. To make the flux unchanged, the current now is $2_I{o} = 0.1A$

Since the rings are superconducting, the induced emf is zero and therefore the flux through each ring must be constant. That is, $\mathcal{E}= I R= 0$ implies $\mathcal{E}=-\frac{d \Phi}{dt}=0$ (Faraday's Law). Now, the flux through a ring must be proportional to its own current. We have that
$\Phi= L I$ where the proportionality coefficient is called self-inductance. Note that because the rings are close to each other, the flux through one of the rings in the initial configuration is $\Phi_{0}= L I_{0}+LI_{0}= 2 L I_{0}.$ The final flux equals for that ring is $\Phi_{1}= L I_{1}.$ Comparing the last two expressions, we obtain $I_{1}=2 I_{0}=0.1~\mbox{A}.$

When we separate two identical superconducting rings far apart, there are two magnetic field vectors tend to the same side. So that, sum vector is greater than the first one 2 times. Thus $I=2I_0=0.1$ A