= Solution
For one wire, <Ampère's circuital law> on a concentric <circle> of radius $r$ gives
$$
2\pi r B(r)=\mu_0I,\qquad B(r)=\frac{\mu_0I}{2\pi r}.
$$
The other wire experiences the <Lorentz force> on its <electric current>, $d\mathbf F=I\,d\boldsymbol\ell\times\mathbf B$. Its direction is perpendicular to this <magnetic field>, so the <force per unit length> is
$$
\boxed{\frac FL=\frac{\mu_0I^2}{2\pi r}.}
$$
The <right-hand rule> gives attraction for parallel <electric currents> in the same direction and repulsion for opposite directions.
For the upper wire, the upward magnetic <force per unit length> balances its weight per unit length, $\lambda g$, where $g$ is the <gravitational acceleration>. Therefore
$$
\boxed{r_{\mathrm{eq}}=\frac{\mu_0I^2}{2\pi\lambda g}.}
$$
This assumes the lower wire is held fixed and both wires remain straight. Within the vertical-motion model, increasing the separation weakens the upward <magnetic force>, so this <equilibrium> is vertically restoring.
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