= Solution
Each long wire lies in the other's <magnetic field>, $B=\mu_0I/(2\pi d)$. The magnetic <Lorentz force> per unit length has magnitude
$$
\boxed{F'=\frac{\mu_0I^2}{2\pi d},\qquad\text{opposite currents repel}.}
$$
For the two-wire circuit's <self-inductance>, a nonzero wire radius is essential. Introduce a radius $a_w\ll d$. Between the wires their <magnetic fields> add. The external <magnetic flux linkage> per unit length through the connecting ribbon is, to leading order in $a_w/d$,
$$
\Phi'=\int_{a_w}^{d-a_w}\frac{\mu_0I}{2\pi}\left(\frac1x+\frac1{d-x}\right)dx
=\frac{\mu_0I}{\pi}\ln\frac{d-a_w}{a_w}.
$$
It follows that the leading thin-wire loop <self-inductance> per unit length is
$$
\boxed{L'_{\rm ext}\simeq\frac{\mu_0}{\pi}\ln\frac d{a_w}.}
$$
It is the loop value for the pair, not the contribution of just one conductor. Corrections from the wire interior depend on the current distribution: for nonmagnetic round wires with uniform direct <electric current>, the pair contributes another $\mu_0/(4\pi)$; for a surface current the internal contribution is zero. The leading logarithm is common to both models.
\b[TODO (source data):] the source gives no wire radius or current-distribution convention. There is no finite radius-independent value of the requested <self-inductance>: taking ideal zero-width wires sends $L'_{\rm ext}$ to infinity. The <force> and the parameterized leading <self-inductance> above are determined, but a unique finite answer requires that additional specification. For example, changing $a_w$ to $a_w/2$ changes the leading result by $\mu_0\ln2/\pi$ even at fixed separation.
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