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I've been asked to calculate the inductance per unit length for two wires or radius $a$ separated by $2d$ where $2d>>a$.

Starting from $\int_{s} B.dS = LI$ Im not sure what surface to take?

For each wire the field at a distance r away is given by $\int_{l} B.dl = \mu_0 I$ and by superposition $B$ in the first integral is their sum. And so $$LI= \int_{s} B_1.dS + \int_{s} B_2.dS $$ where $B_1$ and $B_2$ are the contributions from the two wires.

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    $\begingroup$ An alternative derivation link $\endgroup$ – Sergei Gorbikov Aug 1 '16 at 13:16
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my answer

Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.

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