Solution

Magnetic field between the two wires

Each straight wire produces an azimuthal magnetic field on the outside.
For a wire carrying current I, the field at a distance$\rho$from its center is:

$B = \frac{\mu_0 I}{2\pi\rho} \quad (\rho > r)$

Between the two wires (separation h), the fields add constructively because the opposite currents generate fields in the same direction in that region. If we place wire $1 at x = 0$ and wire 2 at$x = h$, the total field at a point $x \in (r, h-r)$is:

$B(x) = \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi (h-x)}$

Magnetic flux per unit length

The flux through the rectangle of width $h-2r$ and unit height is:

$\Phi' = \int_{r}^{h-r} B(x)\,dx = \frac{\mu_0 I}{2\pi} \int_{r}^{h-r} \left( \frac{1}{x} + \frac{1}{h-x} \right) dx$

Each integral gives $\ln\frac{h-r}{r}$ Adding them:

$\Phi' = \frac{\mu_0 I}{2\pi} \cdot 2\ln\frac{h-r}{r} = \frac{\mu_0 I}{\pi} \ln\frac{h-r}{r}$

Since $h \gg r fwe can approximate$h-r \approx h$

$\Phi' \approx \frac{\mu_0 I}{\pi} \ln\frac{h}{r}$

Inductance per unit length

By definition$L' = \Phi'/I$

$\boxed{L' = \frac{\mu_0}{\pi} \ln\frac{h}{r}}$