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+### Statement
+
+$11.3.11.$ [Insert the problem statement]
+
+### 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$
+
+$\boxed{L' = \frac{\mu_0}{\pi} \ln\frac{h}{r}}$
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]