Electromagnetic induction > Mutual inductance. Inductance of conductors. Transformers > Problem 11.3.11

Statement

$11.3.11$

Find the inductance per unit length of a two-wire line. The line consists of
two parallel straight wires of radius r, the distance between the centerlines
of which are h ≫ r. Through the wires flow equal in modulus, but oppositely
directed currents. There is no magnetic field inside the wires.

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

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