Irodov 3.322: Two Wires, Two Logarithms – Amit & Deepali Sinha Foundation

Problem Irodov 3.322

A double line is made of two long parallel wires carrying a current to a load and back. The distance between the axes of the wires is $\eta$ times the radius of each wire, with $\eta \gg 1$ . Neglecting the field inside the wires, and with permeability equal to one everywhere, find the inductance per unit length of the line.

The double line end on: two wires of radius a whose axes are η a apart (drawn for η near 10), the current out of the page in one and back in the other (sense chosen). — The double line end on: two wires of radius $a$ whose axes are $\eta a$ apart (drawn for $\eta$ near 10), the current out of the page in one and back in the other (sense chosen).

Given

axis separation $= \eta\times$ wire radius
$\eta \gg 1$
field inside the wires neglected
$\mu = 1$

Find

L_1

, the inductance per unit length

Solution

Step 1The field in the plane of the wires

The plane of the two axes, seen edge on between the wires' surfaces, and a strip of it of width dx at the distance x from the left axis. Both wires' fields cross the strip the same way, along B. — The plane of the two axes, seen edge on between the wires’ surfaces, and a strip of it of width $dx$ at the distance $x$ from the left axis. Both wires’ fields cross the strip the same way, along $\vect B$ .

Let each wire have radius $a$ , so the axes are $d = \eta a$ apart, and let the current be $I$ out of the page in the left wire and back into it in the right one. A thin wire far from the other ( $\eta \gg 1$ ) has the field of a straight current outside it, by symmetry circles about its axis with the same magnitude all round, so Ampere’s law on a circle of radius $r$ gives the first line below. At a point between the wires, at the distance $x$ from the left axis, both fields are square to the plane of the axes and, by the right-hand rule, point the same way, so they add:

\begin{align*} B\cdot 2\pi r &= \mu_0 I \\ B(x) &= \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi(d - x)} \tag{1}\end{align*}

Step 2The flux per unit length

The circuit runs along one wire and back along the other, so the flux that links it crosses the strip of the plane between the wires’ surfaces, from $x = a$ to $x = d - a$ ; the field inside the wires is neglected. An element of unit length and width $dx$ carries $B\,dx$ , so with (1):

\begin{aligned} \Phi_1 &= \frac{\mu_0 I}{2\pi}\int_a^{d-a}\left(\frac{1}{x} + \frac{1}{d - x}\right)dx\\ &= \frac{\mu_0 I}{2\pi}\left(\ln\frac{d - a}{a} + \ln\frac{d - a}{a}\right) = \frac{\mu_0 I}{\pi}\,\ln\frac{d - a}{a} \end{aligned}

The two wires contribute equally, as the symmetry of the line requires.

Step 3The inductance per unit length

Dividing the flux by $I$ , and putting $d/a = \eta$ :

L_1 = \frac{\Phi_1}{I} = \frac{\mu_0}{\pi}\,\ln(\eta - 1)

For $\eta \gg 1$ the $1$ beside $\eta$ is negligible, the same order as the corrections already dropped by treating each wire’s field as that of a lone wire. Keeping it is no gain: the exact result for two cylinders, $(\mu_0/\pi)\,\mathrm{arccosh}(\eta/2)$ , lies closer to $\ln\eta$ than to $\ln(\eta - 1)$ :

L_1 \approx \frac{\mu_0}{\pi}\,\ln\eta

(2)

Check

A numerical case: wires 1 mm in radius with axes 10 cm apart, $\eta = 100$ , give

L_1 \approx \bigl(4\times 10^{-7}\text{ H/m}\bigr)\ln 100 \approx 1.8\ \mu\text{H/m}

the familiar microhenry or two per metre of an open two-wire line. Units: $\mu_0$ is H/m and $\ln\eta$ a pure number. The kept form $\ln(\eta - 1)$ and the boxed $\ln\eta$ differ by $\ln\bigl(\eta/(\eta - 1)\bigr)$ , about $1/\eta$ , which is $0.01$ against $\ln 100 = 4.6$ there.

Remark

Compare the coaxial cable, $L_1 = (\mu_0/2\pi)\ln\eta$ : the open line has twice the logarithm because its field is not confined. Each wire’s field fills the space round it and the flux between the wires gathers a logarithm from each, while a coaxial outer conductor kills the field outside and leaves only the inner one’s.