Irodov 3.319: A Logarithm per Metre of Cable – Amit & Deepali Sinha Foundation

Problem Irodov 3.319

A cable is made of two thin-walled coaxial metal cylinders; the outer one has a radius $\eta = 3.6$ times that of the inner one. The medium between them has permeability equal to one. Find the inductance per unit length of the cable.

The cable in cross-section: two thin coaxial metal tubes, of radii a and η a, with μ = 1 between them. — The cable in cross-section: two thin coaxial metal tubes, of radii $a$ and $\eta a$ , with $\mu = 1$ between them.

Given

$\eta = 3.6$
thin walls
$\mu = 1$
equal and opposite currents in the two tubes (assumed)

Find

L_1

, the inductance per unit length

Solution

Step 1The field between the tubes

A coaxial circle of radius r between the tubes, the contour for Ampere's law. — A coaxial circle of radius $r$ between the tubes, the contour for Ampere’s law.

Let the inner tube have radius $a$ and carry the current $I$ , and the outer tube the same current back (assumed: the cable is the go and the return of one circuit). By symmetry $\vect B$ circles the axis with a magnitude depending only on $r$ , so on a coaxial circle of radius $r$ between the tubes Ampere’s law encloses just the inner current:

B\cdot 2\pi r = \mu_0 I

B = \frac{\mu_0 I}{2\pi r},\qquad a < r < \eta a

(1)

Inside the inner tube a circle encloses no current, and outside the outer one it encloses $I - I = 0$ , so the field is zero in both places: it lives only between the tubes.

Step 2The flux per unit length

The circuit is closed along the inner tube and back along the outer one, so the flux that links it crosses a strip of unit length lying in a plane through the axis, from $r = a$ to $r = \eta a$ . The field (1) crosses the strip square to it, and an element of width $dr$ has area $dr$ :

\Phi_1 = \int_a^{\eta a} \frac{\mu_0 I}{2\pi r}\,dr = \frac{\mu_0 I}{2\pi}\,\ln\frac{\eta a}{a} = \frac{\mu_0 I}{2\pi}\,\ln\eta

Step 3The inductance per unit length

The walls are thin, so no field lies inside the metal and every line between the tubes links the whole current once; the inductance per unit length is $\Phi_1/I$ . Numerically,

L_1 = \frac{4\pi\times 10^{-7}\text{ H/m}}{2\pi}\,\ln 3.6

L_1 = \frac{\mu_0}{2\pi}\,\ln\eta \approx 0.26\ \mu\text{H/m}

(2)

With a magnetic medium of permeability $\mu$ between the tubes the field, and with it $L_1$ , would be $\mu$ times larger.

Check

The field energy per unit length gives the same result. The energy density $B^2/2\mu_0$ with (1), integrated over thin coaxial shells of volume $2\pi r\,dr$ per unit length:

W_1 = \int_a^{\eta a}\frac{\mu_0 I^2}{8\pi^2 r^2}\,2\pi r\,dr = \frac{\mu_0 I^2}{4\pi}\,\ln\eta

which is $L_1 I^2/2$ with (2). Units: $\mu_0$ is already H/m, and the logarithm is a pure number.

Remark

The same two tubes have capacitance per unit length $C_1 = 2\pi\varepsilon_0/\ln\eta$ , so the logarithm cancels in the product, $L_1 C_1 = \mu_0\varepsilon_0$ , which is $1/c^2$ . That product sets the speed of a signal along the cable, $1/\sqrt{L_1C_1}$ , so with empty space between the tubes a signal runs along it at the speed of light, whatever their sizes.