Irodov 3.154: Current Spreading Out of an Axis – Amit & Deepali Sinha Foundation

Problem Irodov 3.154

Two thin, perfectly conducting coaxial cylinders of radii $a$ and $b$ , with $a < b$ , each have length $l$ . The space between them is filled with a uniform, poorly conducting medium of resistivity $\rho$ . Ignoring edge effects, find the resistance of the medium between the cylinders.

The thin coaxial cylinders in cross-section: radii a and b, the medium of resistivity ρ between them. Both have length l along the axis, normal to the page. — The thin coaxial cylinders in cross-section: radii $a$ and $b$ , the medium of resistivity $\rho$ between them. Both have length $l$ along the axis, normal to the page.

Given

radii $a < b$
length $l$
resistivity $\rho$
ideal electrodes
edge effects neglected

Find

the resistance

R

of the medium

Solution

Step 1The current density at radius

r

Let a steady current $I$ flow from the inner cylinder to the outer one. Each electrode is ideally conducting, so it is an equipotential, and with the edges neglected the current in the medium is radial and the same all round the axis. All of $I$ crosses every coaxial cylinder of radius $r$ between the electrodes:

j\cdot 2\pi r l = I

By the local form of Ohm’s law the field in the medium is $\rho$ times the current density:

E(r) = \rho j = \frac{\rho I}{2\pi l\,r} \tag{1}

The steady current in the medium with the inner cylinder at the higher potential: radial lines of j, the same number crossing every coaxial surface, so they thin out as 1/r. — The steady current in the medium with the inner cylinder at the higher potential: radial lines of $\vect{j}$ , the same number crossing every coaxial surface, so they thin out as $1/r$ .

Step 2The voltage and the resistance

The voltage between the electrodes is the line integral of (1) along a radius:

V = \int_a^b \frac{\rho I}{2\pi l\,r}\,dr = \frac{\rho I}{2\pi l}\ln\frac{b}{a}

The current $I$ cancels in $V/I$ :

R = \frac{\rho}{2\pi l}\ln\frac{b}{a}

(2)

Check

Dimensions: $\rho/l$ is ohm metres over metres, a resistance, and the logarithm is a pure number. Thin layer: with $b = a + \delta$ and $\delta \ll a$ , $\ln(b/a) \approx \delta/a$ and (2) becomes $\rho\,\delta/(2\pi a l)$ , the familiar $\rho \times$ thickness $/$ area of a flat layer of area $2\pi a l$ .

Remark

The same electrodes, seen as a capacitor with the medium (of permittivity $\varepsilon$ ) between them, have $C = 2\pi\varepsilon_0\varepsilon l/\ln(b/a)$ , and the product $RC = \varepsilon_0\varepsilon\rho$ is free of the geometry: the field lines and the current lines follow the same paths, so the same logarithm appears in both and cancels. Because of the $1/r$ in (1), half of the voltage drops within the radius $\sqrt{ab}$ , close to the inner cylinder, where the current is most crowded.