Irodov 3.159: Leakage Between Two Parallel Wires – Amit & Deepali Sinha Foundation

Problem Irodov 3.159

Two long parallel wires, each with a circular cross-section of radius $a$ , lie in a weakly conducting medium of resistivity $\rho$ . Their axes are a distance $l$ apart, with $a \ll l$ .

A potential difference $V$ is kept between the wires. Find the current density at a point that is the same distance $r$ from both axes.
Find the resistance of the medium per unit length of the wires.

The two wires in cross-section, radius a, axes l apart, in the medium of resistivity ρ; the point P is a distance r from both axes (drawn schematically: the problem has a l). — The two wires in cross-section, radius $a$ , axes $l$ apart, in the medium of resistivity $\rho$ ; the point $P$ is a distance $r$ from both axes (drawn schematically: the problem has $a \ll l$ ).

Given

wire radius $a$
axis distance $l \gg a$
resistivity $\rho$
voltage $V$
$P$ a distance $r$ from both axes
medium fills all space (assumed)

Find

(a)

j

P

, (b) the resistance per unit length

R_1

Solution

Step 1The field of one wire feeding the medium

Let wire 1 feed a current $I_1$ per unit length into the medium and wire 2 take the same current back; take the medium to fill all space. In the steady state $\nabla\cdot\vect j = 0$ and $\vect j = \vect E/\rho$ , so the potential obeys Laplace’s equation and contributions superpose, as in electrostatics. Alone, a long wire spreads its current radially and evenly, so by symmetry, over a coaxial cylinder of radius $r$ and unit length,

j\cdot 2\pi r = I_1 \tag{1}

The field $E = \rho j$ integrated along a radius gives the potential, up to a constant:

\varphi(r) = -\int\frac{\rho I_1}{2\pi r}\,dr = -\frac{\rho I_1}{2\pi}\ln r + \text{const}

Step 2The voltage fixes the current

At a point distances $r_1$ and $r_2$ from the axes of wires 1 and 2, the source $I_1$ and the sink $-I_1$ give together

\varphi = \frac{\rho I_1}{2\pi}\ln\frac{r_2}{r_1}

On the surface of wire 1, $r_1 = a$ and $r_2 \approx l$ since $a \ll l$ , and on wire 2 the roles swap. The difference is the voltage:

V = \frac{\rho I_1}{2\pi}\ln\frac{l}{a} - \frac{\rho I_1}{2\pi}\ln\frac{a}{l} = \frac{\rho I_1}{\pi}\ln\frac{l}{a}

so the current per unit length is

I_1 = \frac{\pi V}{\rho\ln(l/a)} \tag{2}

Step 3The current density at the point

P

Current lines between the wires, circles through both axes, with equipotentials. At P, equally distant from both axes, j is parallel to the line joining them, from the wire feeding the current (+, wire 1) to the wire taking it back (-, wire 2). — Current lines between the wires, circles through both axes, with equipotentials. At $P$ , equally distant from both axes, $\vect j$ is parallel to the line joining them, from the wire feeding the current ( $+$ , wire 1) to the wire taking it back ( $-$ , wire 2).

At $P$ each wire contributes a current density of magnitude $I_1/(2\pi r)$ by (1): away from wire 1 and toward wire 2. Both lie along the lines from the axes to $P$ , at the same angle to the symmetry plane, so the components across the line joining the wires cancel and those along it add. Each has the fraction $(l/2)/r$ along it:

j = 2\cdot\frac{I_1}{2\pi r}\cdot\frac{l}{2r} = \frac{I_1\, l}{2\pi r^2}

Substituting (2):

j = \frac{l\,V}{2\rho\, r^2\ln(l/a)}

(3)

Step 4The resistance per unit length

For a length $h$ of the pair the current is $I_1 h$ , so the resistance is inversely proportional to $h$ , and the constant $R_1 = V/I_1$ characterises unit length: a length $h$ has the resistance $R_1/h$ . From (2):

R_1 = \frac{\rho}{\pi}\ln\frac{l}{a}

(4)

Check

All the current per unit length must cross the symmetry plane. On it a point a height $y$ above the joining line has $r^2 = l^2/4 + y^2$ , and $\vect j$ is normal to the plane, so the flux of the boxed density through a strip of unit length is

\int_{-\infty}^{\infty}\frac{l\,V\,dy}{2\rho\ln(l/a)\,(l^2/4 + y^2)} = \frac{l\,V}{2\rho\ln(l/a)}\cdot\frac{2\pi}{l} = \frac{\pi V}{\rho\ln(l/a)}

which is $I_1$ in (2). Far away, $r \gg l$ , $j$ falls as $1/r^2$ , the law of a two-dimensional dipole. Dimensions: $\rho\ln(l/a)$ is $\Omega\cdot$ m, the unit of a resistance times a length, as $R_1/h$ requires.

Remark

Comparing with the capacitance per unit length of the same two wires in a dielectric, $C_1 = \pi\varepsilon_0\varepsilon/\ln(l/a)$ , the product $R_1 C_1 = \varepsilon_0\varepsilon\rho$ carries no trace of the geometry: the leakage and the capacitance are two readings of the same field pattern.