Irodov 3.169: Current Crowds Toward the Surface – Amit & Deepali Sinha Foundation

Problem Irodov 3.169

A long conductor of circular cross-section with area $S$ is made of a material whose resistivity depends only on the distance $r$ from the axis, as $\rho = \alpha/r^2$ with $\alpha$ a constant. Find:

the resistance of the conductor per unit length;
the electric field strength inside it when it carries a current $I$ .

The cross-section of the conductor, area S, with its axis at the centre; the resistivity depends only on the distance r from the axis. — The cross-section of the conductor, area $S$ , with its axis at the centre; the resistivity depends only on the distance $r$ from the axis.

Given

cross-section area $S$
$\rho = \alpha/r^2$
current $I$

Find

(a) resistance per unit length

R_1

· (b) the field

E

Solution

Step 1The field is axial and uniform over the section

Let the section have radius $a$ , so that $S = \pi a^2$ . In a long straight conductor carrying a steady current the field is along the axis and, by symmetry, depends at most on $r$ . A steady field has zero circulation; around a thin rectangle with its long sides parallel to the axis at radii $r$ and $r'$ , the radial sides contribute nothing, so

E(r)\,\ell - E(r')\,\ell = 0

and the field $E$ is the same at every point of the section. The current density then follows from Ohm’s law in local form, $j = E/\rho$ :

j(r) = \frac{E}{\rho(r)} = \frac{E\,r^2}{\alpha} \tag{1}

Step 2The current carried by the section

The element: a thin ring of radius r and width dr in the section of radius a; every point of it has the same resistivity. — The element: a thin ring of radius $r$ and width $dr$ in the section of radius $a$ ; every point of it has the same resistivity.

A thin ring of radius $r$ and width $dr$ has area $2\pi r\,dr$ , and the resistivity is constant on it. Summing (1) over all such rings from the axis to the surface:

I = \int_0^a j(r)\,2\pi r\,dr = \frac{2\pi E}{\alpha}\int_0^a r^3\,dr = \frac{2\pi E}{\alpha}\cdot\frac{a^4}{4} = \frac{\pi a^4}{2\alpha}\,E

With $a^2 = S/\pi$ the factor in front is $S^2/(2\pi\alpha)$ :

I = \frac{S^2}{2\pi\alpha}\,E \tag{2}

Step 3The resistance per unit length

Over a length $\ell$ the voltage is $E\ell$ , so the resistance of unit length is $E/I$ . By (2):

R_1 = \frac{2\pi\alpha}{S^2}

(3)

Step 4The field for a given current

Solving (2) for the field:

E = \frac{2\pi\alpha I}{S^2}

(4)

Check

Dimensions: $\rho = \alpha/r^2$ makes $\alpha$ an $\Omega\cdot\text{m}^3$ , so $\alpha/S^2$ is $\Omega/\text{m}$ , a resistance per length, and (4) is volts per meter. A bound: every ring conducts worse than the surface layer, whose resistivity $\alpha/a^2$ is the least in the section. A uniform conductor of that resistivity would have

R_1' = \frac{\alpha/a^2}{S} = \frac{\pi\alpha}{S^2},

and (3) is twice this, larger, as it must be.

Remark

By (1) the current density grows as $r^2$ : the axis, where the resistivity is infinite, carries no current at all, and half of the current flows in the outer ring between $a/2^{1/4} \approx 0.84a$ and $a$ . The infinite resistivity on the axis costs nothing, because it sits on a set of zero area.