Irodov 3.266: A Wire's Own Field Makes a Hall Voltage

Problem Irodov 3.266

A straight copper wire of circular cross-section, radius $R = 5.0$ mm, carries a current $I = 50$ A. Copper has $n = 0.9\cdot 10^{23}$ conduction electrons per cm $^3$ . Find the potential difference between the axis of the wire and its surface.

The straight round copper conductor of radius R carrying the current I. — The straight round copper conductor of radius $R$ carrying the current $I$ .

Given

$R = 5.0$ mm
$I = 50$ A
$n = 0.9\cdot 10^{23}$ cm $^{-3}$
carriers electrons, charge $-e$
current density uniform over the section (assumed)

Find

the potential difference between axis and surface

Solution

Step 1The drift of the electrons

Take the current density as uniform over the cross-section (assumed); it is the current over the area. The carriers are electrons of charge $-e$ , so they drift against the current with the speed $u$ fixed by $j = neu$ :

j = \frac{I}{\pi R^2}, \qquad u = \frac{j}{ne} = \frac{I}{\pi R^2 ne} \tag{1}

Step 2The wire’s own field inside it

By symmetry $\vect B$ circles the axis with the same magnitude all round. A circle of radius $r < R$ encloses the fraction $r^2/R^2$ of the current, so Ampere’s law gives

B\cdot 2\pi r = \mu_0 I\,\frac{r^2}{R^2}

B(r) = \frac{\mu_0 I r}{2\pi R^2}

(2)

Step 3The radial field that holds the electrons

The cross-section with the current out of the page, so the electrons drift into it. The wire's own B circles the axis; on an electron the magnetic force F_m points toward the axis, and in the steady state the electric force F_e balances it, so E points toward the axis. — The cross-section with the current out of the page, so the electrons drift into it. The wire’s own $\vect B$ circles the axis; on an electron the magnetic force $\vect F_m$ points toward the axis, and in the steady state the electric force $\vect F_e$ balances it, so $\vect E$ points toward the axis.

An electron moving with velocity $\vect u$ (against the current, along the axis) in the azimuthal $\vect B$ feels $-e\,\vect u\times\vect B$ . By the right-hand rule this force points toward the axis: the wire’s own field pinches the electron stream. Electrons crowd toward the axis until the radial electric field they set up stops any further radial drift:

\begin{aligned} -e\vect E - e\,\vect u\times\vect B &= 0\\ \vect E &= -\vect u\times\vect B \end{aligned}

This field points toward the axis, with the magnitude

E(r) = u\,B(r) \tag{3}

Step 4The potential difference

Since $\vect E$ points toward the axis, the potential rises outward, and the surface is the higher of the two. The rise from axis to surface is the integral of (3) with (2):

\begin{aligned} U = \varphi(R) - \varphi(0) &= \int_0^R u\,\frac{\mu_0 I r}{2\pi R^2}\,dr = u\,\frac{\mu_0 I}{2\pi R^2}\cdot\frac{R^2}{2}\\ &= \frac{\mu_0 I u}{4\pi} \end{aligned}

Substituting $u$ from (1) gives the closed form in the box below. Numerically, with $n = 0.9\times 10^{29}$ m $^{-3}$ and the usual $\mu_0$ and $e$ :

U = \frac{4\pi\times 10^{-7}\times 50^2}{4\pi^2\times(5.0\times 10^{-3})^2\times 0.9\times 10^{29}\times 1.60\times 10^{-19}}

U = \frac{\mu_0 I^2}{4\pi^2 R^2 n e} \approx 0.2 \text{ nV}

(4)

The surface is at the higher potential, the axis at the lower.

Check

Units: $\mu_0 I^2$ is in newtons and $R^2ne$ in C/m, so the ratio is in N $\cdot$ m/C $=$ V. Size: with copper’s resistivity $\rho = 1.7\times 10^{-8}$ $\Omega\cdot$ m and $j = 6.4\times 10^5$ A/m $^2$ from (1), the ohmic field along the wire is

\rho j = 1.7\times 10^{-8}\times 6.4\times 10^{5} \approx 0.01 \text{ V/m},

so the radial voltage, (4), is about $10^{-8}$ of the drop along one metre. The space charge that sets up the radial field, $\varepsilon_0\nabla\cdot\vect E = \varepsilon_0\mu_0 uj$ , is only the fraction $(u/c)^2 \approx 2\times 10^{-26}$ of the electrons’ own charge density $ne$ (with $u \approx 4\times 10^{-5}$ m/s), so the uniform current density assumed in Step 1 is undisturbed. If the carriers were positive, drifting along the current, $q\vect u\times\vect B$ would still point toward the axis, but they would charge the axis positive and the sign of $U$ would reverse.

Remark

This is the Hall effect with the wire’s own field as the applied one: the self-field squeezes the carriers toward the axis, and their charge separation builds the radial voltage. It grows as $I^2$ and as $1/n$ , which is why in a metal it is a fraction of a nanovolt, while in a plasma column carrying a huge current, where both signs of carrier move and no such field can build up to stop them, the same inward magnetic force squeezes the whole column (the pinch effect).