Irodov 3.385: The Wire Field Pulls the Electron Back

Problem Irodov 3.385

A long straight round wire of radius $a$ carries a steady current $I$ . An electron leaves the surface of the wire with a velocity $v_0$ directed perpendicular to the surface. The magnetic field of the current bends its path and turns it back toward the wire. Find the greatest distance from the axis of the wire that the electron reaches.

The round wire of radius a seen side on, carrying the current I, and the electron e leaving its surface with velocity v_0 normal to it. — The round wire of radius $a$ seen side on, carrying the current $I$ , and the electron $e$ leaving its surface with velocity $\vect v_0$ normal to it.

Given

wire radius $a$
current $I$
$\vect v_0$ normal to the surface

Find

the greatest distance

r_{\mathrm{max}}

of the electron from the axis

Solution

Step 1The field and the plane of the motion

Outside the wire the field circles the axis, with the magnitude Amp`ere’s law gives at the distance $r$ (by symmetry $B$ is azimuthal and the same all round):

B \cdot 2\pi r = \mu_0 I \quad\Longrightarrow\quad B = \frac{\mu_0 I}{2\pi r} \tag{1}

The force $-e\,\vect v\times\vect B$ is perpendicular to the azimuthal $\vect B$ , so it has no azimuthal component: the electron stays in the plane through the axis and its starting point. In that plane take $z$ along the current and $r$ from the axis.

Step 2The axial momentum follows the distance

Write the velocity in the plane of the motion and the field along the azimuthal unit vector, and use $\hat{\vect r}\times\hat{\boldsymbol\varphi} = \hat{\vect z}$ together with $\hat{\vect z}\times\hat{\boldsymbol\varphi} = -\hat{\vect r}$ :

\vect v\times\vect B = \left(v_r\,\hat{\vect r} + v_z\,\hat{\vect z}\right)\times B\,\hat{\boldsymbol\varphi} = v_r B\,\hat{\vect z} - v_z B\,\hat{\vect r}

The axial component of the equation of motion, with $B$ from (1), is then

m\frac{dv_z}{dt} = -e\,(\vect v\times\vect B)_z = -e v_r B = -\frac{e\mu_0 I}{2\pi r}\,\frac{dr}{dt}

Both sides are time derivatives. Integrating from the start, where $v_z = 0$ and $r = a$ :

m v_z = -\frac{e\mu_0 I}{2\pi}\,\ln\frac{r}{a} \tag{2}

The electron drifts against the current, and the faster the farther out it gets.

Step 3The turning point

The magnetic force does no work, so the speed stays $v_0$ throughout. The electron is farthest out when $v_r = 0$ , and there all of its speed is axial, $|v_z| = v_0$ . Putting this into (2) at $r = r_{\mathrm{max}}$ :

m v_0 = \frac{e\mu_0 I}{2\pi}\,\ln\frac{r_{\mathrm{max}}}{a} \quad\Longrightarrow\quad \ln\frac{r_{\mathrm{max}}}{a} = \frac{2\pi m v_0}{\mu_0 e I}

r_{\mathrm{max}} = a\exp\!\left(\frac{2\pi m v_0}{\mu_0 e I}\right)

(3)

The path in the plane through the axis, drawn for 2π m v_0/(μ_0 e I) = 1.2. With B out of the page above the wire the electron drifts against I, runs parallel to the wire at r_max with velocity v_0, and returns to the surface. — The path in the plane through the axis, drawn for $2\pi m v_0/(\mu_0 e I) = 1.2$ . With $\vect B$ out of the page above the wire the electron drifts against $I$ , runs parallel to the wire at $r_{\mathrm{max}}$ with velocity $\vect v_0$ , and returns to the surface.

At the turning point the electron moves parallel to the wire, opposite to the current, and it then comes back to the wire along the mirror image of its outward path.

Check

Units: $\mu_0 I$ is $\text{T}\cdot\text{m}$ , so $\mu_0 e I/m$ is $\text{T}\cdot\text{m}\cdot\text{C}/\text{kg} = \text{m/s}$ (as $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ ), and the exponent $2\pi m v_0/(\mu_0 e I)$ is a pure number. Limits: as $I \to 0$ the exponent grows without bound and so does $r_{\mathrm{max}}$ , the electron flying off straight; as $v_0 \to 0$ , $r_{\mathrm{max}} \to a$ . For $I = 10$ A and an electron of $v_0 = 1.0\times10^{5}$ m/s the exponent is $2\pi(0.911\times10^{-30})(10^{5})/((1.257\times10^{-6})(1.60\times10^{-19})(10)) = 0.28$ , so $r_{\mathrm{max}} = 1.3\,a$ .

Remark

The integration works because $-e\,\vect v\times\vect B$ along the axis is $-e B\,dr/dt$ and $B$ depends on $r$ alone: the conserved quantity is the canonical momentum $m v_z - eA_z$ with the wire’s vector potential $A_z = -(\mu_0 I/2\pi)\ln(r/a)$ . The logarithm makes the confinement weak but absolute: no electron leaving the wire radially escapes to infinity, however fast, since $r_{\mathrm{max}}$ is finite for every $v_0$ ; it only grows exponentially with the speed.