Irodov 3.210: The Field Inside a Proton Beam – Amit & Deepali Sinha Foundation

Problem Irodov 3.210

A uniform beam of protons, accelerated through a potential difference $V = 600$ kV, has a circular cross-section of radius $r = 5.0$ mm and carries the current $I = 50$ mA. Find the electric field strength at the surface of the beam, and the potential difference between the surface and the axis of the beam.

A length of the proton beam seen side on: its axis, its radius r and the beam current I. — A length of the proton beam seen side on: its axis, its radius $r$ and the beam current $I$ .

Given

$V = 600$ kV
$r = 5.0$ mm
$I = 50$ mA
uniform beam
protons start from rest (assumed)

Find

the field

E

at the surface, the potential difference

\Delta\varphi

between surface and axis

Solution

Step 1The speed of the protons

A proton accelerated from rest (assumed) through $V$ gains the energy $eV = 0.6$ MeV, far below its rest energy of $938$ MeV, so the motion is non-relativistic:

\begin{align*} \frac{1}{2}mv^2 &= eV \\ v &= \sqrt{\frac{2eV}{m}} \tag{1}\end{align*}

Numerically, with $m = 1.672\times10^{-27}$ kg:

v = \sqrt{\frac{2(1.60\times10^{-19}~\text{C})(6.00\times10^{5}~\text{V})}{1.672\times10^{-27}~\text{kg}}} = 1.07\times10^{7}~\text{m/s}

Step 2The charge density of the beam

The current is the charge per unit volume times the speed times the cross-section, so the beam carries the uniform charge density

\rho = \frac{I}{\pi r^2 v} \tag{2}

Step 3The field inside the beam

By symmetry the field of a long uniform beam is radial and the same all round, so Gauss’s theorem over a coaxial cylinder of radius $x \le r$ and length $L$ gives

\begin{align*} E\cdot 2\pi x L &= \frac{\rho\,\pi x^2 L}{\varepsilon_0} \\ E(x) &= \frac{\rho\,x}{2\varepsilon_0} \tag{3}\end{align*}

At the surface, $x = r$ , with (2) and (1):

E = \frac{\rho\,r}{2\varepsilon_0} = \frac{I}{2\pi\varepsilon_0 r v}

Numerically, with $1/2\pi\varepsilon_0 = 1.8\times10^{10}$ m/F:

E = \frac{(1.8\times10^{10}~\text{m/F})(0.050~\text{A})}{(5.0\times10^{-3}~\text{m})(1.07\times10^{7}~\text{m/s})}

E = \frac{I}{2\pi\varepsilon_0 r}\sqrt{\frac{m}{2eV}} \approx 17~\text{kV/m}

(4)

It points radially outward, away from the axis.

Step 4The potential difference between axis and surface

The field points outward, so the potential falls from the axis to the surface by the integral of (3):

\Delta\varphi = \int_0^r \frac{\rho\,x}{2\varepsilon_0}\,dx = \frac{\rho\,r^2}{4\varepsilon_0} = \frac{I}{4\pi\varepsilon_0 v}

Numerically:

\Delta\varphi = \frac{(9.0\times10^{9}~\text{m/F})(0.050~\text{A})}{1.07\times10^{7}~\text{m/s}}

\Delta\varphi = \frac{I}{4\pi\varepsilon_0}\sqrt{\frac{m}{2eV}} \approx 42~\text{V}

(5)

The axis is the higher of the two.

The field E (top) grows linearly inside the beam and falls as 1/x outside; the potential φ (bottom), measured from the axis, is highest on the axis. — The field $E$ (top) grows linearly inside the beam and falls as $1/x$ outside; the potential $\varphi$ (bottom), measured from the axis, is highest on the axis.

Check

Consistency: for a field growing linearly from zero, the potential drop is the mean field $E/2$ times $r$ , so (5) must equal $Er/2$ , and $(1.68\times10^{4}~\text{V/m})(5.0\times10^{-3}~\text{m})/2 = 42$ V. Dimensions: $I/(\varepsilon_0 v)$ is A/((F/m)(m/s)) = C/F = V.

Remark

The potential difference does not depend on the beam radius: squeezing the same current into a thinner beam raises the field at its surface but leaves the axis-to-surface voltage at $I/4\pi\varepsilon_0 v$ . This space-charge voltage, small against the $600$ kV of acceleration here, is what pushes a dense, slow beam apart; for a proton beam the magnetic attraction between the parallel currents cancels only the fraction $v^2/c^2$ of it, about $10^{-3}$ at this speed.