Irodov 3.406: A Hundred Volts per Turn – Amit & Deepali Sinha Foundation

Problem Irodov 3.406

In a betatron the electrons circulate on an equilibrium orbit of radius $r = 20$ cm. Over a time $\Delta t = 1.0$ ms the magnetic induction at the orbit rises at an essentially constant rate from zero to $B = 0.40$ T. Find the energy an electron gains in each revolution.

A betatron seen along its magnetic field B (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger): the electron on the equilibrium orbit of radius r. — A betatron seen along its magnetic field $\vect B$ (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger): the electron on the equilibrium orbit of radius $r$ .

Given

$r = 20$ cm
$\Delta t = 1.0$ ms
$B$ on the orbit: $0 \to 0.40$ T at a constant rate

Find

the energy

\Delta W

gained per revolution

Solution

Step 1The flux through the orbit

On an equilibrium orbit the betatron condition holds at every moment: the mean induction inside the orbit is twice the induction on it. The flux is that mean times the enclosed area:

\Phi = \pi r^2\langle B\rangle = \pi r^2\cdot 2B_0 = 2\pi r^2 B_0

where $B_0$ is the induction on the orbit at that moment. It rises at a constant rate from zero to $B$ in the time $\Delta t$ , so the flux does too:

\frac{d\Phi}{dt} = 2\pi r^2\,\frac{dB_0}{dt} = \frac{2\pi r^2 B}{\Delta t} \tag{1}

Step 2The energy per revolution

By Faraday’s law the circulation of the eddy field round the orbit is $d\Phi/dt$ , and one revolution carries the electron once round it; the magnetic force does no work. The energy gained in a turn is $e$ times the circulation:

\Delta W = e\,\frac{d\Phi}{dt}

Putting in the rate (1), numerically, in electron volts,

\Delta W = \frac{2\pi\,(0.20)^2\,(0.40)}{1.0\times10^{-3}}\text{ eV} = 100\text{ eV}

\Delta W = \frac{2\pi e r^2 B}{\Delta t} \approx 0.10\text{ keV}

(2)

Check

Units: $r^2B/\Delta t$ is $\text{m}^2\cdot\text{T}/\text{s} = \text{Wb}/\text{s} = \text{V}$ , so $e$ times it is an energy. A cross-check through the whole cycle: at the end $p = erB$ on the orbit, so the electron’s energy is about $pc = (0.20)(0.40)(3.0\times10^{8})$ eV $= 24$ MeV, reached in $24\times10^{6}/100 = 2.4\times10^{5}$ turns; each turn at nearly $c$ takes $2\pi r/c = 4.2$ ns, and $2.4\times10^{5}$ turns take $1.0$ ms, which is $\Delta t$ .

Remark

A hundred electron volts a turn is small, so a betatron needs hundreds of thousands of turns and a millisecond to reach tens of MeV. The gain per turn is set by the flux through the orbit, while the final energy is set by the field on the orbit alone ( $p = erB$ ): the betatron condition, which ties the two together, is what lets one magnet both accelerate the electron and hold it on a fixed circle.