Irodov 3.403: The Betatron's One-Half Rule – Amit & Deepali Sinha Foundation

Problem Irodov 3.403

In a betatron, electrons are accelerated along a circular orbit by a growing magnetic field. Show that the orbit keeps a constant radius provided the magnetic induction at the orbit equals one half of the mean induction over the area enclosed by the orbit (the betatron condition).

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): an electron on a round 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): an electron on a round orbit of radius $r$ .

Given

a round orbit
induction on it
mean induction inside it

Find

show the radius stays fixed if the first is half the second

Solution

Step 1The push along the orbit

Let $r$ be the orbit’s radius and $\Phi$ the magnetic flux through it. By symmetry the eddy field $E$ is tangent to the orbit and the same all round it, so Faraday’s law on the orbit reads, in magnitude,

\oint\vect E\cdot d\vect l = 2\pi r\,E = \frac{d\Phi}{dt}

The tangential force $eE$ changes the magnitude $p$ of the electron’s momentum; Newton’s law in the form $d\vect p/dt = \vect F$ holds at any speed:

\frac{dp}{dt} = eE = \frac{e}{2\pi r}\,\frac{d\Phi}{dt} \tag{1}

The electron on its orbit: the magnetic force F_B points to the centre and bends the path, while the eddy field E, tangent to the orbit, pushes the electron along v (its charge is negative). — The electron on its orbit: the magnetic force $\vect F_B$ points to the centre and bends the path, while the eddy field $\vect E$ , tangent to the orbit, pushes the electron along $\vect v$ (its charge is negative).

Step 2The bend across the orbit

The magnetic force $e v B_0$ , with $B_0$ the induction on the orbit, is perpendicular to the velocity and turns the momentum at the rate $p v/r$ needed for a circle of radius $r$ . Equating the two and cancelling $v$ :

\frac{p v}{r} = e v B_0 \quad\Longrightarrow\quad p = e r B_0 \tag{2}

This must hold at every instant if the electron is to stay on the circle.

Step 3Both at once with

r

fixed

Differentiating (2) in time at constant $r$ and setting the result equal to (1):

e r\,\frac{dB_0}{dt} = \frac{e}{2\pi r}\,\frac{d\Phi}{dt} \quad\Longrightarrow\quad \frac{dB_0}{dt} = \frac{1}{2\pi r^2}\,\frac{d\Phi}{dt}

The flux is the mean induction $\langle B\rangle$ inside the orbit times the area, $\Phi = \pi r^2\langle B\rangle$ . Differentiating it at fixed $r$ and putting it in:

\frac{dB_0}{dt} = \frac{1}{2\pi r^2}\cdot\pi r^2\,\frac{d\langle B\rangle}{dt}

\frac{dB_0}{dt} = \frac{1}{2}\,\frac{d\langle B\rangle}{dt}

(3)

Step 4The betatron condition

The acceleration starts from zero field, with $B_0 = \langle B\rangle = 0$ , and the electron’s starting momentum is negligible. Integrating (3) from that start:

\int_0^{B_0} dB_0' = \frac{1}{2}\int_0^{\langle B\rangle} d\langle B\rangle'

B_0 = \frac{1}{2}\,\langle B\rangle

(4)

So the orbit keeps its radius when the field on it is half the mean field inside it, the betatron condition; strictly, the rates must keep this ratio, (3), and with both fields starting from zero that is the same thing.

Check

Both sides of (3) are inductions per unit time, and $e$ and the speed have dropped out, so the condition holds for a relativistic electron as well as a slow one. A uniform field fails it: with $B_0 = \langle B\rangle = B$ , integrating (1) gives $p = e\Phi/(2\pi r) = erB/2$ , half the momentum (2) demands, so the electron would curl into a circle of radius $p/(eB) = r/2$ , the same halving as a particle in a solenoid that is switched on around it.

Remark

The condition asks for a field that is stronger inside the orbit than on it. The pole pieces of a betatron are shaped for this: the flux through the middle, which does the accelerating, is made large, while the field at the orbit, which does the steering, is held at half the mean. The division of labour is the whole design: one global quantity, the enclosed flux, sets the energy gain, and one local quantity, the field on the orbit, keeps the path from spiralling in or out.