Irodov 3.402: Millions of Turns at Five Volts – Amit & Deepali Sinha Foundation

Problem Irodov 3.402

In a betatron, electrons circulate on an equilibrium orbit of radius $r = 25$ cm. During the acceleration the magnetic flux through the orbit grows at an essentially constant rate $\dot{\Phi} = 5.0$ Wb/s, and the electrons gain the energy $W = 25$ MeV. Find how many revolutions an electron makes during the acceleration, and the distance it travels.

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, through which the flux grows. — 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 $r$ , through which the flux grows.

Given

$r = 25$ cm
$\dot{\Phi} = 5.0$ Wb/s
$W = 25$ MeV

Find

the number of revolutions

N

and the distance

s

Solution

Step 1The energy gained in one revolution

By Faraday’s law the circulation of the eddy field round the orbit equals, in magnitude, the rate of change of the flux through it. One revolution carries the electron once round that circuit, so the work done on it is $e$ times the circulation:

\Delta W = e\oint\vect E\cdot d\vect l = e\,\frac{d\Phi}{dt}

\Delta W = e\dot{\Phi}

(1)

With $\dot\Phi = 5.0$ Wb/s $= 5.0$ V this is $5.0$ eV per turn, the same on every turn, since the flux grows at a constant rate.

Step 2The number of revolutions

The electron collects $W$ in equal steps of $\Delta W$ , so using (1):

N = \frac{W}{\Delta W} = \frac{25\times10^{6}\text{ eV}}{5.0\text{ eV}}

N = \frac{W}{e\dot{\Phi}} = 5.0\times10^{6}

(2)

Step 3The distance travelled

Each revolution is one circumference of the orbit, so with (2):

s = 2\pi\,(0.25\text{ m})\,(5.0\times10^{6})

s = 2\pi r N \approx 7.9\times10^{6}\text{ m} \approx 8\times10^{3}\text{ km}

(3)

Check

Units: $\dot\Phi$ in Wb/s is a voltage, so $e\dot\Phi$ is an energy and $N$ a pure number. A cross-check through the orbit itself: at 25 MeV the electron is ultrarelativistic, $p \approx W/c$ , and the orbit needs $p = eBr$ with $B$ the field on the orbit, so $B \approx 25\times10^{6}/(0.25\cdot3.0\times10^{8}) = 0.33$ T. The betatron condition makes the mean field inside the orbit twice that, so the flux grows to $\pi r^2\cdot 0.67$ T $= 0.13$ Wb, which at $5.0$ Wb/s takes $26$ ms; an electron moving at nearly $c$ covers $7.9\times10^{6}$ m in $s/c = 26$ ms. The two times agree.

Remark

Five electron volts a turn is tiny, yet the betatron needs no radio-frequency cavity and no synchronism at all: the accelerating field is induced uniformly round the whole orbit, and the electron simply goes round several million times. The path, eight thousand kilometres, is longer than the Earth’s radius, packed into a ring half a metre across, which is why the vacuum in a betatron’s chamber has to be very good.