Irodov 3.396: Orbit Spacing Sets the Dee Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.396

The oscillator of a cyclotron runs at the frequency $\nu = 10$ MHz. Protons are accelerated in it, and at the orbit radius $r = 0.5$ m the distance between neighbouring trajectories must be no less than $\Delta r = 1.0$ cm. Find the effective accelerating voltage applied across the dees.

The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field B; r is the orbit radius at which the spacing is asked. — The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field $\vect B$ ; $r$ is the orbit radius at which the spacing is asked.

Given

$\nu = 10$ MHz
$\Delta r \geq 1.0$ cm
$r = 0.5$ m
protons

Find

the effective accelerating voltage

V

Solution

Step 1Resonance fixes the field

A proton circles in the field $B$ with the angular frequency $eB/m$ , whatever its speed, and the dee voltage must reverse in step with it, so the oscillator frequency equals the revolution frequency:

2\pi\nu = \frac{eB}{m} \quad\Longrightarrow\quad B = \frac{2\pi\nu m}{e} \tag{1}

Step 2Energy as a function of radius

On a circle of radius $r$ the force $evB$ is centripetal, which ties the speed to the radius; with (1) the kinetic energy $T$ then depends on $r$ alone:

\begin{aligned} \frac{mv^2}{r} &= evB \quad\Longrightarrow\quad v = \frac{eBr}{m} = 2\pi\nu r, \\ T &= \frac{mv^2}{2} \end{aligned}

T = 2\pi^2 m\nu^2 r^2

(2)

Step 3Spacing of neighbouring orbits

The protons' orbits, growing by a semicircle at each gap crossing; two neighbouring orbits in one dee, one revolution apart, are Δ r apart at the radius r. — The protons’ orbits, growing by a semicircle at each gap crossing; two neighbouring orbits in one dee, one revolution apart, are $\Delta r$ apart at the radius $r$ .

Each crossing of the gap adds $eV$ , so one revolution adds $2eV$ . Over that revolution the radius grows by $\Delta r$ , small against $r$ , so the energy gain is the derivative of (2) times $\Delta r$ :

2eV = \frac{dT}{dr}\,\Delta r = 4\pi^2 m\nu^2 r\,\Delta r \quad\Longrightarrow\quad \Delta r = \frac{eV}{2\pi^2 m\nu^2 r}

The spacing grows with the voltage, so the requirement that it be at least $\Delta r$ is a lower bound on $V$ .

Step 4The voltage

Turning the spacing round for $V$ and inserting $m = 1.672\times10^{-27}$ kg and $e = 1.60\times10^{-19}$ C:

V = \frac{2\pi^2(1.672\times10^{-27})(10\times10^{6})^2(0.5)(0.010)}{1.60\times10^{-19}}\ \text{V}

V \geq \frac{2\pi^2 m\nu^2 r\,\Delta r}{e} \approx 0.10 \text{ MV}

(3)

Check

Energy bookkeeping at $r = 0.5$ m: (2) gives $T = 8.25\times10^{-13}$ J $= 5.2$ MeV, and one revolution adds $2eV = 0.21$ MeV, a fraction $0.040$ of it. Since $T$ goes as $r^2$ , a fractional gain of $2\Delta r/r = 2(0.010)/(0.5) = 0.040$ is the same number. Units: $m\nu^2r\Delta r/e$ is $\text{kg}\cdot\text{m}^2/(\text{s}^2\cdot\text{C}) = \text{J}/\text{C} = \text{V}$ .

Remark

The spacing $eV/(2\pi^2 m\nu^2 r)$ shrinks as $1/r$ : the orbits crowd together toward the rim, so the condition at the largest radius is the one that matters, and it holds with room to spare at every smaller one. A clear gap between the last two orbits is what lets a deflector pick off the outermost one without touching the one inside, which is why the dee voltage, and not only the field and the radius, is a design figure of the machine.