Irodov 3.397: Field and Radius Cap the Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.397

Protons are accelerated in a cyclotron, the largest radius of curvature of their path being $r = 50$ cm. Find:

the kinetic energy of the protons at the end of the acceleration, if the magnetic induction in the cyclotron is $B = 1.0$ T;
the least oscillator frequency at which the protons reach the kinetic energy $T = 20$ MeV by the end of the acceleration.

The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field B; r is the largest radius of the protons' path. — The cyclotron: two dees across a narrow gap, fed by the oscillator, in the uniform field $\vect B$ ; $r$ is the largest radius of the protons’ path.

Given

$r = 50$ cm
(a) $B = 1.0$ T
(b) $T = 20$ MeV
protons

Find

(a) the final kinetic energy · (b) the least oscillator frequency

\nu_{\mathrm{min}}

Solution

Step 1(a) The momentum on the last orbit

The protons' orbits, growing by a semicircle at each gap crossing out to the outermost one, of radius r. — The protons’ orbits, growing by a semicircle at each gap crossing out to the outermost one, of radius $r$ .

On the outermost orbit the magnetic force $evB$ is the centripetal force of the circle of radius $r$ , which fixes the momentum:

\frac{mv^2}{r} = evB \quad\Longrightarrow\quad p = mv = eBr

The kinetic energy follows from the momentum. Numerically, with the proton’s mass $1.672\times10^{-27}$ kg and charge $1.60\times10^{-19}$ C,

T = \frac{p^2}{2m} = \frac{\left[(1.60\times10^{-19})(1.0)(0.50)\right]^2}{2\,(1.672\times10^{-27})}\ \text{J} = 1.91\times10^{-12}\ \text{J}

T = \frac{(eBr)^2}{2m} \approx 12 \text{ MeV}

(1)

Step 2(b) The frequency the oscillator must keep

The oscillator frequency equals the protons’ revolution frequency, the same on every orbit, so on the last orbit the speed is the circumference over the period:

v = 2\pi\nu r, \qquad T = \frac{mv^2}{2} = 2\pi^2 m\nu^2 r^2

With the radius capped at $r$ , a final energy of at least $T$ needs a frequency of at least

\nu = \frac{1}{2\pi r}\sqrt{\frac{2T}{m}} = \frac{1}{2\pi r}\cdot 2\sqrt{\frac{T}{2m}}

Numerically, with $T = 20$ MeV $= 3.20\times10^{-12}$ J,

\nu_{\mathrm{min}} = \frac{1}{\pi\,(0.50)}\sqrt{\frac{3.20\times10^{-12}}{2\,(1.672\times10^{-27})}}\ \text{Hz}

\nu_{\mathrm{min}} = \frac{1}{\pi r}\sqrt{\frac{T}{2m}} \approx 20 \text{ MHz}

(2)

The field must then be raised to keep the protons in step with that frequency:

B = \frac{2\pi\nu_{\mathrm{min}} m}{e} = 1.3 \text{ T}

Check

The two parts agree with each other: at $B = 1.0$ T the revolution frequency is $eB/(2\pi m) = 15.2$ MHz, and $2\pi^2 m\nu^2r^2$ at that frequency gives back $12$ MeV; raising the energy from $12$ to $20$ MeV at the same radius scales the frequency by $\sqrt{20/12} = 1.29$ , to $19.7$ MHz. Units: $\sqrt{T/m}$ is a speed, and a speed over a length is a frequency. The non-relativistic treatment is fair: $20$ MeV is $2$ per cent of the proton’s rest energy, $938$ MeV.

Remark

The final energy $(eBr)^2/(2m)$ is set by the product $Br$ alone, the magnetic rigidity of the last orbit: a cyclotron gains energy only by being larger or by having a stronger magnet, and the dee voltage only decides how many turns it takes. That $2$ per cent mass increase at $20$ MeV is also where the classical cyclotron runs out: the revolution frequency falls as the mass grows and the protons drift out of step with a fixed oscillator, which the synchrocyclotron cures by lowering the frequency during each cycle of acceleration.