Irodov 3.400: Detuning the Drive to Keep Step – Amit & Deepali Sinha Foundation

Problem Irodov 3.400

In a cyclotron the period of revolution of a particle changes as its energy grows, and this harmful effect is removed by slowly varying (modulating) the frequency of the accelerating field. A particle of charge $q$ and mass $m$ circulates in a magnetic field of induction $B$ and gains the energy $\Delta W$ in each revolution. Find the law $\omega(t)$ by which the frequency of the accelerating field must be varied.

A cyclotron seen along its magnetic field B (drawn out of the page): two hollow dees separated by a narrow gap, fed by an alternating voltage from the oscillator. — A cyclotron seen along its magnetic field $\vect B$ (drawn out of the page): two hollow dees separated by a narrow gap, fed by an alternating voltage from the oscillator.

Given

$B$
$\Delta W$ per revolution
$q$
$m$

Find

the law

\omega(t)

of the accelerating field’s frequency

Solution

Step 1The drive must follow the revolution

The particle crosses the gap between the dees twice per revolution, and the field across the gap must reverse between the two crossings so that both accelerate it. So the accelerating field has to oscillate at the particle’s own angular frequency of revolution, at every moment of the acceleration.

A cyclotron seen along B: the particle crosses the gap between the two dees twice per turn, so the field E in the gap must reverse once every half turn. — A cyclotron seen along $\vect B$ : the particle crosses the gap between the two dees twice per turn, so the field $\vect E$ in the gap must reverse once every half turn.

Step 2The revolution frequency at any energy

On a circle of radius $r$ the momentum is $p = qBr$ and the speed is $v = pc^2/W$ , with $W$ the total energy including $mc^2$ . The angular frequency is the speed over the radius:

\omega = \frac{v}{r} = \frac{pc^2/W}{p/(qB)}

\omega = \frac{qBc^2}{W}

(1)

At the start $W = mc^2$ and this is $\omega_0 = qB/m$ , the cyclotron frequency; as the energy grows, $\omega$ falls.

Step 3How fast the energy grows

The energy rises by $\Delta W$ per revolution, and one revolution takes $2\pi/\omega$ . The modulation is slow, so the energy may be treated as rising smoothly at the average rate; using (1) for $\omega$ :

\frac{dW}{dt} = \frac{\Delta W}{2\pi/\omega} = \frac{\Delta W\,\omega}{2\pi} = \frac{\Delta W\,qBc^2}{2\pi\,W}

The right side depends on $W$ only through $1/W$ , so the variables separate.

Step 4The energy as a function of time

Multiplying by $W$ and integrating from $W = mc^2$ at $t = 0$ :

\begin{aligned} \int_{mc^2}^{W} W'\,dW' &= \frac{\Delta W\,qBc^2}{2\pi}\int_0^t dt', \\ \frac{W^2 - m^2c^4}{2} &= \frac{\Delta W\,qBc^2}{2\pi}\,t \end{aligned}

Solving for $W$ and taking $m^2c^4$ out of the root:

W = \sqrt{m^2c^4 + \frac{\Delta W\,qBc^2}{\pi}\,t} = mc^2\sqrt{1 + \frac{qB\,\Delta W}{\pi m^2c^2}\,t} \tag{2}

Step 5The law of the frequency

Putting (2) into (1), the factor $c^2$ cancels against $mc^2$ and leaves $qB/m = \omega_0$ in front:

\omega = \frac{qBc^2}{mc^2\sqrt{1 + qB\,\Delta W\,t/(\pi m^2c^2)}}

\begin{gathered} \omega = \frac{\omega_0}{\sqrt{1 + at}}, \\ \omega_0 = \frac{qB}{m}, \quad a = \frac{qB\,\Delta W}{\pi m^2c^2} \end{gathered}

(3)

The frequency of the accelerating field against time, falling as 1/sqrt(1+at); at the tick t = 3/a it has dropped to half its starting value. — The frequency of the accelerating field against time, falling as $1/\sqrt{1+at}$ ; at the tick $t = 3/a$ it has dropped to half its starting value.

Check

Units: $qB/m$ is $\text{C}\cdot\text{T}/\text{kg} = 1/\text{s}$ , a frequency; $a$ is $(qB/m)\cdot\Delta W/(\pi mc^2)$ , a frequency times a ratio of energies, so $at$ is a pure number. For small $at$ the law gives $\omega \approx \omega_0(1 - at/2)$ ; by then the energy has grown by $W - mc^2 \approx mc^2\,at/2 = \omega_0 t\,\Delta W/(2\pi)$ , which is $\Delta W$ times the $\omega_0 t/2\pi$ revolutions made, as it should be.

Remark

The law depends on time only through $W^2$ growing linearly, which follows from the energy gain per unit time falling as $1/W$ : a more energetic particle goes round more slowly and so visits the gap less often. This is the synchrocyclotron; because the frequency sweep can be followed by only one bunch of particles at a time, it delivers its beam in pulses, not continuously like a cyclotron.