Cyclotron — working principle, frequency, and limitations

Question

Explain the working principle of a cyclotron. Derive the expression for the cyclotron frequency and maximum kinetic energy of accelerated particles. Why can’t a cyclotron accelerate electrons?

(JEE Main 2022, similar pattern — also a favourite in CBSE and NEET)

Solution — Step by Step

A cyclotron uses a combination of a uniform magnetic field and an oscillating electric field to accelerate charged particles in a spiral path. The magnetic field makes the particle move in circles, while the electric field (applied in the gap between two D-shaped electrodes called “dees”) accelerates the particle each time it crosses the gap.

The crucial insight: the time taken for one semicircle is independent of the particle’s speed. As the particle speeds up, it moves in a larger circle but takes the same time — so the electric field frequency can remain constant.

For a charged particle (charge $q$ , mass $m$ ) moving with speed $v$ in a magnetic field $B$ :

qvB = \frac{mv^2}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}

Time for one complete revolution:

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Cyclotron frequency:

\boxed{f = \frac{1}{T} = \frac{qB}{2\pi m}}

This frequency is independent of $v$ and $r$ — this is the resonance condition that makes the cyclotron work.

The maximum radius equals the dee radius $R$ : $R = \frac{mv_{max}}{qB}$ , so $v_{max} = \frac{qBR}{m}$ .

\boxed{KE_{max} = \frac{1}{2}mv_{max}^2 = \frac{q^2B^2R^2}{2m}}

Electrons have extremely small mass ( $9.1 \times 10^{-31}$ kg). For the same energy input, electrons reach relativistic speeds very quickly. At relativistic speeds, the mass increases ( $m = \gamma m_0$ ), which changes the cyclotron frequency — breaking the resonance condition. The particle goes out of sync with the oscillating electric field and stops gaining energy.

Protons and heavier ions remain non-relativistic for much longer, so the cyclotron works well for them.

Why This Works

The magic of the cyclotron lies in the mass-velocity cancellation. A faster particle has a larger radius ( $r \propto v$ ), but it also covers the larger semicircle faster (circumference $\propto v$ ). These effects exactly cancel, making the revolution time $T = 2\pi m/(qB)$ constant. This is what allows a single fixed-frequency oscillator to keep accelerating the particle.

Alternative Method

You can derive the maximum KE by counting the total voltage gained. If the particle crosses the gap $2n$ times (for $n$ full revolutions) with voltage $V$ each time, then $KE = 2nqV$ . But since $n$ depends on $R$ , it is cleaner to use the radius relation directly: $KE = q^2B^2R^2/(2m)$ .

In NEET, the most commonly asked numerical involves finding the cyclotron frequency or maximum KE. Remember that $f$ depends only on $q$ , $B$ , and $m$ — not on the dee radius. The maximum KE depends on $R$ (dee radius), so a bigger cyclotron produces higher energy particles.

Common Mistake

Students often state “the cyclotron cannot accelerate neutral particles.” While true, this is not the limitation the question asks about. The specific limitation for electrons is the relativistic mass increase — not neutrality. Also, some write “electrons are too light to be deflected by the magnetic field.” That is wrong — electrons are deflected just fine; the problem is that they become relativistic too fast for the fixed-frequency oscillator to keep up.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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