= Cyclotron resonance condition
{title2=$\omega_{\mathrm{RF}}=\omega_c=|q|B/m$}
Take an <ion> of <mass> $m$ and <electric charge> $q\ne0$, with <velocity> perpendicular to a uniform <magnetic field> of magnitude $B>0$. In the <classical mechanics> limit, the <Lorentz force> provides the <centripetal acceleration>:
$$
|q|vB=\frac{mv^2}{r},\qquad
\omega_c=\frac vr=\frac{|q|B}{m},\qquad
r=\frac v{\omega_c}=\frac{mv}{|q|B}.
$$
Here $r$ is the local <radius> of curvature and $\omega_c$ is the <cyclotron frequency>, expressed as an <angular frequency>. A half-turn takes
$$
\tau_{1/2}=\frac{\pi r}{v}=\frac{\pi}{\omega_c}=\frac{\pi m}{|q|B}.
$$
Increasing <speed> increases the path length in exactly the same proportion. Drive the gap at <angular frequency> $\omega_{\mathrm{RF}}=\omega_c$, or ordinary <frequency> $f_{\mathrm{RF}}=|q|B/(2\pi m)$. In each half-turn the gap <electric field> reverses once, just as the particle reverses its direction of passage through the gap. The particle therefore receives two accelerating kicks per complete turn, using one oscillation per complete turn. This is the fundamental <resonance>; transit time through the gap and departures from a uniform <magnetic field> have been neglected.
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