Cyclotron energy and orbit radius ($K=|q{|}^2{B}^2{r}^2/(2m)$)

Combining $r=mv/(|q|B)$ with the classical kinetic energy givesHere $E$ denotes kinetic energy, not relativistic total energy. At fixed magnetic field, twice the final radius allows four times the classical kinetic energy. At fixed final kinetic energy, a larger accelerating gap voltage reduces the number of crossings needed, but does not change the required orbit radius.

For an illustrative proton with $B=1.00\mathrm{T}$ and peak gap voltage $V_0=10.0\mathrm{kV}$, take negligible injection kinetic energy, on-crest kicks, a zero-width gap, and no losses. Each crossing adds $10.0\mathrm{keV}$, so $100$ crossings give $1.00\mathrm{MeV}$, using electronvolts. With $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_p=1.67262192595\times 10^{-27}\mathrm{kg}$,The $100$th kick arrives after $99$ intervening half-turns, about $3.247\mu \mathrm{s}$ after the first kick in this idealization. The final two curvature radii differ by $r(1\mathrm{MeV})[1-\sqrt{99/100}]\simeq 0.724\mathrm{mm}$. That difference is a curvature-radius increment, not a general formula for separation of real extracted turns: successive arc centers shift, and extraction also depends on focusing and the gap geometry. These chosen settings are a calculation, not historical measurements.