 Proton exclusion by magnetic rigidity and range

For a concrete independent example, stipulate perpendicular motion, $|q|=e$ with $e$ the elementary charge, uniform magnetic field $B=1.00\mathrm{T}$, incoming gyroradius $r_{\mathrm{in}}=0.100\mathrm{m}$ and outgoing gyroradius $r_{\mathrm{out}}=0.0500\mathrm{m}$. These are invented teaching inputs. Converting SI momentum to electronvolt units givesso $p_{\mathrm{in}}=29.9792458\mathrm{MeV}/c$ and $p_{\mathrm{out}}=14.9896229\mathrm{MeV}/c$. Both a Positron and a proton with these momenta trace the same ideal arcs. Their kinetic energies differ because the relativistic energy-momentum relation givesThe second expression avoids subtracting nearly equal quantities when $pc\ll m{c}^2$. For this calculation stipulate rounded rest energies $m_e{c}^2=0.511\mathrm{MeV}$ and $m_p{c}^2=938.272\mathrm{MeV}$. The Positron interpretation gives $K_{\mathrm{in}}\simeq 29.473\mathrm{MeV}$ and $K_{\mathrm{out}}\simeq 14.487\mathrm{MeV}$; the proton interpretation gives only $0.479\mathrm{MeV}$ and $0.120\mathrm{MeV}$. The corresponding proposed losses of kinetic energy are about $14.985\mathrm{MeV}$ and $0.359\mathrm{MeV}$.

The small proton kinetic energy follows from $K\simeq p^2/(2m_p)$, whereas here the Positron has $pc\gg m_e{c}^2$ and $K\simeq pc-m_e{c}^2$. Consequently magnetic bending alone cannot reject the proton. One needs an independent material constraint: could a proton at the proposed kinetic energy penetrate the plate and then travel at least the observed visible distance in the specified gas? A particle range shorter than that visible distance contradicts the hypothesis. The visible track need not include the stopping point: it already supplies a lower bound on particle range. No numerical particle range is assigned to this teaching example; that would require stopping data for the actual material and gas conditions. The experimental insight is to combine a geometric momentum measurement with an independent test of penetration.