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The beam energy is a total pion energy, so the momentum follows the relativistic relation $p^2{c}^2=E_{\pi }^2-m_{\pi }^2{c}^4$. Taking the standard charged-pion rest energy $m_{\pi }{c}^2=139.57$ MeV,^[ref]For motion perpendicular to a uniform magnetic field, the Lorentz force supplies the transverse momentum change and gives $r=p/(qB)$. With the stipulated $qB=1$ eV s m${}^{-2}$,The unit cancellation leaves length after dividing by $c$ in metres per second. Equivalently, the usual magnetic-rigidity conversion is $p[\mathrm{GeV}/c]\simeq 0.2998B[\mathrm{T}]r[\mathrm{m}]$ for unit charge. The sign of the pion charge chooses the bending direction, while the radius uses its magnitude. Treating the pion as massless would give about $0.831$ m and is not a good approximation at this modest total energy. The production target does not determine the downstream orbit once the pion momentum and magnetic field are specified.