A 5.0 MeV (kinetic energy) proton enters a 0.21 T field, in a plane perpendicular to the field. What is the radius of its path?

The force due to the magnetic field generates a centripetal force on the proton.

F = qvB = mv^2/r

qB = mv/r

qB = p/r

kinetic energy (K) = mv^2/2 = p^2 / (2m)

qB = √ (2mK) / r

r = √ (2mK) / (qB)

r = √ [ (2) (1.67e-27 kg) (5.0 MeV * 1.60e-13 J/MeV) ] / [ (1.60e-19 C) (0.20 T) ]

r = 1.6 m