1. The problem statement, all variables and given/known data A 5.6 MeV (kinetic energy) proton enters a 0.18 T field, in a plane perpendicular to the field. What is the radius of its path? 2. Relevant equations kinetic energy: KE = 1/2 mv^2 velocity: v = sqrt(2KE / m) radius: r = mV/qB 3. The attempt at a solution KE = 5.6 MeV = 8.9722x10^-13 J m(p) = 1.67x10^-27 kg so v = 3.3x10^7 m/s r = (m(p))(v) / (q(e))(B) where m(p) is mass of proton v is velocity as determined by manipulating the KE equation q(e) is the charge of the proton which is the same as that of an electron (except positive) B is the given magnetic field the homework system I'm on, keeps telling me that my answer is wrong. if you see what i'm doing wrong or where i've made a mistake, please let me know.