Question

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?

Answer 1

 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