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

Find velocity from 1/2*mv^2 = 5.0MeV
v^2 =10MeV/m
v= sqrt(10MeV/m)

F=qVXB = F centripetal =(mv^2)/r

e*sqrt(10MeV/m)*B = ((10MeV/m)*m)/r

\[r = \frac{ 10MeV }{eB \sqrt{\frac{ 10MeV }{ m}} }= \frac{ \sqrt{ 10MeV *m} }{ eB }\]