A 5.6 MeV (kinetic energy) proton enters a 0.39 T field, in a plane perpendicular to the field. What is the radius of its path?

the radius of the protons path is r = 0.85 m.

Explanation:

the force due to magnetic fields lead to the cetripetal force, such that:

F = q*v*B = m * (v^2) / r

q*B = m*v/r

then:

r = m*v/q*B

r = p/q*B

then, the kinetic energy of the proton:

K = 1/2*m*v^2 = p^2 / (2*m)

q*B = / sqrt{2*m*K}/r

r = / sqrt{2*m*K} / (q*B)

= / sqrt{2 * (1.67*10^-27) * (5.3*1.60*10^-13) } / (1.60*10^-19*0.39)

= 0.85 m