A5.6 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  

answer would be d.) i hope this

explanation:

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