A 3.5-MeV (kinetic energy) proton enters a 0.39-T field, in a plane perpendicular to the field....

Question

Question

A 3.5-MeV (kinetic energy) proton enters a 0.39-T field, in a plane perpendicular to the field.

Part A: What is the radius of its path?

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Answer 1

Answer #1

r^2 = 2 Km/(q B sin(phi))^2

r^2 = (2 x 3.5 x 10^6 x 1.6 x 10^-19 x 1.67 x 10^-27)/(1.6 x 10^-19 x 0.39)^2

Radius, r = 0.69 m

Comment in case any doubt please rate my answer....

Answer 2

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