A 6.0-MeV (kinetic energy) proton enters a 0.20-T field, in a plane perpendicular to the field. What is the radius of its path? 0.6 m 1.8 m 2.0 m 2.4 m
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0 MeV = 6.0 x \(10^6\) eV = 6.0 x \(10^6\) x 1.6 x \(10^{-19}\) J = 9.6 x \(10^{-13}\) J Mass of proton, m = 1.67 x \(10^{-27}\) kg Show more…
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