Question
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? See Section $23-8$.
Step 1
We know that 1 eV = $1.6 \times 10^{-19}$ J, so 6 MeV = $6 \times 10^{6}$ eV = $6 \times 10^{6} \times 1.6 \times 10^{-19}$ J = $9.6 \times 10^{-13}$ J. Show more…
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