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A cyclotron accelerates protons (mass 1.67 x 10^-27 kg, charge 1.6 x…
Question
A cyclotron accelerates protons (mass 1.67 x 10^-27 kg, charge 1.6 x 10^-19 C) using a uniform magnetic field of magnitude 0.8 T applied perpendicular to the plane of motion. The dees have a maximum usable radius of 0.40 m. Find the maximum kinetic energy (in MeV) that a proton can gain in this cyclotron. (1 MeV = 1.6 x 10^-13 J.)
✓ Verified answer: 4.905389checked by our engine — not a guess
Step-by-step solution
At maximum radius r = mv/(qB), so v = qBr/m.
KE = (1/2) m v^2 = (qBr)^2/(2m).
Plug in: qBr = 1.6e-19 x 0.8 x 0.40 = 5.12e-20.
Square = 2.62144e-39.
Divide by 2m = 2 x 1.67e-27 = 3.34e-27: KE = 2.62144e-39/3.34e-27 = 7.8487e-13 J.
Convert: 7.8487e-13/1.6e-13 = 4.9054 MeV.
Final answer4.9054
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