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JEE Main 2022 · 27 Jul · Shift 2, Physics question 43

A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be :

[use mp=1.6×10−27m_p=1.6\times10^{-27} kg, e=1.6×10−19e=1.6\times10^{-19} C]

Choose one option
Answer and worked solution

Official answer: B

Kinetic energy q2B2r22m=(1.6×10−19)2(0.6)23.2×10−27=2.88×10−12\frac{q^2B^2r^2}{2m}=\frac{(1.6\times10^{-19})^2(0.6)^2}{3.2\times10^{-27}}=2.88\times10^{-12} J =18=18 MeV.

(B)

Topics:Motion in magnetic and electric fields
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