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JEE Main 2022 · 30 Jun · Shift 1, Physics question 43

A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will be :

(Take : e=1.6×10−19e=1.6\times10^{-19} C, mp=1.67×10−27m_p=1.67\times10^{-27} kg)

Choose one option
Answer and worked solution

Official answer: A

Exit speed 2πfR=3.77×1072\pi fR=3.77\times10^7 m s−1^{-1}; K=12mpv2≈1.19×10−12K=\frac12m_pv^2\approx1.19\times10^{-12} J ≈7.4\approx7.4 MeV.

(A)

Topics:Motion in magnetic and electric fields
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