JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 11)
A cyclotron is used to accelerate protons. If the operating magnetic field is $$1.0 \mathrm{~T}$$ and the radius of the cyclotron 'dees' is $$60 \mathrm{~cm}$$, the kinetic energy of the accelerated protons in MeV will be :
$$[\mathrm{use} \,\,\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \,\mathrm{C}$$ ]
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18
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Explanation
$$R = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}$$
$$ \Rightarrow K = {{{B^2}{q^2}{R^2}} \over {2m}}$$
$$ = {{{{(1.6 \times {{10}^{ - 19}})}^2} \times {{0.6}^2}} \over {2 \times 1.6 \times {{10}^{ - 27}}}}$$ J
$$= 18$$ MeV
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