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 :
$$[\mathrm{use} \,\,\mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \,\mathrm{C}$$ ]
Solution
<p>$R = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}$</p>
<p>$\Rightarrow K = {{{B^2}{q^2}{R^2}} \over {2m}}$</p>
<p>$$ = {{{{(1.6 \times {{10}^{ - 19}})}^2} \times {{0.6}^2}} \over {2 \times 1.6 \times {{10}^{ - 27}}}}$$ J</p>
<p>$= 18$ MeV</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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