The maximum velocity to which a proton can be accelerated in a cyclotron of $10\, MHz$ frequency and radius $50\, cm$ is
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The motion of proton in magnetic field will be circular
$\therefore v=2 \pi r f=2 \times 3.14 x 50 \times 10^{-2} \times 10 \times 10^6$
$=3.14 \times 10^7\,ms ^{-1}$
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