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(C) The radius $R$ of a charged particle moving in a uniform magnetic field is given by the formula $R = \frac{mv}{qB}$, where $m$ is the mass, $v$ is

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$\sqrt{3}R$

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(C) The radius $R$ of a charged particle moving in a uniform magnetic field is given by the formula $R = \frac{mv}{qB}$, where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength.Since the kinetic energy $K = \frac{1}{2}mv^2$, we can write $v = \sqrt{\frac{2K}{m}}$.Substituting this into the radius formula: $R = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.This shows that $R \propto \sqrt{K}$.Given that the new energy $K' = 3K$, the new radius $R'$ will be $R' = \sqrt{\frac{K'}{K}} R = \sqrt{\frac{3K}{K}} R = \sqrt{3}R$.

$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes $3$ times the original, the new radius will be

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