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(D) When a charged particle moves in a perpendicular magnetic field, it undergoes uniform circular motion.The magnetic force provides the necessary ce

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$16$

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(D) When a charged particle moves in a perpendicular magnetic field, it undergoes uniform circular motion.The magnetic force provides the necessary centripetal force: $\frac{mv^2}{R} = qvB$.This simplifies to $v = \frac{qBR}{m}$.The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$.Since $q$, $m$, and $R$ are constant, $E \propto B^2$.If the magnetic field $B$ is increased by $4$ times $(B' = 4B)$, the new kinetic energy $E'$ will be:$E' \propto (4B)^2 = 16B^2$.Therefore, $E' = 16E$. The kinetic energy increases by $16$ times.

$A$ proton moving in a perpendicular magnetic field possesses energy $E$. The strength of the magnetic field is increased four times, but the proton is constrained to move in a path of the same radius. The kinetic energy of the proton will increase by: (in $\times$)

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