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(D) When a charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ perpendicularly with velocity $v$,it follows a circular pat

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decreases with increase of specific charge of the particle

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(D) When a charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ perpendicularly with velocity $v$,it follows a circular path.The magnetic force provides the necessary centripetal force: $qvB = \frac{mv^2}{r}$.From this,the radius of the orbit is $r = \frac{mv}{qB}$.The time period $T$ is the time taken to complete one revolution: $T = \frac{2\pi r}{v}$.Substituting the value of $r$,we get $T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$.Since $T = \frac{2\pi m}{qB}$,the time period is independent of the velocity $v$ and the radius $r$ of the orbit.However,$T$ is inversely proportional to the specific charge $(q/m)$.Therefore,as the specific charge $(q/m)$ increases,the time period $T$ decreases.

If a charged particle enters a uniform magnetic field normally with a certain velocity,then the time period of revolution of the particle

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