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(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it undergoes uniform circular motion.The magnetic force prov

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Depends on magnetic field and $(q/m)$ of the particle

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(D) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it undergoes uniform circular motion.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 full circle: $T = \frac{2\pi r}{v}$.Substituting the value of $r$: $T = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}$.Thus,the time period $T$ is independent of velocity $v$ and radius $r$,and depends only on the mass $m$,charge $q$,and magnetic field $B$. Therefore,it depends on the magnetic field and the charge-to-mass ratio $(q/m)$.

$A$ particle of mass $m$ and carrying a charge $q$ enters with a velocity $v$ perpendicular to a uniform magnetic field $B$. The time period of rotation of the particle:

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