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Question

(B) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular m

Vedclass · Accepted Answer

$\frac{2\pi m}{qB}$

Vedclass · Answer

(B) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.Equating the magnetic force to the centripetal force:$qvB = \frac{mv^2}{r}$From this,we can find the velocity $v$ or the radius $r$:$v = \frac{qBr}{m}$The time period $T$ for one complete revolution is the circumference divided by the speed:$T = \frac{2\pi r}{v}$Substituting the expression for $v$:$T = \frac{2\pi r}{(qBr/m)} = \frac{2\pi m}{qB}$Thus,the time taken to complete one revolution is $\frac{2\pi m}{qB}$.

$A$ charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is

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