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A charged particle moves perpendicular to a uniform magnetic field $B$ with speed $v$. The radius of its circular path is:
A$r = \dfrac{qB}{mv}$
B$r = \dfrac{mB}{qv}$
C$r = \dfrac{mv}{qB}$
D$r = \dfrac{qv}{mB}$
Answer & Solution
Correct answer: C. $r = \dfrac{mv}{qB}$
When $\vec{v} \perp \vec{B}$, the magnetic force $qvB$ provides the centripetal force:
$\dfrac{mv^2}{r} = qvB \Rightarrow r = \dfrac{mv}{qB}$.
This is the *cyclotron radius* (gyroradius). The corresponding angular frequency $\omega = \dfrac{qB}{m}$ is the cyclotron frequency, independent of speed — that's why cyclotron particle accelerators work: heavier or slower particles take longer paths but the period stays constant.
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