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A charged particle of mass $\mathrm{m}$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $\mathrm{B}$. The time taken by the particle to complete one revolution is
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The correct answer is:
$\frac{2 \pi m}{q B}$
$\frac{2 \pi m}{q B}$
$\mathrm{m} \omega^2 \mathrm{r}=\mathrm{Bq} \omega \mathrm{r}$
$\omega=\mathrm{Bq} / \mathrm{m}$
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$
$\omega=\mathrm{Bq} / \mathrm{m}$
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$
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