Q219 / 225 · JEE Main 2005 · Physics

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

  • A
    $\frac{2 \pi \mathrm{mq}}{\mathrm{B}}$
  • B
    $\frac{2 \pi q^2 B}{m}$
  • C
    $\frac{2 \pi q B}{m}$
  • Correct answer
    D
    $\frac{2 \pi m}{q B}$

Solution

$\mathrm{m} \omega^2 \mathrm{r}=\mathrm{Bq} \omega \mathrm{r}$ $\omega=\mathrm{Bq} / \mathrm{m}$ $\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$

Hence, the correct answer is option: 'D'.

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