Question

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

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

Answer 1

The correct option is C

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

Explanation: for the correct option is (C).

Step 1:Formula used

For a charged particle of mass m and charge, q travels on a circular path of radius r that is perpendicular to the magnetic field B.

So, $\frac{m{v}^2}{r}=Bqv$

$$\begin{gathered} \frac{mv}{r}=Bq \\ v=\frac{Bqr}{m} \end{gathered}$$

Step 2:

As, the particle completes the one revolution.

Therefore,

$$\begin{gathered} v=\frac{2\mathrm{\pi r}}{t} \\ \frac{2\mathrm{\pi r}}{t}=\frac{Bqr}{m} \\ t=\frac{2\pi m}{qB} \end{gathered}$$

.Hence, The correct option is (C).