Question

A particle of charge $–q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear charge density $+\lambda .$ Then the time period will be

• A. $T=2\pi r\sqrt{\frac{m}{2k\lambda q}}$
• B. $T^2=\frac{4{\pi }^{2}m}{2k\lambda q}{r}^3$
• C. $T=\frac{1}{2\pi r}\sqrt{\frac{2k\lambda q}{m}}$
• D. $T=\frac{1}{2\pi r}\sqrt{\frac{m}{2k\lambda q}}$

Answer 1

The correct option is A $T=2\pi r\sqrt{\frac{m}{2k\lambda q}}$

The electrostatic force acting on the point charge provides the necessary centripetal force for it,

Electrostatic force = Centripetal force

$qE=\frac{m{v}^2}{r}$

$\because \text{Electric field due to an infinite wire}E=\frac{2k\lambda }{r}$

$q\left(\frac{2k\lambda }{r}\right)=\frac{m{v}^2}{r}$

So $v=\sqrt{\frac{2kq\lambda }{m}}$

Now, time period is

$T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{m}{2k\lambda q}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.