JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 15)
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 time period will be given as :
(Consider $$k$$ as Coulomb's constant)
$$T^2=\frac{4 \pi^2 m}{2 k \lambda q} r^3$$
$$T=\frac{1}{2 \pi r} \sqrt{\frac{m}{2 k \lambda q}}$$
$$T=\frac{1}{2 \pi} \sqrt{\frac{2 k \lambda q}{m}}$$
$$T=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$$
Explanation
$$\begin{aligned}
& \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\
& \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\
& \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}}
\end{aligned}$$
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