JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 24)
An electron revolves around an infinite cylindrical wire having uniform linear charge density $$2 \times 10^{-8} \mathrm{C} \mathrm{m}^{-1}$$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is ___________ $$\times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$$. Given mass of electron $$=9 \times 10^{-31} \mathrm{~kg}$$
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Answer
8
Explanation
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$$ \begin{aligned} & e E=\frac{\mathrm{mV}^2}{r} \\\\ & e \cdot \frac{2 \mathrm{~K} \lambda}{r}=\frac{\mathrm{mV}^2}{r} \\\\ & V=\sqrt{\frac{e \cdot 2 \mathrm{k} \lambda}{\mathrm{m}}} \\\\ & =\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}} \\\\ & =8 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned} $$
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