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QJEEM23Ph9027: JEE Main 2023 Physics - Class 12 JEEM23Ph9027 Solution: An electron revolves around an infinite cylindrical wire having …

An electron revolves around an infinite cylindrical wire having uniform linear charge density 2×10−8Cm−12 \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 ___________ ×106 m s−1\times 10^{6} \mathrm{~m} \mathrm{~s}^{-1} . Given mass of electron =9×10−31 kg=9 \times 10^{-31} \mathrm{~kg}

JEE Main 2023 (Online) 10th April Evening Shift Physics - Electrostatics Question 57 English

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JEE Main 2023 (Online) 10th April Evening Shift Physics - Electrostatics Question 57 English Explanation

amp;eE=mV2ramp;e⋅2 Kλr=mV2ramp;V=e⋅2kλmamp;=1.6×10−19×2×9×109×2×10−89×10−31amp;=8×106 m/s\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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