JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Medium
- Numerical
An electron revolves around an infinite cylindrical wire having uniform linear change density \(2 \times 10^{-8}\,Cm ^{-1}\) 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 \(.........\times 10^6\,ms ^{-1}\). Given mass of electron \(=9 \times 10^{-31}\,kg\)

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Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
\(e E=\frac{ mV ^2}{ r }\)
\(e \cdot \frac{2 K \lambda}{ r }=\frac{ mV ^2}{ r }\)
\(V =\sqrt{\frac{e \cdot 2 k \lambda}{ m }}\)
\(=\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}}\)
\(=8 \times 10^6\,m / s\)
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