JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(=9 \times 10^{-31}\,kg\)
- A \(4\)
- B \(2\)
- C \(8\)
- D \(6\)
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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