ExamBro
ExamBro
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 \(=9 \times 10^{-31}\,kg\)

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(C) \(8\)

Step-by-step Solution

Detailed explanation

\(e E=\frac{ mV ^2}{ r }\)

\(e \cdot \frac{2 K \lambda}{ r }=\frac{ mV ^2}{ r }\)

\(V =\sqrt{\frac{e \cdot 2 k \lambda}{ m }}\)

\(=\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}}\)

\(=8 \times 10^6\,m / s\)

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app