Question

An electron revolves around an infinite cylindrical wire having uniform linear charge density \(2×10^{–8}\ Cm^{-1}\) 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 = \(9\times10^{-31}\) kg.

Answer 1

Correct Answer: 8

To find the velocity of an electron revolving around an infinite cylindrical wire with uniform linear charge density, we use the concept of electrostatic force acting as the centripetal force.

Step 1: Calculate Electrostatic Force

The linear charge density is given as \( \lambda = 2 \times 10^{-8} \, \text{C/m} \). The electric field \( E \) due to an infinite line of charge at a distance \( r \) is:

\( E = \frac{\lambda}{2\pi\epsilon_0 r} \)

where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \) is the permittivity of free space.

The electrostatic force \( F \) on the electron (charge \( -e \)) is:

\( F = eE = \frac{-e\lambda}{2\pi\epsilon_0 r} \)

Step 2: Equate Electrostatic and Centripetal Force

The centripetal force required for circular motion is provided by the electrostatic force:

\( \frac{mv^2}{r} = \frac{e\lambda}{2\pi\epsilon_0 r} \)

Simplify to find \( v \):

\( v^2 = \frac{e\lambda}{2\pi\epsilon_0 m} \)

\( v = \sqrt{\frac{e\lambda}{2\pi\epsilon_0 m}} \)

Given: \( e = 1.6 \times 10^{-19} \, \text{C} \) and \( m = 9 \times 10^{-31} \, \text{kg} \).

Step 3: Calculate \( v \) and Verify

Plug in the values:

\( v = \sqrt{\frac{(1.6 \times 10^{-19})(2 \times 10^{-8})}{2\pi(8.85 \times 10^{-12})(9 \times 10^{-31})}} \)

Simplifying further:

\( v \approx 8 \times 10^6 \, \text{m/s} \)

The velocity of the electron is \( 8 \times 10^6 \, \text{m/s} \), which lies within the given range of \(8,8\).