Question

A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then the time period will be given as:

• A. \( T = 2\pi r \sqrt{\frac{m}{2kq}} \)
• B. \( T^2 = \frac{4\pi m r^3}{2kq} \)
• C. \( T = \frac{1}{2\pi r} \sqrt{\frac{m}{2kq}} \)
• D. \( T = \frac{2kq}{m} \)

Answer 1

The Correct Option is A

The electric field \( E \) due to an infinitely long line charge with linear charge density \( +\lambda \) at a distance \( r \) from the line charge is given by:

\[ E = \frac{\lambda}{2 \pi \epsilon_0 r}, \]

where \( \epsilon_0 \) is the permittivity of free space.

Force on the Charged Particle: The force \( F \) acting on the particle due to the electric field is:

\[ F = -qE = -q \left( \frac{\lambda}{2 \pi \epsilon_0 r} \right). \]

Since the particle moves in a circular path, this force provides the centripetal force necessary for circular motion:

\[ F = \frac{mv^2}{r}. \]

Equating the Forces: Setting the electric force equal to the centripetal force:

\[ -q \left( \frac{\lambda}{2 \pi \epsilon_0 r} \right) = \frac{mv^2}{r}. \]

Rearranging gives:

\[ mv^2 = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Finding the Time Period: The velocity \( v \) can also be expressed in terms of the radius and the time period \( T \):

\[ v = \frac{2 \pi r}{T}. \]

Substituting this expression for \( v \) into the equation:

\[ m \left( \frac{2 \pi r}{T} \right)^2 = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Simplifying gives:

\[ m \times \frac{4 \pi^2 r^2}{T^2} = \frac{q \lambda}{2 \pi \epsilon_0}. \]

Rearranging for \( T^2 \):

\[ T^2 = \frac{4 \pi m r^2 \epsilon_0}{q \lambda}. \]

Final Expression: To match the answer choices, if we express \( k = \frac{1}{4 \pi \epsilon_0} \):

\[ T^2 = \frac{4 \pi m r^2}{2 k q}. \]

Thus, the time period is:

\[ T = 2 \pi r \sqrt{\frac{m}{2 k q}}. \]