A particle of charge ' "-q" ' and mass ' "m" ' moves in a circle of radius ' "r" ' around an infinitely long line charge of linear charge density ' `+lambda` '.Then time period will be given as :
(Consider "k" as Coulomb's constant)

Options

A. `T=2 pi r sqrt((m)/(2k lambda q))`
B. `T=(1)/(2 pi)sqrt((2k lambda q)/(m))`
C. `T=(1)/(2 pi r)sqrt((m)/(2k lambda q))`
D. `T^(2)=(4 pi^(2)m)/(2k lambda q)r^(3)`

Answer

To find the time period \( T \) of a particle of charge \( -q \) and mass \( m \) moving in a circle of radius \( r \) around an infinitely long line charge with linear charge density \( +\lambda \), we will follow these steps:

### Step 1: Determine the Electric Field
The electric field \( E \) created by an infinitely long line charge with linear charge density \( \lambda \) at a distance \( r \) from the line charge is given by the formula:

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

### Step 2: Calculate the Electric Force on the Particle
The electric force \( F \) acting on the particle due to the electric field is given by:

\[ F = qE \]

Since the charge of the particle is \( -q \), we have:

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

### Step 3: Set the Electric Force Equal to Centripetal Force
For the particle to move in a circular path, the electric force must provide the necessary centripetal force. The centripetal force \( F_c \) is given by:

\[ F_c = \frac{mv^2}{r} \]

Setting the magnitudes of the electric force and centripetal force equal gives:

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

### Step 4: Solve for Velocity \( v \)
Rearranging the equation to solve for \( v^2 \):

\[ q\lambda = 2 \pi \epsilon_0 mv^2 \]

\[ v^2 = \frac{q\lambda}{2 \pi \epsilon_0 m} \]

### Step 5: Relate Velocity to Angular Velocity \( \omega \)
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:

\[ v = r\omega \]

Substituting \( v \) into the equation gives:

\[ (r\omega)^2 = \frac{q\lambda}{2 \pi \epsilon_0 m} \]

### Step 6: Solve for Angular Velocity \( \omega \)
Taking the square root:

\[ r^2 \omega^2 = \frac{q\lambda}{2 \pi \epsilon_0 m} \]

\[ \omega^2 = \frac{q\lambda}{2 \pi \epsilon_0 m r^2} \]

### Step 7: Find the Time Period \( T \)
The time period \( T \) is related to angular velocity \( \omega \) by:

\[ T = \frac{2\pi}{\omega} \]

Substituting the expression for \( \omega \):

\[ T = 2\pi \sqrt{\frac{m r^2}{q\lambda / (2 \pi \epsilon_0)}} \]

Simplifying gives:

\[ T = 2\pi r \sqrt{\frac{m}{q\lambda / (2 \pi \epsilon_0)}} \]

### Step 8: Substitute Coulomb's Constant
Using \( k = \frac{1}{4\pi \epsilon_0} \), we can express \( \epsilon_0 \) in terms of \( k \):

\[ \epsilon_0 = \frac{1}{4\pi k} \]

Substituting this into the equation for \( T \):

\[ T = 2\pi r \sqrt{\frac{m}{\frac{q\lambda}{(2 \pi)(\frac{1}{4\pi k})}}} \]

This simplifies to:

\[ T = 2\pi r \sqrt{\frac{4\pi k m}{q\lambda}} \]

### Final Expression
Thus, the final expression for the time period \( T \) is:

\[ T = 2\pi r \sqrt{\frac{4\pi k m}{q\lambda}} \]