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

Answer: (A)

$T = 2\pi r \sqrt{\frac{m}{2kq}}$

Answer 2

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:

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