Question

Two protons move parallel to each other, keeping distance r between them, both moving with same velocity , then the ratio of electric and magnetic force of interaction between them is :-

Answer 1

c^2/v^2

Answer 2

The problem asks for the ratio of the electric and magnetic forces between two protons moving parallel to each other with the same velocity $v$, separated by a distance $r$.

1. Electric Force ($F_e$)

The electric force between two protons, each with charge $e$, separated by a distance $r$, is given by Coulomb's Law:

$F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$

This force is repulsive.

2. Magnetic Force ($F_m$)

Each moving proton constitutes a current, and these currents interact magnetically. Consider one proton moving with velocity $\vec{v}$. It produces a magnetic field $\vec{B}$ at the position of the second proton. For a point charge $e$ moving with velocity $\vec{v}$, the magnetic field at a perpendicular distance $r$ is:

$B = \frac{\mu_0}{4\pi} \frac{ev}{r^2}$

The force experienced by the second proton due to this magnetic field is given by the Lorentz force formula: $\vec{F_m} = e (\vec{v} \times \vec{B})$. Since the velocities are parallel and the magnetic field produced by one proton at the location of the other is perpendicular to its velocity, the magnitude of the magnetic force is:

$F_m = evB = ev \left( \frac{\mu_0}{4\pi} \frac{ev}{r^2} \right)$

$F_m = \frac{\mu_0}{4\pi} \frac{e^2v^2}{r^2}$

This force is attractive.

3. Ratio of Electric to Magnetic Force ($F_e/F_m$)

Now, we find the ratio of the magnitudes of the electric and magnetic forces:

$\frac{F_e}{F_m} = \frac{\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}}{\frac{\mu_0}{4\pi} \frac{e^2v^2}{r^2}}$

Cancel common terms ($e^2/r^2$ and $4\pi$):

$\frac{F_e}{F_m} = \frac{1/\epsilon_0}{\mu_0 v^2} = \frac{1}{\mu_0 \epsilon_0 v^2}$

We know that the speed of light in vacuum ($c$) is related to $\mu_0$ and $\epsilon_0$ by the relation:

$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \implies \mu_0 \epsilon_0 = \frac{1}{c^2}$

Substitute this into the ratio:

$\frac{F_e}{F_m} = \frac{1}{(1/c^2) v^2} = \frac{c^2}{v^2}$

The ratio of electric and magnetic force of interaction between them is $c^2/v^2$.