In this article we would be looking at Comparison of Magnetic Force and electric forces Between two Moving Charges

Suppose two charges $q_{1}$ and $q_{2}$ are moving with velocities $v_{1}$ and $v_{2}$ respectively. Let r be the distance between the particles at any instant. Now two forces would be acting on each charges at this particular instant

**a) Electric Force:-** which is given by

$F_{e}=\frac{1}{4\pi \varepsilon {0}}\frac{q{1}q_{2}}{r^{2}}$

**b) Magnetic Forces:- **which is due to the instantanous magnetic field setup by the moving charges and can be calculated as given below

The instantanous magnetic field setup by $q_{1}$ at the point of space where $q_{2}$ is situated is given by

$B=\frac{\mu *{0}}{4\pi}\frac{q*{1}v_{1}sin\theta}{r^{2}}$

where $\theta$ is the angle between $v_{1}$ and r . The magnetic force on the charge $q_{2}$ due to this magnetic field

$F=q_{2}v_{2}B sin\phi$

where $\phi$ is the angle between $v_{2}$ and B

So magnetic force will be

$F_{m}=\frac{\mu_{0}}{4\pi}\frac{q_{1}q_{2}v_{1}v_{2}sin\theta sin\phi}{r^2}$

if $v=v_{2}=v_{1}$ and $\theta =\phi =90^0$ , then

$F_{m}=\frac{\mu_{0}}{4\pi}\frac{q_{1}q_{2}v^2}{r^2}$

So

$\frac{F_{m}}{F_{e}}=\mu_{0}\epsilon_{0}v^2$

We know that

$\mu_{0}\epsilon_{0}=\frac{1}{c^2}$

Where c is the velocity of the light in free space So ,

$\frac{F_{m}}{F_{e}}=\left ( \frac{v}{c} \right )^2$

In general v <<<<c So magnetic force are weaker then electric forces.

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