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# Comparison of Magnetic Force and electric forces Between two Moving Charges

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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