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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