JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Force Between Two Moving Charges

If two charges \[{{q}_{1}}\] and \[{{q}_{2}}\] are moving with velocities v1 and v2 respectively and at any instant the distance between them is r, then .

Magnetic force between them is \[{{F}_{m}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{{{q}_{1}}{{q}_{2}}\,{{v}_{1}}{{v}_{2}}}{{{r}^{2}}}\] .... (i)

and Electric force between them is \[{{F}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]  .... (ii)

From equation (i) and (ii) \[\frac{{{F}_{m}}}{{{F}_{e}}}={{\mu }_{0}}{{\varepsilon }_{0}}{{v}^{2}}\] but \[{{\mu }_{0}}{{\varepsilon }_{0}}=\frac{1}{{{c}^{2}}}\]; 

where c is the velocity of light in vacuum. So \[\frac{{{F}_{m}}}{{{F}_{e}}}={{\left( \frac{v}{c} \right)}^{2}}\]

As \[v<c\] so \[{{F}_{m}}<{{F}_{e}}\]