Two equal point charges q are moving parallel to each other with the same speed v as shown in the given figure. These charges can be same visualised as parallel currents in same direction and hence they will apply magnetic attraction on each other. What will be the magnetic force exerted by one on another? At what speed this magnetic force gets balanced by the electric repulsion between them?

Explanation

We can assume the moving charge as a current element of magnitude qu. Magnetic field on perpendicular bisector of current element is given by:

`dB = (mu_(0))/(4 pi) (idl sin 90^(@))/x^(2) rArr B = mu_(0)/(4pi) (qv)/x^(2) `
Magnetic force on other change:
` F = qvB sin 90^(@) = qv mu_(0)/(4pi) (qv)/(x^(2)) `
` rArr F = mu_(0)/(4 pi) (q^(2)v^(2))/x^(2) `
If this magnetic attraction between charge is balanced by electric repulsion, then we can write the following equation:
`F = mu_(0)/(4 pi) (q^(2)v^(2))/x^(2) = 1/(4 pi epsi_(0)) q^(2)/x^(2) rArr v = 1/sqrt(mu_(0)epsi_(0)) = c`
Hence, the electric and magnetic forces will be balanced only when charges are moving with the speed of light parallel to each other. Also, if two charges are moving parallel to each other with a speed of light then they will apply net were force on each other.