Two protons move parallel to each other with an equal velcity `v = 300 km//s`. Find the ration of forces of magentic and electricla of the protons.

Explanation

Force of magentic interaction `vec(F)_(mag) = e(vec(v) xx vec(B))`
Where, `vec(B) = (mu_(0))/(4pi) (e (vec(v) xx vec(r)))/(r^(3))`
So, `vec(F)_(mag) = (mu_(0))/(4pi) (e^(2))/(r^(3)) [vec(v) xx (vec(v) xx vec(r))]`
`= (mu_(0))/(4pi) (e^(2))/(r^(3)) [(vec(v). vec(r)) xx vec(v) - (vec(v) . vec(v)) xx vec(r)] = (mu_(0))/(4pi) (e^(2))/(r^(3)) (-v^(2) vec(r))`
And `vec(F)_("ele") = e vec(E) = e (1)/(4pi epsilon_(0)) (e vec(r))/(|vec(r)|^(3))`
Hence, `(|vec(F)_(mag)|)/(|vec(F)_("electric")|) = v^(2) mu_(0) epsilon_(0) = ((v)/(c))^(2) = 1.00xx10^(-6)`