Two protons move parallel to each other with an equal velocity `v=300kms^-1`. Find the ratio of forces of magnetic and electric interaction of the protons.

Explanation

`F_(e) = (1)/(4pi in_(0)).(e.e)/(r^(2))`, repulsion
`F_(m) = (mu_(0))/(4pi).(e.e.v.v)/(r^(2))` attraction
`(F_(m))/(F_(e)) = mu_(0) in_(0) v^(2) = (v^(2))/(c^(2))(because c = (1)/(sqrt(mu_(0) in_(0))))`,
`= ((300 xx 10^(3))/(3 xx 10^(8)))^(2) = 10^(-6)`
`F_(m) lt lt F_(e)`
The path of protons will be