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.

Answer

`1.00xx10^-6`

Explanation

The force of magnetic induction is given by
`vecF_(mag)=e(vecvxxvecB) Here vecB=(mu_0)/(4pi) (e(vecvxxvecr))/(r^3)`
`:. vecF_(mag)=(mu_0)/(4pi)xx(e^2)/(r^3)[vecvxx(vecvxxvecr)]`
`=(mu_0)/(4pi)xx(e^2)/(r^3)[(vecv.vecr)xxvecv-(vecv.vecv)xxvecr]`
`=(mu_0)/(4pi)xx(e^2)/(r^3)xx(-v^2vecr)`
`vecF_("ele.")=evecE=e 1/(4piepsilon_0) (evecr)/(r^3)`
`:. |F_(mag)|/|F_("ele.")|=v^2mu_0epsilon_0=(v/c)^2=1.00xx10^-6`