Question

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

Let r be the perpendicular separation between two moving protons and `vecv` be the velocity of motion of each proton. The magnetic field induction due to one moving proton at the another will be
`vecB=(mu_0)/(4pi)(e(vecvxxvecr))/(r^3)`
The force of magnetic interaction is
`vecF_m=e(vecvxxvecB)=e[vecvxx(mu_0)/(4pi)e((vecvxxvecr))/(r^3)]`
`=(mu_0)/(4pi)(e^2)/(r^3)[vecvxx(vecvxxvecr)]`
`=(mu_0)/(4pi)e^2/r^3[(vecv*vecr)vecv-(vecv*vecv)vecr]`
`=(mu_0)/(4pi)e^2/r^3xx(-v^2vecr)`
The force of electric interation is
`vecF_E=evecE=exx(1)/(4piin_0)(evecr)/(r^3)=(1)/(4piin_0)(e^2)/(r^3)vecr`
`(|vecF_m|)/(|vecF_E|)=-v^2mu_0in_0=v^2/c^2=((300xx1000)^2)/((3xx10^8)^2)=10^-6`