Two protons move parallel to the x-axis in opposite directions
at the same speed v (small compared to the speed of light c).
At the instant shown, find the electric and magnetic forces on the upper
proton and compare their magnitudes.
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Concept and Principle:Electric force upper proton due to lower proton is given by Coulomb's law, according to which the force FE experienced by a charge q1 due to another charge q2 at a distance r is given by,FE→=14πε0q1q2r2r^Now the magnetic force is given by,FB→=q(-v→)×B→where B is the magnetic field and v the velocity of the particle.Derivation:Applying Coulomb's law, on upper proton,FE→=14πε0q1q2r2y^q1=q2=qFE→=14πε0q2r2y^ ----→(1)Now to find magnetic force due to lower charge on upper charge, we need magnetic field which is given by,B→=μ04πqv→×r^r2=μ04πqvr2y^Now force due to magnetic field is,FB→=q(-v→)×B→=q(-v x^)×μ04πqvr2y^=-μ04πq2v2r2z^FB→=-μ04πq2v2r2z^ ----→(2)Now comparing the magnitudes |FB→||FE→|=14πε0q2r2μ04πq2v2r2=μ0ε0v2But, μ0ε0=1c2So,|FB→||FE→|=μ0ε0v2=v2c2|FB→||FE→|=v2c2Answer:The ratio of magnitudes is found to be v2c2 ...