Write the expression for the Coulomb’s law.
F→12el=14πε0q1q2s2s^ (I)
Here, F→12el is the electrostatic force between the charges, ε0 is the absolute permittivity, q1,q2 are the charges, s is the distance between the charges and s^ is the unit vector.
Write the equation for the electric force on a charged particle.
F→=E→q (II)
Here, F→ is the electric force, E→ is the electric field and q is the charge.
Write the equation for the electric force on charge q1 due to charge q2 using equation (II).
F→12el=E→(r→1)q1 (III)
Here, E→(r→1) is the electric field at r→1 due to charge q2 .
Write the equation for the electric field at r→1 due to charge q2 .
E→(r→1)=14πε0q2s2s^
Put the above equation in equation (III).
F→12el=14πε0q1q2s2s^
Write the expression for the magnitude of the above force.
F12el=14πε0q1q2s2 (IV)
Write the expression for the magnetic force on a charged particle.
F→=q(v→×B→)
Here, F→ is the magnetic force, v→ is the velocity of the particle and B→ is the magnetic field.
Write the equation for the magnetic force on charge q1 due to charge q2 using the above equation.
F→12mag=q1(v→1×B→(r→1)) (V)
Here, F→12mag is the magnetic force on charge q1 due to charge q2, v→1 is the velocity of charge q1 and B→(r→1) is the magnetic field at r→1 due to charge q2.
Write the equation for the magnetic field at r→1 due to charge q2.
B→(r→1)=μ04π q2s2v→2×s^
Here, μ0 is the absolute permeability and v→2 is the velocity of charge q2 .
Put the above equation in equation (V).
F→12mag=q1(v→1×μ04π q2s2v→2×s^)=q1[(μ04π q2s2)v→1×(v→2×s^)] (VI)
Write the identity of vector triple product.
A→×(B→×C→)=(A→⋅C→)B→−(A→⋅B→)C→
Use the above identity in equation (VI).
F→12mag=q1[(μ04π q2s2)[(v→1⋅s^)v→2−(v→1⋅v→2)s^]] (VII)
The vectors v→1 and v→2 are perpendicular so that their scalar product must be zero.
Write the expression for the scalar product of v→1 and v→2.
v→1⋅v→2=0
Put the above equation in equation (VII).
F→12mag=q1[(μ04π q2s2)[(v→1⋅s^)v→2−0]]=q1[(μ04π q2s2)(v→1⋅s^)v→2]=μ04π q1q2s2(v→1⋅s^)v→2
Write the expression for the magnitude of the above magnetic force.
F12mag=μ04π q1q2s2v1v2=(v1v2)(μ0ε0)14πε0 q1q2s2 (VIII)
Here, F12mag is the magnitude of the magnetic force on charge q1 due to charge q2, v1 is the magnitude of the velocity of charge q1 and v2 is the magnitude of the velocity of charge q2 .
Write the expression for the speed of light.
c2=1μ0ε0
Here, c is the speed of light.
Rewrite the above equation for μ0ε0 .
μ0ε0=1c2
Put the above equation in equation (VIII).
F12mag=(v1v2)(1c2)14πε0 q1q2s2 (IX)
Put equation (IV) in equation (IX).
F12mag=(v1v2)(1c2)F12el (X)
Conclusion:
From equation (VII) and (X), in which case the velocities are not perpendicular, the term (v1v2)F12el becomes larger than F12mag .
Rewrite equation (X) including the case velocities are not perpendicular.
F12mag≤(v1v2)(1c2)F12el
Thus, it is proved that F12mag≤v1v2c2F12el in nonrelativistic domain.