harrylin said:
Can you please elaborate?
Consider two long straight "line charges" with linear charge density λ C/m, separated by distance d. They repel each other with the following force per unit length:
$$f_E = \frac {\lambda^2}{2 \pi \epsilon_0 d}$$
Set those two line charges in motion along their length with speed v (or you move yourself with speed v).
Without taking relativity into account, the moving linear charge density becomes a current: I = λv. The (attractive) magnetic force per unit length between two long straight currents separated by distance d is:
$$f_B = \frac {\mu_0 I^2}{2 \pi d} = \frac {\mu_0 \lambda^2 v^2}{2 \pi d}$$
To find the speed at which the net force is zero, set the two forces equal to each other and solve for v. I'll let you do it.
Taking relativity into account, we work with the charge-current four-vector which has only two components in this situation (letting the lines lie along the x-axis): ##J^\mu = (c \rho, J, 0, 0)## where J is the current density in A/m
2 and ρ is the volume charge density in C/m
3. Between different inertial reference frames, J and ρ are related by a Lorentz transformation exactly the same way as x and t:
$$J^\prime = \gamma (J - v \rho)$$
$$\rho^\prime = \gamma \left( \rho - \frac{vJ}{c^2} \right)$$
In the frame where the line charges are stationary, J = I/A = 0 and ρ = λ/A, where A is the cross-section area of the charges. I'll leave it to you to apply the Lorentz transformation, along with I' = J'A and λ' = ρ'A, to verify that
$$I^\prime = - \gamma I = - \gamma v \lambda$$
$$\lambda^\prime = \gamma \lambda$$
(The "-" sign simply reflects the fact that if you start to move in the +x direction, the line charges move in the -x direction in your reference frame.)
Using these to find v which makes the net force zero, the γ's cancel and we have the same result as before.
