Magnetic Force and Electric Force

Magnetic force and electric force are actually the same thing. We will see using a simple example. Suppose we have a continuous positive line charge with charge density $\lambda_0$, and a continuous negative line charge with charge density $-\lambda_0$. Now, in our inertial frame $S$, the positive line is moving from left to right at speed $v$ while the negative line is moving from right to left, also, at speed $v$. The net current is then $$I = 2\lambda v$$ to the right, where $\lambda$ is the observed line charge density in our frame $S$ $$\lambda = \gamma\lambda_0 = \frac{1}{\sqrt{1-v^2/c^2}}\lambda_0.$$ A point charge $q$ at distance $s$ away is moving at speed $u$ < $v$ to the right. Because the line charge cancel each other, there is no electrical force acting on $q$.

In another inertial frame, $\bar{S}$, which is moving to the right at speed $u$ relative to our original inertial frame. Using the Einstein velocity addition rule, the velocity of the positive and negative lines are $$\bar{v}_{\pm} = \frac{v\mp u}{1 \mp vu/c^2}.$$ Since the speed of the negative line is larger than that of the positive line, the Lorentz contraction of the negative line is more severe than in the positive line. The line charge density of the two lines after applying the Lorentz contraction become $$\bar{\lambda}_{\pm}=\pm(\gamma_{\pm})\lambda_0,$$ where $\gamma_{\pm}$ is given by $$\begin{align} \gamma_{\pm} &= \frac{1}{\sqrt{1-\frac{1}{c^2}(v\mp u)^2(1\mp vu/c^2)^{-2}}}\\ &= \frac{c^2\mp uv}{\sqrt{c^2\mp uv)^2-c^2(v\mp u)^2)}}\\ &= \frac{c^2\mp uv}{\sqrt{(c^2-v^2)(c^2-u^2)}} \\ &= \gamma \frac{1\mp uv/c^2}{\sqrt{1-u^2/c^2}} \end{align}$$ Hence, there is a net negative charge of the system $$\bar{\lambda}_{\text{net}}=\bar{\lambda}_{+}+\bar{\lambda}_{-} = \lambda_0(\gamma_{+}-\gamma_{-})=-\frac{2\lambda vu}{c^2\sqrt{1-u^2/c^2}}.$$ The electric field set up by this net line charge density is given by $$\bar{E} = \frac{\lambda_{\text{net}}}{2\pi\epsilon_0s}$$ The electric force acted on $q$ in $\bar{S}$ is $$\bar{F}=q\bar{E} = \frac{\lambda v}{\pi\epsilon_0c^2s}\frac{qu}{\sqrt{1-u^2/c^2}},$$ perpendicular to $\vec{u}$ since $\bar{E}$ by the net charge is perpendicular to the lines. This force in our original inertial frame $S$ will be $$\begin{align} F &= \sqrt{1-u^2/c^2}\bar{F}\\ &= -\frac{\lambda v}{\pi\epsilon_0c^2}\frac{qu}{s}\\ &= - qu\left(\frac{\mu_0 I}{2\pi s}\right), \end{align}$$ also perpendicular to $\vec{u}$. We recognize this as $$\vec{F} = q\vec{u}\times \vec{B},$$ where $B = \left(\frac{\mu_0 I}{2\pi s}\right)$ is the familiar magnetic field.

We can see that what we call the magnetic force is actually the electric force due to different Lorentz contraction in other inertial frames.