Magnetic force between 2 moving charges If 2 positive charges are moving in the same direction wit

According to "Darwinian Lagrangian - Interacting particles" the force between two charged particles (written in SI units) is
$\begin{array}{rl}\mathbf{F}& =\frac{1}{4\mathrm{\pi <!--\; \pi \; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}_0}\frac{q_1{q}_2}{r^2}\stackrel{^<!--\; ^\; -->}{\mathbf{r}}\\ & +\frac{{\mathrm{\mu <!--\; \mu \; -->}}_0}{4\mathrm{\pi <!--\; \pi \; -->}}\frac{q_1{q}_2}{r^2}\frac{1}{2}\{{\mathbf{v}}_1(\stackrel{^<!--\; ^\; -->}{\mathbf{r}}\cdot <!--\; \cdot \; -->{\mathbf{v}}_2)+{\mathbf{v}}_2(\stackrel{^<!--\; ^\; -->}{\mathbf{r}}\cdot <!--\; \cdot \; -->{\mathbf{v}}_1)-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathbf{r}}[{\mathbf{v}}_1\cdot <!--\; \cdot \; -->{\mathbf{v}}_2+3({\mathbf{v}}_1\cdot <!--\; \cdot \; -->\stackrel{^<!--\; ^\; -->}{\mathbf{r}})(\stackrel{^<!--\; ^\; -->}{\mathbf{r}}\cdot <!--\; \cdot \; -->{\mathbf{v}}_2)]\}\end{array}$
where $q_1$ and $q_2$ are the charges, ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ are the velocities of the two prticles, and $\stackrel{^<!--\; ^\; -->}{\mathbf{r}}$ is the unit vector between them.
The first term is the well-known electric Coulomb force between the two particles.
The second (very long) term is a magnetic force between the two particles. Roughly speaking, this magnetic force is attractive when $q_1{\mathbf{v}}_1$ and $q_2{\mathbf{v}}_2$ are parallel. And it is repulsive when they are antiparallel.
But remember that this Darwinian formula is still an approximation valid only for slow velocities ($v_1,v_2\ll <!--\; \ll \; -->c$). So don't expect correct results when $v_1\to <!--\; \to \; -->c$ or $v_2\to <!--\; \to \; -->c$
Having said that, you see when both $v_1$ and $v_2$ approach $c$, then the magnetic force becomes similar in size to the Coulomb force. (Remember ${\mathrm{\mu <!--\; \mu \; -->}}_0=\frac{1}{{\mathrm{ϵ<!--\; ϵ\; -->}}_0{c}^2}$ to verify this.)

The real question that needs to be asked is this: "according to whom?" The special theory of relativity states that all inertial observers' points of view of the world are equivalent.
According to an observer in the laboratory frame, charges moving at a constant velocity possess both an electric and a magnetic field. The magnetic field of such a moving charge can be calculated without too much difficulty, and it can be shown that if we know the moving charge's electric field $\mathbf{E}$, then
$\mathbf{B}=\frac{\mathbf{v}}{c^2}\times <!--\; \times \; -->\mathbf{E}.$
Thus, an observer in the lab will say that each of the moving charges is affected by force due to both the electric and magnetic fields of the other charge.
However, if we looked at this from the point of view of one of the charges (or perhaps an observer moving with the same speed as the charges), both these charges would appear to be at rest with respect to each other, and therefore there would be no magnetic field there. The charges would interact with a purely electric field, ${\mathbf{E}}^{\prime }$, which is different from $\mathbf{E}$
As I've shown in the linked answer, you can compute the electric and magnetic fields experienced by the second charge due to the first as show that:
$F_E=\mathrm{\gamma <!--\; \gamma \; -->}{F}_E^{\prime }\text{ and}{F}_B=-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\frac{v^2}{c^2}{F}_E^{\prime }⟹<!--\; ⟹\; -->F_{\text{lab}}=\mathrm{\gamma <!--\; \gamma \; -->}(1-<!--\; -\; -->\frac{v^2}{c^2})F_E^{\prime }=\frac{F_E^{\prime }}{\mathrm{\gamma <!--\; \gamma \; -->}},$
where $\mathrm{\gamma <!--\; \gamma \; -->}=1/\sqrt{1-<!--\; -\; -->v^2/c^2}$, and $F_E^{\prime }$ is the Coulomb force in the particle's rest frame.
Using the above relation, you should also be able to prove why imposing the magnetic and electric forces being the same on each of the charges leads to $v=c$. (See the end of the answer I linked.) I doubt it's a coincidence, but I think it needs to be interpreted as saying that the force due to the magnetic field of a moving charge can never be equal to the force due to its electric field or greater than it (since $v$ can never be equal to $c$ for a massive object).
A simple reason for this I can think of is the following: as I've shown above, the force due to the magnetic field is opposite in direction to the force due to the electric field. (Like charges repel, but like "currents" attract.) Now imagine that there could be some constant speed $v$ such that their magnitudes could be equal. This would mean that both these forces would cancel each other out; i.e. there would be some "privileged" speed $v$ where the particle would not experience a net force. But this violates the very idea of an inertial frame! Therefore it must not be possible for any sensible $v$. And indeed, using the formula for $F_{\text{lab}}$ above, you can see that the measured force seems to reduce, but it never goes to zero for, since $v<c$. So I suppose -- for what it's worth -- you could say that this is "due" to special relativity.