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## action = reaction and lorenz force

Newtons third law states that there is a counter force to every force. Unfortunately this doesn't seem to work for moving point charges. The Coulomb force cancels out but
the B-Field of a moving point charge is:
$$\mathbf{B}=\frac{\mu_0}{4\pi}q \frac{\mathbf{v}\times\mathbf{r}}{\left|r\right|^3}$$
And the Lorenz force is
$$\mathbf{F}=q\, \mathbf{v}\times \mathbf{B}$$

Lets assume that the two charges have velocities $\mathbf{v}_1,\mathbf{v}_1$

Therefore the two Lorenz forces are
$$\mathbf{F}_1=k(r)\, \mathbf{v}_1 \times (\mathbf{v}_2 \times\mathbf{r})$$
and
$$\mathbf{F}_2=k(r)\, \mathbf{v}_2 \times (\mathbf{v}_1 \times (- \mathbf{r}))$$
Due to the Jacobi identity the sum of the two forces is not zero
$$\mathbf{F}_1+\mathbf{F}_2=- k(r)\, \mathbf{r}\times(\mathbf{v}_1\times \mathbf{v}_2)$$

What is the solution here? The Pointing vector? Relativity? I think that the basic formulas must be correct for slowly moving charges. So it shouldn't be due to non linear trajectories, neglected acceleration or some such thing.