First, for a charge $q_1$ with position $\vec{s}_1$ and velocity $\vec{v}_1$, the magnetic field caused by the particle at position $\vec{s}_2$, the equation I've found is $$\vec{B} = \frac{\mu_0}{4\pi}\frac{q_1\vec{v}_1\times\hat{r}_{1\to2}}{r^2},$$ where $\vec{r}_{1\to2}=\vec{s}_2-\vec{s}_1$, $r = |\vec{r}_{1\to2}|$, and $$\hat{r}_{1\to2} = \frac{\vec{r}_{1\to2}}{r}.$$
Second, the force caused by a magnetic field on a moving charge $q_2$ with velocity $\vec{v}_2$ is given by
$$\vec{F}_{1\to2} = q_2 \vec{v}_2 \times \vec{B}.$$
So combining them, the force on particle 2 by particle 1 would be
$$\vec{F}_{1\to2}=\frac{\mu_0 q_1 q_2}{4 \pi r^2} \vec{v}_2 \times ((\vec{v_1} \times \hat{r}_{1\to2})).$$
Now here's where I hit a problem: Newton's Third Law predicts $\vec{F}_{1\to2}=-\vec{F}_{2\to1}$. Putting the above equation to both sides and dropping constants gives $$\vec{v}_1 \times (\vec{v}_2 \times \hat{r}_{1\to2}) = -\vec{v}_2 \times (\vec{v}_1 \times \hat{r}_{2\to1}).$$ With how the $\hat{r}$ vectors are defined, $\hat{r}_{2\to1} = -\hat{r}_{1\to2}$, this simplifies to $$\vec{v}_1 \times (\vec{v}_2 \times \hat{r}_{1\to2}) = \vec{v}_2 \times (\vec{v}_1 \times \hat{r}_{1\to2}).$$
Now, this is where I got stuck, because that's not an identity that holds. Indeed, an easy counterexample is $\vec{v}_1=\hat{i}, \vec{v}_2=\hat{k}, \hat{r}_{1\to2}=-\hat{i}$, since
\begin{align} \hat{i}\times(\hat{k}\times-\hat{i})&=\hat{i}\times\hat{j}=\hat{k}, \\ \hat{k}\times(\hat{i}\times-\hat{i})&=\hat{k}\times\vec{0}=\vec{0}. \end{align}
So, what is the equation for magnetic force between two moving charges? Is my derivation incorrect somewhere? Or if I have it right, is momentum conserved by some other force?
