Let a,b be two charged particles. $$\vec{r}_a(0)=\vec{0}$$ $$\vec{r}_b(0)=r\hat{j}$$ $$\vec{v}_a(t)=v_a \hat{i}$$ $$\vec{v}_b(t)=v_b\hat{j}$$
In which both $v_a$ and $v_b$ $<<c$.
Then
$$\vec{E}_{ab}(0)=\frac{q_a}{4\pi \epsilon r^2}\hat{j}$$
$$\vec{B}_{ab}(0)=\frac{\mu q_av_a}{4\pi r^2} \hat{k}$$
$$\vec{E}_{ba}(0)=-\frac{q_b}{4\pi \epsilon r^2}\hat{j}$$
$$\vec{B}_{ba}(0)=\vec{0}$$
Note that $v_a$ and $v_b$ $<<c$ thus a and b almost obey Coulomb's law. Moreover, $\vec{j_i}(\vec{r})=q_i\delta(\vec{r}-\vec{r}_i)\vec{v}_i$ hence BS law can be applied.
Hence
$$\vec{F}_{ab}(0)=q_b(\vec{E}_{ab}+\vec{v}_b \times \vec{B}_{ab})$$ $$=\frac{q_a q_b}{4\pi \epsilon r^2}\hat{j}-\frac{\mu q_av_a v_b}{4\pi r_b^2} \hat{i}$$
But
$$\vec{F}_{ba}(0)=-\frac{q_aq_b}{4\pi \epsilon r^2}\hat{j}$$
Consequently $$\vec{F}_{ab} \ne -\vec{F}_{ba}$$
This result contradict to Newton's 3rd law!! But I cannot find any error... It troubled me.