An example which contradict to Newton's 3rd law?

Question

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.

Answer 1

The details of your analysis are not quite right - that's not what the electric field of a moving charge looks like, for example. This is probably because you haven't learned all the rules of electromagnetism yet. Still, the spirit of your question is hitting at an important point.

Charges do not conserve momentum and don't obey Newton's third law. You have to include the momentum of the electromagnetic field to see conservation laws hold.

There's an accessible discussion in section 8.2 of Griffiths "Introduction to Electrodynamics" if you would like a little more math.

Answer 2

The main thing you have to start out with is to define the current of a single electron using a Delta function:

$j(\vec r')=-e\,\delta(\vec r'- \vec{r}(t)) \dot{\vec{r}}(t)$,

where $r'(t)$ is the position of the particle. Then everything else (Maxwell equations, Biot-Savart law), should work.