Electromagnetic "two-body" problem Consider two electric charges $q_1$ and $q_2$ interacting by means of Lorentz forces:
$$\boldsymbol{F}_{i\to j} = q_j(\boldsymbol{E}_{i}+ \boldsymbol{v}_{j}\times\boldsymbol{B}_{i}), \qquad i,j\in\{1,2\},\text{and}\ i\neq j$$
using Coulomb's law and the Biot–Savart law, we get:
$$\boldsymbol{F}_{1\to 2} = \frac{q_1q_2}{(4\pi \epsilon_0)d^2} \left( \boldsymbol{\hat{u}}_{12}+ \frac{\boldsymbol{v}_2\times(\boldsymbol{v}_1\times \boldsymbol{\hat{u}}_{12})}{c^2}\right)$$
$$\boldsymbol{F}_{2\to 1} = \frac{q_1q_2}{(4\pi \epsilon_0)d^2} \left( \boldsymbol{-\hat{u}}_{12}- \frac{\boldsymbol{v}_1\times(\boldsymbol{v}_2\times \boldsymbol{\hat{u}}_{12})}{c^2}\right)$$
Where we have used that $\mu_0 = 1/(\epsilon_0 c^2)$. Now, it is well known that in the presence of magnetic fields Newton's third law is not fulfilled (since the electromagnetic field is a "third body" that has linear momentum and energy). For this reason, $\boldsymbol{F}_{1\to 2} \neq -\boldsymbol{F}_{2\to 1}$, which can be verified by adding both forces. Using the vector identity $\boldsymbol{a}\times(\boldsymbol{b}\times \boldsymbol{c}) = (\boldsymbol{a}\cdot\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}\cdot\boldsymbol{b})\boldsymbol{c}$, we get that:
$$\boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1} = \frac{q_1q_2}{4\pi \epsilon_0} \left(\frac{(\boldsymbol{v}_2\cdot\boldsymbol{\hat{u}}_{12})\boldsymbol{v}_1-(\boldsymbol{v}_1\cdot\boldsymbol{\hat{u}}_{12})\boldsymbol{v}_2}{d^2c^2}\right)$$
Now my questions are:

*

*It is correct to say, assuming that the particle velocity is small relative to the speed of light, that the electromagnetic field undergoes a change of momentum in the form of radiation given by $\text{d}p_{EM}/\text{d}t = \boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1} \neq 0$? Or do we need to include radiative corrections such as those in the Larmor formula?

*How to demonstrate in a simple way that if the charges move slowly (with respect to light) and are of opposite sign they would eventually approach each other? (it seems clear that if the velocities tend to align then $\boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1}$ tends to zero)

*What about the angular momentum, is there a way to write a simple formula for the evolution of the angular momentum of the system? Does it happen that the electromagnetic field also drags some amount of angular momentum away from the charges?

 A: *

*No. EM field momentum in the vicinity of the particles changes in the opposite direction to momentum of the particles. So the approximate relation would be $\frac{d\mathbf p_{EM}}{dt} = -\mathbf F_{12} - \mathbf F_{21}$. Approximate because forces are not exactly given by the formulae you wrote. Indeed there are terms depending also on particle's accelerations, and all quantities are in the past, not present, due to retardation (assuming purely retarded fields without additional homogeneous solution are the correct description of EM field).


*One variant of this motion was analyzed by J. L. Synge, see my answer here
Does classical electromagnetism really predict the instability of atoms?


*Not sure if the result would be simple, but yes it is possible. For point particles, one can't use the Poynting formulas, but there is the Frenkel theory for point particles with local conservation of momentum (consistent with the Synge model), so extending it to angular momentum should be straightforward. Most probably Synge's system - two point particles with retarded fields - exchanges angular momentum with EM field, just as it exchanges energy.

