How does Newton's third law apply to magnetism? As a magnet passes through a solenoid, currents are induced in the wire with an angular momentum. Since the induced magnetic solenoid repels the magnet linearly, it cannot allow for the conservation of angular momentum. There is another centripetal force that appears to push the electrons in the magnet as it passes through the magnetic solenoid. But again, this cannot account for the conservation of angular momentum?
How can I understand how momentum and energy are conserved in any electromagnetic interaction? Without going into too much mathematical detail, but enough to gain a heuristic understanding?
 A: Without going into too much mathematical detail, the best answer is that the electromagnetic field itself carries angular momentum. The density of the angular momentum of an electromagnetic field can be derived from the Poynting vector
$$\textbf{S}=\frac{1}{\mu_0}\left(\textbf{E}\times\textbf{B}\right)$$
From which the momentum density is given by $\boldsymbol{\mathcal{P}}=\epsilon_0\mu_0\textbf{S}$, and so the electromagnetic angular momentum density is given by
$$\boldsymbol{\mathcal{L}}=\textbf{r}\times\boldsymbol{\mathcal{P}}=\epsilon_0\,\textbf{r}\times\left(\textbf{E}\times\textbf{B}\right)=\epsilon_0\left((\textbf{r}\cdot\textbf{B})\,\textbf{E}-\left(\textbf{E}\cdot\textbf{B}\right)\,\textbf{r}\right)$$
The total angular momentum derived from this density is exactly what cancels the angular momentum of the current in the solenoid.
Energy is not conserved in the interaction since energy is being put in my whatever force is pushing the magnet.
I hope this helped clear up some confusion!
A: Due to the long discussion, let me summarize my conclusions so far.
I assume that the coil axis and the magnet movement are in $z$-direction.
I am quite sure that it is not the electromagnetic field of the magnet. $E$ is very likely to be of type $(E_r,E_\theta, E_\phi)=(0,E_\theta(r,\phi),0)$ meaning that $S$ is of type $(S_r(r,\phi),0,S_\phi(r,\phi))$ (see image) meaning that $\cal L$ is of type $(0,{\cal{L}}_{\theta}(r,\phi),0)$,i.e. the integral around $\theta$ of the angular momentum is zero and locally it is not pointing in $z$ while the angular momentum of the current is. 

Fig. 1 Poynting filed of moving circular current
Furthermore, I am sure that a free electron would not rotate around the magnet but would make some local turning.
Fig. 2 Trajectories of charged particles in the field of a circular current
Hence, to get the circular motion there is obviously a strong interaction of the current with the confinement, i.e. the wire. I have to think about the details, but would expect the coil to turn in the opposite direction.
Finally, it is very important that magnetization itself is angular momentum. In the permanent magnet it is most likely to come from the electron spins, but in a simplified way it may be imagined as a circular current as well. In this simple image the circulating current in the coil is producing an opposite field in the permanent magnet, reducing its magnetization, or in the simplified image, its current and, therefore, its angular momentum. In terms of magnetization and spin, what is happening is the following. The magnetization starts to precess around the local field, so the total magnetization in $z$-direction reduces, due to cylindrical symmetry all new in-plane components sum up to zero.
From this point of view the magnet is already bringing plenty of angular momentum and borrowing some of it to the current.
Similar thinking might help understanding how the free charge gets angular momentum when passing a magnetic field, starting to "circulate" as in Fig. 2.
The list of effects and the corresponding understanding might not be complete, but it gives an idea.
