A particle's dynamics near a solenoid The image below shows a toroidal solenoid, on the $x-y$ plane, carrying a current $I$. At the origin there is a charge $q$ of mass $m$.
At $t=0$ the current begins to decrease in magnitude. How is the particle affected by it? What is the particle's counter action on the solenoid?

At first I thought that by Ampere's law an electric field will act on the particle and accelerate it, and then by Faraday's law the particle will cause the solenoid to recoil. But this doesn't seem to work when the particle's mass is large.   
 A: The toroid's magnetic field is confined to be within the toroid. There is no magnetic field where the charged particle is. When the current is changed then the magnetic field in the toroid changes but it does not change where the particle is. However, that does not mean there is no electric field.
If you consider a loop around the toroid cross-section that goes through the central point, then when you change the current in the toroid, the magnetic flux through that loop will change. Hence there must be a non-zero and time-varying magnetic vector potential.
$$ \oint \vec{A}(t) \cdot d\vec{l} = \int \vec{B}(t) \cdot d\vec{S}$$
Around a toroid you cannot easily evaluate this because the A-field is not constant around the loop, but the sense of it is clear enough from the right-hand rule - if the magnetic field in the toroid circulates anticlockwise when looking from above, then the A-field will point along the z-axis. From there, we know that
$$ \vec{E} = -\frac{\partial \vec{A}}{\partial t},$$
so the electric field acts in the opposite direction to the change in the A-field - for a decreasing current, there will be an E-field in the $+z$ direction.
A positively charged particle will accelerate towards $+z$, which creates a current, which induces a magnetic field that circulates anti-clockwise.
In addition, the particle has its own electric field, but neither the electric of magnetic fields of the particle will produce a net force on an ideal toroid, so the question arises - how can this be reconciled with Newton's third law for an isolated system?
I think the answer lies in considering the electromagnetic momentum density $(\vec{E}\times \vec{H})/c^2$. In order to conserve momentum and therefore satisfy Newton's third law, it must be the case that the integral of the electromagnetic momentum density over all volume, changes in such a way to exactly balance the momentum gained by the particle. This looks like a tricky problem for the geometry described, but we could at least try and establish that things are going in the right direction.
If we consider the charged particle at rest, with the magnetic flux circulating anticlockwise in the toroid, then the Poynting vector is zero outside the toroid and points upwards inside the toroid. As the charge moves (towards +z), the E-field inside the toroid weakens and changes direction to have a growing component in the $\hat{z}$ direction. The Poynting vector magnitude also weakens and it gains a component in the $\hat{R}$ direction. Thus if we ask in what direction is $\partial_t (\vec{E} \times \vec{H})$, then it is towards $-z$.
There is thus a change in the momentum of the electromagnetic field in the opposite direction to the change in momentum of the particle.
