Faraday's Law - recursive? So we know that the EMF is induced by change of flux. The thing that was always confusing me is the following: 


*

*we start changing the magnetic field

*which in turn induces electric field which makes charge carriers move

*this e-field also, in turn, makes another magnetic field that changes

*and the whole process seems to go ad infinitum from there!


As far as I understand, this is the basis for the electromagnetic radiation. But Faraday's equation takes into account only the "first" field that is changing, or so I was led to believe, e.g. when you're calculating the self-impedance of the solenoid, you will only look for the first derivative of the magnetic field caused by the current going through it, not all the subsequent magnetic fields.
Since it's also natural to assume that the Law is valid and my reasoning is wrong, where am I wrong?
 A: The differential form of Maxwell's equation relate the value of the fields at the same instant of time and at the same location.
Your reasoning (or notion) that "this change begats this which begats that..." is leading you astray.
For example, the differential form of Faraday's law (Maxwell-Faraday equation) is
$$\nabla \times \vec E(t) = -\frac{\partial \vec B(t)}{dt} $$
So the curl of the electric field, at one instant of time and at one point, is proportional to the time rate of change of the magnetic field at the same instant of time and at the same point.
Whatever the time rate of change of $\vec B$ is, the (negative of the) curl of $\vec E$ is.
A: The efield in your case is an induced non-time varying field, it does not generate further magnetic fields and hence the process stops at just one generation. 
What you are talking about does happen in em wave radiation, when a time varying electric/magnetic field produces a time varying field and thus the process goes on continually.
A: The below two equations :
$$\nabla\!\times\!\vec{E} = -\frac{\partial \vec{B}}{\partial t} $$
and
$$\nabla\!\times\!\vec{B} = \mu_0\vec{J} + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t} $$
show how electromagnetic waves propagate. The second term in the second equation in particular is necessary for the propagation of electromagnetic waves. In applications to circuit theory often the second term in the second equation is neglected - this is often possible for low frequencies like 50 - 60 Hz. Hence for low frequencies we can often neglect the electromagnetic waves produced by circuits. There is no contradiction.
A: If you have a thin circuit with a total resistance $R$, and place it in an external (changing) $\vec{B}$ field, then there is flux through the ring.
First, there is flux $\Phi_1$ from the external, changing $\vec{B}$ field.  Since that $\vec{B}$ field is changing, there is an emf due to that.
Second, the current from the ring itself produces its own $\vec{B}$ field, so its own flux, $\Phi_2$.  In a quasistatic approximation, you might say that this flux is proportional to the instantaneous current, $I$, through the circuit and denote the proportionality by $\Phi_2=LI$.  If the current is changing, then that flux is also changing, so there is an emf due to that.
If the circuit were moving there might be a third contribution to the change in flux, let's ignore motional emf for now.
So together we have a total emf: $\mathscr E = -d(\Phi_1+\Phi_2)/dt$.  Based on the resistance, we have $$RI=\mathscr E=-d(\Phi_1+\Phi_2)/dt=-d\Phi_1/dt-LdI/dt.$$
This is a differential equation, and the solution depends on how the external $\vec{B}$ field is changing (to get $-d\Phi_1/dt$).  This isn't related to radiation, it's just that you have a differential equation, so you need as an input a whole time dependent function for the external field $\vec{B}=\vec{B}(t)$ and how it is changing (to get $-d\Phi_1/dt$), and what you solve for is a whole function $I=I(t)$ telling you how the current changes.
