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?

• I think it is just a bad idea to think about the $\mathbf{E}$ and $\mathbf{B}$ fields as "causing" each other. They are ultimately both just different aspects of the same electromagnetic field tensor, and they are both "caused" simultaneously by movement of charges (and initial conditions), following the retarded solutions to Maxwell's equations for given current and charge distributions. – Tob Ernack Nov 23 '19 at 5:17

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.

$$\nabla \times \vec E(t) = -\frac{\partial \vec B(t)}{dt}$$

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.

• At the same time, I feel I understand your point, since equation speaks for itself, but at the same time, I have a hard time putting it into my previously described visual scenario. You speak of the time rate of change of B, but which B? All the B's I mentioned earlier, superimposed together in one particular point at one particular moment, or just some particular B? Do they not cause each other but all exist simultaneously? I hope I didn't confused you. – Lajka May 9 '14 at 23:44
• @Lajka, Maxwell's equations hold - period. Since the equations are linear, one could write something like $\nabla \times (\vec E_1 + \vec E_2) = -\frac{\partial}{\partial t}(\vec B_1 + \vec B_2)$ which, due to linearity, is $\nabla \times \vec E_1 + \nabla \times \vec E_2 = -\frac{\partial \vec B_1}{\partial t} - \frac{\partial \vec B_2}{\partial t}$ but I don't see this being interpretable in the way that you're evidently trying to. – Alfred Centauri May 9 '14 at 23:54
• I still don't get it, man, but thanks for trying to help me. This sentence of yours, "Your reasoning (or notion) that "this change begats this which begats that..." is leading you astray.", is too ambivalent for me; could you just elaborate it a little more... why is it leading me astray? – Lajka May 10 '14 at 15:42

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.

• Well, if the current in the circuit is AC, this definitely causes non-stationary magnetic field, and all the conditions for further generation are met, yes? – Lajka May 10 '14 at 19:38
• @Lajka : An AC current would give you a time dependent magnetic field but it would survive only one differential, as soon as you would differentiate it to get an electric field (induced) that field would not be time-dependent. Your question seemed to about simple current configurations so I did not bother to talk about complicated currents which may survive multiple differentials and hence provide time varying induced electric field. – Rijul Gupta May 10 '14 at 19:43

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.

• Ah, so you're saying we're merely approximating by neglecting the second term, but my reasoning was actually ok? – Lajka May 10 '14 at 15:39
• In case of the solenoid you are neglecting the second term - the so-called 'displacement current' - and only considering the magnetic field produced by the actual current. So there is no contradiction. – guru May 11 '14 at 18:11

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.