Your thought train is fine up to Fig.6. But then you get into a needlessly complicated descriptions of signs. You state

"by Faraday’s law an EMF $v_\text{ind}(t)$ is induced across the inductor. To find the polarity of such EMF or voltage, we first find the direction of the current $i_\text{ind}(t)$ such EMF tries to establish, which we determine using Lenz’s law. So while our goal is to find the polarity of the induced EMF"

It seems you are using the word "voltage" and symbol $v_{ind}$ for two different concepts: 1) induced EMF in the ideal inductor - that is due to induced electric field present in the coils; 2) potential drop across the ideal inductor terminals (when moving from one terminal to the other in the designated positive direction) - that is due to the fact that electric field has electrostatic component, and potential drop is integral of this electrostatic field.

This (using "voltage" for two different concepts and getting confused as a result) is common in many people's understanding of electricity, even those with high credentials. I think mostly because many textbooks and teachers do not understand this either, because they don't understand the general concept of electromotive force and its variants. Old papers and textbooks on electricity (pre-WWII) did not suffer from this confusion.

Understanding the difference between EMF and potential difference in physics is important. It also resolves the question of sign of potential drop across the ideal inductor which you are interested in.

Both concepts - EMF and potential drop - are valid and useful, and both depend on the sign convention. The sign convention is that current intensity $i$ is positive when it flows in the designated positive direction in the loop, and emf and potential drops are positive when their effect is to act on current in the circuit element to increase it in that same direction.

*Induced EMF*

Induced emf for the path defined by coils of the inductor is defined as integral of induced electric field over that path.

When we have single inductor of self-inductance $L$ whose interaction with other currents/inductors can be neglected, induced EMF always obeys the equation (whether the inductor is ideal or not):

$$
emf = -L\frac{di}{dt},
$$
the minus sign making sure that emf acts against changes of current. This follows from Faraday's law and the mentioned sign convention for current and emf. It is not a good idea to call this quantity "voltage", but it is often being done, with various variations (electromotive voltage, induced voltage, etc). Although induced EMF has the unit Volt just as voltage in electrostatics has, EMF is a path-dependent quantity and in general one cannot associate two points of space (such as terminals of an inductor) with unique EMF. We need to state the path as well, but because of this, it is always better to say EMF instead. It is also almost never the quantity of interest when measuring in practice; instead, we want to measure potential differences, which do not depend on paths.

*Potential drop*

Potential drop across the inductor has (unfortunately) become a more complicated thing to explain. It is not always the same as EMF, not even in magnitude. Potential drop is integral of the electrostatic part of electric field when going from one terminal to the other in the positive direction, the exact path not being important. This quantity is not, in general, determined merely by induced part of electric field; the value of current $i$, resistance of the inductor $R_i$ and its internal capacitance $C_i$ is also important. So in general, potential drop across inductor cannot be determined from induced EMF alone.

However, in the special case where the inductor is ideal (zero internal resistance $R$, capacitance $C$), there is a simple relation: potential drop is exactly minus induced EMF. This is because electric field in ideal inductor coils has to be zero, and this implies that electrostatic field completely cancels the induced field inside the conductive coils, and that implies integral of electrostatic field from one terminal to the other has to be minus integral of induced field over path from one terminal to the other that goes inside the coils. So, for a perfect inductor, we have

$$
potential~drop = - emf = L\frac{di}{dt}.
$$

This quantity is also often called voltage drop across the inductor, or simply voltage across the inductor. This quantity is useful when writing down the so-called Kirchhoff's voltage law for any closed loop in a lumped model of AC circuit. Then it is simply denoted $V$ or $v$. This is the preferred meaning of the word "voltage"; it is used in electrostatics, and is used also in AC circuits. It is also what we often want to measure in a complicated real circuit via oscilloscope. To make sure we really measure this and not some path dependent quantity like EMF, the probes and wires have to be made with good field insulation (coax cables) and during measurements, we prevent the wires from arranging themselves in loops.

From the above expression all signs are evident; when current increases in positive direction, emf is negative so the potential drop has to be positive, to counteract the induced EMF. When current decreases in that same direction, the induced EMF is positive, so the potential drop is negative to counteract the induced EMF.

What would be different for a real inductor with internal resistance? Here, net electric field must exist in the coil to push against resistance, so we cannot assume effect of potential drop cancels effect of the induced EMF. Let the real inductor be connected directly to a source of varying potential drop (more often called "source of voltage") $V(t)$:

```
---------------
| -> |
| )
(V) ) real inductor
| )
| <- |
---------------
```

Then by assumption, potential drop on the inductor is $V(t)$. We cannot find induced EMF or current $i$ from these assumptions alone.

However, if we assume this real inductor behaves as ideal inductor with self-inductance $L$ with resistor $R$ in series (ignoring capacitance effects), we can use Kirchhoff's second circuital law (for its formulation, see my answer here: Using Faraday's law twice ). The replacement lumped model circuit looks like this:

```
---------------
| -> |
| |
(V) L
| |
| |
| R
| |
| <- |
---------------
```

Now we can write down Kirchhoff's second circuital law for this circuit:

$$
Ri = V - L\frac{di}{dt}.
$$

Here our voltage source contributes to the circuit effective electromotive force of magnitude V, and the other r.h.s. term is induced emf.

From this equation, we can find that potential drop on the real inductor is
$$
V = Ri + L\frac{dI}{dt} = Ri - emf.
$$
So potential drop on the inductor is not the same as induced EMF, not merely due to opposite sign but also due to magnitude. It depends on emf and current $i$ and internal resistance of the inductor $R$. In general it has different phase from EMF, and in special cases it can have the same sign as induced EMF has, something that can't happen with ideal inductor.

This was still a simplification, and a more realistic model would include contribution due to capacitive interactions between the inductor's coils (ideal capacitor $C$ in parallel with the ideal inductor $L$).

## Best Answer

The sign in the case of an inductor is indeed easy to be uncertain about. I would say this is a good illustration of a more general difficulty with signs in physics. The way to get signs right is sometimes not to worry over one equation or another equation of opposite sign, but rather to be clear in your mind about what happens in a simple example case.

I find it very useful to consider the simple circuit with just a resistor and an inductor in it. The voltage around the circuit is zero, so we get the equation

$IR + L dI/dt = 0$

It is easy to see that the sign is correct in this equation, because then we get

$ dI/dt = - (R/L) I$

for which the solution is exponential decay. If we had the opposite sign we would get exponential growth of the current, which is clearly wrong. But you are free to consider the first equation either in the form I wrote it, or in the form

$IR = - L dI/dt$