In comments you said,
The EMF will be induced only after the current starts flowing.
This is not correct. The EMF in an inductor has nothing to do with the magnitude of the current that is flowing (for example, whether it is zero or non-zero).
It only depends on whether the current is changing.
From a physics point of view, this comes from Faraday's Law of Induction:
$$\mathcal{E} = -\frac{d\Phi}{dt}$$
In the inductor, the magnetic flux $\Phi$ is proportional to the current, so we can express this as the constitutive relation of the inductor,
$$V=L\frac{dI}{dt}$$
You can see it's entirely possible for there to be an EMF produced, even if the inductor current is zero, so long as the rate of change of the current ($\frac{dI}{dt}$) is non-zero.
If you connect, at $t=0$, an ideal voltage source to an ideal inductor, the inductor EMF is immediately produced to counter the applied voltage (satisfying Kirchhoff's Voltage Law), and the inductor current immediately starts changing.
The current signal lags the voltage signal if the applied voltage is sinusoidal because the current continues to increase ($\frac{dI}{dt}>0$) for as long as the applied voltage is positive, thus it reaches its peak when the voltage just returns to 0 at the end of the positive half-cycle of the voltage waveform.
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
If there is no resistance in the circuit, the rate of change of current in the circuit does not change, as explained here: inductor back EMF.
Thus, the decrease in the magnitude of the rate of change of current as time progresses must relate to there being resistance in the circuit.
In the circuit, you have $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$.
As time progresses, the current in the circuit increases, so $V_{\rm resistance}$ increases. Because $\mathcal E_{\rm battery}$ stays the same, $\mathcal E_{\rm back} $ must decrease, as does the magnitude of $\frac {dI}{dt}$.
To my knowledge, if there is a back emf, it should oppose the applied emf and the sum of these "emfs" should result in a net emf. Current should only increase if the net emf increases... No?
To understand what is happening one needs to consider where the equation $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$ come from and the use of the terms potential difference and emf.
From what you have written I assume that you have never queried as to what happens in a simple circuit which consists of an ideal battery and a resistor.
In such a situation the potential difference across the battery $V_{\rm battery}$ is equal in magnitude to the potential difference across the resistor $V_{\rm resistor}$ so that $V_{\rm battery}-V_{\rm resistor} =0$.
If the equation is multiplied by the current $i$ then the equation becomes $V_{\rm battery}i-V_{\rm resistor}i =0$ and it perhaps now becomes apparent that the equation is a restatement of the law of conservation of energy for electrical circuits, electrical power output from the battery is equal to electrical power dissipated in the resistor.
The only thing to add is that $V_{\rm battery}$ is also called the emf of an ideal battery, $\mathcal E_{\rm battery}$.
A slightly more complex example is an ideal battery used to power an ideal electric motor.
The steady state condition is that there is no current in the circuit because the emf of the battery, $\mathcal E_{\rm battery}$, is equal in magnitude to the back emf produced by the motor coil rotating in a magnetic field $\mathcal E_{\rm back,motor}$ thus $\mathcal E_{\rm battery} - \mathcal E_{\rm back,motor}=0$.
Now suppose that the motor is made to do some work by you pinching the axle of the motor.
The speed of revolution of the motor decreases and hence the back emf produced by the rotating coil, $\mathcal E_{\rm back,motor,load}$, decreases and what is the net input power, $\mathcal E_{\rm battery}i_{\rm load}-\mathcal E_{\rm back,motor,load}i_{\rm load}$, equal to?
In this case is is equal to the mechanical power output from the motor, the work done per second against the frictional force that you have applied to the axle.
Returning to your question one has to consider a dynamic situation.
Points to note are that connecting a battery to a length of wire will results in a current flowing in the wire.
If the circuit has inductance an emf will be induced because of the changing current (Faraday) and that induced emf will oppose the change producing it (Lenz) which in this case is the changing current.
So connect an ideal battery to and ideal (no resistance) inductor.
At the start even with no current flowing the current must change otherwise there would be no back emf from the inductor and then there is an unnatural situation with a battery being connected to a piece of wire with no resistance and no current is flowing. Now we come to the situation which you are unsure about.
Two equal and opposing emf, $\mathcal E_{\rm battery}$ and $\mathcal E_{\rm back}$, and yet the current changes and noting that if there is no change in the current $\mathcal E_{\rm back}=0$.
$\mathcal E_{\rm battery}i-\mathcal E_{\rm back}i$ relates two powers.
$\mathcal E_{\rm battery}i$ is the instantaneous power delivered by the battery.
So what is $\mathcal E_{\rm back}i$?
It is the rate of change of the magnetic energy stored in the inductor.
So in this situation the instantaneous electrical power delivered by the battery is equal to the instantaneous rate of increase in magnetic magnetic energy stored by the inductor.
In a time $\Delta t$ the battery delivers electrical energy equal to $\mathcal E_{\rm battery}i\Delta t$.
During that time the magnetic energy stored by the inductor changes by $\frac 12 L(i+\Delta i)^2 - \frac 12 Li^2 = Li\Delta i + \mathcal O \Delta i^2$ and you will note that $Li\Delta i = L \frac{\Delta i}{\Delta t} i \Delta t= \mathcal E_{\rm back} i \Delta t \,(=\mathcal E_{\rm battery}i\Delta t)$.
Perhaps do not think of the emfs as battling it out for domination rather that one emf is the source of electrical energy and the other emf is the sink (user) of electrical energy and they are equal because energy is a conserved quantity.