Doubt in the mode of action of inductor in the RL circuit

Question

I feel that my initial question was slightly incoherent and not precise. So instead of making addendum to the question which is making it lengthier, I am editing it completely. So, sorry about that.

As in comments, people are suggesting to incorporate the resistance in the circuit. But the problem still continues.

Suppose we have an RL circuit with dc power suply of emf $\mathscr{E}$. Then we know from the differential equation obtained by $KVL$ that the induced emf by the inductor goes from $\mathscr{E}$ to 0 and the current acquires the steady value from $0$ to $\frac{\mathscr{E}}{R}$.

I am not able to physically visualize why the induced emf decreases. At $t=0$, before closing the switch there is no current. $\frac{dI}{dt}=0$. When we close the switch, the induced emf equals the applied voltage thus no current at $t=0^+$.

But now, why the induced emf has to decrease. If it does not decrease, then the current in the circuit will remain $0$, and $\frac{dI}{dt}=0$, which the inductor likes. If induced emf decreases say by $\Delta V$, then the current will tend to flow in the circuit, thus $\frac{dI}{dt}>0$, so instantaneously the inductor has to induced more emf (by an amount $\Delta\mathscr{E}$) to resist this change. So, inductor has to maintain the constant induced voltage $\mathscr{E}$, so as not to change the rate of current flow.

If we see from the perspective of differential equation. At $t=0$, induced emf by inductor is $\mathscr{E}$. Thus no current. But by the Faraday's law, this induced emf corresponds to a particular $\frac{dI}{dt}$ in the cicuit. Thus current increases in the circuit. By $KVL$, some voltage drop will occur across $R$, thus induced emf by inductor decreases so as to satisfy $KVL$ and thus $\frac{dI}{dt}$ decreases, and this continues till all the voltage will be across resistance.

But this analysis suggests that induced voltage governs the rate of current flow in the circuit and also the induced emf changes so that net voltage in the circuit is $0$ (to satisfy $KVL$). So, the guiding force is the $KVL$.

But Faraday's law does opposite. It considers the rate of flow of current as the cause and induced emf as the effect. I think that this is the point where intuition does not match with the mathematical result and essentially $KVL$ escapes us from the intuitive picture of the action of inductor.

Answer 1

But Faraday's law does opposite. It considers the rate of flow of current as the cause and induced emf as the effect.

Both Faraday’s law and Kirchoff’s laws express non-causal relationships. Neither of them is written as a cause and effect. They are relationships that hold at all times.

In contrast, consider the retarded potentials: $$\varphi(\mathbf{r},t) = \dfrac{1}{4\pi \varepsilon_0} \int \dfrac{\rho(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|} dV'$$$$\mathbf{A}(\mathbf{r},t) = \dfrac{\mu_0}{4\pi} \int \dfrac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|} dV'$$Notice that $\varphi$ and $\mathbf{A}$ at time $t$ are determined by $\rho$ and $\mathbf{J}$ at time $t_r<t$. That time ordering is essential for a cause and effect relationship. Causes always come before effects, so the fact that $t_r<t$ means that $\rho$ and $\mathbf{J}$ are unambiguously the causes and $\phi$ and $\mathbf{A}$ are the effects.

There is no similar temporal ordering of Faraday’s law: $$\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$Both sides happen at the same time, so neither side can claim to cause the other. The word for Faraday’s law is “induces” and either side can be said to induce the other, so “induce” does not mean “cause”.

Now, let’s look at the RL circuit. Upon closing the switch the battery’s EMF must be distributed across the resistor and the inductor. As you correctly analyzed it immediately goes across the inductor. However, your analysis goes wrong when you say:

If it does not decrease, then the current in the circuit will remain $0$, and $\frac{dI}{dt}=0$, which the inductor likes.

An inductor with a voltage across it does not like $\frac{dI}{dt}=0$. An inductor requires at all times that $\frac{dI}{dt}=V/L$. So when $V\ne 0$ it demands $\frac{dI}{dt} \ne 0$. An inductor with a voltage across it does not oppose a change in current, it requires a change in current.

With that understanding, it is clear that the current increases, leading to an increasing voltage drop across the resistor, a decreasing voltage drop across the inductor, and a reduced rate of increase of the current.