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enter image description here

Consider the example shown above. Here we have connected an active inductor, in which the initial current is "I", to a resistance, with the help of a switch and close the switch at t=0. We know and it is given in almost all the textbooks that the current and magnetic flux through the inductor will slowly decay in this case to zero.

We also know from Faraday's Law of EMI that nature doesn't like the time variation of magnetic flux at any point in space. To maintain the same flux in the inductor, with the help of non-conservative induced electric field, it induces such a current into itself such that the uniformity of flux can be maintained.

Now the question comes is if the tendency of an inductor is to maintain the same amount of flux through itself, then the flux through it should not change at all and it should remain the same till eternity, independent of the fact that it is connected to a resistance, but mathematically and also from energy conservation point of view, we know that the current decays! and the magnetic flux also decays to zero!

So, the question is, how does it happen? Where does the tendency of an inductor to maintain the same flux go in this case? Can somebody help me with a microscopic explanation that how the current and magnetic flux change in the above circuit? and why do they change?

I am not expecting a mathematical or energy-conservation-related argument as answers to my question.

I am expecting microscopic explanation in the circuit (i.e. what exactly happening inside the wires and resistance) and an intuitive explanation to understand the phenomena.

Kindly help. Thank you.

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  • $\begingroup$ There's a problem with the circuit that you've drawn - since the inductor is in series with the switch, the inductor current must be zero when the switch is open and so $I_0 = 0$. $\endgroup$ – Alfred Centauri Jun 20 at 16:44
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    $\begingroup$ @AlfredCentauri, I imagine it is a SPDT switch and it was connected to something else (not shown) for t < 0. $\endgroup$ – The Photon Jun 20 at 16:48
  • $\begingroup$ Here's one way to implement a circuit that works like you describe. $\endgroup$ – Alfred Centauri Jun 20 at 16:55
  • $\begingroup$ Thanks @AlfredCentauri. I did not know that part, so thanks for correcting me. And, my question is not primarily on this point. $\endgroup$ – Devansh Mittal Jun 20 at 16:58
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The inductor cannot win!

When the current changes so does the magnetic flux linked with the inductor, an emf is induced which produces a current in opposition to the changing current producing it - Faraday and Lenz.

You get an endless sequence if the inductor stopped the current changing, no emf would be induced and there would be no opposition to any changing current so the current can change and then the induced emf produces a current which stops the current changing so the current can change . . . . . . . . . .

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Objects with mass tend to resist (due to inertia) changes in their velocity. But that doesn't mean that nothing ever experiences acceleration. It's the same story here.

When we say "an inductor doesn't like changes in current" or "an inductor opposes changes in current" we don't mean that the current through the inductor can't change at all. We only mean that there must be an energy transfer for the change to happen.

In your example, the stored magnetic energy in the inductor must be dissipated as heat in the resistor in order for the inductor current to change.

I am not expecting a mathematical or energy-conservation-related argument as answers to my question.

I am expecting microscopic explanation in the circuit (i.e. what exactly happening inside the wires and resistance) and an intuitive explanation to understand the phenomena.

When current flows from the inductor through the resistor, a voltage is developed across the resistor, according to Ohm's law. Because of the way they're connected, this exact same voltage appears across the inductor. Since all it takes to change the current through the inductor is a voltage across its terminals, we now know the current of the inductor will change at a rate proportional to that voltage.

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  • $\begingroup$ "Since all it takes to change the current through the inductor is a voltage across its terminals, we now know the current of the inductor will change at a rate proportional to that voltage." This is the kind of answer I am expecting, but still I do not exactly understand what you mean to say above, in your edit. I feel you have the exact answer to my question, so I request you to kindly help me with greater explanation with greater detail. Thanks in advance. $\endgroup$ – Devansh Mittal Jun 20 at 16:59
  • $\begingroup$ @DevanshMittal, to explain it exactly would be to write down the differential equations. Which I assume you don't want since you said you are "not expecting a mathematical ... argument". $\endgroup$ – The Photon Jun 20 at 17:01
  • $\begingroup$ True. I can write the differential equations and prove that the current will exponentially decay with time with some time constant. That I understand and agree. Logically and mathematically I am convinced, but my intuition is somehow not satisfied. Your analogy of inertia and block is very very good. BUT, if I can understand it directly in the considered example itself then it will further help me. Thank you. $\endgroup$ – Devansh Mittal Jun 20 at 17:03
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    $\begingroup$ Search the site. There are many questions about how resistors work microscopically. Probably a few about how inductors work, but that is honestly harder to explain. But trying to go from the microscopic explanation of how the components work to why the current in the circuit behaves as it does is like trying to explain the ecology of a forest by explaining the chemical process of photosynthesis. $\endgroup$ – The Photon Jun 20 at 17:44
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    $\begingroup$ Sure. Thanks a lot for time and patience. $\endgroup$ – Devansh Mittal Jun 20 at 18:20
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First, here's an example of a circuit that works as you describe. enter image description here

We say that the switch has been open a long time which is to say that, before the switch closes at $t=0$, the circuit is in DC steady state (all voltages and currents are constant in time).

Since the current through the inductor is constant, the magnetic flux threading the inductor is constant and so, there is zero inductive emf. This means that the voltage across the inductor terminals is zero and the voltage across the current source is just the voltage across the resistor

$$v_L = 0,\,v_S = v_R = I_0\cdot R,\quad t < 0$$

The instant after the switch closes, the current through the inductor is still $I_0$ but now, the voltage across the inductor (left-most terminal marked with the $+$ symbol) is

$$v_L(0+) = -v_R(0+) = -I_0\cdot R$$

This is just KVL; the voltage across the closed (ideal) switch is zero (by definition) and so the sum of the voltages across the inductor and resistor must be zero.

Thus, there must be an inductor emf that is equal in magnitude to the voltage across the inductor. And it follows, since the inductor emf is non-zero, that the magnetic flux threading the inductor must be changing. When you work out the sign, it further follows that flux is decreasing, i.e., the inductor current is decreasing.

As the current decreases, the voltage across the resistor decreases which means that the magnitude of the induced emf also decreases which means that the current decreases at a slower rate. The result is the familiar exponential decrease in inductor current.

Microscopically, the (ideal) conductor making up the inductor can sustain no electric field within. Thus, if there is an applied voltage across the inductor (as there must be if there is current through the resistor in parallel with the inductor), there must be an induced electric field that precisely cancels (within the conductor) the applied field. This induced electric field is necessarily associated with a changing current through the inductor.

As an aside, when the inductor is made of non-ideal conductor with non-zero resistance, there is actually some electric field inside the conductor consistent with (microscopic) Ohm's law $\vec j = \sigma \vec E$. In this case, the inductor emf cannot precisely cancel the applied electric field within the conductor. However, we can model this with an ideal inductor in series with an ideal resistor $r_L$ to represent the non-zero resistance of the inductor windings. The terminal voltage of the inductor is then higher than the induced emf due to the additional voltage drop across $r_L$.

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