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I have spent a lot of time thinking about Faraday's law and common explanation as given on wikipedia does not sit well with me. I may be confused. My concern is entirely with the wording of this explanation From wiki:

"The increase in current causes a back EMF (voltage) across the inductor due to Faraday's law of induction which opposes the change in current."

So the above says that polarity of emf is opposite the change in current. I interpret this as the induced voltage across the inductor will have a polarity such that it would create a current opposing the change in current which produced it. This makes sense as the induced voltage exists because the inductor tries to maintain its magnetic field. For increasing current then, as more current is flowing from source to ground, the current (call it I') which would oppose this change flows from ground to source. The voltage which would be associated with this I' is in the opposite direction of the actual induced voltage, because the strengthening magnetic field due to increasing current actually causes the inductor to push positive current from ground to source, which is really sending positive charge into the source-side of the inductor. So, like we see in the figure below, the induced voltage appears on the source side of the battery. The same as the changing current. Which agrees with V = Ldi/dt. So why do they explain it as "emf opposes change in current" ? Seems to me that emf has same polarity as change in current. Can someone please offer some clarification?

To put it another way, looking at the situation below where the the switch closes, current begins to flow so di/dt > 0, then we know induced voltage V > 0. So from the way we define positive current in the circuit, and the equation V = Ldi/dt, the induced voltage should be positive on the source side (inductor side 1 as shown below). This agrees with the figure but not with the wording of the explanation. Because the change in current is positive, current is increasing from source-> ground, the opposing current which OPPOSES this change would be flowing from the ground->source. The voltage which would be associated with this opposing current would then be positive on the ground side relative to the source side, but this is opposite what we see in the figure below and the equation V = Ldi/dt. So why would they put this explanation on wikipedia? Which should be the most clear and eloquent explanation of this mathematical relationship?

From Wikipedia Flyback Diode page

Thank you for taking the time to read and respond!

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  • \$\begingroup\$ An inductor is a source of EMF so, to oppose the increase in current in a clockwise direction, it generates a voltage with its top terminal +ve wrt ground, i.e. it tries to send current around the loop in an anticlockwise direction. An inductor does not like change, it strives to counteract changes to what it sees as the status quo. If the voltage source is turned down to zero instantaneously while a current, I, is flowing clockwise, the inductor will generate an emf with its lower terminal +ve wrt its top terminal - trying to maintain the clockwise current flow. \$\endgroup\$
    – Chu
    Dec 17, 2021 at 16:42
  • \$\begingroup\$ @chu - you said this: If the voltage source is turned down to zero instantaneously while a current, I, is flowing clockwise, the inductor will generate an emf with its lower terminal +ve wrt its top terminal - trying to maintain the clockwise current flow. - if the applied voltage instantly drops to zero, the current remains stable exactly at the value it was AND, there is no back-emf required to sustain that. \$\endgroup\$
    – Andy aka
    Dec 17, 2021 at 17:54
  • \$\begingroup\$ electronics.stackexchange.com/questions/470171/… \$\endgroup\$
    – G36
    Dec 17, 2021 at 18:47
  • \$\begingroup\$ "from the way we define positive current in the circuit, and the equation V = Ldi/dt" - That's not right. the correct formula is V = -Ldi/dt hyperphysics.phy-astr.gsu.edu/hbase/electric/induct.html \$\endgroup\$ Dec 17, 2021 at 20:05
  • \$\begingroup\$ @BruceAbbott are you referring to applied voltage (V) or back emf (V)? \$\endgroup\$
    – Andy aka
    Dec 18, 2021 at 9:40

2 Answers 2

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Because the change in current is positive, current is increasing from source-> ground ...

I'm not sure you can ascribe a polarity to a current change.

The wikipedia quote says that the back EMF opposes the current change. I think it might be clearer to say that the back EMF opposes the voltage that is making the current change.

General language is not good at describing physical phenomena accurately, and I think the wiki quote is guilty of some sloppiness.

The important thing to note is that the effect of the voltage generated by the current change will be to reduce the change you were trying to make, whether the current is increasing or decreasing. When you are trying to apply Faraday's Law of Induction, if you misplace a minus sign (as I invariably do), just remember that the effect of the EMF generated is to reduce the change that the current was making.

Something that irks me in such descriptions is the due to Faraday's law of induction bit. The back EMF is not due to Faraday's Law of Induction, it's just described by Faraday's Law. Physical phenomena happen because they do, not because we describe what happens quantitatively and then call that description a Law. But I digress/rant.

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@Javaming, here's the roadmap of the problem - debunking ideas first: -

enter image description here

enter image description here

The above is for Javamang so that he understands why I took the main path (below) of explaining how an inductor works (he wasn't happy in comments that I'd done this) but, there's no point trying to fix someone's incorrect ideas without debunking them first so, that is why I'm explaining this to him.


I interpret this as the induced voltage across the inductor will have a polarity such that it would create a current opposing the change in current which produced it.

No, that's not correct. No extra current is created at all: -

  • There is an induced (back) emf that exactly equals the applied voltage (ideal inductor)
  • An ideal inductor is, for this example, one with no DC resistance
  • And, the induced voltage is not across the inductor but in series with it.

So, that induced emf is in series with the inductor and, we know that if there is no net voltage across an ideal inductor (due to the applied voltage and back-emf being equal) then the current that flows through the inductor is not defined by this approach: -

schematic

simulate this circuit – Schematic created using CircuitLab

We can also say that when both sides of an inductor have the same voltage, we can regard the inductance as having zero impedance hence, using ohms law, the current flow is this: -

$$I = \dfrac{0 \text{ volts}}{0\text{ }\Omega}\hspace{1cm}\text{ i.e. indeterminate}$$

Leaving us with just the expected current flow due to \$V = L\dfrac{di}{dt}\$

So why would they put this explanation on wikipedia? Which should be the most clear and eloquent explanation of this mathematical relationship?

So, the quote from wiki: -

"The increase in current causes a back EMF (voltage) across the inductor due to Faraday's law of induction which opposes the change in current."

It's fundamentally wrong - the back emf has no part in defining the current that flows into an ideal inductor when a source voltage is applied.

Because the change in current is positive, current is increasing from source-> ground, the opposing current which OPPOSES this change would be flowing from the ground->source.

This is flawed as mentioned above - there is no extra opposing current. You need to think about things in a different way.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Dec 20, 2021 at 17:33

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