Do all inductors produce 1 weber after one second when 1 volt DC is applied?

Question

A definition of magnetic flux (the weber) is stated here as: -

If you take a loop of superconducting wire, and apply 1V to this wire during 1s, then the magnetic flux inside this loop will have changed by 1Wb. Note that this is true regardless of size or shape of the loop, and regardless of the matter that's inside the loop! In practice it holds true enough even when the wire is not superconductive, as long as its resistance is low enough to cause only a negligible voltage drop at resulting current.

I believe the above definition is true but I'm prepared to have this belief reset. As an aside this is a basic form of Faraday's law i.e. voltage = rate of change of flux.

So, a big coil (or a little coil) both produce the same flux after one second when 1 volt DC is applied. But what about when the coil is two closely wound turns?

With closely wound turns, the coil inductance is proportional to the square of the number of turns so, 2 turns produces 4 times the inductance and accordingly the rate of rise of current (when voltage is applied) reduces by 4.

This is embodied in the other well-know formula, \$V = L\dfrac{di}{dt}\$.

Given also that the definition of inductance is flux per amp, we can re-arrange this so that flux = inductance x current and, because inductance has risen by 4 with current reducing by 4, it appears that the flux produced by a 2-turn coil (after one second) is exactly the same as the flux produced by a single-turn coil.

You could extend this to as many #turns as you want providing these turns are closely coupled so basically you could say (as per the title): -

All inductors produce 1 weber after one second when 1 volt DC is applied

Now Faraday's law states that \$V = -N\dfrac{d\Phi}{dt}\$

And this is where I'm starting to have a contradiction.

Faraday's law is about induction i.e. the rate of change of flux coupling through \$N\$ turns produces a terminal voltage that is \$N\$ times higher than that for one turn. It works the other way around also; if one volt were applied for one second then the total flux produced by a two turn coil would be half that produced by a single turn coil.

Where am I going wrong in my thinking?

Answer 1

My stab at it (revised). The original block quote :

If you take a loop of superconducting wire, and apply 1V to this wire during 1s, then the magnetic flux inside this loop will have changed by 1Wb.

With qualifications that this is independent of size, shape. material ... but with no qualification about the number of turns. This leads to:

Wb = V * s ... eq1

It says nothing about the current flowing in the turn (or turns) and leaves unanswered whether an N turn coil obeys
Wb = V * s ... eq1a
or
Wb = V * s * N ... eq1b
or even
Wb = V * s / N ... eq1c

Note the definition of Weber

The weber is the magnetic flux that, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second

(yes from Wiki but that links to a primary reference) so it is the flux related to 1 V-s explicitly in a single turn. A crucial difference of phrasing absent from the linked page...

A second turn in the same field would be an independent voltage source. This brings the definition in line with eq1c because 1 Weber is the flux related to 1V-S per turn.

So my (revised!) understanding of the original quote is

If you take a loop of superconducting wire, and apply 1V per turn to this wire during 1s, then the magnetic flux inside this loop will have changed by 1Wb.

This supports Andy's understanding of Faraday's Law expressed in the question - to keep the rate of change of flux constant, you need to keep the voltage per turn constant. Alternatively, if you halve the voltage per turn you will indeed halve the rate of change of flux.

It also leads to the modification in Eq1 of the linked webpage. Which then leads logically to his final equation

H = Wb * turns / A
or
Wb = H * A / turns

This originally made me suspicious, because one normally sees flux as proportional to ampere-turns, so amperes/turn looked ... unfamiliar. The reason is that the inductance already contains a turns-squared term :
L = Al * n^2 (where Al is called "specific inductance" and is a constant for a particular geometry and material)
H = Al * turns^2

Substituting for inductance brings us back to the familiar ampere-turns
Wb = Al * A * turns
which is a more convenient form for some purposes in inductor design.

Answer 2

Compared to a one turn inductor, a two turn inductor has 4 times the inductance.

Therefore the current of a two turn inductor will be 1/4 that of a one turn inductor after 1s.

The flux is proportional to the number of turns and the current. So the flux with 1/4 the current and 2 times the turns will be half that of a one turn inductor.

Magnetic fields generated by multiple sources add together linearly. If the flux generated by one loop of loop is one webber. Then the flux generated by two loops having the same current must be two webbers.

Flux is not proportional to inductance. Flux must be proportional to current and number of turns because electric and magnetic fields add linearly.

As for the units...
Henries = Wb / A is dimensionally eqivalent to Wb / A / Turn (because Turns is a unitless quantitly).

Answer 3

Points go to Brian but, I think, after such lengthy meanderings, my thoughts needs mentioning. My fundamental misunderstanding was that I believed the following formula applied to any inductor irrespective of turns: -

Inductance is total flux per amp

Many web sites state the above (without much clarification) but the real truth is: -

Inductance per turn is total flux per amp

This fixed my thinking.

If two closely packed turns are used, then inductance increases 4 times and, for a fixed DC voltage, the rate in which current builds is quartered compared to the single turn scenario.

So, from this, the inductance per turn is now \$2L\$ (where L is the single turn value)

And \$2L = \dfrac{\Phi}{I/4}\$ or \$\Phi=\dfrac{2LI}{4}\$ i.e. half the amount for a single coil.

And this now (thankfully) ties in with Faraday's Law (\$V = -N\dfrac{d\Phi}{dt}\$)

With twice the number of turns and a fixed 1 volt applied voltage, the rise in flux in one second is half that for a single turn inductor.

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Another way of looking at it (more in line with Brian's answer) is to think about ampere turns (magneto motive force). The idea here is that you convert the ampere turns into the equivalent of a single coil scenario: -

1. Inductance of the equivalent single turn reverts back to L (not 4L)
2. Current was I/4 (for 2-turns) but ampere-turns make it I/2

Hence \$L = \dfrac{\Phi}{I/2}\$ or \$\Phi = \dfrac{LI}{2}\$