What's different in a transformer core when coupling high & low power?

[cmb]

This is one of those 'oh, I am not sure I know the answer' kind of thoughts.

The strength of magnetic flux in a transformer core is a function of volts per turn, Faraday figured that one out.

So, if I have a light load on a secondary (let's say it is pure ohmic for now, let's not confuse it) and turn up the AC volts on the primary, hey, there it is, voltage 'V' and next to no current going through.

Then I put a heavy ohmic load on the secondary and the volts get pulled down and so I have to crank up the power, more current, to get back to my voltage 'V' across the whole coil.

First case is some lower power with some given voltage across each turn, and second is a higher power at the same voltage across each turn but more current.

Here is my 'duh' moment; the magnetic flux is the same, because that is defined as volts per turn. We have this magnetic core 'transmitting' power from one side to another, but the magnetic domains in the middle would never know there was any difference in the power transmitted.

One might casually assume that if one were to 'look into' the detail of power being transported from one place to another then there would be 'something different' happening within that conduction space, wherein the power is being transported, between different power settings.

Is there? What is it, if magnetic flux is a function of volts only?

 

[anorlunda]

https://en.wikipedia.org/wiki/Poynting's_theorem

 

[cmb]

https://en.wikipedia.org/wiki/Poynting's_theorem

But Poynting relates to electric charges 'outside' the transformer core, right?

I'm asking what is going on 'within' the magnetic core that is different between power transfer magnitudes?

Maybe the question doesn't make sense. Sure I know what the equations say, I was hoping for a 'note of wisdom' than equations. I guess it is a philosophical POV I was coming to it at; that transfer of EM energy is so 'not like what we know' that there isn't really an analogy (like 'more water in a pipe' or such), and that it's just how it is?

 

[vela]

I think you need to consider the magnetic field created by the current in the secondary, the subsequent induced emf in the primary, and how the circuit driving the primary responds.

 

[anorlunda]

But Poynting relates to electric charges 'outside' the transformer core, right?

Not right. Fields outside the core caused by charges inside the core. It applies.

 

[Dale]

Summary:: Faraday's law says magnetic induction field is down to volt turns, not current, so what transmits 'power' in a magnetic core?

First case is some lower power with some given voltage across each turn, and second is a higher power at the same voltage across each turn but more current.

I don’t think this is correct. Both the current and the voltage should be higher.

 

[cmb]

I don’t think this is correct. Both the current and the voltage should be higher.

Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!

Yet, if the rate of change of flux is zero, then the magnitude of the flux is proportional to the current in the current loop, rather than the voltage. (Biot-Savart).

It's really the dichotomy of Faraday versus Biot-Savart. Hence, my question. Sure, I get all the equations and explanations one at a time, but it just seems odd to me if one stops to think about those two sides of the magnetic core story.

As for charges outside a magnetic core affecting the charges inside, it is an interesting idea I'd like to try to picture that, but how does this apply, in physical terms, to a ferrite with a very high bulk resistance? Why would this not be different for ferrites with low bulk Ohmic resistances versus those with high resistances? If that were true then would the inductive behaviour not be a function of core Ohmic conductivity as well as permeability?

 

[Dale]

Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!

You are forgetting the self inductance. The voltage is proportional to $di/dt$. So as you increase the voltage you also increase the current.

 

[hutchphd]

You are forgetting the self inductance. The voltage is proportional to $di/dt$. So as you increase the voltage you also increase the current.

Consider for simplicity the case of a perfectly flux linked transformer with primary and secondary. Then$$N_p\dot{\phi}=V_p\text{ and }N_s\dot{\phi}=V_s$$ So the volts per turn is a characteristic number and that will be unchanging for the (ideal) transformer. Notice core losses and actual resistance will make this slightly untrue but an imperfect flux linkage just changes the numbers.
Of course the loading of the secondary will determine the actual primary and secondary currents flowing in steady state AC configuration. But for an ideal transformer the heat will end up in the secondary load resistor I believe

 

[cmb]

You are forgetting the self inductance. The voltage is proportional to $di/dt$. So as you increase the voltage you also increase the current.

For a given fixed load, yes.

But I am asking what happens to the core for different loads versus currents but a given fixed voltage.

 

[cmb]

Consider for simplicity the case of a perfectly flux linked transformer with primary and secondary. Then$$N_p\dot{\phi}=V_p\text{ and }N_s\dot{\phi}=V_s$$ So the volts per turn is a characteristic number and that will be unchanging for the (ideal) transformer. Notice core losses and actual resistance will make this slightly untrue but an imperfect flux linkage just changes the numbers.
Of course the loading of the secondary will determine the actual primary and secondary currents flowing in steady state AC configuration. But for an ideal transformer the heat will end up in the secondary load resistor I believe

Thanks, yes, exactly.

The thought has come up because of the following scenario; I am working with the latest LDMOS transistors, that can withstand >65:1 VSWR, and a tuneable load that can 'set' such a mismatch if I actually wanted to.

There is a 'legacy' impedance matching 9:1 transformer in there between the two (I don't think it is necessary but it is there!), and as I can basically set what I like either side of it, the question is what is best load to set for minimum losses in that transformer for a given power?

It seems to me that if I set the load so that it pulls the two turn transformer to (say) 50V/T and 1A then I get 100W through but a flux that is a function of the 50V/T. Just for the sake of numbers let's say it is 13.56MHz in a 1cm^2 xs core, so the flux will be 92 Gauss, which for 61 material would be a loss of about 400mW/cc according to data sheet.

But if I set the load so that it is 10V/T but drawing 20A, then I get through 400W but a flux of 46 Gauss. This is a power loss of about 50mW/cc.

I end up transmitting 4 times MORE power 'through the core' for LESS core loss, about an 8th in fact.

If one can set the load arbitrarily, a low impedance load will always give a much lower power loss in the core, it seems?

What is the 61 material core doing differently in those two scenarios?

 

[Dale]

For a given fixed load, yes.

But I am asking what happens to the core for different loads versus currents but a given fixed voltage.

The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.

 

[cmb]

The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.

Could you please help me understand this.

If the load on the secondary is one ohm the relationship would be one amp in the coil for each volt across it.

If it is 10 ohms then are you suggesting it is still one amp for each volt across the coil?

Is this something to do without of phase currents, or something else?

 

[hutchphd]

The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.

This is true but the question had to do with the magnetic flux deep in the core. After some thought I believe it does not change for a fixed input . The phase between current and voltage will be driven by output loading

 

[cmb]

This is true but the question had to do with the magnetic flux deep in the core. After some thought I believe it does not change for a fixed input . The phase between current and voltage will be driven by output loading

Yes, thanks, I'm coming to the same sort of conclusion.

Analogies aren't the greatest things, but I think the best analogy is like a tug-o-war rope, you can pull on either end lightly or really strongly. The rope itself can conduct small amounts of power (work x distance) or large amounts of power, the rope itself doesn't know but is put under tension. I think magnetic domains themselves don't have 'tension' as such so just exhibit the net balance.

The consequence is that the phases of current become increasingly displaced in phase for increasing loads.

As for the losses in the ferrite, if the rope doesn't stretch and lose work that way (i.e. magnetic domains don't 'stretch' like rope, that's where the analogy ends) the actual tension doesn't matter.

The actual power loss in the material we can measure is like a damper attached to the middle of the rope and the other end fixed to a stationary point on the ground. As voltage is like displacement and current like the tension, the damper will absorb the displacement not the power (displacement x tension).

Does this sound like a reasonable stab at the scenario we're thinking about?

 

[hutchphd]

I will not comment on your anology (we all have our own pictures in our heads you can deal with yours) but the big losses in cores are due to real induced currents in laminated steel. And there are other problems when saturation impends because the domains all get aligned. And of course the hysteresis comes from the domain structure and mechanical pinning thereof. One need to be careful about where the energy actually goes, if that matters.

 

[cmb]

I will not comment on your anology (we all have our own pictures in our heads you can deal with yours) but the big losses in cores are due to real induced currents in laminated steel. And there are other problems when saturation impends because the domains all get aligned. And of course the hysteresis comes from the domain structure and mechanical pinning thereof. One need to be careful about where the energy actually goes, if that matters.

Of course, saturation and steel excepted. These carry with them some very specific physics related to their material type.

I'm thinking primarily here of 'more ideal' transformers, like RF frequencies in ferrites, so no currents (albeit not the tiny currents in high resistance ferrite grains) and fields are milliTesla and far from saturation.

 

[etotheipi]

we all have our own pictures in our heads

true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now

 

[Dale]

true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now

Hey! Me too (now)

 

[hutchphd]

My panda always rides a unicycle. He's talented.

 

[vela]

I'm thinking primarily here of 'more ideal' transformers, like RF frequencies in ferrites, so no currents (albeit not the tiny currents in high resistance ferrite grains) and fields are milliTesla and far from saturation.

Wouldn't your basic question about currents and voltages apply equally to a transformer with no core, i.e. just two coils? It seems to me like focusing on what's going on in the core is just an unnecessary complication.

 

[Dale]

OK, so consider a simple circuit with two loops. The primary loop has a voltage source $v_p(t)= V \cos(\omega t)$ and an inductor $L_p$, and the current through the primary loop is $i_p(t)$. The secondary loop has an inductor $L_s$, and a resistive load $R$, and the current through the secondary loop is $i_s(t)$ with the voltage across the secondary loop being $v_s(t)$. The primary and secondary are perfectly coupled with mutual inductance $M=\sqrt{L_p L_s}$.

$$v_p(t) = L_p i_p'(t) + M i_s'(t)$$ $$v_s(t)=L_s i_s'(t) + M i_p'(t)$$ $$v_s(t)=R i_s(t)$$

Solving we get

$$v_p(t) = V \cos(\omega t)$$ $$i_p(t)= -\frac{V L_s}{R L_p} \cos(\omega t) + \frac{V}{L_p \omega} \sin(\omega t)$$ $$v_s(t)= V \sqrt{\frac{L_s}{L_p}} \cos(\omega t)$$ $$i_s(t)=\frac{V}{R}\sqrt{\frac{L_s}{L_p}}\cos(\omega t)$$

Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!

As you can see, changing the load does change the current in both the primary and the secondary. It does not change the voltage in the primary since that is fixed by the supply, and it also does not change the voltage in the secondary, which surprised me.

So with a fixed voltage supply, the additional power is delivered as additional current.

 

[cmb]

So with a fixed voltage supply, the additional power is delivered as additional current.

Of course it is. That was never in question.

So, then, what is the answer to the question?

There is no physical manifestation I am aware of in the magnetic core other than to the voltage, because EMF (volt/turns) is rate of change of flux.

If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?

So long as I keep good volt/turns that works well for the core's loss characteristics, I can put in and draw out low impedance power and make the core as small as I like, it seems!? I'm now picturing putting 10kW through a diddy 35mm toroid! The only question on size seems to be if one can wind wire of sufficient ampacity around the core.

 

[cmb]

true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now

I'm trying to perform the serious business of making the most efficient kW class RF amplifier I can make into the smallest space. I'm unclear on the origin of your comment. Could we keep 'off-topic and humour' to at least until after there is an answer to the question?

 

[Dale]

There is no physical manifestation I am aware of in the magnetic core other than to the voltage, because EMF (volt/turns) is rate of change of flux.

There is the current too.

If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?

I am not sure why you think the current being different is not something different.

So long as I keep good volt/turns that works well for the core's loss characteristics, I can put in and draw out low impedance power and make the core as small as I like, it seems!? I'm now picturing putting 10kW through a diddy 35mm toroid! The only question on size seems to be if one can wind wire of sufficient ampacity around the core.

And what would happen if you put a large current through that 35 mm toroid?

You seem to know that the current is different and yet want to ignore the current and be confused as a result. Current is rather important in general.

Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!

Don’t forget $L = \Phi/I$. You cannot ignore current simply because it doesn’t show up in one single equation.

 

[cmb]

I'm not following sorry.

What is the physical manifestation of more or less coil current in the core? That is my question. I don't think there is any.

The current is 'coil current' and only coil current, in the windings, not in the core.

Does the core do anything differently with more or less current in the windings? This has been the only question.

There isn't any 'current' in the core, only magnetic flux which is a function of voltage only.

Is that right, or does the 'core' behave in some different way with more or less current in the windings?

 

[Dale]

More current gives a larger self induced flux. As I said above $L=\Phi/I$. Your “look ma no current” is only one equation out of many. You cannot separate the core from the windings they go together. Without the windings there isn’t a core.

 

[cmb]

I think I understand your confusion.

I mentioned nothing about having a fixed inductance.

If one puts on more or less turns, one can set the impedance to anything one wants.

I am discussing what the difference in the core material is between one condition where there is (say) 10V/turn and 1A, versus 10V/turn and 100A in the windings? Put on whatever loads and numbers of turns you want to, to set those conditions, the inductance and impedance are not fixed here in my question.

 

[Dale]

I never said they were fixed. That is why I put them as variables.

Regardless of whether the inductance is fixed or not there is inductance and the equation $L=\Phi /I$ holds

 

[cmb]

I never said they were fixed. That is why I put them as variables.

Regardless of whether the inductance is fixed or not there is inductance and the equation L=Φ/I holds

In that case, my bad for wholly confusing the picture by referring to ohmic loads.

If I add a capacitance to the load on the secondary, the circuit impedance changes. The current in the coils goes up while the volt/turn can remain the same.

More current in the coils but the same volts/turn. What happens to the flux? Can we funnel endless VAr via the core, ever increasing coil current, with no effect on it so long as the volt/turns is constant?

 

[Dale]

If I add a capacitance to the load on the secondary, the circuit impedance changes.

This is starting to feel like a moving target.

More current in the coils but the same volts/turn. What happens to the flux?

$\Phi=LI$ for self inductance. $\Phi_1=L_1 I_1 \pm M I_2$ for mutual inductance with $M=k\sqrt{L_1 L_2}$

 

[cmb]

This is starting to feel like a moving target.

It's totally stationary from post #1.

I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.

 

[agrumpyoldphysicstec]

It's totally stationary from post #1.

I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.

Only one question and only two answers:
1) The short answer: Nothing changes in the core.
2) The rather longer answer is that when the secondary current increases (which would tend to cancel out the flux in the core), the current in the primary increases (thus keeping the flux the same). This is illustrated in the attachment for the very simple case of a DC source connected to the primary (obviously a situation that cannot be allowed to continue).

 

[Dale]

It's totally stationary from post #1.

Your last post was the first one that mentioned a capacitative load. You also didn’t mention a variable inductance until post 28. Even if those things were in your mind and intended, I am telling you that to me it feels like a moving target.

I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.

Not sure why you are still wondering. It is easy enough to calculate $\Phi=LI$ from what I posted earlier for a resistive load and $k=1$. Or you can change the load to a complex impedance and use phasors to simplify the calculations.

The full differential eq results for $k\ne 1$ are particularly interesting, but too big to post. That might be a good one to use phasors on.

 

[hutchphd]

Only one question and only two answers:
1) The short answer: Nothing changes in the core.
2) The rather longer answer is that when the secondary current increases (which would tend to cancel out the flux in the core), the current in the primary increases (thus keeping the flux the same). This is illustrated in the attachment for the very simple case of a DC source connected to the primary (obviously a situation that cannot be allowed to continue).

I agree wholeheartedly.

If I add a capacitance to the load on the secondary, the circuit impedance changes. The current in the coils goes up while the volt/turn can remain the same.

If we redo @Dale solution for abitrary (complex) impedance $Z$ on the secondary, the primary complex current is $$I_p=\frac {V_p} { Z_{in}}$$
$$=\frac {V_pL_s} {ZL_p}\left[ 1+\frac Z {j\omega L_s}\right]$$
Where J is for the EE imaginary.
So there are two terms in quadrature. The second term is the response of the primary coil with no secondary current. The first term is the added current supplied by the source to counteract the "Lenz Law" response of the secondary if current flows (at 90deg phase to primary current).
So the the magnetic flux is unchanged for any secondary impedance, reactive or not.

 

[cmb]

This is quite a weird thread. After positing in my first post that the flux is the same, irrespective of the current in the coils, we've now come full circle with everyone agreeing with each other but not my original post? (Notwithstanding me creating a misleading aspect by referring to ohmic loads, sorry about that)

It must be me.

So, on to the mystery that I was alluding to from the outset.

Is no-one remotely curious that somehow the primary and secondary coils 'communicate' their currents to each other by influencing and affecting the way each other behaves, yet the behaviour of the thing in between them shows no differences?

I mean, if I send some message to someone, then that message has some tangible effect on the transmission medium in between me and them.

But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.

 

[hutchphd]

This is quite a weird thread. After positing in my first post that the flux is the same, irrespective of the current in the coils, we've now come full circle with everyone agreeing with each other but not my original post? (Notwithstanding me creating a misleading aspect by referring to ohmic loads, sorry about that)

It must be me.

So, on to the mystery that I was alluding to from the outset.

Is no-one remotely curious that somehow the primary and secondary coils 'communicate' their currents to each other by influencing and affecting the way each other behaves, yet the behaviour of the thing in between them shows no differences?

I mean, if I send some message to someone, then that message has some tangible effect on the transmission medium in between me and them.

But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.

The "thing" in between them can just as easilly be air as magnetic material. This is no more mysterious than radio waves. Of course it is no less mysterious than radio waves!
I will say it took me a while to show myself analytically that the "deep fields" were rigorously unaffected. But in the final analysis it is one charge in primary wire talking to one charge in secondary wire.
I liked the question very much and it made me think through a few loose ends. Thanks for the question

 

[rude man]

It may help to consider the relation magnetomotive force mmf = flux times reluctance. In your toroid, mmf = $N_1 i_1 + N_2 i_2$ with "1" referring to the primary, N = no. of turns and i the current in each winding. Also, mmf = reluctance times flux. Flux is constant thruout the toroid.

With no secondary load $i_1 = emf/\omega L_1$ and $i_2 = 0$. Flux times reluctance = $N_1 i_1$.

When a load is applied to the secondary, each winding contributes to mmf the same way: $\Sigma Ni = mmf$. But now the current in both windings increases but with the secondary current 180 out of phase with the primary (ignoring the magnetizing current). So we really have 2 sources of flux which cancel each other except for the remanent magnetizing current. Flux therefore does not change.

This is all with zero resistance in both windings and no leakage inductance. If that is not the case then phase shifts between primary and secondary voltage (but not emf) occur but the flux is still unaffected, its magnitude and phase solely determined by the emf (not terminal voltage) applied to the primary coil, i.e. by the magnetizing current.

The magnetic field serves to transfer energy from the generator to the primary to the secondary without changing its own energy level.

Think of two monochromatic laser beams 180 out of phase. Each carries energy but the sum is always zero everywhere in the net beam.

 

[Dale]

But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.

I just disagree that ignoring the currents is “analyzing the physics”. I fail to see the point of wondering why there isn’t a difference when there is one that you are aware of but deliberately ignoring.

Also, greater current does increase flux in general.

 

[Charles Link]

One thread that might be helpful is https://www.physicsforums.com/threa...ormer-schematic-attached.999796/#post-6457990 See post 10.

See post 15, 16 https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ See also post 18. I think the late Jim Hardy gave a very good response here.

I'm still on a learning curve with transformer power and efficiency, but the OP @cmb might find these threads useful.

 

[hutchphd]

Also, greater current does increase flux in general.

But I hadn't really thought through the fact that the induced current in the secondary was coincident in phase with an induced additional countervailing current in the primary even though they produce flux that cancels. It explains why that additional current makes both primary and secondary heat for real rersistive windings, but with no increased flux. Perhaps I should have known all this but apparently I had forgotten it (if ever I knew it).

 

[DaveE]

First, I'm too lazy and busy to do the real physical treatment of this, you know, with equations and such. But, I think the way to approach this is to keep in mind that the transformer is a coupled inductor. In the ideal case everything about the core can be modeled as a shunt magnetizing inductance in parallel with the windings.

There will be some magnetizing current which flows that is only due to the excitation from the applied voltage (usually on the primary). The magnetizing current has nothing to do with the secondary current or power delivered (again, ideal model), but there is some reactive "power" associated with it. The value of the core is to redirect the flux in the windings to increase coupling and increases the shunt inductance. It doesn't really change that flux, nor does that flux change the core (again, ideal case).

Then your question about power flow becomes fairly trivial, the ideal transformer part of the model just changes the ac impedance seen by the source. The ideal transformer doesn't deliver, or process, power any more than a superconducting wire does. But it does have a shunt inductive component added, even in an ideal model.

So, your question really revolves around the non-idealities of the transformer, which is necessarily messy. Nevertheless, you can still mostly treat the core as a separate thing primarily related to the voltage excitation of the inductor.

 

[Dale]

So, I recalculated the circuit using phasors for a general complex impedance load, $Z$, and a general mutual inductance $M=k\sqrt{L_p L_s}$, $0 \le k \le 1$.

$$i_p=\frac{j V \left(j Z+\omega L_s\right)}{\omega L_p \left(-j Z+k^2 \omega L_s-\omega
L_s\right)}$$ $$ i_s=-\frac{j k V \sqrt{L_p L_s}}{L_p \left(-j Z+k^2 \omega L_s-\omega L_s\right)} $$

So then the flux on the primary side is $$L_p i_p + M i_s = -\frac{j V}{\omega }$$ and the flux on the secondary side is $$L_s i_s + M i_p = -\frac{k V Z L_s}{\omega \sqrt{L_p L_s}
\left(\left(k^2-1\right) \omega L_s-j Z\right)}$$

So interestingly the flux on the primary does not depend on the load $Z$ nor on the transformer's geometry factor $k$. But the flux on the secondary side does depend on both.

 

[hutchphd]

So interestingly the flux on the primary does not depend on the load Z nor on the transformer's geometry factor k.

This is a very satisfactory result. In the limit of perfect coupling (k=1) the two fluxes are equal and independent of Z and there is no change in actual internal fields
As soon as there is imperfect coupling and the cancellation of the "out of phase" field no longer obtains. The OP must take care to realize that H field in space will now respond everywhere to changes in the the loading even though the "primary flux" is unchanged. (We sort of cook the books here by our definitions of inductance).

 

[Charles Link]

For $k= 1$ the $H$ and $B$ fields are the same for both the primary and secondary windings, but using $L_p=C N_p^2$, and $L_s=C N_s^2$, with $M=C N_p N_s$, the result will be that the flux "links", (what @Dale is calling flux) obey $\Phi_s=\frac{N_s}{N_p} \Phi_p$ , where $\Phi_p=N_p B A$ and $\Phi_s=N_s B A$ represents the flux links. This essentially says $V_s=\frac{N_s}{N_p} V$.

Note: The magnetomotive force $MMF=\oint H \cdot dl=NI$, so that the $H$ or $B$ that is generated is proportional to the number of turns $N$, but the flux links for the self inductance will then be proportional to $N^2$.

The flux links for mutual inductance will be proportional to $N_p N_s$, because the current and the windings in one creates the $H$ and $B$ in the other, and the number of links is proportional to the number of windings of the other.

In addition, in this ideal case you should also get $|N_s i_s| \approx |N_p i_p|$, to a very good approximation if $Z$ is small,(a large load), and that is indeed the case. I agree with @Dale 's solution.

 

[cmb]

The "thing" in between them can just as easilly be air as magnetic material. This is no more mysterious than radio waves. Of course it is no less mysterious than radio waves!
I will say it took me a while to show myself analytically that the "deep fields" were rigorously unaffected. But in the final analysis it is one charge in primary wire talking to one charge in secondary wire.
I liked the question very much and it made me think through a few loose ends. Thanks for the question

You're sort of simultaneously concluding but also re-opening my original dilemma.

I accept, of course, that photons mediate the electromagnetic field. As counter intuitive as it seems, in the absence of apparent photons, magnetics must also use photons as their mediators, including permanent magnets. Where one would look to see such photons, between two moving permanent magnets, I struggle to imagine.

This is probably the same question, actually. I don't really know the answer to that and if I did I'd probably know the answer to this.

The thing is that I tended to have assumed that the flux in the core of an inductor/transformer is the manifestation of the mediating photons between the two coils, just like measuring a flux of photons between two radio transmitting/receiving antenna. It seems there must be more photons being mediated between the coils than 'just' those associated with the magnetic flux?

[I think it is reasonable to assume that in a high Q (>400), low conductivity (>10^8 ohm.cm), ferrite toroid that k is as close to 1 as can be measured and there are no electrical currents present as can be measured within it.]

 

[Charles Link]

I think I see what @cmb is puzzled by, and that is basically the "magic" of the transformer, that it can adjust its primary current when the load increases, and the magnetic flux in the core basically stays the same, with the primary mmf $N_p I_p$ balancing the secondary mmf $N_s I_s$.

With the "links" of post 40, the upper limit to the secondary and primary currents seems to be governed by how much current the copper wire can handle, rather than by any changes that may be happening in the iron core due to the increased loading.

Perhaps there is something that is present in the calculations, in addition to this, but with the balance $|N_p I_p|=|N_s I_s |$, it seems that higher powers are readily accommodated by the transformer core, at least when $k \approx 1$. I think this is what the OP has been asking several times throughout this thread. See also https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ post 17, where Jim Hardy talks about this self-balancing. @cmb Perhaps you would find this entire thread good reading, where the operation of the transformer is discussed in detail.

Perhaps it would be useful to take the limit in the equations of post 43 for $Z \rightarrow 0$, when $k \neq 1$, but I haven't figured out the implications yet.

 

[hutchphd]

You're sort of simultaneously concluding but also re-opening my original dilemma.

I understand your dilemma but the answer is simple.
The magnetic field is a calculational fiction. We have shown that for ana idealized transformer with perfect coupling (and no relativistic effects incidentally) the internal field does not change because it is instantly compensated. But the observable quantities (the currents) do change with more power and just happen to cancel the field changes. Charges interact with other charges. The fact that our fiction behaves a certain way is not fundamental: we set it up this way.
This is, to quote @etotheipi, an argument about the picture in your head and whether the baby panda occasionally becomes invisible...

 

[jartsa]

The thing is that I tended to have assumed that the flux in the core of an inductor/transformer is the manifestation of the mediating photons between the two coils, just like measuring a flux of photons between two radio transmitting/receiving antenna. It seems there must be more photons being mediated between the coils than 'just' those associated with the magnetic flux?

So there is a very small energy associated with the turning motion of the microscopic magnets in the core. And at each moment that energy must be increased by a large amount by the primary coil and decreased by equally large amount by the secondary coil.

So it must be so that the two coils exert opposing torques on the microscopic magnets. The coil that is winning the wrestling match is losing energy, the loser is gaining energy.

Now I try to get into photons, hmm, somehow virtual photons are able to transfer energy, like when I spin a ball, the ball and my hand exchange virtual photons, and the ball gains energy. So I guess now we could ask what is different in the empty space between my hand and a ball spun by my hand, when the ball is a one with small rotational inertia , compared to a case when the ball has a large rotational inertia.

In the latter case more angular momentum is exchanged, so I guess in the latter case the group of virtual photons has more angular momentum. Right?

Hmm actually that last paragraph is messed up in many ways. Like for example it's the ball with the smaller rotational inertia that gains more energy per some amount of angular momentum.

 

[alan123hk]

If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?

Yes, the magnetic field inside the transformer core will not change, but when the power flow changes, in addition to the inevitably changing current, there is indeed another thing that will change. I think you should be able to find the answer from the Poynting vector.

https://en.wikipedia.org/wiki/Poynting_vector