Mutual inductance in a transformer

[axim]

TL;DR Summary

In a transformer, the secondary coil length is changed but the number of windings stays constant (inductance of secondary changes). Flux created by primary stays the same (open load), why does the secondary voltage change?

Hello!
I have a question regarding the mutual inductance of two coils in a transformer.

In the formulas for linked coils $M=k*\sqrt{(L_1*L_2)}$ where $k=1$ for a perfect coupling.

I wanted to check my understanding about mutual inductance and see how it can be determined. When I vary the secondary coil length I'm getting into trouble.

My approach:
To determine the mutual inductance, I found many examples where one coil $L_2$ of length $l_2$ with $N_2$ windings (I refer to it as output) is surrounded by a coil $L_1$ of length $l_1$ with $N_1$ windings (input). The radius is $r$.
With the assumption that length $l_1>=l_2$ and $l >>r$ the simplification is done that the flux from $L_1$ will be nearly constant inside the solenoid and pass completely through $L_2$. The surface area entering the coils should be $A$ and is equal (they have basically the same radius). This example seems to be similar to a transformer (flux created by a current flowing through $L_1$ is guided completely through $L_2$ and will induce a voltage here). No current flow in $L_2$, open load.
Can I use this approach here? For better understand I put a sketch below.

Because of $l>>r$ the flux generated inside $L_1$ is $μ*N_1*I_1/l_1$. As it passes through $L_2$ completely it causes the mutual induction $M_{21}=μ*N_2*N_1*A*/l_1$. This is valid for any length $l_2$ as long as it is shorter than $l_1$.

$L_1$ and $L_2$ have same length:
In case $l=l_1=l_2$, the inductance values are: $L_1=μ*N_1^2*A/l$ and$L_2=μ*N_2^2*A/l.$ I see that $M_{21}=\sqrt{(L_1*L_2)}$, same as in the transformer formula, good.
If $N_1=2*N_2$ then $L_1=4*L_2$, that matches a regular 2:1 transformer operation:
The input voltage will cause a magnetizing current (input divided by impedance of $L_1$), which will cause the self inductance of $L_1$ with the same voltage opposed to the input. The induced output at $L_2$ will be half of the input voltage (As $N_2*N_1$ from $M_{21}$ divided by $N_1^2$ from $L_1$ substituted = $0.5*N_1^2/N_1^2 = 0.5$ and I know $L_1$ counteracted the input voltage). The output is open so no current can flow in $L_2$.
Or in shorter words: $M_{21}$ is $0.5*L_1$ and both see the same flux.

$L_2$ made shorter (my problem starts here):
Now, lets assume the coil $L_2$ is made shorter, but still keeps the same amount of windings as before. It is now basically the input coil $L_1$ cut in half, so it only has the length $l_2=l_1/2$. If the windings are kept the same and the length is now halved, the inductance has doubled (same $N_2$ but half length).

As $L_2$ is still completely within $L_1$ (even more now with the half length), the flux it sees from $L_1$ is still the same (entry/exit fringes neglected), and it passes through the same amount of windings. $M_{21}=μ*N_2*N_1*A*/l_1$ - no change to the equation above, same flux and same number of windings. Now $L_1=2*L_2$ instead of $4*L_2$ as the $N_2/l_2$ part of the $L_2$ inductance has changed.
But, according to the formula, the mutual inductance will stay the same.

This means: The input voltage at $L_1$ will still create the same magnetizing current and self inductance as before (no change here). As the mutual inductance stays the same, also the same voltage should be induced at $L_2$ (same number of windings as before).
Now here I have a discrepancy.

$L_1$ has stayed the same, it will have the same magnetizing current and according to my understanding create the same flux in the circuit. The inductance of $L_2$ has changed but not its windings.

Now with $l_2=l_1/2$ there is a discrepancy: $M_{21} != \sqrt{(L1*L2)}$
$\sqrt{(L_1*L_2)} = \frac{μ*N_1*N_2*A} {0.707*l_1}$
Now because of the changed length the output is $\sqrt{2}$ higher, which does not match my own derivation where the mutual inductance stayed the same.

If I simulate in Spice and link two coils $L_1$ and $L_2$ where $L_1=2*L_2$ with $k=1$, without load for $1V$ input I get $0.707V$ output, which matches the equation $M=\sqrt{(L_1*L_2)}$. When following the scheme described above it should be $0.5V$ only.

Where is the mistake?
Summarized: The flux created at $L_1$ is always the same and should pass completely through $L_2$ so I expect it will always see the same flux. If the number of windings in $L_2$ stays the same according to my derivation above the induced output voltage should be the same. But as the inductane of $L_2$ is different the voltage is different, matching $\sqrt{(L_1*L_2)}$. If the voltage induced is number of windings multiplied with the flux change $e=N*\frac{dΦ}{dt}$, and neither windings nor flux have changed at $L_2$, why is the induced voltage seen different?

Some help to clear this up would be appreciated a lot!

 

[berkeman]

I've tried reading your post, but have trouble following a lot of it.

First, with a ferrous core, the flux is guided by that core. So changing the length of one of the windings and keeping the same number of turns should not affect the magentizing inductance. It will likely affect the leakage inductance some, but I don't think that is your question.

Could you be getting confused by using the equation for inductance of a coil that is not wound on a high-$\mu$ ferrous core? In that case, the inductance of that air-core coil could well depend on the length of the coil...

 

[berkeman]

BTW, one key concept in transformer construction is the tradeoff between magnetizing inductance and leakage inductance for each coil winding and the combination of them.

When you say "adjust the length of the coil", this is probably not what you mean, but if you take a 2-layer wound coil and double its length to get all of the windings against the core, you will increase the magnetizing inductance for that winding and decrease its leakage inductance. That's because for a multi-layer coil, some of the flux from the outer layers does not couple into the core, and is not contributing to the magnetizing inductance and the mutual coupling to other windings on the core.

 

[axim]

I've tried reading your post, but have trouble following a lot of it.

First, with a ferrous core, the flux is guided by that core. So changing the length of one of the windings and keeping the same number of turns should not affect the magentizing inductance. It will likely affect the leakage inductance some, but I don't think that is your question.

Could you be getting confused by using the equation for inductance of a coil that is not wound on a high-$\mu$ ferrous core? In that case, the inductance of that air-core coil could well depend on the length of the coil...

Possibly, I'm not sure. Thanks for reading and trying to understand the long post!
I'll try to boil it down to a more simple example:
Two coils are wound on the same core. Now both have the exact same amount of windings but different length. For this exercise I will assume no leakage because of the core.
In air, the inductance of the primary is lower (windings more spaced), inductance of the secondary is higher (windings more close).

My expectation is that even with the core enhancing the inductance the primary would still have a lower inductance.

This creates the discrepancy:
If I apply an input voltage on coil 1 (load on coil 2 open) there will be a magnetizing current and flux. So I have a ratio $Φ_1/i_1$. If I apply the same voltage on coil 2 (load on coil 1 open) there will be a lower current (as the inductance is higher) and a ratio $Φ_2/i_2$. But the ratios are different, coil 2 with the high inductance will create more flux per current (e.g. half length means double inductance, means half current but divided by half length the same flux).

And as the flux of one coil - because of the core - is completely guided through the other coil, this would mean that a current $i$ from coil 1 would cause a voltage in coil 2 but the same current in coil 2 would create a different flux and therefore voltage in coil 1. Meaning the mutual inductance $M_{12}$ and $M_{21}$ would be different. But that can't be true.
I hope this was understandable, sorry for the lengthy explanation.

 

[Baluncore]

When I vary the secondary coil length I'm getting into trouble.

You are assuming the core gives 100% coupling. The inductance is then proportional to the square of the number of turns, independent of the coil area or the length of the winding over the core.

The area and the length of the winding term, A/l, change the solenoid inductance only when the magnetic core is absent, and assumed to be replaced with non-magnetic air.

 

[axim]

Ah! That matches with what berkeman was suspecting. Thanks for identifying and clarifying my mistake here.
It was a little bit confusing because even some webpages used examples like that (coil in coil with air core) to explain transformer-like behaviour and the mutual inductance. And I was not sure if I could apply it, but thought it would be possible as one coil completely covered the second one, so I tried to transfer it (and for the "normal" case it even worked).

So my understanding is that both coils now appear with the same inductance which is dependend now almost only on the number of windings, and the geometric difference does not matter as the flux is guided by the core. If that is correct I really learned something today which kept me busy some time.

 

[Baluncore]

If that is correct I really learned something today which kept me busy some time.

The two ends of a long air-core solenoid are far apart, so are poorly coupled. That coil has lower inductance than a shorter coil, or one wound on a magnetic core.

To get 50% coupling in air, between a primary and a secondary single turn, you overlap the two turns to share 50% area, or you move them apart to allow half the flux to leak. That is the origin of the area/length term.

To get 50% coupling with a magnetic core, you must split the core lengthwise, then wind one coil about only half of the core flux path.

 

[Charles Link]

This looks very similar to something I just computed in another thread but with a transformer that has an air gap. See https://www.physicsforums.com/threads/inducing-emf-through-a-coil-understanding-flux.940861/page-5 posts 154 and 155.

In any case, I think the flux just depends on the number of turns $N$ and the loop length $l$ of the torus. If I am computing it properly, it doesn't matter, at least to first order, whether you spread the $N$ loops around the whole torus or bunch them together. With a ferromagnetic core that is a ring, you use the loop length $l$ of the torus, rather than the length that the windings are spread in computing the $B$.

Note: This comes from the integral form of the Maxwell equation for $H$ with a transformer coil of $N$ turns: $\oint H \cdot dl=NI$. The $H$ is assumed to be made uniform around the whole core, independent of how the loops are wrapped=whether close together, or far apart. The uniform $H$ is a property that the ideal transformer with a core has, (when there is no air gap). In any case, you always get a uniform $B$ with an ideal transformer, air gap or no air gap. (Note also that the $H$ from the windings is $H =\frac{NI}{l}$ and is uniform around the whole core, gap or no gap).

In computing inductance, we thereby have $L= \frac{\Phi}{I}=\frac{\mu N^2 A }{l}$ where $l$ is the length of the torus loop and $A$ is the cross-sectional area of the core, independent of the spread of the $N$ windings. (Here we are assuming the windings are wrapped close to the core, but the spread doesn't matter). This formula works for both the primary and secondary. Note: $\Phi$ is the flux links: $\Phi=N \mu H A =\frac{N^2 \mu I A }{l}$.

Note for both the primary and secondary, the use of $l_1$ or $l_2$ in the inductance formula in the case of a toroidal type transformer is incorrect.

Edit: Note that $l$ will depend on the radial distance, but to a good approximation we can use the loop length $l$ around the torus that is measured at the center of the cross-sectional area.

 

[Charles Link]

I am still hoping the OP returns to see the input I have for him in post 8, where the length $l$ in his inductance formula needs to be the length around the torus, rather than the distance over which the windings are spread.

 

[Baluncore]

I am still hoping the OP returns to see the input I have for him in post 8, ...

@axim if you are watching.
@Charles Link
Google bots will index it, and over the next decade, many others will read it and benefit.
We can't take it with us when we go, so thank you for leaving your wisdom here for recycling, where others will appreciate it.

 

[axim]

I am still hoping the OP returns to see the input I have for him in post 8, where the length $l$ in his inductance formula needs to be the length around the torus, rather than the distance over which the windings are spread.

@Baluncore @Charles Link Sorry I must have overlooked the mail and got logged out in the browser. Thanks for the additional information I will have a look at it and think about it tonight, don't want to miss that input.

 

[Charles Link]

@axim This transformer problem is actually a very interesting problem. It also turns out that when current $I_s$ flows in the secondary coil, (which is 90 degrees out of phase with the original current in the primary with a resistive load in the secondary), that the primary coil will balance that result which would otherwise create its own magnetic field with extra current of its own, so that $|N_s I_s |=|N_p I_p |$ for this additional current, so that the magnetic field (and the voltage) stays the same, load or no load, in an ideal transformer, and is found from the keep-alive current (no load from the secondary coil) in the primary coil which is often designated as $I_{po}$.

Note that it is the product of the turns $N$ and the current $I$ which determines the magnetic field strength , (as well as any magnetic field that would have occurred if it wasn't balanced by the primary's response), just as in your problem above, independent of the spacing that occurs in the windings of the coil.

 

[axim]

@axim This transformer problem is actually a very interesting problem. It also turns out that when current $I_s$ flows in the secondary coil, (which is 90 degrees out of phase with the original current in the primary with a resistive load in the secondary), that the primary coil will balance that result which would otherwise create its own magnetic field with extra current of its own, so that $|N_s I_s |=|N_p I_p |$ for this additional current, so that the magnetic field (and the voltage) stays the same, load or no load, in an ideal transformer, and is found from the keep-alive current (no load from the secondary coil) in the primary coil which is often designated as $I_{po}$.

Yes that balancing was also what started me puzzling in the first place and got me into the topic. I started by noticing that for high inductor values, just a few mA ($I_{po}$) are sufficient to stop the current from flowing through the "shorted" source coil - so I knew that a load causing several Ampere of current would cause a voltage drop in the source which would be many times higher than the input voltage. So I started to calculate the sequence of this events, which worked for the default examples until I came to the mutual inductance topic here.
Actually I just noticed I had already read parts of the topic mentioned above - https://www.physicsforums.com/threads/inducing-emf-through-a-coil-understanding-flux.940861/page-5 - but could only understand parts of it.
I mean that's always the case, some effects are shown or seem to appear obvious - but when having a look at the details everything gets complex. Does not matter how simple it seems in the first place. But usually I learn something and will have a better understanding when revisiting the topics later.

What I understood from your post #8 above and other linked posts is that $H$ in the transformer is seen uniform and only depends on the number of turns and the current. This comes from the Maxwell equation relating a current through a conductor to a magnetic field $\oint H\cdot dl$. Which for my understanding is the current integrated along the wire, giving the resulting $H$ .
Now here without a core the result would not be uniform at all to my understanding, instead the field would be stronger the closer we are to the wire. In the center there would be an overlap of the contributions from all sides, but still the result is lower. This is at least my basic understanding when trying to use Biot-Savart.
But - here is the part where I need to improve my knowledge for the future - because of the core the field is guided and uniform. This is now a vague statement, but probably what I need to explore next.

 

[Charles Link]

because of the core the field is guided and uniform. This is now a vague statement, but probably what I need to explore next.

Perhaps the chief reason behind this is the magnetic surface currents that occur, whether in reality or simply mathematically, where the magnetic surface current per unit length is $\vec{K}_m=\vec{M} \times \hat{n}/ \mu_o$. These will be on the outer surface of the core and in the same direction as the currents in the coil, but perhaps several hundred times stronger, and the result is that the magnetization $M$ tends to be uniform around the entire core, even when the windings of the primary or secondary coils are limited in their extent.

See also https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/ to get a good description of how the surface current model of magnetization compares to the pole model.

 

[Charles Link]

This comes from the Maxwell equation relating a current through a conductor to a magnetic field ∮H⋅dl. Which for my understanding is the current integrated along the wire, giving the resulting H .

The integral here $\oint H \cdot dl =NI$ is not integrated along the wire. Instead, it is integrated through/around the inside of the core. The current is that which passes through the plane of the disc whose boundary is the loop through the core. (Note also that this includes currents in the secondary coil, and how the primary will then supply additional current to offset the current in the secondary, keeping $H$ and thereby $B$ the same, with $|N_p I_p|=|N_s I_s |$, and the $N_p I_{po}$ thereby determining the $H$ and $B$).

Note: This is different from the integral that is done to compute the voltage of one of the coils, where the $V= \int E_c \cdot dl$ does go along the wire of the coil.

 

[axim]

I created a small drawing after a short clarification (thanks to Charles!).
Maybe it will help someone who will find this later.
$\vec H$ and $d \vec l$ always point in the same direction so that $\oint H \cdot dl$ will simplify to just $HL$ with $L$ being the torus length.

 

[Charles Link]

It may be worth mentioning that the currents in the formula $\oint H \cdot dl=\sum N_i I_i$ are often sinusoids, so that $H$ will also be a sinusoid, i.e. $I_i=I_i(t)=\bar{I}_{i} \cos(\omega t+\phi_i)$, etc., and $H=H(t)=\bar{H} \cos(\omega t+\phi_H)$.

In addition, the voltage (in both primary and secondary) leads the primary sinusoidal keep-alive current $I_{po}$ by 90 degrees, so that with a resistive load in the secondary, the currents $I_s$ and $I_p$, where $N_s I_s=-N_p I_p$, are also 90 degrees ahead of the keep-alive current $I_{po}$.

(Note that the current in the secondary coil doesn't generate any net magnetic field of its own because any current in the secondary is offset by additional current in the primary coil, with no change in the net magnetic field from that which is created by the keep-alive current in the primary coil).