19
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When we consider two inductors that are magnetically coupled, we sometimes talk about mutual inductance. As an example, consider two inductors (such as 9 μH and 4 μH) that are 100% coupled and produce magnetic fields that are aiding. We calculate the mutual inductance (M) like this: -

$$M = \sqrt{L_1\times L_2} \hspace{1cm}=\hspace{1cm} \sqrt{9\text{ }\mu H\times 4\text{ }\mu H} \hspace{1cm}=\hspace{1cm} 6\text{ }\mu H$$

So, a 100% coupled 9 μH and 4 μH inductor have a mutual inductance of 6 μH. But what does the 6 μH represent? I can see that 6 μH is the geometric mean of 9 μH and 4 μH but is that all?

Using "M", we can find the net inductance using this formula: -

$$L_{NET} = L_1 + L_2 + 2\sqrt{L_1\times L_2}$$

And, in the simple example above, we get 9 μH + 4 μH + (2 x 6) μH = 25 μH.

But we don't need M to calculate \$L_{NET}\$. We can calculate it more directly with this formula: -

$$L_{NET} = \left(\sqrt{L_1}+\sqrt{L_2} \right)^2\hspace{1cm}=\hspace{1cm} \left(3+2\right)^2\text{ }\mu H\hspace{1cm} = \hspace{1cm}25 \mu H$$

  • So, does M (mutual inductance) have a numerical (or any other) useful meaning?
  • Does it represent something that is important to us?

Is it only of (some) use when calculating the net inductance of two coupled inductors?


Update Sept 10th

Given the answers so far, I'm sorry to say that I still find M to be fairly meaningless given that it translates into more useful terms like N (turns ratio) and k (coupling factor) in every situation the current answers have raised.

If someone can come up with an EE example where M is clearly a benefit over more pragmatic variables, it would be good.

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  • 12
    \$\begingroup\$ Andy asks an electromagnetism-related question? O.o The world is falling apart or he got his account stolen. \$\endgroup\$ Sep 8, 2021 at 9:34
  • 4
    \$\begingroup\$ @RohatKılıç - the back-story is that I have never found M to be useful and don't understand why it appears to have such prominence. I think it's an "emperor's new clothes" situation. \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 9:37
  • 2
    \$\begingroup\$ Very good question and +1 for the character to ask it. \$\endgroup\$ Sep 8, 2021 at 10:33
  • 7
    \$\begingroup\$ Actually, this is a very good question. Thinking about it, I never required an M value as a design parameter (or goal) for a, say, transformer design, never needed to measure it as well, and never seen in the datasheet of a transformer or a coupled inductor. So, why do we need this then? \$\endgroup\$ Sep 8, 2021 at 11:52
  • 2
    \$\begingroup\$ +1 right on. @Rohat Kılıç says what I was going to say, exactly. \$\endgroup\$
    – john
    Sep 8, 2021 at 22:40

8 Answers 8

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If you are writing mesh equations for a circuit containing a transformer (or coupled inductors), use of mutual inductance leads to a simple formulation for the potential difference across a branch containing one of the inductors.

For example if \$V_1\$ is the voltage across the branch formed by inductor L1 and \$V_2\$ is the voltage across L2, with mutual inductance \$M\$ between the two inductors, then

$$V_1 = L_1\frac{dI_1}{dt}-M\frac{dI_2}{dt}$$ and $$V_2 = L_2\frac{dI_2}{dt}-M\frac{dI_1}{dt}$$

Or, in the frequency domain $$v_1 = j\omega L_1 i_1 - j\omega M i_2$$ and $$v_2 = j\omega L_2 i_2 - j\omega M i_1.$$

Of course there's no information lost if you substitute \$M=k\sqrt{L_1L_2}\$, but the equation has a simpler form if you condense that into one term instead of three terms with a square root to be calculated.

Also, the first three or four pages I found by googling (that was part of the question, right?) all define \$M\$ first, and then define \$k\$ in terms of \$M\$. Presumably this was the historic way the terms were developed as outlined in user4574's answer. So one benefit of defining the mutual inductance is to give you something from which to define the coupling factor \$k\$, if you prefer to formulate your equations in terms of \$k\$.

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  • 1
    \$\begingroup\$ Yes. And the coupling factor provides you a relative measure (0..100%) which is more insightful as a gauge for efficacy (transformers) or isolation (wires/traces), whereas mutual inductance -as you write- is used to solve coupled networks. Neither is superfluous. \$\endgroup\$
    – P2000
    Sep 9, 2021 at 6:32
  • 2
    \$\begingroup\$ @Andyaka, It doesn't matter what circuit. If you're doing mesh analysis (sorry, I wrote "node" initially, it's fixed now) the first step is write a set of KVL equations for your circuit. The second step is use the constitutive equations for the components to eliminate \$v\$ variables from the equations, leaving only \$i\$ variables. This expression for \$v\$ of one branch in terms of \$i\$ in terms of both branches of the transformer allows you to do that very simply. \$\endgroup\$
    – The Photon
    Sep 9, 2021 at 14:52
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    \$\begingroup\$ If you had a description of your transformer in terms of \$k\$ and turns ratio, you would end up using that to calculate \$M\$ and plug it in to the equations, even if you didn't think to give that term the name \$M\$ at the time. \$\endgroup\$
    – The Photon
    Sep 9, 2021 at 14:53
  • 3
    \$\begingroup\$ And not just textbooks. Circuit simulators use M as a basic element, EM simulators (e.g. COMSOL) use M to model coupling, and wireless inductive links are characterized and modelled like the eqns in the above answer. \$\endgroup\$
    – P2000
    Sep 9, 2021 at 23:19
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    \$\begingroup\$ Hi photon, I'm not sure but may be this example helps to understand, why M is more useful than k, for any given configuration of two coils, if we need to find the value of K experimentally, then from your equation we need V2, I2, I1, L1 and L2. While for M we don't need L1 anymore and hence we reduced one variable , and there may be certain application where we actually don't need L1. \$\endgroup\$
    – user215805
    Sep 10, 2021 at 12:53
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I have never found M to be useful and don't understand why it appears to have such prominence.

One reason M enjoys so much prominence is because it was part of Maxwell's original publication on electrodynamics. The first known appearance of M in literature that I am aware of is right in Maxwell's 1865 publication "A dynamical theory of the electromagnetic field" (See page 468).

Is it only of use when calculating the net inductance of two coupled inductors?

Maxwell lays it out pretty clearly in his publication (although his terminology is a bit archaic).

For two circuits (inductors) A and B. The electromagnetic momentum belonging to A is...

L*x + M * y

The electromagnetic momentum belonging to B is...

M*x + N * y

The equation for the current x in A is...

v = R * x + d/dt(L * x + M * y)

The equation for the current y in B is...

n = S * y + d/dt(M * x + N * y)

where v and n are the electromotive forces (voltages), x and y are the currents, and R and S are the resistances in A and B respectively.

The implication of the equation is that the voltage on one inductor consists of three components.

  1. An I *R resistive part.
  2. An inductance * di/dt part from itself
  3. A mutual-inductance * di/dt from the current in the other inductor.

So basically the point of mutual inductance is that if I change the current in one inductor then it causes a voltage in the other one, which is proportional to M and the rate of change in current. This is of course the basis for a transformer.

Also, note that inductance is typically proportional to the square of the number of turns, which is why M is proportional to the square root of the inductance of each winding.

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  • \$\begingroup\$ I'm not so sure about your final sentence. \$M=\sqrt{L_1\times L_2}\$ so, M is proportional to inductance i.e. it's the geometric mean of two coupled inductors when k=1. It's interesting that Maxwell used M. \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 14:46
  • 1
    \$\begingroup\$ @Andyaka I clarified which inductance I was talking about. M is proportional to the square root of the inductance of each winding. \$\endgroup\$
    – user4574
    Sep 8, 2021 at 14:56
  • \$\begingroup\$ Mutual inductance IS inductance. \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 19:41
  • \$\begingroup\$ M appears to just obfuscate basic and more fundamentally important circuit elements. It seems to be transferable to other variables so, it's only use appears to be shorthand but, if that is the case then why is it held up as being important or particularly useful? \$\endgroup\$
    – Andy aka
    Sep 11, 2021 at 15:50
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    \$\begingroup\$ @Andyaka sometimes it's not obfuscation but abstraction, e.g. when solving diff.-eqns. to find Ls and M from V and I, then M (an abstraction or effect) will be found with Ls, before determining k from it. Then M models a physical effect without necessarily being a corpuscular thing with turns etc... \$\endgroup\$
    – P2000
    Sep 11, 2021 at 20:17
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It may be true that M is not important in transformers or coupled inductors.

If someone can come up with an EE example where M is clearly a benefit over more pragmatic variables, it would be good.

1. Here is an example where M could be useful.

Consider an example of two multi-turn air-core coils is placed at a distance in a wireless charging application. If you need to design the wireless charging coils to have desired performance/characteristics you need to calculate M and it becomes useful. (Image sources: 1 and 2)

enter image description here

Let's say the gap between two coils varies between 4cm - 6cm, the maximum diameter of the coils should be less than 10 cm. The design goal is to optimize the coil parameters such as diameter, the number of turns, turn pitch, etc to achieve the best power transfer efficiency at a given output power. (Formula image source.)

Now the output power and efficiency are:

enter image description here

You need to first calculate mutual inductance using Elliptic integrals e.g. follows: (Formula image source.)

enter image description here
enter image description here

With this calculation of M (and of course known ways of calculating coil resistances), you can implement a complete optimization problem as a function of coil geometric parameters and distances.

In this example, you cannot use the coupling coefficient directly as the value of coupling depend on the distance and coil parameters in a non-linear manner.

2. Meaning of M

Although we write as \$M=k \sqrt{L_1 L_2}\$, in fact, the fundamental physical quantity is \$M\$ because: (Image source)

enter image description here

\$k\$ is a derived coefficient to get an understanding of coupling percentage. It would be more appropriate to express \$k=M/ \sqrt{L_1 L_2}\$, at least in the above example.

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  • \$\begingroup\$ Unfortunately the link that might allow me to analyse what your equations might actually be hides behind a researchgate login page. Neither have you explained why M is a fundamental physical quantity and why k is derived from M. I mean, just throwing in a few formulas that express k in terms of M, L1 and L2 doesn't prove or substantiate anything. Basically you have no justified why M is useful. Don't get me wrong - I'm sure you are right (just as everyone with an answer is probably right) but, nothing is shaking the ground underneath my feet regards the usefulness of M. \$\endgroup\$
    – Andy aka
    Sep 24, 2021 at 13:50
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    \$\begingroup\$ why M is useful: because the example in the answer needs M for the stipulated objective. Can you provide a way to achieve this without calculating/measuring M? \$\endgroup\$
    – Pojj
    Sep 24, 2021 at 15:13
  • \$\begingroup\$ I don’t see an example in your answer and I don’t see a specific formula. Show me both please. I need something that is clear and not ambiguous. If you want to paraphrase what method is used to directly solve M then that’s a start. Assertions are not enough. \$\endgroup\$
    – Andy aka
    Sep 24, 2021 at 19:13
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    \$\begingroup\$ Can you show how you can calculate k directly in this example? \$\endgroup\$
    – Pojj
    Sep 24, 2021 at 20:37
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    \$\begingroup\$ Well, I am giving an EE example that M is useful. If you don't agree with it, please provide the alternative way of solving the problem without M. \$\endgroup\$
    – Pojj
    Sep 25, 2021 at 5:23
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What precisely is the numerical (or any) meaning of mutual inductance

So, a 100% coupled 9 μH and 4 μH inductor have a mutual inductance of 6 μH. But what does the 6 μH represent?

So, does M (mutual inductance) have a numerical (or any other) useful meaning?

As we know, for a two-inductor system, the total voltage in each coil is:

\$v_1(t) = L_1 i_1'(t) \pm M i_2'(t) \tag 1\$

\$v_2(t) = L_2 i_2'(t) \pm M i_1'(t) \tag 2\$

Let's assume one coil is kept open-circuited while we apply a time-varying current to the other coil. Then, separately, the previous equations simplify to:

\$v_1(t) = \pm M i_2'(t) \tag 3\$

\$v_2(t) = \pm M i_1'(t) \tag 4\$

The plus-minus signs represent the fact that the voltage induced by mutual induction (i.e. on one coil due to the other coil) depends on the relative spatial orientation of the two windings, i.e. it depends on the position of the dots. Of course it also depends on the reference direction for both currents and the reference polarity for both voltages.

Now suppose the rate of change of the current in one coil is 1 amp per second. Then, ignoring units the previous equations separately simplify to:

\$v_1(t) = \pm M \tag 5\$

\$v_2(t) = \pm M \tag 6\$

From the two above equations (5) and (6), the take-away is: the mutual inductance between two inductors tells you the volts induced in one open-circuited coil when we (linearly, i.e. at a constant rate) change the current in the other coil by one ampere in one second. Of course, if such change in current is not of one amp, which is in general the case, then we must use eqs. (3) and (4).

So for the two-inductor system of your question, using eqs. (5) and (6), the \$M\$ = 6 μH means that 6 volts are induced in one (open-circuited) coil, when we linearly change the current in the other coil by 1 ampere in 1 second (even if the coupling coefficient is not 100% as your example). More generally (eqs. (3) and (4)), the \$M\$ = 6 μH means that, when we linearly change the current in one coil by \$\Delta I\$ amperes in 1 second, \$6 \cdot \Delta I\$ volts are induced in the other (open-circuited) coil.

For example, consider the following circuit, in which \$L_1\$ = 7 H, \$L_2\$ = 4 H and \$M\$ = 3 H. (I know those numbers are big, but it doesn't matter in this discussion.) Notice that from \$t_1\$ = 1 s to \$t_2\$ = 2 s (so the change in time is \$\Delta t\$ = 1 s), the current in \$L_1\$ changes from \$I_1\$ = 0.5 A to \$I_2\$ = 1 A, thus the change in current is \$\Delta I\$ = +0.5 A, therefore the voltage induced in \$L_2\$ is \$M \cdot \Delta I = 3 \cdot 0.5 = 1.5\$ volts.

enter image description here

Figure 1. Image source: own.


[...] and, is it useful

It allows us to use circuit equations (equations involving currents and voltages only, which are scalar functions of scalar variables) instead of directly using Maxwell's equations (equations involving electric and magnetic fields, which are vector functions of vector and scalar variables, which are harder to solve).

Note that many circuit analysis textbooks (Sadiku, Dorf, Hayt, Irwin, Nilsson, Thomas, Van Valkenburg...) start by defining the mutual inductance, then introduce the coupling coefficient by defining it as a ratio of two quantities measured in units of henrys (mutual inductance and the geometric mean of the self-inductances, where the three inductances can be derived from Maxwell's equations). It's kind of similar to how power factor is defined as a ratio of two quantities measured in watts (active power and apparent power, both of which are independently defined). (Yes, I know apparent power is measured in volt-amperes, but that's exactly equivalent to watts; we use VA for apparent to not confuse it with active power.)

As you've shown in part 3 of your answer, we can formulate the circuit equations in terms of the mutual inductance and the self-inductances, or, in terms of the coupling coefficient (which depends on the mutual inductance and self-inductances) and the turns ratio (which depends on the self-inductances). Either way is just as useful in the sense that they allow us to still work with circuit variables (voltages and currents) only.

Perhaps one way (using \$k\$ and \$N\$) is more practical than the other (using \$M\$ and \$L_1\$ and \$L_2\$, as you showed) because it yields shorter equations, but that doesn't make the latter less useful. And for that matter, we could substitute \$k = M/\sqrt{L_1 L_2}\$ and \$N = \sqrt{L_2/L_1}\$, to see that our equations still depend on \$M\$, \$L_1\$ and \$L_2\$.

Here's another way I look at this. If we change the current in one coil, we find out a voltage is induced in the other open-circuited coil, in other words the induced voltage is directly proportional to the rate of change of current: \$v_2(t) \propto i_1'(t)\$. The mutual inductance simply allows us to write this proportionality as an equation: \$v_2(t) = M i_1'(t)\$. I would say this is analogous to how Georg Ohm found out that, in some devices under some circumstances, the voltage across the device is directly proportional to the current through the device (\$v(t) \propto i(t)\$); then he introduced (static/DC) resistance as the proportionality constant, to allow us to write the proportionality as an equation: \$v(t) = R i(t)\$. So the mutual inductance is what allows us to relate changing currents to induced voltages, that's why it's useful.


representative (of something)

Qualitatively (in words only, without numbers), the mutual inductance between two coils is the ability of one inductor to induce a voltage across the other inductor, and vice versa.

For a quantitative explanation, I'll repeat it: the mutual inductance between two inductors tells you the volts induced in one open-circuited coil when we linearly change the current in the other coil by one ampere in one second.

The above interpretations of mutual inductance are analogous to self-inductance. Qualitatively, the self-inductance of a coil is the ability of a time-varying current in the coil to induce a voltage across the coil. Quantitatively, the self-inductance of a coil tells you the volts induced in the coil when we linearly change the current in the coil by one ampere in one second.


or important?

\$M\$ (and \$L_1\$ and \$L_2\$) is just as important as \$k\$ (and \$N\$) for setting up the circuit equations. You can use one or the other.


Does it represent something that is important to us?

What is important to you? Do you consider resistance important? Because we can write circuit equations in terms of conductance \$G\$ instead of resistance \$R\$, just like we can write the circuit equations in terms of \$k\$ instead of \$M\$.


Is it only of (some) use when calculating the net inductance of two coupled inductors?

As far as I know, the only purpose of calculating the net inductance is to simplify inductors connected in series or in parallel. But we can use \$M\$ even when the inductors aren't neither in series nor in parallel, as shown in the following circuit.

enter image description here

Figure 2. Image source: own.


Part 2 - you don't need to use M (coupling less than 100%)

You have said that using \$k\$ is simpler than \$M\$. Do you really think doing the procedure you showed in part 2 of your answer is simpler than just using \$M\$? In your procedure (shown as an image), you had to 1) manipulate the self-inductances and mutual inductance; 2) convert the inductances of the new coupled inductors into turns; 3) combine the turns; 4) convert the equivalent turns to an equivalent inductance. Don't you think it's easier to use \$M\$ instead, since you don't need to do any conversion nor manipulation?

Furthermore, even if you considered your method of part 2 to be simpler, I think you can't apply it for inductors that aren't in aiding or opposing series or parallel connections, such as the those in figure 2 of this answer.


Edit

Here's another way some people (like me) look at this \$M\$-vs-\$k\$ dispute.

As we know, (static/DC) resistance is defined as the ratio \$R = V_{ab}/I\$. We can use electromagnetic theory to find an expression for the resistance in terms of the material of the conductor (using the resistivity \$\rho\$ or conductivity \$\sigma\$) and the geometry of the conductor (using the length \$L\$, the surface area \$S\$, etc.) For example, if we want to find the resistance of a cylindrical conductor, and if we assume the conductor has uniform electric field \$\mathbf E\$ and uniform current density \$\mathbf J\$, then we can find the resistance as follows. The voltage is:

\$\begin{align} V_{ab} &= \displaystyle\int_a^b {\mathbf E} \cdot {\mathrm d} {\mathrm l} \\ &= \displaystyle\int_a^b E {\mathrm d} l \cos{(\theta)} && \text{expand dot product} \\ &= \displaystyle\int_a^b E {\mathrm d} l && \text{electric field is parallel to length vector} \\ &= E \displaystyle\int_a^b {\mathrm d} l && \text{electric field is independent of length} \\ &= E L && \text{integrate the differential length} \end{align}\$

the current is:

\$\begin{align} I &= \int_S {\mathbf J} \cdot {\mathrm d} {\mathbf S} \\ &= \int_S J {\mathrm d} S \cos{(\theta)} && \text{expand dot product} \\ &= \int_S J {\mathrm d} S && \text{current density is parallel to surface vector} \\ &= J \int_S {\mathrm d} S && \text{current density is independent of area} \\ &= J S && \text{integrate the differential surface} \end{align}\$

and substituting the two previous results for the current and voltage into \$R = V_{ab}/I\$, we get:

\$\begin{align} R &= \dfrac{EL}{JS} \\ &= \dfrac{EL}{(\sigma E) S} && \text{substitute $J = \sigma E$} \\ &= \dfrac{L}{\sigma S} && \text{simplify} \\ &= \dfrac{\rho L}{S} && \text{substitute $\sigma = 1/\rho$, if desired} \end{align}\$

which is a well-known formula. (End of example.)

For capacitance, we define the capacitance between two separated conductors/plates as ratio of the electric charge stored on either conductor, to the voltage or potential difference across the two conductors (\$C = Q/V_{ab}\$); then we can use Gauss' law and the voltage equation \$V_{ab} = \int_a^b {\mathbf E} \cdot {\mathrm d} {\mathrm l}\$ to express the capacitance in terms of the dielectric and its geometry only; in this way we get formulas such as \$C = \epsilon S/d\$ for parallel-plate capacitor.

For self-inductance, we define the self-inductance of a conductor as the ratio of the total magnetic flux linkages to the current which it they link (\$L = \lambda/I\$); then we can use Ampère's law and the magnetic flux equation \$\Phi = \int_S {\mathbf B} \cdot {\mathrm d} {\mathrm S}\$ to express the inductance in terms of the material and its geometry only; in this way we get formulas such as \$L = \mu_0 N^2 S/(2 \pi R)\$ for toroidal inductor.

In this way we can find the resistances, capacitances and inductances of conductors. That's how it's done for power transmission lines (e.g. read the textbooks by Glover & Sarma [chapter 4] or Grainger & Steveson [chapters 4 and 5] on power systems analysis).

Similar to the above properties (resistance, capacitance, inductance), the mutual inductance can be derived from equations of electromagnetic theory, so what's why we say it's a "fundamental property". On the other hand, we conceive the coupling coefficient \$k\$ simply as a dimensionless factor defined in terms of fundamental properties, specifically as the ratio of mutual inductance to the geometric mean of the self-inductances.

Even though calculating \$R\$, \$C\$, \$L\$ and \$M\$ involves electromagnetic theory/Maxwell's equations, those constants are considered as given/known in circuit analysis, so that we don't have to ever use Maxwell's equations again.

There's nothing wrong on only using \$k\$ (and \$N\$) in the circuit equations. If you find ways to always express the circuit equations in terms of \$k\$ without \$M\$, that's fine, whether the resulting expressions are shorter or longer. Simply put, I and other people think of \$k\$ as a derived quantity from \$M\$, so that saying \$M\$ is useless is like saying \$k\$ is also useless (which obviously is not true).

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  • \$\begingroup\$ Thank you - that gives me something to work on. \$\endgroup\$
    – Andy aka
    Sep 26, 2021 at 18:00
  • \$\begingroup\$ The part with the mutual inductances is what I was aiming for in my (deleted) answer, but didn't get the words for it. +1 \$\endgroup\$ Sep 27, 2021 at 6:58
  • \$\begingroup\$ 2) convert the inductances of the new coupled inductors into turns, which requires to know the inductance factor - that is incorrect, it is sufficient to recognize that they will be identical in both coils and, no I don't think it's easier to use M instead. More to come later... \$\endgroup\$
    – Andy aka
    Sep 27, 2021 at 11:03
  • \$\begingroup\$ You are the only person to use the term "useless" on this Q and A page. Maybe I have used a similar (but not exactly the same) term somewhere that you are paraphrasing? \$\endgroup\$
    – Andy aka
    Sep 28, 2021 at 7:33
  • \$\begingroup\$ @Andyaka Re: “You are the only person to use the term "useless" on this Q and A page […]” // In your first comment of the question post, you said “I have never found M to be useful”. I would say that is the same as saying “I have always found M to be useless”. Or do you think not? \$\endgroup\$
    – alejnavab
    Sep 28, 2021 at 15:05
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Part 1 - you don't need to use M (coupling = 100%)

This statement came from the first posted answer (now deleted): -

Try analysing any circuit with coupled inductors without using M

If two inductors are coupled 100% series aiding, you can convert the inductances into "turns" by assuming that they share the same core (100% coupled). Consider this from the question: -

As an example, consider two inductors (such as 9 μH and 4 μH) that are 100% coupled and produce magnetic fields that are aiding.

You can arbitrarily assume that the inductance factor (\$A_L\$) is 1 μH per turn.

Then, you have this: -

  • 9 μH has 3 turns
  • 4 μH has 2 turns

And, the total number of turns is 5 hence, the net inductance is 5² μH = 25 μH.

In other words, you do this: -

$$L_{NET} = (\sqrt{L_1}+\sqrt{L_2})^2$$

Mister mutual inductance was not needed nor harmed!


Part 2 - you don't need to use M (coupling less than 100%)

Another comment suggested that 100% coupling was a special case. Here's the same example with 60% coupling. I've done it pictorially because I think it's more meaningful to see the steps: -

enter image description here

If I were to calculate net inductance using the formula in the question I'd get the same answer.

To answer my own question; so far, M appears to be fairly meaningless, not particularly useful and not representative of anything significantly important. Coupling factor, k, on the other hand, appears to be much more fundamental and useful in the understanding coupled inductors.


Part 3 - you don't need to use M (isolated windings with any degree of coupling)

In @user4574's answer this was said: -

So basically the point of mutual inductance is that if I change the current in one inductor then it causes a voltage in the other one, which is proportional to M and the rate of change in current.

The inference I take here is that M is useful, but I have my doubts when one considers the importance of k, the coupling factor...

In the answer supplied by @ThePhoton this formula was given "for a circuit containing a transformer (or coupled inductors)": -

$$V_2 = L_2\frac{dI_2}{dt}+M\frac{dI_1}{dt}$$

This was also mentioned in a comment by @P2000: -

@ThePhoton points out, if you attempt a similar example as the one above but where the current or voltage are NOT shared (i.e. only flux is coupled) you'll end up calculating an M to solve equations.

But this just doesn't appear to be true.

Consider the \$V_2\$ formula directly above. It contains two terms; the first term is when there is an output current on the "secondary" winding and there is a resultant volt-drop but, the main term to concentrate on is that one that includes M. So, in the case when there is no secondary current the \$V_2\$ formula becomes this: -

$$V_2 = M\frac{dI_1}{dt}$$

Knowing that \$L_1\frac{dI_1}{dt}\$ can be equated to the "primary" voltage \$V_1\$, we can manipulate the simplified \$V_2\$ equation to this: -

$$V_2 = M\cdot \frac{L_1}{L_1}\frac{dI_1}{dt}\hspace{1cm} =\hspace{1cm} M\frac{1}{L_1}\cdot V_1$$

Then, because \$M = k\sqrt{L_1 L_2}\$, the above formula can be rewritten: -

$$V_2 =k\sqrt{L_1 L_2}\frac{1}{L_1}\cdot V_1\hspace{1cm} =\hspace{1cm}k\cdot V_1\cdot\sqrt{\frac{L_2}{L_1}}$$

And, the square root of the inductance ratio is well-known to be the turns ratio, N, hence: -

$$V_2 =k\cdot V_1\cdot N$$

Can anyone explain why this derived formula isn't simpler, more useful and more intuitive compared to the formula that uses M (as proposed by @ThePhoton and @user4574)?

Anyone who knows stuff about transformers will recognize it as the basic primary-secondary voltage relationship.

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    \$\begingroup\$ Sure, in a few particular cases you can solve the circuit without more than "Elecronics for Dummies". I thought it was a serious question. I broke my rule of not answering something that google can answer. \$\endgroup\$
    – Tesla23
    Sep 8, 2021 at 10:21
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    \$\begingroup\$ @Tesla23 it is a serious question. If you have an example where M gives a better way then please do reveal. Can google truly answer this question? \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 10:25
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    \$\begingroup\$ @user215805 to know M you need to know k and, if you know k, you don't need to know M. You can calculate directly but, as I said, please leave an example. \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 11:32
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    \$\begingroup\$ @Andyaka I've ever only noticed that M is just k, with situational units (Henries) applied. So I take your points. I've only one added thought. The form of your summing equation is easy to explain. Since the inductance is proportional to N^2, to sum them you need to take their square root first to map from L-space back to N-space so that you can add the turns and then square the result to go from N-space back to L-space, so to speak. It may help a beginner to understand why that equation does what it does. \$\endgroup\$
    – jonk
    Sep 8, 2021 at 19:09
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    \$\begingroup\$ @P2000 I'm not saying that M cannot be used for anything rather, I'm saying that there are better ways that are more pragmatic and useful to EEs. You don't need to recognize or calculate M to solve coupled inductor problems where voltage and current are not shared. \$\endgroup\$
    – Andy aka
    Sep 9, 2021 at 13:13
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Inductance tells you the voltage a current change induces in a given inductor. Mutual inductance tells you the voltage a current change in one inductor induces in the other inductor.

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    \$\begingroup\$ Maybe you can give an example of Mutual inductance tells you the voltage a current change in one inductor induces in the other inductor? \$\endgroup\$
    – Andy aka
    Sep 8, 2021 at 10:36
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Mutual inductance can be computed (if not always analytically, numerically) from the sole geometry of the coils (which need not be solenoids, can have arbitrary shapes and orientation) using Neumann's formula:

Neumann's formula

Can you do the same with k from first principles? Can you compute k, given the geometry of the coils, without invoking the quantities used to define M (magnetic flux or vector potential)?

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  • \$\begingroup\$ Using Biot Savart you can and, I believe your formula originates in the same place. \$\endgroup\$
    – Andy aka
    Sep 27, 2021 at 6:21
  • \$\begingroup\$ @Andyaka but you still need to go through magnetic flux to get there. B is prop to I, then flux of B is prop to I (oh, look!) and then the time derivative of the flux is prop to the time derivative of I. Anyway, your formula for k is...? \$\endgroup\$ Sep 27, 2021 at 12:29
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Andy's observation has good merit. It is clear to me that,

"\$M\$ is a single letter representation for \$k\sqrt{L_{1}L_{2}}\$ and really nothing more."

So why does \$M\$ for mutual inductance exist. I speculate that back in the day a place holder for an unknown inductive quantity was required until a solution was found.

Is \$M\$ useful or important? Well, I suppose that if using \$M\$ adds clarity or reduces clutter in how a problem is solved then the answer is yes. If not then No. I guess I am sidestepping the question. But in reality, \$M\$ is not necessary as Andy has clearly demonstrated in his answer.

Does \$M\$ or \$k\sqrt{L_{1}L_{2}}\$ have any sort of meaning?

I start with the meaning of inductance as derived from Faraday and Lenz. These folks gave us:

$$v_{\phi}=-\frac{d\phi}{dt}$$

If the magnetic flux \$\phi\$ is created by current in a wire (single loop), then the equation may be modified into:

$$v_{\phi}=-\frac{d\phi}{di}\frac{di}{dt}$$

For several loops (turns)\$N\$: $$v_{\phi}=-N\frac{d\phi}{di}\frac{di}{dt}$$ \$-N\frac{d\phi}{di}\$ is the ability for current to create the flux within its region of influence. It is this ability that is called inductance \$L\$. Really it is self-inductance. $$L=-N\frac{d\phi}{di}$$

If there is a second loop of wire nearby sitting in the changing flux of the first, then clearly by Faraday/Lenz, a voltage will be induced across the second loop of wire.

This is not new, EE folks should be familiar with this. So then consider \$M\$. The two loops of wire can be easily envisioned with a proportion (\$k\$) of the flux from the first loop flowing through the second loop. The inductance equation can be written: $$L_{12}=-N_{1}\frac{d\phi_{2}}{di_{1}}=-kN_{1}\frac{d\phi_{1}}{di_{1}}=kL_{1}$$

$$L_{21}=-N_{2}\frac{d\phi_{1}}{di_{2}}=-kN_{2}\frac{d\phi_{2}}{di_{2}}=kL_{2}$$

These coupling inductances represent the ability of a current in one loop to establish a magnetic flux in another loop. This is where the mutual inductance is defined as the geometric mean of the coupling inductance:

$$M=\sqrt{L_{12}L_{21}}=k\sqrt{L_{1}L_{2}}$$

So here is the meaning:

Mutual inductance is the geometric mean of the coupling inductances.

This averaging tends to hide the meaning of the coupling inductances.

So in the end I support Andy's answer to his own question.

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