# Finding Inductance and Mutual Inductance

• Ntip

#### Ntip

TL;DR Summary
I am trying to solve for inductance through FEA simulation. I and doing eddy current simulations so I can have H field, B field, and current density available
I am trying to solve for inductance through FEA simulation. I and doing eddy current simulations so I can have H field, B field, and current density available.

I am also trying to find an equation for the equivalent inductance with multiple conductors. I know that for two conductors it is Leq = L1 + L2 +/- M12 +M21. If L1=L2 and M12=M12, it simplifies to L=2L1+/-2M12. I'm trying to find how to expand this out to maybe 4 conductors.

Well, that is interesting. What software are you using ?
Do you have a specific question ?
The inductance of each element is the sum of it's self inductance + the mutual inductance with all other elements of the circuit conductor. Since the M of segment a with b is the same as b with a, you can generate a triangular matrix, then double it.

Only two element interactions are required to compute the matrix.
The equations needed for computation of M are given in chapters 3 to 7 of the book by Frederick W. Grover. (1973) "Inductance Calculations. Working formulas and tables". Chapter 4, is only 3 pages. It describes the method of analysis and matrix construction.

You are in for plenty of 2D or 3D vector computations, and some Atanh() functions for the wires at odd angles.

• DaveE
I'm using Ansys Maxwell 3D. I can also do it in Ansys Maxwell 2D to eliminate the thickness component for simplicity.

My question is more in general I guess. I'm not really to sure how to calculate inductance from the parameters that I have. Ideally, I would like to tie the inductance calculation back to Maxwells equations. I've been looking online and haven't found it which surprised me. I have the material properties, current I, current density J, magnetic field H field, and magnetic flux density B.

As far as the mutual inductance...let's look at the case of 4 parallel conductors and say that their current directions alternate. In this way, the effects of their mutual inductance should be subtractive. If my understand is correct, it is subtractive because the magnetic field from the adjacent conductor is in the opposite direction when is passes through the other conductor and by vector addition the B-field inside of the conductor is decreases which decreases the tendency to opposite the change in current in the conductor (inductance). So if we have four parallel conductors with current in the opposite directions, the equivalent inductance would be

L1 + L2 + L3 + L4 - M12 - M21 - M23 - M32 - M34 - M43

If we assume all geometries between the conductors are the same, the coupling M12, M21, M23, M32, M34, M43 would all be the same. so if L = L1 = L2 = L3 = L4, and M = M12, we would have

Leq = 4*L -6*M.

That doesn't make sense to me because does that it makes it look like if I put say 1,000,000 in parallel, I could possibly get a negative inductance. Of course 1,000,000 is just an exaggeration to show my point but still. Any suggestions?

So if we have four parallel conductors with current in the opposite directions,...
You appear to be describing two return circuits or transmission lines.
If my understand is correct, it is subtractive because the magnetic field from the adjacent conductor is in the opposite direction when is passes through the other conductor and by vector addition the B-field inside of the conductor is decreases which decreases the tendency to opposite the change in current in the conductor (inductance).
Reduce the circuit to many straight elements, all vectors and all pointing along the circuit in the direction of reference current flow. For a return circuit, the parallel elements will be running in opposite directions, so the sign of M and the induced voltage will also be reversed.

Where the two currents flow in opposite directions, the magnetic fields will sum in the area between the wires, while they tend to cancel outside. Where currents are in the same direction, the magnetic fields will tend to cancel in the area between the wires while they sum outside.

Yes, I understand that the induced voltage is reversed which is what leads to a negative coupling (-M). What I'm saying is that in the case of 2 conductors in the case of subtractive mutual inductance, it makes sense that the equivalent inductance is 2L-2M or n(L+M) for n=2. But if you have 4 conductors it turns out to be 4L-6M. Six conductors would be 6L- 6M - M45 - M54 - M56 - M65 = 6L - 10M. The equation seems to equate to n(L-M)-(n-2)M for even numbers of n. So if n is really large, you would end up negative inductance which wouldn't be the case. So I'm wondering what's really happening with the B field inside of the conductors. It seems that even if the magnitude of current in each conductor is the same, the magnitude of mutual inductance M must be reduced for n>2 compared to n=2 so that Leq does will always be positive.

If you consider filaments, you eliminate internal inductance.

It is insufficient to simply specify parallel.
You must also specify distribution of elements in space as GMD.
N conductors can pass through a fixed sectional area of a ;
or can be alternated with separation d across a flat plane of width d·(n-1) .

If you present a fixed voltage step of V to your model, and then measure di/dt ;
you can compute L = V · dt / di

I think the easiest way to calculate the self inductance is to use the magnetic field energy. If you have a single conductor, you can solve for the B-field everywhere. Then the total magnetic field energy is $E = \int \frac{B^2}{2 \mu_0} dV$, where the integral is over all space. Then you can calculate the inductance from $E = \frac{1}{2}L I^2$. When you have multiple conductors, you can calculate the individual inductance of each conductor separately. I'm not sure how you would go about calculating the mutual inductance in this way.

It is insufficient to simply specify parallel.
You must also specify distribution of elements in space as GMD.
N conductors can pass through a fixed sectional area of a ;
or can be alternated with separation d across a flat plane of width d·(n-1) .

Ok so yes in additional to the conductor geometries being the same, separation is also the same which is what would allow M12=M21=M23=... So let's say these are all parallel conductors with alternating current direction, Leq would normally be Leq=L/n where n is the number of conductors. Considering the subtractive mutual inductance, I would get $$\frac {(L-M)-(n-2)M} {n}$$.

That's the equation that I get doing 4 and 6 conductors anyway but it wouldn't hold as n get's really large as that would lead to a negative. If the convention that I mentioned in the posts above to determine Leq is correct, then it seems like M must get smaller to prevent Leq from going negative in the case where n is really large. Am I missing something with my logic? Or is it unclear what I'm trying to say/ask?

Am I missing something with my logic? Or is it unclear what I'm trying to say/ask?
You are missing something in your logic, and failing to clearly specify the circuit.

You cannot have equal spacing between 4 or more parallel elements, it is geometrically impossible beyond the equilateral triangle. Your separations between elements do not hold in 3D space.

How many circuits do you have? The elements that are parallel are only the currents. Until you connect the ends to define one circuit with two terminals, the voltage summation is undefined.

A folded wire makes two close parallel elements. That has low inductance, NOT negative inductance. If the fields of the elements cancel, there is no field and so, no inductance.