can someone explain mutual inductance?

Self-inductance is defined as L=V/(dI/dt). Similarly, mutual inductance M between coils 1 and 2 is defined as M_21 = M_12 = V_1/(dI_2/dt) = V_2/(dI_1/dt). This definition requires that the induced voltage be measured in an open circuit, and that the current ramp is slow enough that capacitive effects have a chance to die away.

Say you have two coils with inductance L_1 and L_2 and mutual inductance M. If you wire them in series aiding, the inductance becomes L_1 + L_2

• 2M, because dI/dt is the same in both coils, so besides the self-inductance contribution, you via mutual inductance, you get the current in L_1 contributing to the field in L_2 and vice versa, which is where the 2M comes from.

If you short out L2, then instead of M producing an open-circuit voltage, it produces the current that cancels out the net magnetic field in L2, which is I_1*(M/L_2) That current, which changes linearly just like I_1, in turn produces an open circuit voltage in L_1 of I_2*M**2/L_2. So the net inductance of the first coil becomes

L_1' = L_1 - M**2/L_2.

These things are easier to express in terms of the coupling coefficient k, which is defined as

k = M/sqrt(L_1*L_2)

and is always less than 1.

Mutual inductance is perfectly well defined in these situations, but it's much easier to calculate when L_2 doesn't perturb the fields due to L_1, which requires that L_2 be open-circuited, just as your manual suggests.

Cheers

Phil Hobbs

Mutual inductance is essentially the voltage you see in one loop of wire when you change the current through another loop of wire.

I had trouble with it for years before I ran into the transformer equation

V1=3D L1.dI1/dt +M.dI2/dt

V2=3D M. dI1/dt + L2.dI2/dt

M is less than or equal to root L1.L2. . Transformers usually give you an mutual inductance that is more then 99% of the geometric mean of the primary and secondary inductance.

-- Bill Sloman, Nijmegen

I was doing something basic not to long ago and some of this is currently at the top of my head.

M = Mutual Inductance.

Is basically the amount of inductance that is shared between multiple windings and works with the coefficient number of K which can range from lowww (.xxx) up to 1.0. Because there is virtually no such thing as perfect coupling between 2 or more windings, the K is used in the calculations and averages from .7 to .99

so M = K*sqrt(Lpri*LSec) which is the amount of L (inductance) that is shared between the windings.

If you would like a text on it, I think I have something kicking around on the PC..

Hoped that did something for you.

Okay, this helps me understand where I'm falling short.

I'm not very experienced with FEMM, so I don't know for sure, but it appears to me that I can't really do this in FEMM. Because it models a coil as an "axisymmetric problem," I'm only looking at a cross-section of the wire and it looks like this problem can't be solved this way. If I have an open coil, then current is going to zero at the ends and have a maximum somewhere in the middle. Unfortunately this means I can't do the problem I'm trying to do as FEMM only works in two dimensions.

I have some more thoughts on this but I need to try some calculations first and confirm my suspicions. Thanks for the nudge.