# Mutual inductance of transformer with coils of different area and length

## Main Question or Discussion Point

Hello,

I am trying to derive the mutual inductance between the two coils of a transformer. I want to do this for two coils with different lengths and different cross-sectional areas (assuming that one side of the transformer core is thicker than the other). The parameters are therefore:

I1 = current in coil 1
I2 = current in coil 2
L1 = length of first coil
L2 = length of second coil
A1 = Cross sectional area of coil 1
A2 = Cross sectional area of coil 2
N1 = Number of turns in coil 1
N2 = Number of turns in coil 2

To calculate the mutual inductance, I leave the second coil open-circuited and drive a current into the first coil, so I get the flux density:

B11 = μ(N1)(I1)/L1

The total flux generated is therefore:

∅11 = (B1)(A1) = μ(N1)(I1)(A1)/L1

Assuming that all flux generated in coil 1 also links coil 2 (no flux leakage with coupling coefficient = 1) , we can say that ∅21 = ∅11.

Therefore, the voltage induced in coil 2:

V2 = (N2)d(∅21)/dt = μ(N1)(N2)(A1)/L1 d(I1)/dt

Therefore, I can concluded that the mutual inductance M21 = μ(N1)(N2)(A1)/L1

Now, repeating the steps above but with coil 1 open-circuited while driving a current I2 into the coil 2 gives the following:

B22 = μ(N2)(I2)/L2
∅22 = (B2)(A2) = μ(N2)(I2)(A2)/L2

And assuming all of ∅22 links coil 1, ∅12 = ∅22 and we get

V1 = (N1)d(∅12)/dt = μ(N1)(N2)(A2)/L2 d(I2)/dt

And this expression gives M12 = μ(N1)(N2)(A2)/L2

I know that this isn't correct because both M21 and M12 are supposed to be equal, but my derivation gives different results.

In the simple case where the coil area and lengths are the same, the above derivation does yield mutual inductances that are the same. When these are made different, I cannot seem to get the expressions for the mutual inductances to match.

Any help / comments much appreciated.