# Computing mutual inductance given only magnetic fields

I am experimenting with numerically computing the mutual inductance between a number of coils in the presence of various magnetic materials.

Traditionally, the mutual inductance is found by determining the flux from one coil that passes through the other, then normalizing by current.

$$M_{21}=\frac{N_2 \phi_2}{I_1}$$

To find the flux, you'd take the integral of the magnetic field over the area of the coil.

$$\phi=\iint \vec B \cdot d \vec A$$

However, given real coil geometry, it would be somewhat difficult to compute what the area of the coil would be, and might require a very fine mesh to do accurately.

Intuitively, since mutual inductance is related to how coupled the magnetic fields between two coils are, is it possible to compute it given just the $$B$$ fields from each coil (and the current that caused them)?

Something along the lines of

$$\frac1{I_1I_2}\iiint\vec{B_1}\cdot\vec{B_2}dV$$

to find the amount the fields are "aligned" over all space...?

If my dimensional analysis is correct, this would give units of $$H^2A^{-2}m^{-1}$$, which obviously isn't just Henries. I haven't seen this done anywhere, so is there a fundamental problem regarding this idea? If not, what might be the next step regarding turning this into Henries?

• Your integral has units of $\text H^2\text{/m}$.
– Puk
Mar 16, 2023 at 2:48
• Looks like I forgot I had already included $1/A^2$ when doing the units for the integral part, so I double counted it with the $1/I_1I_2$ at the start. So yes – the units for that whole last expression is $H^2/m$. Mar 16, 2023 at 7:24

You have the right intuition. The magnetic energy (assuming linear materials) is given by $$W = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2.$$ If you have linear materials, you can calculate $$W$$ as $$\iiint{\frac{|\vec B|^2}{2\mu}} dV=\iiint{\frac{|\vec B_1|^2}{2\mu}} dV+\iiint{\frac{|\vec B_2|^2}{2\mu}} dV+\iiint{\frac{\vec B_1\cdot \vec B_2}{\mu}} dV.$$ The first two terms are just $$\frac{1}{2}L_1I_1^2$$ and $$\frac{1}{2}L_2I_2^2$$ respectively, so the mutual inductance is $$M= \frac{1}{I_1I_2}\iiint{\frac{\vec B_1\cdot \vec B_2}{\mu}} dV.$$