Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I'm trying to understand how permanent magnets can be modeled in finite element methods. FEMM's manual, here, states that it treats a PM as a solenoid with the same surface current as the PM, and then solves the field from there, which I understand. However, they use the coercivity $H_c$ in place of the magnetization $M$ (I might be dropping factors of $\mu$ here.) to calculate this surface current, i.e., $H_c (\hat{m} \times \hat{n})$.

The argument given is that this is the amount of current you would need to surround the magnet in (in opposite direction) to drive the field to zero (this is the definition of coercivity), which is the same field you get with no current at all. Therefore, the surface current of the original PM can be physically cancelled out with $-H_c (\hat{m} \times \hat{n})$, so this must be it's actual value.

My problem is that when we model a permanent magnet in absence of any external currents, we are at the $H=0$ point of the $B-H$ diagram (i.e., $B = B_r$), and this is not the same physical point. Intuitively I would want to use $B_r$ in place of $M$, modulo some factors of $\mu$ again.

I know that this makes assumptions about linearity of the magnet, but I'm not exactly sure where they creep in. Thanks in advance for any input.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

I think the answer is that you can't directly measure $B$ inside the material, but you can figure out $H$. So you need to know how $B$ depends on $H$ for calculation purposes. For so-called linear demagnetization curve permanent magnets, we can linearize the $B-H$ curve in the second quadrant, and write

$H(B) = \nu B - H_c$

where $\nu = \frac{H_c}{B_r}$ is the reluctivity (inverse of permeability). Note that if this were a linear material without hysteresis (e.g. a soft iron core), you would just have

$H(B) = \nu B$.

If you apply Ampere's law you get $\nabla \times H = J_f$. For our permanent magnet, the extra term $-\nabla \times H_c$ in the first equation is equivalent to adding an extra bound current term to the second equation. The bound current only has support on the surface, so we treat the permanent magnet like a soft iron core surrounded by a surface current of value $-H_C \hat {n}\times \hat{m}$.

I found the following documents useful: http://www.permagsoft.com/english/assets/applets/DemagnetisingCurve.pdf http://www.permagsoft.com/english/assets/applets/CompEngl.pdf

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.