How does one define the magnetic $\mathbf{H}$-field for a space entirely filled with a uniform magnetization $\mathbf{M}$? I know this sounds silly/trivial, but hear me out. In all of the following examples, let the uniform magnetization be denoted $\mathbf{M}$.

Consider three cases:

Case 1: Consider a uniformly magnetized sphere. It can be shown that the $\mathbf{B}$ and $\mathbf{H}$ fields inside are:

$$\mathbf{B}=\frac{2}{3}\mu_0\mathbf{M},~~~~~\mathbf{H}=\frac{\mathbf{M}}{3} \tag{1}$$

If you were go into this material and shrink down, you would effectively "see" a space of uniform magnetization.

Case 2: Consider a huge uniformly magnetized flat plate/plane (sort of like the parallel-plate capacitor from electrostatics). The bound-current, $\nabla\times\mathbf{M}$, is identically zero within the volume, and the bound-surface, $\mathbf{M}\times\hat{n}$ (where $\hat{n}$ is the surface-normal), is zero-except for at the very small edges far far away, where it can be essentially ignored. However, we do have a "magnetic surface-charge" term, $-\mathbf{M}\cdot\hat{n}$, which is non-zero on both the top and bottom surface. This leads to the following interior fields:

$$\mathbf{B}=0,~~~~~\mathbf{H}=-\mathbf{M} \tag{2}$$

Once again, if you shrink down and go into this material, you will "see" a space of uniform magnetization.

Case 3: Consider a uniformly magnetized cylinder which is very skinny. The "surface magnetic charges" on the top and bottom face are so small and far away that they contribute nothing to the field inside. However, now the bound surface-current, $\mathbf{M}\times\hat{n}$, is non-zero along the outer-face of the cylinder. This configuration leads to the following fields:

$$\mathbf{B}=\mu_0 \mathbf{M},~~~~~~\mathbf{H}=0 \tag{3}$$

And, as in the previous two cases, if you shrink down and go into the cylinder, you will "see" a uniformly magnetized space.

Since each of these configurations give different answers for the interior fields - fields (1), (2), and (3) - I do not think I could begin with any one of these configurations and take "limiting cases" or "limiting parameters" in order to calculate the fields within a space entirely filled with uniform magnetization. So then how do we calculate the field within a space of uniform magnetization?

  • $\begingroup$ Is this not simply a statement that, in order to solve the Poisson equation, one must supply boundary conditions? $\endgroup$
    – Aaron
    Dec 8, 2016 at 5:17
  • $\begingroup$ Well, I guess it is, since there are no free currents, but even with that in mind I can't figure out the proper "boundary conditions". The equations for $\mathbf{H}$ are $\nabla\cdot\mathbf{H}= -\nabla\cdot\mathbf{M}$ and $\nabla\times\mathbf{H}=0$. If one naively assumes "Oh hey look $\nabla\cdot\mathbf{M}=0$ everywhere, so $\mathbf{H}=0$!", then you get an answer which does not agree at all with the 3 limiting cases which I discussed above. $\endgroup$ Dec 8, 2016 at 5:28
  • $\begingroup$ You should think of the boundary of the object itself as the boundary, and the $\mathbf{H}$ value it takes there as the boundary conditions. $\endgroup$
    – Aaron
    Dec 8, 2016 at 5:53
  • $\begingroup$ @Aaron Ahh, that makes sense. So then you're saying each of those three cases I mentioned in my question would not be applicable to the problem at hand, because of them would require me to introduce boundary conditions which are not present in the original problem - correct? By the way, if you have this all figured out, please feel free to write an answer. :) $\endgroup$ Dec 8, 2016 at 6:23

1 Answer 1


The issue at hand is that the original problem does not provide enough information to give a unique solution. A region $R = \mathbb{R}^3$ with uniform magnetization merely imposes $\nabla\cdot\mathbf{H} = 0$ and $\nabla\times\mathbf{H} = 0$, which is equivalent to $\nabla^2 \phi_m = 0$ in $R$ where $\nabla\phi_m = \mathbf{H}$. However, in order to guarantee a unique solution, one must introduce boundary conditions to the Poisson equation. By assigning different boundary conditions, you can recover the three different cases you have written. Another way to think about this is that we can always add a constant to $\mathbf{H}$, and this will still satisfy the differential equations. Only the boundary conditions determines this constant.

Now, if you want to go in reverse and figure out what the correct boundary conditions are for the three cases you mentioned, the easiest way is to treat the object itself as your region $R$, and let the magnetic field $\mathbf{H}$ on the boundary $\partial R$ be the boundary condition. Then, you can take the limit $R \rightarrow \mathbb{R}^3$ in an appropriate manner to recover the appropriate boundary conditions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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