Permanent ferromagnets

Let us consider the magnetic field generated by a distribution of permanent ferromagnets. Let us suppose that the magnets in question are sufficiently ``hard'' that their magnetization is essentially independent of the applied field for moderate field strengths. Such magnets can be treated as if they contain a fixed, specified magnetization ${\bfm M}({\bfm r})$.

Let us assume that there are no true currents in the problem, so that ${\bfm j}_t = 0$. Let us also assume that we are dealing with a steady state situation. Under these circumstances Eq. (3.121) reduces to

$$\nabla\wedge{\bfm H} = 0.$$ (554)

It follows that we can write

$${\bfm H} = - \nabla\phi_m,$$ (555)

where $\phi_m$ is called the magnetic scalar potential. Now, we know that

$$\nabla\!\cdot\!{\bfm B} = \mu_0 \nabla\!\cdot\!({\bfm H} + {\bfm M}) = 0.$$ (556)

Equations (3.134) and (3.135) combine to give

$$\nabla^2\phi_m =- \rho_m,$$ (557)

where

$$\rho_m = -\nabla\!\cdot\!{\bfm M}.$$ (558)

Thus, the magnetostatic field ${\bfm H}$ is determined by Poisson's equation. We can think of $\rho_m$ as an effective magnetic charge density. Of course, this magnetic charge has no physical reality. We have only introduced it in order to make the problem of the steady magnetic field generated by a set of permanent magnets look formally the same as that of the steady electric field generated by a distribution of charges.

The unique solution of Poisson's equation, subject to sensible boundary conditions at infinity, is well known:

$$\phi_m({\bfm r}) = \frac{1}{4\pi}\int \frac{\rho_m({\bfm r}')}{ \vert{\bfm r} - {\bfm r}'\vert}\,d^3{\bfm r}'.$$ (559)

This yields

$$\phi_m({\bfm r}) =-\frac{1}{4\pi}\int \frac{\nabla'\!\cdot\!... ... M}({\bfm r}')}{\vert{\bfm r} - {\bfm r}'\vert}\,d^3{\bfm r}'.$$ (560)

If the magnetization field ${\bfm M}({\bfm r})$ is well behaved and localized we can integrate by parts to obtain

$$\phi_m({\bfm r}) =\frac{1}{4\pi}\int {\bfm M}({\bfm r}')\!\c... ...\frac{1}{\vert{\bfm r} - {\bfm r}'\vert}\right)\,d^3{\bfm r}'.$$ (561)

Now

$$\nabla'\!\left(\frac{1}{\vert{\bfm r} - {\bfm r}'\vert}\righ... ...\nabla\!\left(\frac{1}{\vert{\bfm r} - {\bfm r}'\vert}\right),$$ (562)

so our expression for the magnetic potential can be written

$$\phi_m({\bfm r})= - \frac{1}{4\pi}\nabla\!\cdot\!\int\frac{{... ...M}({\bfm r}')} {\vert{\bfm r} - {\bfm r}'\vert}\,d^3{\bfm r}'.$$ (563)

Far from the region of non-vanishing magnetization the potential reduces to

$$\phi_m({\bfm r}) \simeq -\nabla\!\left(\frac{1}{4\pi r}\righ... ...,d^3{\bfm r}'\simeq \frac{{\bfm m}\!\cdot{\bfm r}} {4\pi r^3},$$ (564)

where ${\bfm m} = \int{\bfm M}\,d^3{\bfm r}$ is the total magnetic moment of the distribution. This is the scalar potential of a dipole. Thus, an arbitrary localized distribution of magnetization asymptotically produces a dipole magnetic field whose strength is determined by the net magnetic moment of the distribution.

It is often a good approximation to treat the magnetization field ${\bfm M}({\bfm r})$ as a discontinuous quantity. In other words, ${\bfm M}({\bfm r})$ is specified throughout the ``hard'' ferromagnets in question, and suddenly falls to zero at the boundaries of these magnets. Integrating Eq. (3.137) over a Gaussian pill-box which straddles one of these boundaries leads to the conclusion that there is an effective magnetic surface charge density,

$$\sigma_m = {\bfm n}\!\cdot\!{\bfm M},$$ (565)

on the surface of the ferromagnets, where ${\bfm M}$ is the surface magnetization, and ${\bfm n}$ is a unit outward directed normal to the surface. Under these circumstances Eq. (3.139) yields

$$\phi_m({\bfm r})= - \frac{1}{4\pi}\int_V\frac{\nabla'\!\cdot... ...\bfm r}')\!\cdot\!d{\bfm S}'}{\vert{\bfm r} - {\bfm r}'\vert},$$ (566)

where $V$ represents the volume occupied by the magnets and $S$ is the bounding surface to $V$. Here, $d{\bfm S}$ is an outward directed volume element to $S$. It is clear that Eq. (3.145) consists of a volume integral involving the volume magnetic charges $\rho_m = -\nabla\!\cdot\!{\bfm M}$ and a surface integral involving the surface magnetic charges $\sigma_m = {\bfm n}\!\cdot\!{\bfm M}$. If the magnetization is uniform throughout the volume $V$ then the first term in the above expression vanishes and only the surface integral makes a contribution.

We can also write ${\bfm B} = \nabla\wedge{\bfm A}$ in order to satisfy $\nabla\!\cdot\!{\bfm B} = 0$ automatically. It follows from Eqs. (3.121) and (3.122) that

$$\nabla\wedge{\bfm H} = \nabla\wedge({\bfm B}/\mu_0 - {\bfm M}) = 0,$$ (567)

which gives

$$\nabla^2{\bfm A} =- \mu_0 \,{\bfm j}_m,$$ (568)

since ${\bfm j}_m = \nabla\wedge{\bfm M}$. The unique solution to Eq. (3.147), subject to sensible boundary conditions at infinity, is very well known:

$${\bfm A}({\bfm r}) = \frac{\mu_0}{4\pi}\int \frac{{\bfm j}_m({\bfm r}')} {\vert{\bfm r}- {\bfm r}'\vert}\,d^3{\bfm r}'.$$ (569)

Thus,

$${\bfm A}({\bfm r}) = \frac{\mu_0}{4\pi}\int \frac{\nabla'\we... ... M}({\bfm r}')} {\vert{\bfm r}- {\bfm r}'\vert}\,d^3{\bfm r}'.$$ (570)

If the magnetization field is discontinuous it is necessary to add a surface integral to the above expression. It is straightforward to show that

$${\bfm A}({\bfm r}) = \frac{\mu_0}{4\pi}\int_V \frac{\nabla'\... ...}({\bfm r}')\wedge d{\bfm S}'}{\vert{\bfm r}- {\bfm r}'\vert}.$$ (571)

It is clear that the above expression consists of a volume integral involving the volume magnetization currents ${\bfm j}_m = \nabla\wedge{\bfm M}$ and a surface integral involving the surface magnetization currents ${\bfm J}_m = {\bfm M} \wedge {\bfm n}$ (see Eq. (3.132)). If the magnetization field is uniform throughout $V$ then only the surface integral makes a contribution.