2.1.3 Magnetostatic energy

The origin of domains still cannot be explained by the two energy terms above. Another contribution comes from the magnetostatic self-energy, which originates from the classical interactions between magnetic dipoles. For a continuous material it is described by Maxwell's equations

$$\mathrm{div~}\mathbf{D} = \rho$$ (2.6)

$$\mathrm{div~}\mathbf{B} = 0$$ (2.7)

$$\mathrm{curl~}\mathbf{E} = -\frac{\partial \mathbf{B} }{dt}$$ (2.8)

$$\mathrm{curl~}\mathbf{H} = \frac{\partial \mathbf{D}}{dt} + \mathbf{j} \quad.$$ (2.9)

In our magnetostatic problem, we do not have any electric fields $\mathbf{E}$ or free currents $\mathbf{j}$. Thus, there are two remaining equations

$$\mathrm{div~}\mathbf{B} = 0$$ (2.10)

$$\mathrm{curl~}\mathbf{H} = 0 \quad.$$ (2.11)

The magnetic induction $\mathbf{B}$ is given by $\mathbf{B}=\mu_0 (\mathbf{H} + \mathbf{M})$. A general solution of () is given by

$$\mathbf{H}=-\nabla U \quad,$$ (2.12)

where $U$ is the magnetic scalar potential. Inserting the expressions for $\mathbf{B}$ and $\mathbf{H}$ in () gives

$$\Delta U_\mathrm{in} = \mathrm{div~}\mathbf{M}$$ (2.13)

inside magnetic bodies and

$$\Delta U_\mathrm{out} = 0$$ (2.14)

outside in air or vacuum.

These equations have to be solved with the boundary conditions

$$U_\mathrm{in}=U_\mathrm{out}, \qquad \frac{\partial U_\mat... ... U_\mathrm{out}}{\partial n}=\mathbf{M} \cdot \mathbf{n} \quad$$ (2.15)

on the surface of the magnet to obtain $U$ and derive from it $\mathbf{H}$. $\mathbf{n}$ is the unit normal to the magnetic body, taken to be positive in outward direction.

In micromagnetics, the magnetization distribution $\mathbf{M}(\mathbf{r})$ is given. With relation () the magnetic scalar potential can be calculated from the magnetization distribution. The demagnetizing field $\mathbf{H}_\mathrm{ms}$ is then obtained by using ().

Finally the magnetostatic energy is given by

$$E_\mathrm{ms}=-\frac{1}{2} \mu_0 \int_V \mathbf{M} \cdot \mathbf{H}_\mathrm{ms} \,d{^3r}\, \quad.$$