1.2.1.3 Magnetostatic energy

By taking the first-order variation of the free energy functional (1.50), one has:

$$\delta F_ =-\int_{\Omega} \frac{1}{2} \mu_0 M_s \delta\textbf{{m}}\cdot{\... ... \frac{1}{2} \mu_0 M_s \textbf{{m}}\cdot\delta{\mathbf{H}_\text{m}} dV\quad.$$ (1.61)

The above two integral term are identical as stated by the reciprocity theorem [4,5] and then the latter equation can be rewritten in the following form:

$$\delta F_ =-\int_{\Omega} \mu_0 M_s {\mathbf{H}_\text{m}}\cdot \delta\textbf{{m}} dV \quad.$$ (1.62)