View Single Post
PotatoMerrick is offline
Nov13-12, 12:14 PM
P: 1

I'm currently reading material on micromagnetics. In these papers, authors define a quantity called the demagnetizing energy ([itex]E_d[/itex]) as
[tex] E_d = -\frac{1}{2} \int_V \vec{m} \cdot \vec{H}_d\;dV [/tex]
where [itex]\vec{m}[/itex] is the internal magnetization of a material sample of volume [itex]V[/itex] and [itex]\vec{H}_d[/itex] is the demagnetising field. The demagnetizing field itself is defined as the negative derivative of the demagnetizing energy with respect to the material magnetisation, i.e.
[tex] \vec{H}_d = -\frac{dE_d}{d\vec{m}} [/tex]
My problem is that I would like to know how to derive [itex]\vec{H}_d[/itex] by taking the derivative of [itex]E_d[/itex] with respect to [itex]\vec{m}[/itex]. This is as far as I have got (and I'm not too sure that this is correct)
[tex]\frac{dE_d}{d\vec{m}} = \frac{d}{d\vec{m}} \left( -\frac{1}{2}\int_V \vec{m}\cdot\vec{H}_d\;dV \right)[/tex]
[tex]\frac{dE_d}{d\vec{m}} = -\frac{1}{2}\int_V \frac{\partial}{\partial\vec{m}}\left(\vec{m}\cdot\vec{H}_d\right)\;dV=-\frac{1}{2}\int_V\frac{\partial\vec{m}}{\partial \vec{m}} \cdot \vec{H}_d + \vec{m}\cdot\frac{\partial \vec{H}_d}{\partial \vec{m}}\;dV=-\frac{1}{2} \int_V \vec{H}_d\;dV - \frac{1}{2}\int_V \vec{m}\cdot\frac{\partial \vec{H}_d}{\partial \vec{m}}\;dV[/tex]
Could some kind soul please give me some pointers as to how to proceed and/or explain to me where I'm going wrong?
Phys.Org News Partner Physics news on
Sensitive detection method may help impede illicit nuclear trafficking
CERN: World-record current in a superconductor
Beam on target: CEBAF accelerator achieves 12 GeV commissioning milestone