Relationship of the Demagnetizing Energy to the Demagnetizing Fieldby PotatoMerrick Tags: electro magnetism, magnet force, vector analysis 

#1
Nov1312, 12:14 PM

P: 1

Hello,
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? 


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