How is the Total Energy of a Magnetic Field in Matter Calculated?

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The total energy of a magnetic field in matter is calculated using the formula \(\frac{\mu H^2}{2}\), but when determining the energy spent on magnetizing the material, one must subtract the magnetic field energy \(\frac{B^2}{2}\) and the dipolar interaction \(-\vec{M} \cdot \vec{B}\). A key point of confusion arises regarding why the self-energy density of the magnetic field is expressed as \(\frac{H^2}{2}\) rather than \(\frac{B^2}{2}\). This distinction is rooted in the definitions of H, B, and \(\mu\), as well as the differing interactions of matter with magnetic versus electric fields. Ultimately, the choice of equations varies based on the system of units used, such as cgs, where the expressions differ significantly.
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The total energy of the magnetic field in the matter is \frac{\mu H^2}{2}, I want to calculated the energy that is being spent as a the work on magnetizing the material, so I need to subtract the energy of the magnetic field itself \frac{B^2}{2} and the dipolar interaction -\vec{M} \cdot \vec{B}, however here is the problem $$ \frac{\mu H^2}{2} - \frac{B^2}{2} = \frac{\mu H^2}{2} - \frac{(\mu H)^2}{2} < 0 $$ for \mu &gt; &gt; 1.

Why the energy of magnetic field itself is given by \frac{ H^2}{2} and not by \frac{ B^2}{2}?
 
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That equation is for the energy density of the field, not the total energy.
And it doesn't matter if you write \frac{\mu H^2}{2} or \frac{B^2}{2\mu} because B = \mu H
 
That all was implied.

The question is why should we take the self energy density of the magnetic field as H^2/2 and not B^2/2, unlike the electric field where it is E^2/2 and not D^2/2.
 
Why do you keep writing \frac{H^2}{2} instead of \frac{\mu H^2}{2}?
Anyway, That's due to the way H, B and \mu are defined. And because matter often interacts with magnetic fields in a way that's opposite to how it interacts with electric fields.
If you use different definitions, the equations look different.
e.g. in the cgs system the equations are \frac{B^2}{8\pi} and \frac{E^2}{8\pi}
 
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