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Energy of magnetic field

  1. Feb 14, 2014 #1
    The total energy of the magnetic field in the matter is [itex] \frac{\mu H^2}{2} [/itex], 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 [itex] \frac{B^2}{2} [/itex] and the dipolar interaction [itex] -\vec{M} \cdot \vec{B} [/itex], 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 [itex] \mu > > 1 [/itex].

    Why the energy of magnetic field itself is given by [itex] \frac{ H^2}{2} [/itex] and not by [itex] \frac{ B^2}{2} [/itex]?
  2. jcsd
  3. Feb 15, 2014 #2
    That equation is for the energy density of the field, not the total energy.
    And it doesn't matter if you write [itex]\frac{\mu H^2}{2}[/itex] or [itex]\frac{B^2}{2\mu}[/itex] because [itex]B = \mu H[/itex]
  4. Feb 15, 2014 #3
    That all was implied.

    The question is why should we take the self energy density of the magnetic field as [itex] H^2/2 [/itex] and not [itex] B^2/2 [/itex], unlike the electric field where it is [itex] E^2/2 [/itex] and not [itex] D^2/2 [/itex].
  5. Feb 16, 2014 #4
    Why do you keep writing [itex]\frac{H^2}{2}[/itex] instead of [itex]\frac{\mu H^2}{2}[/itex]?
    Anyway, That's due to the way H, B and [itex]\mu[/itex] 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 [itex]\frac{B^2}{8\pi}[/itex] and [itex]\frac{E^2}{8\pi}[/itex]
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