# Magnetostatic energy density — derivation without introducing induchance?

I was looking for a derivation of the expression for the energy density at any point in a static magnetic field. I do know that it is $\dfrac {1}{2 \mu_0}\left|\vec{B}\right|^2$ -- I was just wondering if there was a derivation that could be built up the way one derives the energy density $\dfrac {\epsilon_0}{2}\left|\vec{E}\right|^2$ at any point in an electric field, by considering the energy needed to build up a 'source' charge bit by infinitesimal bit.

Whatever proof I have come across seems to bring inductance into the picture -- is there a way of doing it without that? I ask because the corresponding proof for the electric field does not seem to need the definition of capacitance anywhere...

Thanks...

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Try taking two parallel sheets of current, finding their force on each other, and calculating the mechanical work required to move them farther or closer. –  Ben Crowell Jun 16 '13 at 19:34
@BenCrowell the comments are meant to be used to comment on the question, rather than to provide an answer. –  Larry Harson Jun 16 '13 at 21:14
@LarryHarson: I could be mistaken, but I think it's pretty common that when you have something that's not a full answer, but may be helpful, you post it as a comment. –  Ben Crowell Jun 16 '13 at 23:19
@BenCrowell it's tolerated but not encouraged for reasons given in answering in comments –  Larry Harson Jun 17 '13 at 0:33
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## 1 Answer

Have you considered Poynting's theorem? It originates from the general derivation of the energy density contained in the electric and magnetic fields (starting from the Lorentz force and the definition of work).

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Hi @sybtc -- in the Wikipedia article you've mentioned, I could find the explanation for $\vec{E}\cdot\vec{J}$ part of Poynting's theorem, but $|\vec{B}|^2/2\mu_0$ as energy density, or at least $\dfrac {\vec{B}}{\mu_0}\cdot\dfrac{\partial \vec{B}}{\partial t}$ as the density of reactive power, seems to be assumed. Hence my question... –  Avijit Mar 18 '13 at 18:49
I did the derivation following the text from Griffiths. I did not read the Wiki article. The steps you take are: write down the definition of work using the Lorentz force (magnetic part of force does not contribute since it is oriented perpendicular to the displacement of a charge). Then replace the charge by a charge density, take this density together with the velocity vector to form the current. Then apply Maxwell's equations to replace the current by it's equivalent fields. –  sybtc Mar 18 '13 at 18:52
Hi @sybtc -- thanks for pointing to Griffiths, I'm trying to digest what the text says there. –  Avijit Mar 18 '13 at 19:21
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