# 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...

-
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
This website starts with a proof by by introducing inductance, then considers a second proof that doesn't require inductance. farside.ph.utexas.edu/teaching/em/lectures/node84.html –  David 16 mins ago

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
You can certainly do the derivation without resorting to the inductance. When it comes about to consider the contributions of induced electromotive forces $V_i$ you can consider that $$V_i=-\frac{\Phi}{dt}$$ where $Phi$ is the flux of the magnetic field $B$ through the surface of the circuit you're considering. You can find a derivation according to this line of reasoning in the text from Stratton.