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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 $$u_B=\dfrac {1}{2 \mu_0}\left|\mathbf{B}\right|^2,$$ I was just wondering if there was a derivation that could be built up the way one derives the energy density, $$u_E=\dfrac {\epsilon_0}{2}\left|\mathbf{E}\right|^2$$ 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.

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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
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@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 Jul 22 at 18:26

4 Answers 4

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Magnetic Field



We set $$|\textbf{B}|=B$$ to simplify the notation and assume $$\textbf{B}=\mu_0\textbf{H}$$ Energy density contained in magnetic field: $$\begin{aligned}du_B&=H{\cdot}dB\\&=\frac{1}{\mu_0}B{\cdot}dB\end{aligned}$$ We build magnetic source bit by infinitessimal bit then the field $B$ is increased bit by infinitessimal bit $dB$ until $B$ (from $0$ to $B$), then the energy density contained in the field is

$$\begin{aligned}u_B&=\frac{1}{\mu_0}\int_0^B{B}{\cdot}dB\\&=\frac{1}{2\mu_0}B^2\\&=\frac{1}{2\mu_0}|\textbf{B}|^2\end{aligned}$$

Hope this will answer your question
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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

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.

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It's a reasonable question, and the answer is: one can't prove it, without introducing induction.
Consider a conducting loop with no current. Then someone starts creating a current in it, using, for example, a battery. The question is: why should we perform work to create this current, if we know that magnetic force $$ \mathbf {F} = \frac{q}{c}[\mathbf{v},\mathbf{B}]$$ is always perpendicular to the charge's displacement (thus it doesn't perform work at all)? The answer for this is that when current appears, magnetic field changes, thus electric field is created. Electric field acts upon charges with the force $$ \mathbf {F} = q\mathbf{E}$$ and because of that we must take an effort to create the current.
Moreover, it holds true in general case. We must $\textit {always}$ work against electric field in order to create magnetic field. In some sense, it is meaning of magnetic energy.
The best way to introduce magnetic and electric energy is Poynting theorem. It states: $$\mathbf{E}\cdot\mathbf{j}+\frac{\partial u}{\partial t}=-\triangledown\mathbf{S}$$ where $u=\frac{E^2}{8\pi}+\frac{B^2}{8\pi}$, $\mathbf{j}$ is current density and $\mathbf{S}$ is Poynting vector. I strongly recommend you to read a corresponding Feynman's lecture, you will defenetly get the sense of these things.

$\textbf{PS:}$ I used CGS system here, I hope it didn't bug you very much.

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