# Magnetostatic energy density — derivation without introducing inductance?

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.

-

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}

-
This is completely useless - you've just defined the symbol $u_B$. How does it relate to energy at all? – Emilio Pisanty Sep 30 '15 at 14:17
And by the way I do not define $u_B$ in this case. $u_B$ is given by the problem from Kyle Kanos – aharisjo Nov 27 '15 at 6:44
$u_B$ is energy density. To find energy, say $E_B$, you can do it by integrating $u_B$ through out the volume $E_B=\int u_BdV$ – aharisjo Nov 27 '15 at 6:47

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.

-

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).

-

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 $$V_i=-\frac{d\Phi}{dt}$$ 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.

-