Magnetic energy

The energy stored in a magnetic field is

$W_m=\frac{1}{2}{\int }_v\mu H^{2}dv\text{(J)},$

with magnetic energy density

$w_m=\frac{1}{2}\mu H^2=\frac{1}{2}\mathbf{B}\cdot \mathbf{H}{\text{(J/m}}^3\text{)}.$

Like Electrostatic energy, this energy isn’t localized on currents or magnets — it’s stored in the field that fills space.

Derivation via inductor

Build up the current through an inductor of inductance $L$ from $i=0$ to $i=I$ in time $t$. The voltage required is $v=Ldi/dt$, instantaneous power $p=vi=Lidi/dt$. Total energy:

$W_m={\int }_0^{t}pdt={\int }_0^{t}Li\frac{di}{dt}dt={\int }_0^{I}Lidi=\frac{1}{2}L{I}^2.$

The $\frac{1}{2}$ factor catches everyone — same origin as in $W_e=\frac{1}{2}C{V}^2$: the field builds gradually, so the average $v$ during charging is $\frac{1}{2}{V}_{\text{final}}$.

Generalization

Specialize to a solenoid of length $l$, $N$ turns, cross-section $S$, current $I$. Field inside: $B=\mu NI/l$, so $H=NI/l$. Inductance $L=\mu N^{2}S/l$. Stored energy:

$W_m=\frac{1}{2}L{I}^2=\frac{1}{2}\cdot \frac{\mu N^{2}S}{l}\cdot I^2.$

Rewrite using $H=NI/l$ and volume $v=Sl$:

$W_m=\frac{1}{2}\mu H^2\cdot Sl=\frac{1}{2}\mu H^2\cdot v.$

So inside the solenoid, the energy density is $\frac{1}{2}\mu H^2$ — uniform across the solenoid interior.

This expression $w_m=\frac{1}{2}\mu H^2$ turns out to be universally valid, not just for solenoids: at every point in space with a magnetic field, the local energy density is $\frac{1}{2}\mu H^2$. Integrating over all space recovers the total.

Worked example: coaxial cable

Inner radius $a$, outer $b$, length $l$, current $I$, permeability $\mu$ between conductors. The field is $H_{\varphi }=I/(2\pi r)$ for $a<r<b$, zero elsewhere (approximately).

$W_m={\int }_v\frac{1}{2}\mu H^{2}dv=\frac{1}{2}\mu {\int }_0^l{\int }_0^{2\pi }{\int }_a^b\frac{I^2}{(2\pi r{)}^2}rdrd\varphi dz=\frac{\mu I^{2}l}{4\pi }\ln \frac{b}{a}.$

Cross-check with $W_m=\frac{1}{2}L{I}^2$ and $L=(\mu l/2\pi )\ln (b/a)$: yes, $W_m=\frac{1}{2}\cdot (\mu l/2\pi )\ln (b/a)\cdot I^2=(\mu I^{2}l/4\pi )\ln (b/a)$. Match.

Duality with electric

Electrostatic and magnetic energy densities are dual:

	Electric	Magnetic
Density	$\frac{1}{2}ϵE^2=\frac{1}{2}\mathbf{D}\cdot \mathbf{E}$	$\frac{1}{2}\mu H^2=\frac{1}{2}\mathbf{B}\cdot \mathbf{H}$
Circuit form	$\frac{1}{2}C{V}^2$	$\frac{1}{2}L{I}^2$
Material constant	$ϵ$	$\mu$

In an electromagnetic wave both densities coexist, with average values equal — half the wave’s energy is “electric,” half is “magnetic.” The propagating energy flux is the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{H}$.

Why factor 1/2

Same explanation as for electrostatic energy: bring up the current (or charge) gradually. Early bits of current don’t see the full final field — they see whatever field has been built so far, which is proportional to the current. So work done on each $di$ is $i(Ldi/dt)dt=Lidi$, integrated to $\frac{1}{2}L{I}^2$. The half is built into the integral, not added by hand.