Energy Stored in Magnetic Fields

The relay coil of Self-Inductance ended its worked example holding one joule that had to go somewhere when the switch opened. This article answers the question that example left standing: where, physically, was that joule the microsecond before? Not in the electrons — their kinetic energy in a copper winding is nanojoules. Not in "the circuit" as an accounting abstraction. The joule was in space: distributed through the volume around the windings, at a density of $B^2/2\mu_0$ joules per cubic metre, wherever the magnetic field reached. Cut the current and the field collapses inward, handing its energy back to the charges; that reflux is the kilovolt kick. The argument is a ladder with three rungs, and climbing it is the business of this article. Rung one is circuit bookkeeping: the work done driving current into an inductor against its back-EMF is $W = \tfrac12 LI^2$ — Lenz's opposition, integrated. Rung two rewrites the same energy in terms of sources and potential, $W = \tfrac12\int \mathbf{J}\cdot\mathbf{A}\,\mathrm{d}\tau$ , honest but still clinging to the currents. Rung three lets go: one integration by parts relocates every joule into the field itself, $W = \frac{1}{2\mu_0}\int B^2\,\mathrm{d}\tau$ , integrated over all space, no reference to any circuit at all. Three expressions, one number, escalating ontology. The third rung matters far beyond magnets. Paired with electrostatics' $\varepsilon_0E^2/2$ , it says the electromagnetic field is a physical system with its own energy inventory — the premise on which Poynting's Theorem will build energy flow, and without which light carrying energy across vacuum (Maxwell's Equations in Vacuum, four articles hence) would be unintelligible. It also settles a famous verbal paradox: magnetic forces do no work, yet a 6-megajoule MRI magnet is plainly storing something. Both statements are true, and this article is where they stop conflicting.

Key takeaways

Charging an inductor against its back-EMF banks work: $W = \tfrac12 LI^2$ . The energy is recoverable — this is storage, not dissipation.
The same energy, rewritten via sources: $W = \tfrac12\int\mathbf{J}\cdot\mathbf{A}\,\mathrm{d}\tau$ ; and relocated into the field: $W = \frac{1}{2\mu_0}\int_{\rm all\ space} B^2\,\mathrm{d}\tau$ — energy density $u_B = B^2/2\mu_0$ , no circuit mentioned.
The full electromagnetic density is $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ ; at practical field strengths magnetism wins by four orders of magnitude ( $1.5$ T stores $9.0\times10^5$ J/m³; air-breakdown $E$ manages 40 J/m³) — why energy tech that stores in fields stores magnetically.
The coax computed by energy reproduces Self-Inductance's flux answer exactly, and adds the piece flux methods miss: the internal inductance $\mu_0/8\pi$ per metre, geometry-independent.
"Magnetic forces do no work" survives: $\mathbf{B}$ brokers energy between mechanical and electrical accounts without ever paying from its own; the stored energy was deposited by electric fields fighting induced EMFs during ramp-up.
KEY IDEA: the work done against Lenz's opposition while establishing a current is stored in space at density $B^2/2\mu_0$ — the field, not the circuit, is the reservoir.

The core insight

Establishing a current stores the work done against the back-EMF in the magnetic field itself: $W = \tfrac12 LI^2 = \tfrac12\int\mathbf{J}\cdot\mathbf{A}\,\mathrm{d}\tau = \frac{1}{2\mu_0}\int B^2\,\mathrm{d}\tau$ — one energy, three descriptions, with the field-density form $u_B = B^2/2\mu_0$ the fundamental one.

Where this lives today

Superconducting magnetic energy storage (SMES) is this article commercialized with no moving parts: persistent coils bank grid energy at $B^2/2\mu_0$ densities and return it in milliseconds, providing frequency stabilization and bridge power where response time beats capacity. At the opposite end of duty cycle, every tokamak is a magnetic energy warehouse before it is anything else — ITER's toroidal-field system stores about 41 GJ, four orders beyond the MRI of worked example 2, and its quench-protection engineering (detection in milliseconds, energy extraction to massive dump resistors) is the direct scale-up of the hospital's vent duct. Fusion's central-solenoid "flux swing" is even a working use of the recoverable character of $\tfrac12LI^2$ : energy parked in the field is withdrawn to drive the plasma current. The everyday deployment is smaller and everywhere: the energy-transfer inductors of every buck, boost, and flyback converter shuttle microjoules-to-millijoules of $B^2/2\mu_0$ into and out of ferrite tens of thousands to millions of times per second — a phone charger is a magnetic-field energy pump with a duty cycle. And the density comparison of worked example 2 quietly runs battery-adjacent debates: field-based storage (magnetic or electric) trades energy density away for power density and cycle life, which is why supercapacitors and SMES bracket, rather than replace, chemistry. The field-energy concept itself, meanwhile, does its deepest work out of sight: every antenna-design tool and every optical simulation integrates $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ as routinely as a bank sums accounts — the third rung of the ladder, treated as infrastructure.

Origins · 1 minWilliam Thomson (Lord Kelvin) and how this was discovered

William Thomson (Lord Kelvin)

1824–1907 · Glasgow, Cambridge, the Atlantic cable

Thomson — Kelvin from 1892, the first scientist ennobled for physics — is the figure through whom energy became the organizing currency of physical theory, and magnetic energy specifically got its first honest accounting. In the early 1850s, fresh from co-founding thermodynamics (his absolute temperature scale, the Joule–Thomson collaboration), he computed the "mechanical value" of distributions of electricity and magnetism: the work required to assemble a magnetic configuration, expressed as a volume integral over the field region — the direct ancestor of this article's $\int B^2/2\mu_0\,\mathrm{d}\tau$ , written before Maxwell's equations existed. His method of images, his analogy machine (heat flow ↔ electrostatics) and his insistence that fields carry energy shaped the young Maxwell decisively. Kelvin's engineering face made the same points at industrial scale: his telegraph theory ( $RC$ signal retardation) and instrument patents made the 1866 Atlantic cable work, minting him the fortune that funded a 69-year Glasgow professorship. The kelvin honors the temperature scale; this chapter honors the ledger — assembling fields costs work, and the work is stored where the field is.

History

1847

— Helmholtz's conservation-of-energy treatise puts electromagnetic processes inside one universal ledger

1853

— William Thomson computes the mechanical value of magnetic distributions: field energy as a volume integral

1865

— Maxwell's dynamical theory assigns energy density to the electromagnetic field itself

1884

— Poynting (and Heaviside, independently) derive the flow theorem the density demands

1911

— Onnes's superconductivity makes lossless magnetic storage physical: persistent currents

2035

— ITER-era tokamaks inventory tens of gigajoules in toroidal fields — quench protection as civil engineering

The physical picture

Watch a superconducting magnet being energized. For forty minutes a power supply pushes current against the coil's back-EMF — real work, megajoules of it, metered at the wall. Then the supply is disconnected, a superconducting switch closes the loop, and the magnet persists: no source, no dissipation, field at full strength, for years. The megajoules are still there — the magnet can be de-energized through a resistor and every joule recovered as heat. During those years of persistence, where does the inventory sit? There is no current change, no EMF, no power flow. Only the field remains — and the field is exactly where the energy is: $B^2/2\mu_0$ per cubic metre, roughly a megajoule per cubic metre at MRI strength, sitting in the bore and the windings' throat like compressed spring-stuff.

Figure

Where the joule lives: not in the wire, but distributed through the field volume at density B squared over 2 mu0

Two sanity checks calibrate the picture. Scale: the density $B^2/2\mu_0$ at one tesla is $4.0\times10^5$ J/m³ — the kinetic energy of a car at highway speed, per cubic metre of field. Electric fields cannot compete in air: dielectric breakdown caps $E$ near $3\times10^6$ V/m, where $\varepsilon_0E^2/2$ is a feeble 40 J/m³. The four-orders-of-magnitude gap is why grid-scale electromagnetic energy storage (SMES), fusion-reactor energy inventories, and every pulsed-power bank that matters are magnetic technologies. (The fields tie only when $E = cB$ — a proportion nature reserves, suggestively, for light; Maxwell's Equations in Vacuum will cash that observation.) Honesty: who deposited the energy, given the theorem — standing since the Lorentz-force chapter — that magnetic forces do no work? The ledger of Lenz's Law already answered. During ramp-up, the changing flux induces an electric field (an Induced Electric Field, exactly the species of two articles ago) that opposes the current; the power supply pushes charges through that electric field, and it is this electric force that does the work, joule by joule, at rate $-\mathcal{E}I = LI\,\mathrm{d}I/\mathrm{d}t$ . The magnetic field orchestrates — it is the flux whose change makes the induced $\mathbf{E}$ — but the work is done against $\mathbf{E}$ , and the deposit lands in the field configuration. At constant current: no induced $\mathbf{E}$ , no work, inventory frozen. The theorem survives untouched; only the slogan "magnetic fields have no energy" (which was never the theorem) dies.

The ladder, rung by rung

Rung one is the charging ledger, inherited directly from Self-Inductance. Raising a current against the back-EMF $\mathcal{E} = -L\,\mathrm{d}I/\mathrm{d}t$ requires the source to deliver power $P = -\mathcal{E}I = LI\,\mathrm{d}I/\mathrm{d}t$ ; integrate from zero:

W = \int_0^I L\,I'\,\mathrm{d}I' = \tfrac12 L I^2 \qquad (1)

where: $W$ — the work banked (joules), fully recoverable on ramp-down (run the integral backwards: the collapsing flux drives the current, and the coil returns the energy — the flyback diode's 1.0 J was exactly this account being emptied through a resistor). Note what (1) does not say: where the energy is. $L$ is a terminal quantity; the ledger knows deposits and withdrawals, not warehouses. For several coupled circuits the same charging audit (performed in Mutual Inductance, recalled here rather than re-derived) generalizes to $W = \tfrac12\sum_{ij}M_{ij}I_iI_j$ with $M_{ii} = L_i$ — a quadratic form whose positivity gave that article its coupling bound. Rung two dresses the ledger in field clothing. Write $LI = \Phi = \oint\mathbf{A}\cdot\mathrm{d}\boldsymbol\ell$ (Self-Inductance's equation 5), so $W = \tfrac12 I\oint\mathbf{A}\cdot\mathrm{d}\boldsymbol\ell$ ; promote the filament to a distribution ( $I\,\mathrm{d}\boldsymbol\ell \to \mathbf{J}\,\mathrm{d}\tau$ ):

W = \tfrac12\int \mathbf{J}\cdot\mathbf{A}\,\mathrm{d}\tau \qquad (2)

where: the integral runs over the current-carrying region only — the energy appears as sources sitting in their own potential, the exact magnetic twin of electrostatics' $\tfrac12\int\rho V\,\mathrm{d}\tau$ . Same virtues (computable, general), same philosophical defect: it looks like the energy lives in the wires. The electrostatic chapter faced the identical ambiguity and resolved it by integration by parts; we now do the same.

Derivation ARelocating the energy — from currents-in-potential to B squared over 2 mu0

The goal is the third rung: $W = \frac{1}{2\mu_0}\int_{\rm all\ space}B^2\,\mathrm{d}\tau$ , derived from equation (2), with the energy density $u_B = B^2/2\mu_0$ read off. The strategy is one vector-calculus integration by parts: trade $\mathbf{J}$ for $\nabla\times\mathbf{B}$ (Ampère), then shuffle the derivative from $\mathbf{B}$ onto $\mathbf{A}$ with a product rule, paying a surface term. The watch-point is that surface term: the whole claim "the energy is in the field, everywhere" rests on it dying at infinity, so we audit its decay honestly rather than waving it away. Magnetostatics assumed throughout (steady currents; the time-dependent generalization is Poynting's, deferred).

Step 1. Eliminate the current via Ampère's law, $\mu_0\mathbf{J} = \nabla\times\mathbf{B}$ :

W = \frac{1}{2\mu_0}\int (\nabla\times\mathbf{B})\cdot\mathbf{A}\,\mathrm{d}\tau.

Step 2. Deploy the product rule $\nabla\cdot(\mathbf{A}\times\mathbf{B}) = \mathbf{B}\cdot(\nabla\times\mathbf{A}) - \mathbf{A}\cdot(\nabla\times\mathbf{B})$ , i.e. $(\nabla\times\mathbf{B})\cdot\mathbf{A} = \mathbf{B}\cdot(\nabla\times\mathbf{A}) - \nabla\cdot(\mathbf{A}\times\mathbf{B}) = B^2 - \nabla\cdot(\mathbf{A}\times\mathbf{B})$ , using $\nabla\times\mathbf{A} = \mathbf{B}$ :

W = \frac{1}{2\mu_0}\int B^2\,\mathrm{d}\tau \;-\; \frac{1}{2\mu_0}\oint_{\mathcal{S}} (\mathbf{A}\times\mathbf{B})\cdot\mathrm{d}\mathbf{a},

the divergence term converted to a surface integral over the boundary $\mathcal{S}$ of the integration volume.

Step 3. Audit the surface term as $\mathcal{S}$ recedes to infinity. For any localized current distribution, the far field is at worst dipolar: $B \sim 1/r^3$ and $A \sim 1/r^2$ , so the integrand falls as $1/r^5$ while the surface grows as $r^2$ — the term dies as $1/r^3$ . Gone, and fast. Why the audit matters: at finite radius the split between the volume and surface pieces is genuinely different — take the volume to be just the wire and virtually all of (2)'s energy sits in the "surface" term. Only over all space is the assignment unambiguous. The field form is not merely another formula; it is a claim about where energy resides, and the claim is licensed precisely by the vanishing of this term.

Step 4. Read off the result and its density:

W = \frac{1}{2\mu_0}\int_{\rm all\ space} B^2\,\mathrm{d}\tau, \qquad u_B = \frac{B^2}{2\mu_0}. \qquad (3)

Positive everywhere, quadratic in the field, and — decisive point — evaluable without knowing which currents made the field, or whether any still exist. Together with the electrostatic $\tfrac12\varepsilon_0E^2$ , the total electromagnetic density is $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ — the inventory whose local conservation law (with the flow term $\mathbf{S}$ that this density demands) is the opening theorem of the Conservation laws chapter (Poynting's Theorem, deferred by name). Inside magnetizable matter the density is amended to $\tfrac12\mathbf{H}\cdot\mathbf{B}$ , with the honest hysteresis fine print owned by Magnetic Energy in Materials (E5). The surface-term decay goes as $1/r^3$ , confirmed by tracing $A \sim 1/r^2$ and $B \sim 1/r^3$ explicitly. Dimensional check: $\mathrm{T}^2/(\mathrm{T{\cdot}m/A}) = \mathrm{J/m^3}$ ✓. The solenoid consistency check below and worked example 1 confirm (3) against (1) numerically.

The mandatory consistency check, in two lines: a long solenoid ( $B = \mu_0nI$ inside, zero outside) filled with its own density gives $W = \frac{(\mu_0nI)^2}{2\mu_0}\,A\ell = \tfrac12(\mu_0n^2A\ell)I^2 = \tfrac12LI^2$ ✓ — rung three reproducing rung one through Self-Inductance's own $L$ . Worked example 1 runs it with the bench numbers. The deeper test is a geometry where the field is not uniform, and where the field method can see something the flux method missed. That is the coax — and settling it discharges two promissory notes issued by the previous article.

Go deeper · Derivation B · 2 minThe coax by energy — flux method confirmed, and the internal inductance it could not see

Derivation BThe coax by energy — flux method confirmed, and the internal inductance it could not see

The goal is double: recompute the coaxial cable's inductance per unit length from $\int B^2/2\mu_0$ , confirming Self-Inductance's flux-method answer $\frac{\mu_0}{2\pi}\ln(b/a)$ — two independent routes, one number, the chapter's standing habit — and then capture the contribution that article explicitly deferred: the energy stored inside the metal of the inner conductor, which no flux ribbon between the conductors can see. The strategy is to integrate the density shell by shell, in the gap and then in the metal (current spread uniformly over the inner conductor's cross-section). Watch the internal piece: its answer will be independent of every geometric parameter — a small classic.

Step 1. The gap contribution, $a < s < b$ , where $B = \mu_0I/2\pi s$ (Ampère, full current enclosed):

\frac{W_{\rm gap}}{\ell} = \int_a^b \frac{1}{2\mu_0}\left(\frac{\mu_0 I}{2\pi s}\right)^2 2\pi s\,\mathrm{d}s = \frac{\mu_0 I^2}{4\pi}\int_a^b\frac{\mathrm{d}s}{s} = \frac{\mu_0 I^2}{4\pi}\ln\frac{b}{a}.

Step 2. Equate to $\tfrac12(L/\ell)I^2$ and read off:

\frac{L_{\rm ext}}{\ell} = \frac{\mu_0}{2\pi}\ln\frac{b}{a}

— identical to the flux-ribbon answer, as it must be: for $b/a = 7$ , the same $0.39\ \mathrm{\mu H/m}$ . Two methods, two different integrals (one over a ribbon of flux, one over a volume of density), one physical constant. The cross-check the previous article promised is closed.

Step 3. Now the piece the ribbon could not reach: inside the inner conductor, $s < a$ , uniform current density gives $B(s) = \frac{\mu_0 I s}{2\pi a^2}$ (Ampère with partial enclosure — the fat-wire interior of E4):

\frac{W_{\rm int}}{\ell} = \int_0^a \frac{1}{2\mu_0}\left(\frac{\mu_0 I s}{2\pi a^2}\right)^2 2\pi s\,\mathrm{d}s = \frac{\mu_0 I^2}{4\pi a^4}\int_0^a s^3\,\mathrm{d}s = \frac{\mu_0 I^2}{16\pi}.

Step 4. Read off the internal inductance:

\frac{L_{\rm int}}{\ell} = \frac{2W_{\rm int}}{I^2\ell} = \frac{\mu_0}{8\pi} \approx 5.0\times10^{-8}\ \mathrm{H/m},

independent of the wire's radius — the $a^4$ in the density integral cancels the $a^4$ in the field's normalization. Every round wire carrying uniformly distributed current, whatever its gauge, adds the same 50 nH/m of internal inductance. Two comments earn their keep. First, this is the flux method's blind spot made visible: field lines inside the metal link only part of the current, and the honest flux-linkage bookkeeping (weighting each line by the fraction linked) is precisely what the energy integral performs automatically — the energy method's characteristic advantage. Second, the fine print: at high frequency the skin effect expels current (and field) from the conductor's interior, and $L_{\rm int}$ melts away — one reason cable inductance is quoted "at DC" or "at $f$ ", with the frequency machinery deferred to the wave chapter's treatment of conductors. The gap integral gives $0.39\ \mathrm{\mu H/m}$ at $b/a = 7$ , agreeing with the flux-ribbon result. The $s^3$ integral and the $a^4$ cancellation that make $L_{\rm int}/\ell = \mu_0/8\pi$ geometry-independent have been checked explicitly.

Figure

The storage scoreboard: magnetic density crushes air-limited electric density by four orders of magnitude

Limiting cases

Case	Result	Meaning
uniform-field volume (solenoid bore)	$W = \frac{B^2}{2\mu_0}\times\mathrm{volume}$	density × volume; reproduces $\tfrac12LI^2$ exactly
$1/s$ field across an annulus (coax gap)	$W \propto \ln(b/a)$	the logarithm of every annular geometry, now in energy form
inside a round conductor	$L_{\rm int}/\ell = \mu_0/8\pi$ , always	geometry-free 50 nH/m; flux methods miss it, energy methods don't
high frequency (skin effect)	$L_{\rm int} \to 0$	current expelled from the metal takes its stored energy with it
$E$ and $B$ both present	$u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$	equal shares only when $E = cB$ — the proportion light will turn out to obey
practical air-gap storage	$u_B/u_E \sim 2\times10^4$	breakdown caps $E$ ; magnetic storage wins — SMES, pulsed power
magnetizable media	$u = \tfrac12\mathbf{H}\cdot\mathbf{B}$ (linear)	the honest in-matter form; hysteresis losses owned by E5
persistent superconducting loop	$W$ frozen with the field	inventory without traffic: storage needs no ongoing work

Common trapMagnetic forces do no work — and the field still stores megajoules. Both are true.

The apparent contradiction dissolves once "the magnetic field has energy" stops being misread as "the magnetic force does work." The force theorem is kinematic and exact: $q\mathbf{v}\times\mathbf{B} \perp \mathbf{v}$ , always, so the magnetic force on a charge never changes its speed. The storage statement is about the configuration: assembling a current distribution requires work — done by electric fields (the induced EMFs of ramp-up) against the charges — and that work is inventoried in space at $B^2/2\mu_0$ . During any later transaction (a motor turning, a magnet lifting), the magnetic field again brokers: it redirects, converts, and routes energy between mechanical and electrical accounts, while the actual work along every charge's path is traceable to electric fields (including motional-EMF bookkeeping in moving conductors). Misassigning the worker leads to real errors — most famously overcounting in magnet-lifting problems. When an energy audit confuses you, follow the electrons: whatever pushed one along its velocity was an electric force.

Worked examples

1 · The bench solenoid's inventory, audited twice (confidence builder)

The solenoid of Self-Inductance ( $n = 1000$ turns/m, $\ell = 20$ cm, $r = 1.5$ cm, so $L = 0.178$ mH) carries $I = 2.0$ A. Find the stored energy by the circuit ledger and by the field density, and compare.

Step 1. Circuit ledger: $W = \tfrac12LI^2 = 0.5 \times 1.78\times10^{-4} \times 4.0 = \mathbf{0.36\ mJ}$ .

Step 2. Field route. Interior field: $B = \mu_0 n I = 1.26\times10^{-6} \times 1000 \times 2.0 = 2.5\times10^{-3}\ \mathrm{T}$ . Density: $u_B = B^2/2\mu_0 = (2.51\times10^{-3})^2 / (2 \times 1.26\times10^{-6}) = 2.5\ \mathrm{J/m^3}$ .

Step 3. Volume: $A\ell = \pi(0.015)^2 \times 0.20 = 1.41\times10^{-4}\ \mathrm{m^3}$ . Inventory: $W = u_B \times A\ell = 2.5 \times 1.41\times10^{-4} = \mathbf{0.36\ mJ}$ ✓ — rung three equals rung one to every displayed digit, with the ideal-solenoid caveat (exterior field neglected) costing only the same end-corrections that the $L$ formula itself already ignores.

Step 4. Read the density for scale: 2.5 J/m³ at 2.5 mT. Density scales as $B^2$ , so an MRI's 1.5 T — six hundred times the field — stores $360{,}000$ times the density. Which is example 2.

2 · The MRI magnet: megajoules in the bore (exam level)

A 1.5 T superconducting MRI magnet has $L = 50$ H and operates in persistent mode at $I = 500$ A. Find the stored energy, the peak energy density, and assess the quench stakes (liquid helium's volumetric latent heat ≈ 2.6 kJ per litre).

Step 1. Inventory: $W = \tfrac12LI^2 = 0.5 \times 50 \times (500)^2 = \mathbf{6.3\ MJ}$ — the energy of about 1.5 kg of TNT, held silently, sourcelessly, in the field of a hospital instrument.

Step 2. Where it sits: peak density $u_B = B^2/2\mu_0 = (1.5)^2/(2 \times 1.26\times10^{-6}) = \mathbf{9.0\times10^{5}\ J/m^3}$ — nearly a megajoule per cubic metre through the bore and winding region, consistent with several cubic metres of high-field volume holding the 6.3 MJ.

Step 3. The quench scenario — the warning box of Self-Inductance at industrial scale. If any winding segment goes normal (resistive), the 6.3 MJ starts dissipating in the magnet. Dumped into the helium bath alone it would boil $6.3\times10^6 / 2.6\times10^3 \approx \mathbf{2400\ litres}$ of liquid helium — the entire cryostat, explosively, which is why every MRI room has a quench vent duct to the roof. Real quench protection (heaters that deliberately spread the normal zone, dump resistors that extract energy to an external mass) is engineering built entirely on this article's two formulas: $\tfrac12LI^2$ for the stakes, $B^2/2\mu_0$ for where the reservoir presses.

Step 4. The electric comparison that motivates the whole technology: to match $9.0\times10^5$ J/m³ electrostatically requires $E = \sqrt{2u/\varepsilon_0} = 4.5\times10^{8}$ V/m — $150\times$ beyond air breakdown, achievable in no macroscopic gap. Note the number $4.5\times10^8 = c \times 1.5$ : matching densities demands exactly $E = cB$ , the ratio that free electromagnetic waves — and only they — maintain everywhere. The chapter is three articles from explaining why.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §7.2.4 · Jackson, Classical Electrodynamics, 3rd ed., §5.16–5.17 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 7 · Zangwill, Modern Electrodynamics, §13.6. The solenoid energy $\tfrac12\mu_0 n^2I^2A\ell$ is consistent with $L=\mu_0 n^2A\ell$ ; the coupling bound $M^2\le L_1L_2$ follows by completing the square (treated in Mutual Inductance); and $\sqrt{4\times9}=6\,\mathrm{mH}$ . The constant $\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}$ is the standard CODATA value. The coupled-pair energy and $M^2\le L_1L_2$ derivation are treated in Mutual Inductance. Worked numbers: bench solenoid (0.178 mH, 2.0 A): $W$ = 0.36 mJ both routes ( $B$ = 2.5 mT, $u_B$ = 2.5 J/m³, vol $1.41\times10^{-4}$ m³); MRI ( $L$ = 50 H, 500 A): $W$ = 6.25 MJ, $u_B$ (1.5 T) = $8.95\times10^5$ J/m³, quench $\approx$ 2400 L LHe at 2.6 kJ/L; density comparison $u_B/u_E \approx 2.2\times10^4$ at 1.5 T vs 3 MV/m; equal-density condition $E = cB = 4.5\times10^8$ V/m at 1.5 T; internal inductance $\mu_0/8\pi = 5.0\times10^{-8}$ H/m.

Builds on: Self-Inductance · Faraday's Law of Induction · Electromotive Force · Lenz's Law

Leads to: Mutual Inductance · Maxwell's Equations in Vacuum · The Displacement Current

Is the energy REALLY in the field, or is that just one bookkeeping choice among several?

In magnetostatics, strictly, it is a choice: all three rungs give the same total for any complete system, and no static experiment can probe where the joules "sit" — only totals and differences are measured. The field assignment earns its privileged status dynamically: once fields change, energy demonstrably moves through empty space (radiation reaches you from the Sun across vacuum containing no $\mathbf{J}$ and no circuits), and only the local density $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ with its flow $\mathbf{S}$ supports a local conservation law — energy accounted continuously, everywhere, with no teleporting. General relativity then makes the assignment binding: energy density gravitates locally, so the field's energy has to be somewhere. The honest summary: statics permits agnosticism; electrodynamics and gravity vote decisively for the field.

If magnetic forces do no work, what physically pushes the energy in during ramp-up?

The induced electric field — the species built two articles ago. While the current rises, the changing flux threads the winding with an azimuthal $\mathbf{E}$ opposing the current (Lenz). The power supply drives each electron against that electric force, and electric forces along velocities are exactly what "doing work" means. Rate: $(-\mathcal{E})I = LI\,\mathrm{d}I/\mathrm{d}t$ ; integrated, $\tfrac12LI^2$ . The magnetic field's role is architectural — it is the growing flux whose change generates that $\mathbf{E}$ — but every joule crosses from the supply into the field through electric force acting on charge. Run time backwards for discharge: the collapsing flux reverses the induced $\mathbf{E}$ , which now pushes the charges, returning the inventory.

Why is the internal inductance the same 50 nH/m for every wire, thick or thin?

A scaling conspiracy worth seeing once. Inside a wire of radius $a$ with uniform current density, the field at radius $s$ is $B \propto Is/a^2$ . Squaring and integrating over the cross-section: $\int_0^a (s/a^2)^2\,s\,\mathrm{d}s = a^4/4a^4 = 1/4$ — the $a$ 's cancel exactly. Physically: a thicker wire has weaker interior field (same current, more spread) but more volume storing it, and the two effects trade off perfectly for the $s^3$ integrand. The result, $\mu_0/8\pi$ per metre, is one of electromagnetism's few geometry-free constants. It is also fragile: raise the frequency and the skin effect crowds the current into a surface layer, gutting the interior field — internal inductance is a DC-and-low-frequency possession that RF engineers correctly ignore.

Can I always get the stored energy back, or is some of it lost like in a resistor?

The field inventory itself is fully recoverable — that is the deep difference between $\tfrac12LI^2$ and $I^2Rt$ . Ramp a lossless (superconducting) coil up and down and the supply's ledger closes exactly; the persistent MRI magnet holds its megajoules for years without topping up. Losses in real transactions come from the plumbing: winding resistance (Joule heat during every ramp), eddy currents induced in nearby metal, and — in ferromagnetic cores — hysteresis, where the magnetization's history-dependence eats a fixed area of the $B$ – $H$ loop per cycle (Hysteresis and Magnetic Energy in Materials, E5, own that story). Engineering folklore states it cleanly: air gaps store energy, iron guides flux, resistance burns money.

How does a fixed density B²/2μ0 explain magnetic forces, like two attracting coils?

Through the oldest trick in energy methods: force is the gradient of stored energy at fixed sources or fixed fluxes. Move two attracting coils closer and the total $\int B^2/2\mu_0$ changes; the force is $F = -\mathrm{d}W/\mathrm{d}x$ (with the standard care about what is held constant — currents held fixed by sources flip the sign bookkeeping via the co-energy). Faraday's beautiful visual for it: field lines carry tension along themselves and pressure sideways, both of magnitude $B^2/2\mu_0$ — attracting coils are linked by stretched lines pulling them together. That line-tension picture is quantified by the Maxwell stress tensor, the field-momentum machinery of the E7 chapter (The Maxwell Stress Tensor), where this article's density becomes one diagonal entry of a bigger object.

Why do power supplies for big magnets need "dump resistors," and how big must one be?

Because de-energizing means relocating the full $\tfrac12LI^2$ somewhere on purpose, at a controlled voltage. Open the circuit abruptly and Self-Inductance's kick chooses the somewhere for you (an arc). A dump resistor $R_d$ across the magnet sets both the discharge clock $\tau = L/R_d$ and the peak terminal voltage $IR_d$ — a genuine design squeeze: fast extraction (big $R_d$ ) means high voltage stressing the winding insulation; gentle voltage means slow extraction, dangerous if a quench is spreading. ITER-class systems resolve it with counterpulsed breakers and resistor farms sized in tens of gigajoules — civil-engineering-scale versions of the flyback-diode-vs-zener trade from Self-Inductance's seed-circuit discussion. Sizing rule of thumb: the resistor mass must absorb $\tfrac12LI^2$ within its allowed temperature rise — for the worked MRI, 6.3 MJ into steel at ~0.5 kJ/(kg·K) with a 100 K rise means ≈ 130 kg of resistor.

Does the B²/2μ0 formula work inside magnetic materials too?

With amendments that matter. For linear media the density generalizes to $u = \tfrac12\mathbf{H}\cdot\mathbf{B} = B^2/2\mu$ — note $\mu$ , not $\mu_0$ : at fixed $B$ , iron stores less energy per volume than vacuum, which is precisely why gapped inductor cores concentrate their stored energy in the air gap (vacuum is the better spring; iron is the better duct). For ferromagnets driven around their hysteresis loop, no single-valued density exists at all: the work to reach a given $B$ depends on the path, and each full cycle deposits the loop's enclosed area $\oint H\,\mathrm{d}B$ as heat. The clean statements and their proofs are E5 property (Magnetic Energy in Materials, Hysteresis); this article's vacuum formula is the foundation they amend.