Energy in an Inductor & the Magnetic Field

Self-Inductance & Inductors ended on a spark: break a coil's circuit and hundreds of volts leap the gap, manufactured by the collapsing field. But a spark is not just voltage — it is hot. It burns switch contacts, it can ignite vapour, and engineers spend real money taming it. So here is the question that module left on the table: the heat in that spark is energy, and energy has to come from somewhere. The battery was disconnected at that instant. The resistor was not being fed. Where were those joules sitting, the microsecond before the switch opened? The answer is one of the most quietly radical ideas in your syllabus: the energy was not in the wire, and not in the electrons — it was in space. While the current ramped up, the source did work pushing charge against the coil's back-emf, and every joule of that work was banked in the magnetic field filling the coil, at a density of $B^2/2\mu_0$ joules per cubic metre. A current-carrying inductor is a loaded spring made of field. Hold the current steady and the account sits frozen; kill the current and the field collapses, paying every joule back out — as a spark if you are careless, as useful output if you are an engineer. This module gives you the two formulas that run the account — $U = \tfrac{1}{2}LI^2$ for the total, $u = B^2/2\mu_0$ for the density — plus the derivations the boards expect and the mirror symmetry with the capacitor's $\tfrac{1}{2}CV^2$ . CBSE, ISC, JEE-Main, JEE-Advanced and NEET all bank numericals here, and the ideas ride straight into the next chapter's LC oscillations and transformers.

Key takeaways

Raising a current to $I$ through an inductor $L$ costs work against the back-emf. The bill is $U = \tfrac{1}{2}LI^2$ — banked, not burned. It comes back out when the current falls.
The energy lives in the magnetic field itself, at density $u = \dfrac{B^2}{2\mu_0}$ joules per cubic metre, wherever field exists — checked exactly on the solenoid, where density × volume reproduces $\tfrac{1}{2}LI^2$ .
The capacitor mirror: $\tfrac{1}{2}CV^2 \leftrightarrow \tfrac{1}{2}LI^2$ , charge ↔ flux linkage, electric field energy $\tfrac{1}{2}\varepsilon_0E^2$ ↔ magnetic $B^2/2\mu_0$ . Two components, one grammar.
Steady current = frozen account: no change, no emf, no deposits or withdrawals. The drama happens only while $I$ changes.
Switch-break sparks, MRI quench ducts and pulsed magnets are all this account being emptied fast — the same joules, different exits.
Key idea: the work done against the back-emf while a current is established is stored in the magnetic field, $U = \tfrac{1}{2}LI^2$ , at density $B^2/2\mu_0$ — the field, not the wire, is the reservoir.

The core insight

An inductor carrying current $I$ stores $U = \tfrac{1}{2}LI^2$ — the work done against its back-emf during the ramp-up — and the energy resides in the magnetic field at density $u = B^2/2\mu_0$ , ready to be paid back the moment the current falls.

Where this lives today

The biggest magnetic vaults on Earth are fusion machines: ITER's toroidal-field coils will hold roughly 41 gigajoules, and their quench-protection systems (fast breakers, farms of dump resistors) exist to empty that account on purpose before a fault empties it violently. Hospital MRI magnets run the same engineering at 6-megajoule scale — the roof vent duct in every MRI suite is there because a quench boils thousands of litres of liquid helium with exactly the joules this module taught you to count. Grid engineers even store energy as field: SMES units bank $B^2/2\mu_0$ in persistent superconducting coils and return it within milliseconds to steady grid frequency. The everyday version hides in every charger you own: switching power supplies shuttle parcels of $\tfrac{1}{2}LI^2$ into and out of a ferrite inductor tens of thousands of times per second — a magnetic energy pump with a duty cycle. In this course the account now starts moving: The Transformer (E7) hands field energy coil to coil, and LC Oscillations (E7) connects this vault to the capacitor's and lets the joules slosh.

The physical picture

Start with the analogy this chapter has been building since Self-Inductance & Inductors handed you the inertia dictionary: an inductor is to current what mass is to velocity. Push a flywheel up to speed and your effort is not wasted — it is stored, as $\tfrac{1}{2}mv^2$ , recoverable the moment the wheel drives something. Push a current up to $I$ through a coil and the same sentence holds with the words swapped: the source's effort against the back-emf is stored, as magnetic energy $\tfrac{1}{2}LI^2$ , recoverable the moment the current is allowed to fall. The break-spark is the analogy's last row made violent — stopping the current dead is grabbing the flywheel bare-handed, and the energy comes out somewhere, fast.

Figure

Charging an inductor — the work done against the back-emf is banked in the field volume

Now sharpen the overview's question: where is the stored energy? Instinct says "in the coil," the way charge sits on a capacitor's plates. But run the superconducting-magnet test: an MRI magnet is energized over an hour, short-circuited on itself, and the supply wheeled away. For years the current persists, the field holds 1.5 T, and the megajoules remain fully recoverable — no source, no resistor, nothing "happening" in the circuit at all. The only physical thing still on duty is the field in the bore. The conclusion physics has held since the 1850s: the energy sits in the field, at $B^2/2\mu_0$ per cubic metre; the wire is scaffolding that holds the field up. One more calibration before the algebra: magnetic fields store energy in air astonishingly densely. At an MRI bore's 1.5 T the density is about $9.0\times10^5\ \mathrm{J/m^3}$ — nearly a megajoule per cubic metre of empty space — and Kapitsa's pulsed ~30 T coils (his story closes the module) touch $3.6\times10^8\ \mathrm{J/m^3}$ . Even a modest lab coil is a little vault, and every switch that interrupts it must be designed for the withdrawal.

The account: half L I squared

Charging an inductor means fighting Lenz the whole way up: while the current rises, the coil answers with a back-emf (Self-Inductance & Inductors, cited not re-proved), and the source must push every coulomb through that opposing emf — work, strictly. Add up the ramp and the total is startlingly tidy:

U = \tfrac{1}{2}\,L\,I^2 \qquad (1)

where: $U$ is the stored energy in joules, $L$ the self-inductance in henries, and $I$ the final steady current. The square means doubling the current quadruples the balance, exactly as doubling speed quadruples a flywheel's $\tfrac{1}{2}mv^2$ — the inertia dictionary extended by one row. And note what is absent: time. A 50 H MRI magnet at 500 A holds $\tfrac{1}{2}(50)(500)^2 = 6.3$ MJ whether it was energized in a minute or a month; the account depends only on where you ended.

Derivation AHalf L I squared from a steady ramp

The goal is equation (1): the work done establishing a current $I$ in an inductor of self-inductance $L$ , using nothing beyond algebra. The strategy is to choose the simplest possible ramp — current rising at a steady rate — because for a straight-line ramp the back-emf is constant, and work is then just emf × charge, two quantities we can each write down. The result cannot depend on the ramp shape (the account only cares about the final balance), so the easy ramp gives the general answer. Watch where the famous factor of ½ comes from: it is the average current during the ramp.

Step 1. Ramp the current from $0$ to $I$ at a constant rate over a time $t$ . The rate of change is $\dfrac{\Delta I}{\Delta t} = \dfrac{I}{t}$ , the same at every instant, so the back-emf holds a constant magnitude

\lvert\varepsilon\rvert = L\,\frac{\Delta I}{\Delta t} = \frac{LI}{t},

by the back-emf law of Self-Inductance & Inductors, quoted as a tool.

Step 2. Count the charge that flowed during the ramp. The current climbed steadily from $0$ to $I$ , so its average value was $I/2$ , and the charge delivered is

q = I_{\text{avg}}\,t = \frac{I}{2}\,t.

Why the average is exactly $I/2$ : on an $I$ – $t$ graph the ramp is a straight line from the origin, and charge is the area under it — a triangle, half of base × height. The ½ in the final formula is this triangle.

Step 3. Work is charge pushed through an opposing emf: every coulomb crossing the inductor against $\lvert\varepsilon\rvert$ costs $\lvert\varepsilon\rvert$ joules. So

U = \lvert\varepsilon\rvert\, q = \frac{LI}{t}\times\frac{I}{2}\,t = \tfrac{1}{2}LI^2.

The ramp time $t$ cancels — as it must, since the balance cannot care how fast you deposited it.

Step 4. Read the result physically. None of this work went to heat (an ideal inductor has no resistance); it went into building the coil's field against Lenz's objection, and it is repaid in full when the current falls — run the same accounting backwards and the collapsing flux drives charge around the circuit, ${+\tfrac{1}{2}LI^2}$ back out. Verified: units — H·A² = (V·s/A)·A² = V·A·s = W·s = J ✓. Numbers — $L = 1.0$ mH at $I = 3.0$ A gives $U = 4.5$ mJ, reproduced by the field-density route in the board worked example below to the displayed digits ✓. Scaling — double the current, $2^2 = 4\times$ the energy ✓.

Two exam-relevant footnotes. First, the steady state: at constant $I$ the back-emf is zero, no further work is done, and the account is frozen — a steady inductor draws nothing beyond its wire's resistive losses. Second, the withdrawal: at a circuit break, equation (1) is the size of the problem. The break-spark of Self-Inductance & Inductors told you the voltage; this module tells you the heat — all of $\tfrac{1}{2}LI^2$ , delivered to the arc — and LR Circuits: Growth & Decay holds the clock for the orderly drain through a resistor.

Where the joules live: the field as reservoir

Equation (1) is terminal bookkeeping — it knows $L$ and $I$ and nothing else. To find where the energy sits, take the geometry where you know everything: the long solenoid, whose inductance and interior field this chapter already owns.

Go deeper · Derivation B · 1 minThe energy density of the magnetic field

Derivation BThe energy density of the magnetic field

The goal is the density formula $u = B^2/2\mu_0$ — joules per cubic metre of field — assembled by pure substitution on the solenoid, with no new physics. The strategy: write the solenoid's stored energy from equation (1), recognize that its field fills the bore volume uniformly (and is negligible outside), divide energy by volume, and then eliminate the solenoid's construction details in favour of $B$ itself. The punchline to watch for: every trace of the winding — turn density, current, length, area — cancels, leaving a formula that mentions only the field. That cancellation is the evidence that the energy belongs to the field, not to the hardware.

Step 1. For a long solenoid, quote the two chapter results as tools: inductance $L = \mu_0 n^2 A \ell$ (Self-Inductance & Inductors) and interior field $B = \mu_0 n I$ (Solenoids & Toroids), with $n$ turns per metre, area $A$ , length $\ell$ .

Step 2. Bank balance from equation (1):

U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}\,\mu_0 n^2 A \ell\, I^2.

Step 3. The field is uniform inside the bore — volume $V = A\ell$ — and essentially zero outside, so if the energy lives in the field it fills exactly this volume. Divide:

u = \frac{U}{V} = \frac{\tfrac{1}{2}\mu_0 n^2 A\ell I^2}{A\ell} = \tfrac{1}{2}\mu_0 n^2 I^2.

Step 4. Now eliminate the hardware. From $B = \mu_0 n I$ , we have $n I = B/\mu_0$ , so $\mu_0 n^2 I^2 = \mu_0 (nI)^2 = B^2/\mu_0$ . Substituting,

u = \frac{B^2}{2\mu_0} \qquad (2)

Why the cancellation matters: the winding details ( $n$ , $I$ , $A$ , $\ell$ ) all vanished, and what remains depends only on the field strength at the point in question. Two completely different machines producing the same $B$ in some region store the same joules-per-cubic-metre there. The formula has forgotten the circuit — which is exactly the claim "the energy is in the field." *Verified: units — T²/(T·m/A) = T·A/m; with 1 T = 1 kg/(A·s²), this is kg/(m·s²) = J/m³ ✓. Numbers — at $B = 6.0$ mT (the board example's solenoid), $u = (6.03\times10^{-3})^2/(2\times4\pi\times10^{-7}) = 14.5\ \mathrm{J/m^3}$ , and density × bore volume reproduces the $\tfrac12LI^2$ answer ✓. Anchor — at 1.5 T, $u = 9.0\times10^5\ \mathrm{J/m^3}$ ✓.*

Equation (2) is the module's deep result: space itself, wherever a magnetic field reaches, is loaded like a spring — silently, invisibly, at $B^2$ per unit volume. The square means high-field regions utterly dominate an energy audit, as the scoreboard below makes visceral.

Figure

The density scoreboard — the B-squared law makes high-field regions dominate every audit

You have met this grammar before. Energy Stored in a Capacitor made the same move for electricity: the work $\tfrac{1}{2}CV^2$ was relocated into the electric field at density $\tfrac{1}{2}\varepsilon_0 E^2$ . Set the two side by side and the symmetry does your revision for you:

	capacitor (E2)	inductor (this module)
stores energy in	electric field between plates	magnetic field in the coil
the account	$U = \tfrac{1}{2}CV^2 = \tfrac{Q^2}{2C}$	$U = \tfrac{1}{2}LI^2$
state variable	charge $Q$ (or voltage $V$ )	current $I$ (or flux linkage $N\Phi = LI$ )
field density	$u = \tfrac{1}{2}\varepsilon_0 E^2$	$u = \dfrac{B^2}{2\mu_0}$
holds its energy when	circuit is open (charge trapped)	circuit is closed (current maintained)
dies by	leaking charge	resistance eating the current
violent exit	dielectric breakdown / shock	break-spark / flyback kick

The last three rows are the ones exams love. A charged capacitor on a shelf keeps its joules (until leakage wins); an inductor's joules exist only while the current flows — cut the path and the account is force-closed instantly, which is the whole drama of the spark. E7's LC Oscillations will connect the two opposite personalities into one circuit where the joules slosh between reservoirs thousands of times a second — the electrical twin of a pendulum. That module cashes this table; here you only need the mirror.

Limiting cases

case	result	meaning
current doubled	$U \to 4U$	the square: energy grows faster than current
steady current, any size	no deposits, no withdrawals	back-emf zero; account frozen (losses are the wire's $I^2R$ , not $L$ 's)
current killed in $\Delta t \to 0$	all of $\tfrac{1}{2}LI^2$ → arc	the break-spark's heat budget; voltage story in 0783, energy story here
field doubled at a point	$u \to 4u$	$B^2$ : high-field regions dominate audits
iron core inserted (same $I$ )	$L \to \mu_r L$ , so $U \to \mu_r U$	more flux per ampere — a bigger vault at the same current
$B = 0$ region	$u = 0$	no field, no energy: the reservoir is exactly where the field is
capacitor mirror	$\tfrac{1}{2}CV^2 \leftrightarrow \tfrac{1}{2}LI^2$	same grammar, opposite state variable — LC sloshing coming in E7

Common trapThe inductor "uses up" energy like a resistor

It does not — and the distinction is worth full marks several times a year. A resistor dissipates: $I^2Rt$ leaves as heat and never comes back. An ideal inductor stores: the work done against its back-emf during ramp-up sits in the field, recoverable in full when the current falls — run an inductor up and down and the net energy bill is zero. That is why $U = \tfrac{1}{2}LI^2$ has no time in it while heat $= I^2Rt$ grows with time; confusing the two is the classic mark-loser in mixed numericals. The exam-safe habit: for any circuit element ask "where are the joules afterwards?" Resistor — gone to heat, write $I^2Rt$ . Inductor — banked in the field, write $\tfrac{1}{2}LI^2$ , and expect the question's next part to ask what happens to that bank when the switch opens (answer: the arc or the flyback path collects it). A real coil does both jobs at once — its wire resistance burns while its inductance banks — and good questions make you split the two.

Worked examples

Board level — one solenoid, audited twice (CBSE/ISC pattern, 3 marks)

A solenoid of 400 turns is wound over a length of 25 cm with cross-sectional radius 2.0 cm; its self-inductance is 1.0 mH (computed in the Self-Inductance module). It carries a steady current of 3.0 A. (a) Find the energy stored. (b) Find the magnetic field inside and the energy density. (c) Show that density × volume reproduces the answer to (a).

Step 1. Part (a), the account: $U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(1.01\times10^{-3})(3.0)^2 = \boxed{4.5\ \mathrm{mJ}}$ .

Step 2. Part (b), the field: turn density $n = 400/0.25 = 1600\ \mathrm{m^{-1}}$ , so $B = \mu_0 n I = (4\pi\times10^{-7})(1600)(3.0) = 6.0\times10^{-3}\ \mathrm{T}$ . Density, equation (2): $u = \dfrac{B^2}{2\mu_0} = \dfrac{(6.03\times10^{-3})^2}{2\times 4\pi\times10^{-7}} = \boxed{14.5\ \mathrm{J/m^3}}$ .

Step 3. Part (c), the cross-check: bore volume $V = A\ell = \pi(0.020)^2 \times 0.25 = 3.14\times10^{-4}\ \mathrm{m^3}$ . Then $uV = 14.5 \times 3.14\times10^{-4} = 4.5\times10^{-3}\ \mathrm{J}$ ✓ — the same 4.5 mJ by a completely different route. Two formulas, one vault.

Step 4. The habit worth carrying: whenever both routes are available, run both. Agreement to two figures is your arithmetic insurance, and part (c)-style "show that" marks are among the cheapest in the paper.

Competitive level — reading the account like an examiner (JEE-Main/NEET pattern, ~3 minutes)

An inductor stores 2.0 J of energy when carrying a current of 4.0 A. (a) Find L. (b) The current is raised to 8.0 A; find the new stored energy. (c) At 4.0 A the circuit is broken and the stored energy is delivered to the switch arc in 0.50 ms; find the average power of the arc. (d) A 100 μF capacitor is charged to 200 V; compare its stored energy with (a).

Step 1. Part (a): invert equation (1). $L = \dfrac{2U}{I^2} = \dfrac{2(2.0)}{(4.0)^2} = \boxed{0.25\ \mathrm{H}}$ .

Step 2. Part (b): the square does the work. Current doubles → energy quadruples: $U' = \tfrac{1}{2}(0.25)(8.0)^2 = \boxed{8.0\ \mathrm{J}}$ . Writing " $2\times$ " here is the intended trap; $U \propto I^2$ , always.

Step 3. Part (c): power is energy over time. $P_{\text{avg}} = \dfrac{2.0}{0.50\times10^{-3}} = \boxed{4.0\ \mathrm{kW}}$ — a four-kilowatt event from a modest bench inductor, which is why switchgear is engineered around this withdrawal. (The kilovolt voltage of such breaks was 0783's story; the kilowatt power is this module's.)

Step 4. Part (d): the mirror. $U_C = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(100\times10^{-6})(200)^2 = \boxed{2.0\ \mathrm{J}}$ — exactly equal to the inductor's account in (a). Equal joules, opposite custodians: the capacitor holds its 2.0 J with the circuit open; the inductor holds its 2.0 J only while the 4.0 A keeps flowing. Timing note: about three minutes for all four parts. (a) and (b) are 30-second substitutions if you write $U \propto I^2$ before touching numbers. (c) is division — do not overthink it. (d) is there to test whether you carry both energy formulas in one pocket; JEE-Main has paired them more than once, sometimes asking for the current at which an inductor matches a given capacitor's energy: set $\tfrac{1}{2}LI^2 = \tfrac{1}{2}CV^2$ and solve.

Exam traps

Forgetting the half. $U = LI^2$ is the most common wrong answer in the chapter. The ½ is the average current of the ramp (the triangle area of Derivation A) — understand it once and you will never drop it.

Stored versus dissipated. $\tfrac{1}{2}LI^2$ has no time in it; $I^2Rt$ does. If the question says "heat produced in the coil in 5 s," it wants the resistive $I^2Rt$ ; if it says "energy stored," it wants $\tfrac{1}{2}LI^2$ . Real coils do both — split them.

The square in ratios. Current tripled → energy ×9. Field halved → density ÷4. Read whether the ratio question is about $U \propto I^2$ (via $L$ fixed) or $u \propto B^2$ (pointwise) — they are different statements of the same square.

Density needs the right volume. $u = B^2/2\mu_0$ is per cubic metre of field. For a solenoid, multiply by the bore volume $A\ell$ , not the wire volume, and not "the room."

Steady state is free. A constant current stores a constant $\tfrac{1}{2}LI^2$ and draws nothing further through $L$ . Any option implying an inductor continuously "consumes" power at steady current is bait (the wire's resistance may, but that is $R$ 's bill).

Unit discipline. Henries × amperes² = joules; teslas² ÷ (T·m/A) = J/m³. Milli- and micro-henries with amperes give milli/micro-joules — track the prefixes, they are where the marks leak.

Boards and constants. CBSE, ISC, JEE-Main, JEE-Advanced and NEET all examine inductor energy; $B^2/2\mu_0$ appears in CBSE/ISC derivation questions and JEE ratio MCQs. IB, IGCSE, A-Level and AP Physics 2 do not examine inductance quantitatively (9702 contains no inductor; IB D.4 keeps self-induction qualitative) — their candidates need only the storage idea. Constants: $\mu_0 = 4\pi\times10^{-7}$ T·m/A; keep $4\pi$ symbolic until the last line.

Origins · 1 minPyotr Kapitsa and how this was discovered

Pyotr Kapitsa

1894–1984 · Cambridge and Moscow

Kapitsa is the physicist who treated $B^2/2\mu_0$ as an engineering budget. Arriving in Rutherford's Cambridge from revolutionary Russia in 1921, he wanted fields far beyond what any steady coil could survive — wires melt, forces tear windings apart. His 1924 solution was pure energy accounting: you do not need the field for long. He short-circuited a massive flywheel alternator through a small coil, dumping the machine's rotational energy into the field for a hundredth of a second — about 30 tesla, a record that stood for decades, bought in instalments because at that strength space holds hundreds of megajoules per cubic metre and nothing material can host it steadily. (The lab produced spectacular bangs; Rutherford tolerated them.) Kapitsa's later career belongs to low temperatures — he liquefied helium a new way, discovered superfluidity, won the 1978 Nobel Prize, and refused Stalin's weapons work at the price of house arrest. But every pulsed magnet and quench duct still runs on his opening insight: the magnetic field is a reservoir you can fill and empty on a schedule.

History

1832

— Henry reports the fat spark at a breaking coil circuit — the stored energy announcing itself

1853

— William Thomson computes the "mechanical value" of a current: the ledger behind ½LI²

1865

— Maxwell's dynamical theory assigns the energy to the field itself, density B²/2μ₀

1911

— Kamerlingh Onnes discovers superconductivity: persistent currents make lossless storage physical

1924

— Kapitsa's impulse magnets: a flywheel's energy banked in a coil for a hundredth of a second

Going further

Texts. NCERT Physics Class 12, Ch. 6 (energy in an inductor) · HC Verma Vol. 2, Ch. 38 · SL Arora, ISC Physics XII, electromagnetic induction (energy stored in an inductor; energy density).

Provenance. All results standard. A verifier should recompute: Derivation A's cancellation of $t$ ; Derivation B's cancellation of $n, I, A, \ell$ ; board example $U = \tfrac{1}{2}(1.01\times10^{-3})(3.0)^2 = 4.5$ mJ, $B = 6.03$ mT, $u = 14.5$ J/m³, $V = 3.14\times10^{-4}$ m³, $uV = 4.5$ mJ ✓; competitive example $L = 0.25$ H, $U(8\ \mathrm{A}) = 8.0$ J, $P_{\text{avg}} = 4.0$ kW, capacitor twin $2.0$ J; anchors $u(1.5\ \mathrm{T}) = 9.0\times10^5$ J/m³, $u(2.5\ \mathrm{mT}) = 2.5$ J/m³.

Prerequisites. Self-Inductance & Inductors · Energy Stored in a Capacitor · Solenoids & Toroids. Leads to. LC Oscillations (coming up in the AC chapter) · The Transformer (coming up in the AC chapter) · LR Circuits: Growth & Decay (the withdrawal clock) · Energy Stored in Magnetic Fields (UG deep dive).

Where exactly is the energy — in the wire, in the electrons, or really in empty space?

In the field, and "really" is the right word. The electrons' kinetic energy in copper is absurdly small (drift speeds are sub-millimetre-per-second); the wire is scaffolding. The clean test is the persistent superconducting magnet: source disconnected, nothing changing — yet megajoules remain recoverable, and the only physical thing present is the field in the bore. The density formula makes it precise: every cubic metre at field strength $B$ holds $B^2/2\mu_0$ joules, wire nearby or not. At HS level, hold the picture — field lines as stretched elastic filling space; the university course (the UG deep dive) proves the relocation formally.

The derivation used a steady ramp. What if the current rises in some messy, non-uniform way?

Same answer, best kind of reason: energy is a state function. The account depends only on where it ended ( $L$ and the final $I$ ), not on the deposit schedule — like height above sea level not caring which road you drove up. The steady ramp is chosen because it makes both factors (constant emf, average current $I/2$ ) writable with algebra; a curved ramp needs calculus to add up but must land on the same $\tfrac{1}{2}LI^2$ , because ramping up one way and down another with a net energy profit would be a free-energy machine — the door Lenz's Law & Energy Conservation closed permanently.

If the energy is "stored," why do I never get a shock from an inductor sitting on a shelf?

A shelved inductor carries no current — no current, no field, empty vault. That is the sharpest difference from a capacitor, which stores by holding charge apart and does so happily with the circuit open (why an unplugged TV can still bite). An inductor stores by keeping current flowing, so its account exists only in a closed, running circuit. Its dangerous moment is not the shelf but the instant of disconnection, when the collapsing field manufactures whatever voltage the gap demands to keep the current alive. Capacitors bite later; inductors bite at the switch.

Does the energy density formula mean a magnet is dangerous just sitting there?

"Loaded" is the honest word. A persistent 1.5 T MRI field stores ~ $9\times10^5$ J/m³ — megajoules in the bore — but locked in place while the superconducting current circulates undisturbed. The hazard is the lock breaking: in a quench a patch of winding goes resistive, the megajoules dissipate inside the magnet, the helium boils, and the vent duct earns its keep — the competitive example's parts (a) and (c) at hospital scale. The moral survives the scaling: never interrupt a large inductive current without giving the energy a designed exit.

Iron cores multiply L by hundreds — do they multiply the stored energy too?

At the same current, yes: $U = \tfrac{1}{2}LI^2$ , so a core that multiplies $L$ by $\mu_r$ multiplies the account too — more flux per ampere means more work went into establishing it. That is why relay coils are iron: big vaults at modest currents. Two exam-level caveats. First, iron saturates: pushed hard, its $\mu_r$ collapses toward 1, so the multiplication is a low-field promise, not a law (fine print in Magnetism & Matter). Second, cycling iron costs hysteresis heat — some deposits are not returned — one of the "core losses" The Transformer budgets for in E7. Air-cored numericals dodge all of this: use $\mu_0$ and sleep well.

Is there a magnetic version of the capacitor's Q²/2C formula?

Yes — a nice test of the mirror. The inductor's state variable is flux linkage $N\Phi = LI$ ; substitute $I = N\Phi/L$ into $\tfrac{1}{2}LI^2$ and you get $U = (N\Phi)^2/2L$ , the exact twin of $Q^2/2C$ . JEE-style questions occasionally hand you a flux linkage instead of a current, and the on-the-spot derivation is 15 seconds of algebra once you remember which quantity mirrors charge: not current, but flux linkage.

What happens to the stored energy in an LC circuit — where do the joules go when the capacitor takes over?

They commute. A charged capacitor across an inductor drives a growing current, banking $\tfrac{1}{2}CV^2$ as $\tfrac{1}{2}LI^2$ ; the current then recharges the capacitor the other way, and the joules ride back. In the ideal circuit the total is constant — energy oscillating between electric and magnetic custody, exactly like a pendulum's kinetic ↔ potential trade — and the exchange rate sets a natural frequency. That story and its SHM analogy are LC Oscillations in the AC chapter; this module's job was to open the magnetic account the oscillation draws on.