In short

When a current I flows through an inductor of inductance L, the inductor stores energy

\boxed{\;U \;=\; \tfrac{1}{2}\,L\,I^{2}\;}

in its magnetic field. This is the exact magnetic analogue of the capacitor formula U = \tfrac{1}{2}CV^{2}: where a capacitor stores energy in its electric field between the plates, an inductor stores energy in its magnetic field inside and around the coil.

The energy is not located in the wire. It lives in the volume of space where \vec{B} is non-zero, with a magnetic energy density

\boxed{\;u \;=\; \dfrac{B^{2}}{2\mu_{0}}\;}

(in joules per cubic metre). Integrating this density over the field region reproduces \tfrac{1}{2}LI^{2} — two views of the same stored energy.

Current in an LR circuit (a battery of EMF \varepsilon pushed through resistance R in series with inductance L):

  • Growth (switch closed at t=0, starting from I=0):
I(t) \;=\; \dfrac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right),\quad \tau \;=\; \dfrac{L}{R}
  • Decay (battery replaced by a short at t=0, starting from I_{0}):
I(t) \;=\; I_{0}\,e^{-t/\tau}

The time constant \tau = L/R has units of seconds and sets the pace of both. After one \tau the current is 63% of its final value (during growth) or 37% of its initial value (during decay); after five \tau it is within 1% of the steady state.

Why it matters. The energy stored in an inductor is how a defibrillator at AIIMS delivers a 200 J shock, how the spark plug in your cousin's Honda Activa fires on every combustion stroke, how ISRO's pulsed-power experiments at Sriharikota dump megajoules of magnetic energy into a load in a microsecond, and why switching off the inductor of an electric motor without a flyback diode fries whatever is nearby.

A CRT television in a 1990s Indian household had a large fat coil inside it called the flyback transformer. Every time the picture refreshed — sixty times a second — that coil had its current yanked to zero in about a microsecond. And every time, a spark the size of a grain of rice jumped across the switch. The spark was not a defect. It was the inductor's stored magnetic energy, refusing to vanish when the current was cut, finding the fastest path to ground it could — which, in that microsecond, was through the air.

An inductor looks inert. It is, after all, just a coil of copper. But once a current is flowing through it, there is a pool of energy sloshing around it, invisible, carried entirely by the magnetic field threading the coil. Switch the current off and that pool does not politely disappear; it insists on being spent. The whole subject of power electronics — every inverter in a solar rooftop in Rajasthan, every driver circuit in an LED streetlight, every spark plug in every two-stroke autorickshaw — is the art of controlling that pool: storing energy in an inductor, then releasing it on cue. This chapter is about how big the pool is, where it actually lives, and how fast it fills and empties.

Why an inductor stores energy at all

Start with a plain question. Hook a 12 V battery to a resistor, and the battery does work on the charges as they flow through the circuit. That work becomes heat in the resistor. What happens to the same 12 V battery if the "resistor" is replaced by an ideal inductor?

Ideal means zero resistance. No heat. But the current cannot jump from zero to infinity — the inductor fights the change in current by producing a back-EMF (Faraday's law applied to the flux the current makes through the coil's own turns). While the current is building up, the battery pushes charge against that back-EMF. The battery is doing work. Since there is no resistor to soak up the energy, and since the inductor itself has no dissipation, the work must be going somewhere. It is going into the magnetic field.

That is the whole argument. An inductor stores energy because building up a current requires doing work against the induced EMF, and that work has to be accounted for somewhere. Let us turn the argument into a formula.

Deriving U = ½LI² from first principles

Take an inductor of self-inductance L connected to a battery with enough EMF to drive a current I through it, with negligible resistance. At an instant when the current is i and is growing at rate di/dt, the inductor opposes the battery with a back-EMF

\varepsilon_{L} \;=\; L\,\dfrac{di}{dt}

Why: this is the definition of self-inductance — the induced EMF in a coil is proportional to the rate of change of current through it. See Self-Inductance and Mutual Inductance for the derivation from Faraday's law.

To push charge dq = i\,dt through the inductor against this back-EMF, the battery must do work

dW \;=\; \varepsilon_{L}\,dq \;=\; L\dfrac{di}{dt}\cdot i\,dt \;=\; L\,i\,di

Why: work done against an EMF equals the EMF times the charge it moves through it — exactly as work in mechanics is force times distance. Here the EMF plays the role of "force per unit charge" and dq plays the role of "distance".

All of that work goes into the magnetic field (there is no resistor to turn it into heat, and the inductor is not moving). So the energy stored when the current reaches its final value I is the total work:

U \;=\; \int_{0}^{I} L\,i\,di \;=\; L\,\cdot\,\dfrac{i^{2}}{2}\bigg|_{0}^{I}
\boxed{\;U \;=\; \tfrac{1}{2}\,L\,I^{2}\;}

Why: integrating L\,i\,di from 0 to I gives \tfrac{1}{2}L I^{2}. This is a pure consequence of the back-EMF being proportional to di/dt — exactly the same integration that gives \tfrac{1}{2}CV^{2} for a capacitor (where the stored charge is proportional to voltage).

The formula has three things worth noticing:

  1. The energy grows as I^{2}, not I. Doubling the current quadruples the stored energy. Tripling it stores nine times as much. This is why the spark from a large industrial coil can be viciously large — the energy rises steeply with current.
  2. The dependence on L is linear. An inductor with twice the inductance stores twice the energy at the same current.
  3. It does not depend on the voltage or the EMF of the source. Two different sources that both push current I through the same L store the same energy. Only the final current and the inductance matter.

Energy lives in the magnetic field, not in the wire

The formula U = \tfrac{1}{2}LI^{2} tells you how much energy is stored. It does not tell you where it is stored. The surprising answer, which drops out the moment you calculate it for a concrete geometry, is: the energy is in the empty space around the wire, wherever the field is non-zero, not in the copper.

Take a long solenoid of length \ell, cross-section A, with n turns per metre, carrying a steady current I. From the solenoid and toroid chapter, the magnetic field inside is uniform with magnitude

B \;=\; \mu_{0}\,n\,I

and is essentially zero outside (for an ideal long solenoid). The self-inductance is

L \;=\; \mu_{0}\,n^{2}\,A\,\ell

Substitute into U = \tfrac{1}{2}LI^{2}:

U \;=\; \tfrac{1}{2}\,\mu_{0}\,n^{2}\,A\,\ell\,I^{2} \;=\; \tfrac{1}{2}\,(\mu_{0}\,n\,I)^{2}\,\dfrac{A\,\ell}{\mu_{0}} \;=\; \dfrac{B^{2}}{2\mu_{0}}\,\cdot\,(A\,\ell)

Why: pull out a factor of (\mu_{0}nI)^{2} = B^{2} and what is left is A\ell/\mu_{0}. The quantity A\ell is the volume of the solenoid interior — exactly the region where the field is non-zero.

Dividing both sides by the volume V = A\ell, you get the magnetic energy density — energy per unit volume of field:

\boxed{\;u \;=\; \dfrac{B^{2}}{2\mu_{0}}\;}

The units check out: B in teslas gives B^{2} in T², and \mu_{0} in T·m/A combined with a denominator factor of 2 gives J/m³. One tesla is strong. Squaring it gives one T² = 1 J·m⁻³·(4π×10⁻⁷)⁻¹·2 ≈ 4×10⁵ J/m³, which is a lot of energy for a cubic metre of field — explaining why high-current-density experimental magnets at institutions like TIFR Mumbai and the IUAC Delhi need serious containment.

We derived u = B^{2}/(2\mu_{0}) for a solenoid, but the result is general: any magnetic field, anywhere in space, carries energy at this density. Integrate u over all space where \vec{B} \neq 0 and you recover \tfrac{1}{2}LI^{2} for any geometry. This is the deeper picture of what is stored in an "inductor": the coil is just a convenient device to set up a region of field; the energy is in the field itself.

The electric analogue is exact. A capacitor's energy density is u_{E} = \tfrac{1}{2}\varepsilon_{0}E^{2}, and both expressions have the same dimensional structure: a square of the field divided by twice the vacuum constant of the relevant medium.

The LR circuit — current that takes time to turn on

Now pair the inductor with a resistor. This is the simplest realistic circuit: an EMF source \varepsilon, a resistor R, and an inductor L, all in series, with a switch. Real inductors have some resistance in their windings; real power-supply loads have some inductance. The LR circuit is the archetype.

The circuit diagram:

Circuit diagram of a series LR circuit with a battery and a switch A battery of EMF epsilon is connected through a switch to a resistor R and an inductor L in series. Current i flows clockwise. + − ε S R L i
An LR circuit. Close the switch S and the EMF $\varepsilon$ pushes current $i$ clockwise through $R$ and $L$. The inductor opposes the rise, so the current does not reach $\varepsilon/R$ instantly — it grows gradually.

Growth of the current

Close the switch at t = 0, with I(0) = 0. Apply Kirchhoff's voltage law around the loop. Going in the direction of the current, the battery contributes +\varepsilon, the resistor drops IR (voltage falls in the direction of current), and the inductor drops L\,dI/dt (back-EMF opposes increasing current). The sum is zero:

\varepsilon \;-\; IR \;-\; L\,\dfrac{dI}{dt} \;=\; 0

Rearrange into the standard form:

L\,\dfrac{dI}{dt} \;+\; IR \;=\; \varepsilon

Why: this is a first-order linear ODE in I(t). The same mathematical form governs a charging capacitor, a falling object with linear drag, or Newton's law of cooling. The general strategy is: find the steady state, then add an exponential decay to it.

Steady state. After a long time, if anything settles, it does so with dI/dt = 0. Plug that in: IR = \varepsilon, so I_{\infty} = \varepsilon/R. Call this the final current — it is exactly what you would get if the inductor were absent (because once the current is steady, the inductor stops producing back-EMF).

Transient. Write I(t) = I_{\infty} + i(t) where i(t) is the departure from steady state. Substitute:

L\,\dfrac{d(I_{\infty} + i)}{dt} \;+\; (I_{\infty} + i)\,R \;=\; \varepsilon

Since I_{\infty} is constant, d(I_{\infty})/dt = 0. And I_{\infty}R = \varepsilon cancels the right side. What is left:

L\,\dfrac{di}{dt} \;+\; i\,R \;=\; 0 \quad\Rightarrow\quad \dfrac{di}{dt} \;=\; -\dfrac{R}{L}\,i

Why: splitting I into steady state plus transient turns the non-homogeneous ODE into a homogeneous one. The homogeneous equation is the one whose solution is a pure exponential — standard fare for first-order linear systems.

The solution to di/dt = -(R/L)\,i is an exponential decay:

i(t) \;=\; A\,e^{-t/\tau},\quad \tau \;\equiv\; \dfrac{L}{R}

Why: if a quantity's rate of change is proportional to its own negative, the quantity decays exponentially. The constant \tau = L/R has units of seconds (henries over ohms — check: [L] = V·s/A, [R] = V/A, so [L/R] = s).

Fix A using the initial condition I(0) = 0. Since I(t) = I_{\infty} + i(t), at t = 0 we need I_{\infty} + A = 0, so A = -I_{\infty} = -\varepsilon/R. Putting it together:

I(t) \;=\; \dfrac{\varepsilon}{R} \;-\; \dfrac{\varepsilon}{R}\,e^{-t/\tau}
\boxed{\;I(t) \;=\; \dfrac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right)\;}

Why: at t=0 the bracket is 0, so I=0 as required. As t \to \infty, the exponential vanishes and I \to \varepsilon/R, the steady state. At t = \tau, the bracket equals 1 - 1/e \approx 0.632: the current has risen to about 63% of its final value.

The voltage across the inductor, V_L = L\,dI/dt, is the mirror image — it jumps to \varepsilon at t = 0 and decays exponentially to 0. Differentiating I(t):

V_{L}(t) \;=\; L\,\dfrac{dI}{dt} \;=\; L\,\cdot\,\dfrac{\varepsilon}{R}\,\cdot\,\dfrac{1}{\tau}\,e^{-t/\tau} \;=\; \varepsilon\,e^{-t/\tau}

(using \tau = L/R, so L/(R\tau) = 1).

Watch the current and the inductor voltage fill in over time:

Animated: LR growth curves for current and inductor voltage A red dot tracing I(t)/Iinfty = 1 − exp(−t/τ) and a dark dot tracing V_L(t)/epsilon = exp(−t/τ) over five time constants. At t = τ, I reaches 0.632 and V_L drops to 0.368. At t = 5τ, I is within 1% of 1. τ 2τ 3τ 4τ 5τ 0 0.37 0.63 1 time (in units of τ = L/R) fraction of final value
Red: $I(t)/I_\infty = 1 - e^{-t/\tau}$ — the current filling up. Dark: $V_L(t)/\varepsilon = e^{-t/\tau}$ — the inductor voltage draining away. The two sum to 1 at every instant, because the voltage across $L$ plus the voltage across $R$ always equals $\varepsilon$. At $t = \tau$ the current is 63% done; at $t = 5\tau$ it is within 1% of steady state.

Decay of the current

Now suppose the current is already I_{0} and you short-circuit the battery out (say, flip the switch to connect the inductor directly to the resistor without the EMF source). At t = 0:

L\,\dfrac{dI}{dt} \;+\; IR \;=\; 0 \quad\Rightarrow\quad \dfrac{dI}{dt} \;=\; -\dfrac{R}{L}\,I

Same homogeneous equation as before, with initial condition I(0) = I_{0}:

\boxed{\;I(t) \;=\; I_{0}\,e^{-t/\tau}\;}

Why: without a battery, there is nothing to hold the current up. The stored magnetic energy drives the current through the resistor, where it dissipates as heat, until the field collapses completely. The current decays at the same rate \tau = L/R it built up at — a symmetry of first-order linear systems.

During decay, the inductor is the source, pushing current through the resistor. The voltage across the resistor is V_R = IR = I_{0}R\,e^{-t/\tau}. The voltage across the inductor is the opposite: V_L = -V_R = -I_{0}R\,e^{-t/\tau} (it has flipped sign, because now the inductor is driving the current, not resisting it).

Where the stored energy goes

Integrate the power dissipated in the resistor during the full decay from I_{0} to 0:

W_{\text{dissipated}} \;=\; \int_{0}^{\infty} I^{2}R\,dt \;=\; \int_{0}^{\infty} I_{0}^{2}R\,e^{-2t/\tau}\,dt
= \; I_{0}^{2}R\,\cdot\,\dfrac{\tau}{2} \;=\; I_{0}^{2}R\,\cdot\,\dfrac{L}{2R} \;=\; \tfrac{1}{2}L\,I_{0}^{2}

Why: the time integral of e^{-2t/\tau} from 0 to ∞ is \tau/2. Plug in \tau = L/R and the Rs cancel. The dissipated heat exactly matches the energy we started with in the inductor — no energy is lost, nothing is gained, it just converts from magnetic field to heat.

This is a beautiful consistency check. The \tfrac{1}{2}L I^{2} formula was derived from the back-EMF during current build-up. The decay calculation is independent — it only uses Ohm's law and the exponential solution — and yet it returns the same \tfrac{1}{2}LI_{0}^{2} for the stored energy. Two calculations, two agreements: the energy really is \tfrac{1}{2}LI^{2}.

Worked examples

Example 1: The defibrillator pulse

A portable defibrillator at a Bangalore cardiac clinic delivers a 200 J shock by discharging an inductor. The inductor has L = 40 mH and a peak current of I_{0} = 100 A, discharged into the patient's chest (treated as a resistance R = 50\ \Omega). Find (a) the energy stored just before discharge, (b) the time constant of the discharge, and (c) the peak inductor voltage at t = 0.

Schematic of a defibrillator inductor discharging through a patient An inductor with L = 40 mH and stored current 100 A connected in a loop to a patient's chest represented as a 50 ohm resistor. Current decays exponentially. L = 40 mH R = 50 Ω (patient) I₀ = 100 A
At $t=0$ the inductor is already charged to 100 A. The energy stored in its magnetic field dumps into the 50 Ω patient-chest load as an exponentially decaying current pulse.

Step 1. Energy stored.

U \;=\; \tfrac{1}{2}\,L\,I_{0}^{2} \;=\; \tfrac{1}{2}\,\times\,40\times 10^{-3}\,\times\,(100)^{2} \;=\; \tfrac{1}{2}\,\times\,0.040\,\times\,10^{4} \;=\; 200\ \text{J}

Why: plug directly into U = \tfrac{1}{2}LI^{2}. 40 mH is 0.04 H; 100 A squared is 10⁴ A². The product is 400, halved is 200.

Step 2. Time constant of the discharge.

\tau \;=\; \dfrac{L}{R} \;=\; \dfrac{0.040}{50} \;=\; 8\times 10^{-4}\ \text{s} \;=\; 0.8\ \text{ms}

Why: \tau = L/R. In 0.8 ms the current drops to 37% of its starting value; in about 4 ms it is essentially zero. That matches the millisecond-duration pulse a defibrillator needs to deliver.

Step 3. Peak inductor voltage at t = 0.

During the decay, V_{R} = IR at every instant, and by Kirchhoff's voltage law the inductor supplies that voltage. At t = 0:

V_{L}(0) \;=\; I_{0}\,R \;=\; 100 \times 50 \;=\; 5000\ \text{V}

Why: at the instant of discharge, the full voltage I_{0}R appears across the inductor as it drives current through the load. This is why defibrillator circuits need high-voltage insulation even though the battery inside runs on a few volts — the inductor multiplies the instantaneous voltage dramatically when switched.

Result: U = 200 J; \tau = 0.8 ms; peak V_{L} = 5 kV.

What this shows: The power delivered at the peak is I_{0}^{2}R = 10^{4} \times 50 = 5\times 10^{5} W — half a megawatt, for less than a millisecond. Over the full decay the total energy delivered is exactly 200 J, matching \tfrac{1}{2}LI_{0}^{2}. A defibrillator is an inductor discharge dressed up as a medical device.

Example 2: Growth in an LR circuit

An electromagnet at NTPC's Vindhyachal thermal plant in Madhya Pradesh is modelled as an inductor with L = 2.5 H and internal resistance R = 10\ \Omega, connected to a 60 V DC supply. The magnet is switched on at t = 0. Find (a) the steady-state current, (b) the time constant, (c) the current at t = 0.25 s, and (d) the energy stored once steady state is reached.

Plot of I(t) for an LR circuit reaching steady state Current rises from 0 to its final value of 6 A along 1 − exp(−t/τ) with τ = 0.25 s. Horizontal dashed line at 6 A marks the asymptote. 0 6 A 3.8 A τ = 0.25 s 2τ 5τ current I time (τ, 3.8 A)
The current grows exponentially from 0 towards 6 A. At one time constant, it has covered 63% of the gap — reaching 3.79 A. By $5\tau$ it is within 1% of the asymptote.

Step 1. Steady-state current. Once dI/dt = 0, the inductor contributes no voltage, so Ohm's law alone governs:

I_{\infty} \;=\; \dfrac{\varepsilon}{R} \;=\; \dfrac{60}{10} \;=\; 6\ \text{A}

Step 2. Time constant.

\tau \;=\; \dfrac{L}{R} \;=\; \dfrac{2.5}{10} \;=\; 0.25\ \text{s}

Why: a quarter of a second is slow on the scale of a light switch, which is why industrial electromagnets like this audibly hum into life instead of snapping on instantly — the current takes time to rise because of the large inductance.

Step 3. Current at t = 0.25 s (which equals \tau).

I(\tau) \;=\; I_{\infty}\,(1 - e^{-1}) \;=\; 6\,\times\,(1 - 0.368) \;=\; 6 \times 0.632 \;=\; 3.79\ \text{A}

Step 4. Energy stored at steady state.

U \;=\; \tfrac{1}{2}\,L\,I_{\infty}^{2} \;=\; \tfrac{1}{2}\,\times\,2.5\,\times\,(6)^{2} \;=\; \tfrac{1}{2}\,\times\,2.5\,\times\,36 \;=\; 45\ \text{J}

Why: the magnetic field surrounding the magnet now holds 45 joules of energy. If you were to suddenly open the supply switch with no flyback path, that 45 J would try to discharge through whatever air gap the switch opened — causing an arc. Real industrial electromagnets always have a freewheeling diode or snubber circuit across them for exactly this reason.

Result: I_{\infty} = 6 A; \tau = 0.25 s; I(\tau) = 3.79 A; U = 45 J.

What this shows: Two separate time scales — the electrical one \tau = L/R that sets how fast the current rises, and the energy integral \tfrac{1}{2}LI^{2} that sets how much energy is ultimately stored. Both scale linearly with L; both have clean physical meanings; and together they describe everything you need to know about an LR transient.

Example 3: Energy density inside a solenoid

A solenoid in an ISRO pulsed-power test facility has 4000 turns per metre, cross-section A = 25\ \text{cm}^{2}, length \ell = 50 cm, and carries I = 20 A. Find (a) the magnetic field inside, (b) the magnetic energy density, (c) the total stored energy by integrating density over volume, and verify that it matches \tfrac{1}{2}LI^{2}.

Step 1. Magnetic field.

B \;=\; \mu_{0}\,n\,I \;=\; 4\pi\times 10^{-7}\,\times\,4000\,\times\,20 \;=\; 4\pi\times 10^{-7}\times 8\times 10^{4} \approx 0.1005\ \text{T}

Why: inside an ideal long solenoid the field is uniform and B = \mu_{0}nI independent of position. About 0.1 T is a modest field — a refrigerator magnet is around 0.005 T, a typical MRI scanner is 1.5 to 3 T.

Step 2. Energy density.

u \;=\; \dfrac{B^{2}}{2\mu_{0}} \;=\; \dfrac{(0.1005)^{2}}{2\times 4\pi\times 10^{-7}} \;=\; \dfrac{0.01010}{2.513\times 10^{-6}} \approx 4020\ \text{J/m}^{3}

Step 3. Total stored energy by integrating density.

U \;=\; u\,\times\,(\text{volume}) \;=\; 4020\,\times\,(25\times 10^{-4})\,\times\,0.50 \;=\; 4020\,\times\,1.25\times 10^{-3} \approx 5.03\ \text{J}

Why: the field is uniform inside and zero outside (ideal solenoid), so the total energy is just density × interior volume. Cross-section 25 cm² = 25×10⁻⁴ m²; multiplied by length 0.5 m gives 1.25×10⁻³ m³.

Step 4. Cross-check against \tfrac{1}{2}LI^{2}. Self-inductance of the solenoid:

L \;=\; \mu_{0}\,n^{2}\,A\,\ell \;=\; 4\pi\times 10^{-7}\,\times\,(4000)^{2}\,\times\,25\times 10^{-4}\,\times\,0.50
= 4\pi\times 10^{-7}\,\times\,1.6\times 10^{7}\,\times\,1.25\times 10^{-3} \approx 0.0251\ \text{H}
U \;=\; \tfrac{1}{2}\,L\,I^{2} \;=\; \tfrac{1}{2}\,\times\,0.0251\,\times\,(20)^{2} \;=\; \tfrac{1}{2}\,\times\,0.0251\,\times\,400 \approx 5.03\ \text{J}

Why: both methods give 5.03 J, to within rounding. This is the promised consistency of the two viewpoints — the energy is either "in the inductor" (reckoned by \tfrac{1}{2}LI^{2}) or "in the volume of field" (reckoned by u = B^{2}/(2\mu_{0})), but it is the same energy.

Result: B \approx 0.10 T; u \approx 4.0 kJ/m³; total U \approx 5.0 J from both methods.

What this shows: The magnetic field itself is the repository. An ISRO pulsed-power rig that boasts "100 kJ of stored magnetic energy" is literally describing a large volume filled with a strong field, computed by integrating B^{2}/(2\mu_{0}) over the apparatus.

Common confusions

If you have followed the derivations so far and can compute both stored energy and LR transients, you have everything a JEE problem can demand. What follows is for readers who want to see the argument two levels deeper: where the energy density u = B^{2}/(2\mu_{0}) comes from at the level of Maxwell's equations, and how energy balance plays out for a general LR loop, not just a solenoid.

Magnetic energy density from Ampère's law

The derivation in the main body relied on a specific geometry (the solenoid) and then claimed the formula u = B^{2}/(2\mu_{0}) is general. Here is why. Start with Ampère's law for steady currents, \oint \vec{B}\cdot d\vec{\ell} = \mu_{0}I_{\text{enc}}.

When you build up current slowly through a circuit, the work done per unit time to push charges against the induced EMF in a small filament of the field is

\dfrac{dW}{dt} \;=\; -\varepsilon\,I \;=\; I\,\dfrac{d\Phi}{dt}

Integrating from zero current to final current, and summing over all filaments of the field, one arrives at the general result

U \;=\; \dfrac{1}{2\mu_{0}}\int_{\text{all space}} B^{2}\,dV

This integral, evaluated over the region where \vec{B} is non-zero, equals \tfrac{1}{2}LI^{2} for any shape of current loop. The key insight: the field, not the wire, is the repository. A field exists in space independent of the coil that created it; its energy exists in space too.

The LR loop with mutual inductance

Two coupled coils, coil 1 with self-inductance L_{1}, coil 2 with L_{2}, mutual inductance M, carrying currents I_{1} and I_{2}. The total magnetic energy is

U \;=\; \tfrac{1}{2}L_{1}I_{1}^{2} \;+\; \tfrac{1}{2}L_{2}I_{2}^{2} \;+\; M\,I_{1}I_{2}

The third term is the coupling energy — it represents the energy that arises because the field of coil 1 threads coil 2 and vice versa. It can be positive or negative depending on the relative winding directions.

Since U \geq 0 for all I_{1}, I_{2}, it follows (treat U as a quadratic form and require the discriminant to be non-negative) that

M^{2} \;\leq\; L_{1}L_{2}

which is the mathematical origin of the coupling coefficient k = M/\sqrt{L_{1}L_{2}} with |k| \leq 1. Every real transformer on the Indian grid — the NTPC Singrauli generator step-up transformer, the 33-to-11 kV distribution transformer on a pole outside your gali — has a coupling coefficient k close to but not equal to 1 (typically 0.95–0.98). The small shortfall from perfect coupling is called the leakage inductance, and accounting for it is the single biggest complication in designing real transformers.

Why \tau = L/R has dimensions of time

A dimensional analysis that doubles as a sanity check. Inductance L is measured in henries: 1 H is 1 V·s/A (from V = L\,dI/dt: volts divided by amps-per-second). Resistance R is ohms: 1 Ω is 1 V/A (from V = IR). So:

[L/R] \;=\; \dfrac{\text{V}\cdot\text{s}/\text{A}}{\text{V}/\text{A}} \;=\; \text{s}

The same dimensional argument works for the RC time constant \tau = RC. The pattern — a storage element divided by a dissipative element — sets the natural time scale of a first-order linear system. In mechanics the analogue is m/b (mass over linear-drag coefficient); in thermal systems it is C_{\text{thermal}}/k_{\text{thermal}} (heat capacity over thermal conductance).

The RLC circuit, briefly

If you put a capacitor in series with L and R instead of just R, the circuit equation becomes second order:

L\,\dfrac{d^{2}q}{dt^{2}} \;+\; R\,\dfrac{dq}{dt} \;+\; \dfrac{q}{C} \;=\; \varepsilon(t)

which is exactly the equation for a damped harmonic oscillator (with L as mass, R as damping, 1/C as spring constant, q as displacement). The stored energy oscillates between the inductor's magnetic field (\tfrac{1}{2}LI^{2}) and the capacitor's electric field (q^{2}/2C), slowly bleeding away through the resistor. This is the subject of LCR Series Circuit and Resonance, and every tuning circuit in every radio built since the invention of radio rests on it.

Where this leads next