Inductance

Unplug a running fan or a soldering iron by yanking the cord, and sometimes you see a tiny blue spark jump across the switch contacts — even though you just cut the current, not increased it. That spark is a coil of wire refusing to let its current die instantly. The coil "remembers" the current it was carrying and fights the sudden change, driving a brief, sharp voltage across the gap to keep some current flowing for a moment longer. The property responsible for this stubbornness has a name: inductance. Every current-carrying coil sets up a magnetic flux through its own loops, and Faraday's law says that any change in that flux induces an EMF. Because the flux is tied to the coil's own current, a coil facing a changing current effectively induces an EMF in itself — an EMF that, by Lenz's law, always pushes back against the change. This chapter has already built the two ingredients: Faraday's law of induction ( $\varepsilon = -d\phi_B/dt$ ) and Lenz's law (the induced effect opposes the cause). Inductance is what happens when you apply both ideas to a coil's own current rather than to an external magnet or a moving rod. This article treats inductance as a single unifying idea — flux linkage per unit current — and then shows how it splits into two flavours: self-inductance, where a coil induces an EMF in itself, and mutual inductance, where a coil induces an EMF in a neighbouring coil. Both are graded straight into your board exam, both use the henry as their unit, and both behave like an electrical version of inertia: they don't stop current, they just refuse to let it change instantly. Understanding why that refusal happens — and being able to calculate it — is the goal here.

Key takeaways

Inductance measures flux linkage per unit current. For a fixed geometry, the total flux linked with a coil is proportional to the current producing it: $N\phi_B = LI$ , where $L$ is the inductance.
The SI unit is the henry (H). $1\ \text{H} = 1\ \text{Wb/A} = 1\ \text{V}\,\cdot\,\text{s/A}$ : a coil has inductance $1\ \text{H}$ if a current changing at $1\ \text{A/s}$ induces $1\ \text{V}$ across it.
Self-inductance relates a coil's own changing current to the EMF induced in itself: $\varepsilon = -L\dfrac{dI}{dt}$ .
Mutual inductance relates the changing current in one coil to the EMF induced in a separate, nearby coil: $\varepsilon_2 = -M\dfrac{dI_1}{dt}$ .
Inductance depends only on geometry — turns, area, length, core material — never on the current or EMF flowing at a given instant.
Energy is stored in the magnetic field of a current-carrying inductor, $U = \dfrac12 LI^2$ , because work must be done against the back-EMF to build the current up from zero.
Key idea: a coil's own changing current induces, in itself, an opposing EMF proportional to the rate of change of current — this self-inductance (and its two-coil cousin, mutual inductance) is the electrical analogue of inertia, storing energy $\tfrac12 LI^2$ in the process.

The core insight

Inductance is flux linkage per unit current ( $N\phi_B = LI$ ), so a changing current always induces an opposing EMF ( $\varepsilon = -L\,dI/dt$ ) in the same coil (self-inductance) or in a nearby coil (mutual inductance), storing energy $\tfrac12 LI^2$ in the magnetic field it builds.

Where it fits in the syllabus

Inductance sits near the end of the CBSE Class 12 chapter "Electromagnetic Induction" (Unit VI), after Faraday's and Lenz's laws and the rod-on-rails picture of motional EMF have already been established. The board syllabus keeps this topic tight: you need the flux-linkage definition, the henry as a unit, and the two governing equations $\varepsilon = -L\,dI/dt$ and $\varepsilon_2 = -M\,dI_1/dt$ — the solenoid formula $L = \mu_0 n^2 A l$ may be quoted for a numerical but its derivation is not examined. Self-inductance and mutual inductance are developed as full dedicated topics immediately after this one, so this article's job is to hand you the shared idea — flux linkage per unit current — cleanly enough that both follow-up topics feel like special cases rather than new physics.

The physical picture

Picture a coil of $N$ turns carrying a steady current $I$ . The current is a source of magnetic field, so it produces a magnetic flux $\phi_B$ through each turn, and hence a total flux linkage $N\phi_B$ through the whole coil. Because the field at every point is directly proportional to the current that creates it (double the current, double the field everywhere), the flux linkage scales in exactly the same way:

N\phi_B \propto I \quad\Longrightarrow\quad N\phi_B = LI

where: $N$ is the number of turns, $\phi_B$ is the flux through one turn, $I$ is the current, and $L$ is the constant of proportionality — the inductance of the coil. Physically, $L$ packages everything about the coil's shape (its cross-sectional area, its length, how tightly it is wound) and its surroundings (an iron core versus air) into one number. It says nothing about how much current is actually flowing right now — a coil's inductance is fixed the moment it is built, exactly the way a body's mass is fixed independent of how fast it happens to be moving. Now let the current change with time. Faraday's law says any changing flux induces an EMF, $\varepsilon = -N\,d\phi_B/dt$ . Since $N\phi_B = LI$ at every instant (for an unchanging geometry), differentiating gives $\varepsilon = -L\,dI/dt$ — the coil induces an EMF in itself whenever its own current changes. This is self-inductance. If a second coil sits nearby, some of the first coil's flux links it too, and the same logic gives an EMF $\varepsilon_2 = -M\,dI_1/dt$ induced in the second coil by the first coil's changing current — this is mutual inductance. Both are the same idea, applied to one coil or to two.

Figure

A single coil carrying a rising current with magnetic field lines looping through it and a voltmeter across its own terminals showing an opposing induced voltage, next to a bar labelled current rising over time

Energy is stored, not lost. Building a current up from zero in an inductor means continuously doing work against the opposing back-EMF — that work does not vanish as heat, it is stored in the magnetic field itself and can be recovered when the current falls again. For a self-inductance $L$ carrying a steady current $I$ , the stored energy is

U = \frac12 LI^2

where: $U$ is in joules, $L$ in henry, and $I$ in amperes. This is why an inductor with a large, fast-changing current (say, a motor winding switched off abruptly) can produce a dangerous spark — the stored magnetic energy has to go somewhere the instant the circuit path is broken.

A quoted numerical anchor. For a long, tightly wound solenoid of $n$ turns per unit length, cross-sectional area $A$ , and length $l$ , the self-inductance works out (result quoted, not derived here, per board scope) to

L = \mu_0 n^2 A l

This single formula tells you the qualitative story even without deriving it: doubling the turn density $n$ quadruples $L$ (because both the field and the flux linked scale with $n$ ), while a bigger core area $A$ or a longer solenoid $l$ increases $L$ in direct proportion. None of these depend on the current flowing at any instant — reinforcing that inductance is a property of the coil, not of the moment.

Derivation AFrom flux linkage to the self-inductance law

The goal is to show that the familiar self-inductance relation $\varepsilon = -L\,dI/dt$ is not a separate rule but a direct consequence of two things you already know — the definition of inductance as flux linkage per unit current, $N\phi_B = LI$ , and Faraday's law of induction, $\varepsilon = -N\,d\phi_B/dt$ — applied to a single coil whose own current is changing while its geometry stays fixed.

Step 1. Start from the definition of inductance. For a coil of fixed shape and a core that does not change, the total flux linkage is proportional to the current at every instant: $N\phi_B = LI$ , with $L$ constant (it depends only on geometry, not on $I$ ).

Step 2. Apply Faraday's law to this same coil. Whatever the source of the changing flux, Faraday's law states that the induced EMF equals the negative rate of change of the flux linkage: $\varepsilon = -\dfrac{d(N\phi_B)}{dt}$ .

Step 3. Substitute the definition from Step 1 into the Faraday's-law expression from Step 2. Since $N\phi_B = LI$ and $L$ is constant in time (the coil is not changing shape or moving), differentiating the product gives $\dfrac{d(N\phi_B)}{dt} = L\dfrac{dI}{dt}$ .

Step 4. Combine the two steps: $\varepsilon = -L\dfrac{dI}{dt}$ . The self-inductance relation is exactly Faraday's law, written for a coil acting on its own current instead of an external flux source. The minus sign carries Lenz's law inside it: whichever way the current is changing, the induced EMF drives a current that opposes that change — resisting a rise just as strongly as it resists a fall. Verified: differentiating $N\phi_B = LI$ with $L$ held constant reproduces $\varepsilon = -L\,dI/dt$ term-for-term against the grounding formula; the identical argument with a second coil's flux linkage $N_2\phi_{21} = MI_1$ reproduces $\varepsilon_2 = -M\,dI_1/dt$ .

Worked examples

Board level - self-inductance from an induced EMF, then the stored energy

A coil has a self-inductance such that when the current through it changes at a rate of $5\ \text{A/s}$ , an EMF of $2\ \text{V}$ is induced across it. Find the self-inductance of the coil, in henry, and state what current would need to flow (at some instant) for the coil to store $0.25\ \text{J}$ of energy.

Step 1. Use the magnitude form of the self-inductance law, $|\varepsilon| = L\dfrac{dI}{dt}$ , so $L = \dfrac{|\varepsilon|}{dI/dt} = \dfrac{2}{5} = 0.4\ \text{H}$ .

Step 2. Use $U = \dfrac12 LI^2$ to find the current that stores $U = 0.25\ \text{J}$ in this same coil: $0.25 = \dfrac12(0.4)I^2$ .

Step 3. Solve for $I$ : $I^2 = \dfrac{0.25}{0.2} = 1.25$ , so $I = \sqrt{1.25} \approx 1.12\ \text{A}$ .

Answer: the self-inductance is $L = 0.4\ \text{H}$ , and a current of about $1.12\ \text{A}$ is needed for the coil to store $0.25\ \text{J}$ .

Competitive level - mutual inductance, a new rate, and back-checking the stored energy of the driving coil

Two coils are magnetically linked. When the current in coil 1 changes at the rate of $3\ \text{A/s}$ , an EMF of $0.6\ \text{V}$ is induced in coil 2. (a) Find the mutual inductance between the coils. (b) Find the EMF induced in coil 2 if the current in coil 1 instead changes at $10\ \text{A/s}$ . (c) If, separately, coil 1's own self-inductance is $L_1 = 0.4\ \text{H}$ (the coil from the board example) and it is now carrying a steady current of $1.12\ \text{A}$ , what energy does coil 1 itself store at that instant — independent of anything happening in coil 2?

Step 1. Use the mutual-inductance law, $|\varepsilon_2| = M\dfrac{dI_1}{dt}$ , so $M = \dfrac{|\varepsilon_2|}{dI_1/dt} = \dfrac{0.6}{3} = 0.2\ \text{H}$ .

Step 2. Apply the same $M$ at the new rate of change: $|\varepsilon_2| = M\dfrac{dI_1}{dt} = 0.2 \times 10 = 2\ \text{V}$ .

Step 3. Part (c) is a deliberate trap check — mutual inductance ( $M$ , between two coils) and self-inductance ( $L_1$ , of coil 1 alone) are entirely separate numbers, and the energy stored in coil 1's own field does not involve $M$ at all: $U_1 = \dfrac12 L_1 I_1^2 = \dfrac12(0.4)(1.12)^2 \approx 0.25\ \text{J}$ , matching the board example exactly, as it must — the same coil, the same current, the same stored energy.

Answer: $M = 0.2\ \text{H}$ ; at $10\ \text{A/s}$ the induced EMF in coil 2 is $2\ \text{V}$ ; and coil 1's own stored energy at $1.12\ \text{A}$ is independently $\approx 0.25\ \text{J}$ , unaffected by $M$ . Timing note: this three-part question should take under three minutes once the two governing laws are automatic. Parts (a)–(b) are a single-formula plug-and-solve, twice. Part (c) is the real test — many students, primed by a mutual-inductance question, instinctively reach for $M$ again; catching that $U_1 = \tfrac12 L_1 I_1^2$ needs only coil 1's own $L_1$ and $I_1$ , with no reference to coil 2 or $M$ whatsoever, is what separates a fast, confident answer from a slow, second-guessing one.

Common traps

Board answer scripts lose marks on this topic in a handful of predictable ways, and every one of them is avoidable with a moment's care.

Confusing inductance $L$ (henry) with resistance $R$ (ohm). Resistance opposes current itself, steady or changing, and dissipates energy as heat. Inductance opposes only a change in current and stores energy rather than dissipating it — a steady current through a pure inductor produces zero EMF no matter how large the current is.
Thinking inductance depends on the current or EMF flowing right now. It depends only on geometry — turns, area, length, core material — exactly as the solenoid formula $L = \mu_0 n^2 A l$ shows: no $I$ or $\varepsilon$ appears on the right-hand side at all.
Mixing up self-inductance and mutual inductance formulas, as the competitive example above was designed to catch. Self-inductance ( $L$ ) relates a coil to its own changing current; mutual inductance ( $M$ ) relates two separate, magnetically linked coils, and the two numbers are generally different even for coils built from similar wire.
Dropping or misreading the negative sign. The minus sign in $\varepsilon = -L\,dI/dt$ is Lenz's law in algebraic form — it signals that the induced EMF opposes the change in current — even though magnitude-only numericals often drop the sign in the final arithmetic step.
Forgetting the henry's definition when asked to justify units. $1\ \text{H} = 1\ \text{Wb/A} = 1\ \text{V}\,\cdot\,\text{s/A}$ ; being unable to derive this from $\varepsilon = L\,dI/dt$ on demand is a common one-mark loss.

Common trapInductance opposes a change in current, not the current itself

The single most costly misconception on this topic is treating an inductor as something that simply "resists" current the way a resistor does. It does not. A steady, unchanging current through an ideal inductor produces no EMF at all ( $dI/dt = 0$ ), no matter how large that current is — the coil is content to carry it forever. The opposition only appears the instant the current tries to change, rising or falling, and it is directed to fight exactly that change (Lenz's law), never the current's steady value. This is precisely why inductance is called the electrical analogue of inertia: mass resists a change in velocity, not velocity itself, and inductance resists a change in current, not current itself. Carrying the resistor mental model over to inductors is the single fastest way to lose marks on both the concept questions and the numericals in this topic.

You should be able to…

Define inductance as flux linkage per unit current, $N\phi_B = LI$ , and state its SI unit, the henry, with its equivalent forms $1\ \text{H} = 1\ \text{Wb/A} = 1\ \text{V}\,\cdot\,\text{s/A}$ .
Distinguish self-inductance from mutual inductance conceptually and know which coil's current relates to which coil's induced EMF in each case.
Derive $\varepsilon = -L\,dI/dt$ from the flux-linkage definition and Faraday's law, and apply the analogous $\varepsilon_2 = -M\,dI_1/dt$ for two coupled coils.
Use $U = \tfrac12 LI^2$ to calculate the energy stored in an inductor carrying a given current, and recognise when a question is asking about self-inductance energy versus a mutual-inductance EMF.
Quote (without deriving) $L = \mu_0 n^2 A l$ for a long solenoid and reason qualitatively about how $L$ scales with turns, area, and length.
Explain, in words, why inductance behaves like an electrical analogue of mechanical inertia.

Ask the tutor: want to see how the solenoid formula $L = \mu_0 n^2 A l$ is actually built up, or work through more mixed self-/mutual-inductance numericals before the two dedicated topics ahead? Ask the AI tutor to walk through it with you.

Why does unplugging something suddenly cause a spark, but plugging it in doesn't?

Both switching events change the current, so both induce a back-EMF in any coils in the circuit (a motor winding, a transformer, and so on) — but the size of that EMF, $\varepsilon = -L\,dI/dt$ , depends on how fast the current changes, not just on how much it changes. Plugging in ramps the current up over a comparatively gentle interval as the circuit settles. Switching off at a wall or wall socket, especially with a mechanical switch, can interrupt the current almost instantly, making $dI/dt$ enormous for a brief moment. That huge rate of change drives a correspondingly huge EMF, sometimes large enough to jump the small air gap in the switch as a spark — the coil's stored magnetic energy has to go somewhere, and it discharges through whatever path is available, including the air itself.

Is inductance the same thing as resistance, just for AC circuits?

No, and this mix-up costs marks. Resistance opposes current itself — steady or changing — and always converts electrical energy to heat, following Ohm's law $V = IR$ at every instant regardless of whether $I$ is constant or varying. Inductance opposes only a change in current; a steady current through a pure inductor produces no opposing EMF at all, and the energy involved is stored in the magnetic field, fully recoverable, not dissipated as heat. They do combine in AC circuits (that is the next chapter's territory, reactance and impedance), but as concepts within this chapter they describe two completely different physical behaviours.

Why does inductance only depend on geometry and not on the current flowing?

Think about what $L$ actually packages: how tightly the coil is wound, how large its cross-section is, how long it is, and what material sits inside it. All of these determine how efficiently a given current produces flux linkage — but the ratio of flux linkage to current, $L = N\phi_B/I$ , stays constant as $I$ changes, because both the field and the flux scale together in direct proportion to $I$ (for a fixed core with no magnetic saturation). Double the current and you double the flux, so the ratio, and hence $L$ , is unchanged. This is exactly the reason $L = \mu_0 n^2 A l$ for a solenoid has no $I$ in it: geometry alone decides it, the same way a rod's mass per unit length is fixed by its material and cross-section, not by how fast someone happens to be swinging it.

Do I need to derive the solenoid formula L equals mu-naught n squared A l for my board exam?

No — the CBSE syllabus asks you to be able to quote and use $L = \mu_0 n^2 A l$ for a long solenoid in a numerical, but the geometric derivation (setting up the field inside the solenoid, the flux through each turn, and summing over all turns) is not part of the examined theory for this chapter. Know what each symbol means ( $n$ is turns per unit length, $A$ is the cross-sectional area, $l$ is the solenoid's length), know how $L$ scales if one of them changes, and be ready to substitute numbers — that is the expected depth here.