Self-inductance

Switch off a large electromagnet and, for an instant, a fat blue spark can leap across the switch contacts — far hotter and more violent than anything the battery voltage alone could produce. The coil, robbed of the current that was sustaining its magnetic field, fights back. It manufactures its own emergency voltage, sometimes hundreds of volts, purely to keep the current flowing a little longer. This stubbornness has a name: self-inductance, the property by which a coil's own changing current induces an EMF in itself. The idea traces back to Michael Faraday's discovery that a changing magnetic flux induces an EMF in any loop it threads. What is easy to miss on first hearing is that a coil's magnetic field always threads its own turns as well as any neighbouring coil. So the moment the current in a coil starts to rise or fall, the coil's own flux linkage changes, and by Faraday's law the coil induces an EMF in itself — with no second coil required anywhere in sight. Heinrich Lenz's law then fixes the sign: this self-induced, or "back," EMF always opposes the very change that created it, so a coil behaves like electrical inertia, resisting any attempt to speed up or slow down the current through it. This article builds that single physical idea into a working toolkit: the definition $\phi_B = LI$ and its consequence $\varepsilon = -L\,dI/dt$ , the SI unit henry, the quoted (not derived) solenoid formula $L = \mu_0 n^2 A l$ , the energy $U = \tfrac12 LI^2$ stored in an inductor's magnetic field, and two fully worked numbers — a solenoid's self-inductance and a coil's back EMF as its current is ramped down — that show exactly how examiners expect these formulas to be used.

Key takeaways

Self-inductance is a one-coil effect. A coil's own changing current changes its own flux linkage, inducing an EMF in itself — no second coil is needed.
Definition: flux linkage is proportional to current in the same coil, $\phi_B = LI$ , where $L$ is the self-inductance, a purely geometric property of the coil.
Faraday plus the definition gives the working law: $\varepsilon = -L\dfrac{dI}{dt}$ , the self-induced (back) EMF.
Lenz's law fixes the sign: the back EMF always opposes the change in current — it fights a rising current and props up a falling one — never the current's own steady value.
SI unit is the henry (H): $1\ \text{H}$ means a current changing at $1\ \text{A/s}$ induces $1\ \text{V}$ of back EMF.
Solenoid formula (quoted, not derived at this level): $L = \mu_0 n^2 A l$ , with turns density $n = N/l$ — geometry alone decides $L$ .
Energy is stored in the magnetic field itself: $U = \tfrac12 LI^2$ , which is why doubling the current quadruples the stored energy.
Key idea: a coil's own changing current changes its own flux linkage ( $\phi_B = LI$ ), inducing a self-EMF $\varepsilon = -L\,dI/dt$ that Lenz's law forces to oppose the change, giving the coil an "electrical inertia" whose stored energy is $U = \tfrac12 LI^2$ .

The core insight

Self-inductance $L$ is the flux-linkage-per-unit-current of a single coil ( $\phi_B = LI$ ), so a changing current $I(t)$ induces a self-opposing back EMF $\varepsilon = -L\,dI/dt$ in that same coil, and the work done against this back EMF while building up the current is stored as magnetic energy $U = \tfrac12 LI^2$ .

Where it fits in the syllabus

Self-inductance sits at the end of the CBSE Class 12 "Electromagnetic Induction" chapter (Unit VI), right after Faraday's and Lenz's laws, and immediately before mutual inductance — the two named forms of inductance the syllabus lists explicitly. You are expected to define $L$ as flux linkage per unit current in the same coil, use $\varepsilon = -L\,dI/dt$ to find induced/back EMF, know the henry as its SI unit, and quote (never derive) the solenoid result $L = \mu_0 n^2 A l$ for plug-in numericals. The full geometric derivation of that solenoid formula is explicitly kept out of scope at this level, as is any AC-generator or transformer machinery — those belong to the next chapter, Alternating Current, where self-inductance reappears as reactance $X_L = \omega L$ .

The physical picture

Picture a coil of wire carrying current $I$ . That current is a source of magnetic field, and by Ampere's law the field pattern it produces is completely fixed by the coil's own current and geometry — nothing else. Some of that field's flux inevitably passes back through the coil's own turns, because the coil sits inside the very field it created. For a fixed shape and fixed number of turns, this self-flux is exactly proportional to the current driving it:

\phi_B = LI

where: $\phi_B$ is the magnetic flux linkage through the coil (weber, Wb), $I$ is the current in that same coil (ampere, A), and $L$ is the self-inductance (or coefficient of self-induction), a constant fixed entirely by the coil's geometry — number of turns, cross-sectional area, length, and any magnetic core material — never by how much current happens to be flowing at a given instant. Read physically, $L$ is "how much flux this coil manages to trap in itself for every ampere it carries" — a bigger, tighter, more-turns coil traps more flux per ampere and so has a larger $L$ . Now let the current change with time — as it always does when a switch is thrown or an AC source drives the coil. The flux $\phi_B = LI(t)$ changes in step, and Faraday's law says any changing flux linkage induces an EMF:

\varepsilon = -\frac{d\phi_B}{dt} = -L\frac{dI}{dt}

where: $\varepsilon$ is the self-induced EMF (volt, V) appearing across the coil, and $dI/dt$ is the rate at which the current is changing (A/s). The minus sign is Lenz's law made explicit: the induced EMF always drives a current that opposes the change producing it. If $I$ is increasing, $\varepsilon$ opposes the rise — it is often called the back EMF because it pushes back against whatever is trying to increase the current. If $I$ is falling, $\varepsilon$ reverses and tries to sustain the current instead of letting it die instantly — this is exactly the mechanism behind the spark you see when a large inductive circuit (like an electromagnet or a relay coil) is switched off: the coil suddenly demands a very large $\varepsilon$ to keep some current flowing as $dI/dt$ spikes to a huge negative value.

Figure

A coil carrying a current with its own magnetic field lines looping back through its own turns, and a graph beside it showing current ramping up over time with a back EMF pulse that opposes the rise

SI unit. Self-inductance is measured in the henry (H): from $\varepsilon = -L\,dI/dt$ , a coil has $L = 1\ \text{H}$ if a current changing at $1\ \text{A/s}$ induces a back EMF of exactly $1\ \text{V}$ across it. Typical laboratory coils have $L$ of a few millihenry to a few henry; a compact air-core solenoid, as the first worked example below shows, often lands in the millihenry range.

Solenoid self-inductance. For a long solenoid of $N$ turns, length $l$ , cross-sectional area $A$ , and turns density $n = N/l$ , the standard result (quoted here, not derived, as the syllabus asks) is:

L = \mu_0 n^2 A l

where: $\mu_0 = 4\pi \times 10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space. This confirms that $L$ is fixed purely by geometry: more turns packed per metre (larger $n$ , and note it enters squared), a fatter cross-section $A$ , or a longer solenoid $l$ all raise $L$ . Slipping a magnetic core such as iron inside the solenoid raises $L$ further still, because the core's much larger permeability effectively replaces $\mu_0$ in the formula — this is why practical inductors used in electronics are almost always wound on iron or ferrite cores rather than left as plain air-core coils.

Stored energy. Building up a current against a back EMF costs work, and that work does not vanish — it is stored as magnetic energy in the field around the coil, recoverable when the current is later allowed to fall:

U = \frac{1}{2}LI^2

Because $I$ appears squared, doubling the current does not double the stored energy — it quadruples it. This quadratic dependence is the single most commonly tested subtlety of this formula, and it is examined directly in Worked example 2 below.

Derivation AEnergy stored in a current-carrying inductor

The goal is to show, using nothing beyond the definition of work and the self-inductance relation $\varepsilon = -L\,dI/dt$ , that the total work done in building a current up from zero to $I$ in an inductor of self-inductance $L$ is $U = \tfrac12 LI^2$ , and that this work is stored as recoverable magnetic energy rather than dissipated. This matters because it is the energetic partner of the back-EMF idea: the back EMF is not just a nuisance that resists change, it is doing real work that ends up stored in the field, exactly the way a stretched spring stores the work done in stretching it.

Step 1. Identify the work done against the back EMF in a small time interval. At any instant, the current is $I$ and the self-induced EMF has magnitude $|\varepsilon| = L\,dI/dt$ . To push the current forward against this opposing EMF, an external source must supply electrical power $P = |\varepsilon|\,I = LI\,\dfrac{dI}{dt}$ at that instant — this is simply voltage times current, the definition of electrical power.

Step 2. Convert instantaneous power into a small amount of work. In a short time $dt$ , the work done is $dW = P\,dt = LI\,\dfrac{dI}{dt}\,dt = LI\,dI$ . Every joule of this work is supplied by whatever source (battery, generator) is forcing the current to rise, and every joule of it goes into overcoming the coil's self-opposition — none of it is lost to resistance here, since we are isolating the inductive effect alone.

Step 3. Sum this work as the current climbs from $0$ to its final value $I$ . Adding up $dW = LI\,dI$ over all the small steps from current $0$ up to current $I$ gives the total work $W = L\displaystyle\int_0^I I'\,dI' = L\left[\dfrac{I'^2}{2}\right]_0^I = \dfrac{1}{2}LI^2$ . (For a Class 12 argument this sum can equally be pictured as the area of a triangle on a graph of "opposing voltage-related quantity" against current, rising linearly from $0$ to $LI$ — the same $\tfrac12 \times \text{base} \times \text{height}$ result.)

Step 4. Interpret the result as stored, recoverable energy. This total work $U = \tfrac12 LI^2$ does not heat the coil or radiate away — it is stored in the magnetic field that the current has built up in the space around and inside the coil. Proof that it is truly stored, not dissipated, is that if the circuit is later opened slowly through a resistor, this same energy reappears as heat in that resistor as the current (and the field) collapses back to zero — energy conservation, exactly as Lenz's law demands.

Worked examples

Board level - self-inductance of an air-core solenoid

Question: A solenoid has $500$ turns, length $0.5\ \text{m}$ , and cross-sectional area $2\times10^{-3}\ \text{m}^2$ , with no magnetic core (so use $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ ). Find its self-inductance.

Step 1. Find the turns density: $n = N/l = 500/0.5 = 1000\ \text{turns/m}$ .

Step 2. Write down the formula to use: $L = \mu_0 n^2 A l$ , quoted directly since its derivation is beyond board-level scope.

Step 3. Substitute the numbers: $L = (4\pi\times10^{-7}) \times (1000)^2 \times (2\times10^{-3}) \times (0.5)$ .

Step 4. Simplify the numerical part before bringing in $\pi$ : $(1000)^2 = 10^6$ ; then $10^6 \times 2\times10^{-3} = 2\times10^{3}$ ; then $2\times10^{3} \times 0.5 = 1\times10^{3}$ .

Step 5. Multiply by $4\pi\times10^{-7}$ : $L = 4\pi\times10^{-7} \times 1\times10^{3} = 4\pi\times10^{-4} \approx 1.257\times10^{-3}\ \text{H}$ .

Answer: The self-inductance of the solenoid is approximately $1.26\times10^{-3}\ \text{H}$ , about $1.26\ \text{mH}$ .

Competitive level - back EMF and energy change as current is ramped down

Question: A coil of self-inductance $0.2\ \text{H}$ carries a current that decreases uniformly from $6\ \text{A}$ to $2\ \text{A}$ in $0.1\ \text{s}$ . Find the magnitude of the induced (back) EMF, and the energy stored in the coil at the initial and final currents. Hence find how much energy was released from the field during this fall.

Step 1. Find the rate of change of current: $\dfrac{dI}{dt} = \dfrac{2 - 6}{0.1} = \dfrac{-4}{0.1} = -40\ \text{A/s}$ , so the magnitude is $40\ \text{A/s}$ .

Step 2. Find the magnitude of the induced EMF using $|\varepsilon| = L\left|\dfrac{dI}{dt}\right| = 0.2 \times 40 = 8\ \text{V}$ .

Step 3. Find the energy stored at the initial current using $U = \tfrac12 LI^2$ : $U_i = \tfrac12 (0.2)(6)^2 = 0.1 \times 36 = 3.6\ \text{J}$ .

Step 4. Find the energy stored at the final current the same way: $U_f = \tfrac12(0.2)(2)^2 = 0.1\times 4 = 0.4\ \text{J}$ .

Step 5. Subtract to find the energy released by the collapsing field: $\Delta U = U_i - U_f = 3.6 - 0.4 = 3.2\ \text{J}$ . This $3.2\ \text{J}$ is exactly what would appear as heat in any resistor across which the coil is discharged, or partly as the spark energy if the circuit is broken abruptly.

Answer: The induced back EMF has magnitude $8\ \text{V}$ ; the stored energy falls from $3.6\ \text{J}$ to $0.4\ \text{J}$ , releasing $3.2\ \text{J}$ from the magnetic field. Timing note: this problem packs three separate formula applications ( $\varepsilon = L\,dI/dt$ , then $U=\tfrac12LI^2$ twice) into one question, which is exactly how competitive exams like JEE Main test self-inductance — budget about $2$ minutes total, and do the two energy calculations back-to-back so the $0.1 \times (\text{square})$ pattern stays fresh rather than re-deriving it twice.

Common traps

Thinking self-inductance needs two separate coils — it is a property of a single coil interacting with its own changing flux; mutual inductance is the two-coil case, and the two are easy to conflate under exam pressure.
Forgetting that the back EMF opposes the change in current, not the current itself — a steady, unchanging current (even a very large one) produces zero self-induced EMF, since $dI/dt = 0$ .
Misapplying $L = \mu_0 n^2 A l$ by substituting total turns $N$ where turns density $n = N/l$ belongs, or forgetting that $n$ is squared — both errors are extremely common and both change the answer by large factors.
Assuming energy stored is proportional to $I$ rather than $I^2$ — doubling the current quadruples the stored energy, not doubles it, as Worked example 2 shows numerically.
Dropping or mishandling the sign in $\varepsilon = -L\,dI/dt$ — for a Class 12 numerical, always compute the magnitude from $|dI/dt|$ and reason about direction separately using Lenz's law in words, rather than chasing signs through the algebra.

You should be able to…

Define self-inductance as flux linkage per unit current in the same coil, $\phi_B = LI$ , and explain in words why it depends only on geometry.
Use $\varepsilon = -L\,dI/dt$ to calculate the magnitude of the back EMF induced by a changing current, and use Lenz's law in words to reason about its direction.
State the SI unit of self-inductance (the henry) and explain what it physically means.
Quote and correctly substitute into the solenoid self-inductance formula $L = \mu_0 n^2 A l$ , being careful to use turns density $n = N/l$ , not total turns $N$ .
Use $U = \tfrac12 LI^2$ to find the magnetic energy stored in a coil at a given current, and compute the energy released as current falls between two values.
Explain why self-inductance is often called the electrical analogue of mechanical inertia.

Ask the tutor: want to work through a few more solenoid and back-EMF numericals, or see how self-inductance turns into reactance $X_L = \omega L$ in the next chapter on Alternating Current?

Why does switching off a big electromagnet sometimes cause a spark?

When you open the switch, you are forcing the current to drop from some large value to zero in a very short time, which makes $dI/dt$ enormous and negative. Since $|\varepsilon| = L|dI/dt|$ , even a modest $L$ can produce hundreds of volts for that brief instant, because the coil is desperately trying to keep the current flowing (Lenz's law). That huge momentary voltage is often enough to ionise the air across the switch gap, producing a visible spark or arc — this is also why relay and motor coils in real circuits are fitted with a "flyback diode" to give the current a safer path to decay through instead of arcing.

Is self-inductance the same thing as resistance? Both seem to oppose current.

No, and the difference is important. Resistance opposes the current itself, all the time, converting electrical energy irreversibly into heat via $I^2R$ . Self-inductance opposes only changes in the current, and the energy involved is not lost — it is stored in the magnetic field as $U = \tfrac12LI^2$ and can be recovered later when the current falls. A steady current flows through an ideal inductor with zero opposition (no back EMF), whereas the same steady current through a resistor is dissipating heat continuously.

Why do we just quote L = mu_0 n squared A l instead of deriving it, if it is on the syllabus?

CBSE deliberately keeps the derivation out of scope at Class 12 level, because deriving it rigorously needs the formula for the magnetic field inside a long solenoid combined with careful flux-linkage bookkeeping across all $N$ turns, which is more machinery than the chapter wants to spend time on. You are expected to understand what the formula tells you physically (geometry alone fixes $L$ ) and to be fluent at substituting numbers into it, exactly as in Worked example 1 — the derivation itself may appear as background reading but will not be asked as a "derive" question in a CBSE board paper.

Does the direction of current matter for how much self-inductance a coil has?

No. Self-inductance $L$ depends only on the coil's geometry and core material, never on the direction or even the magnitude of the current at any instant — reversing the current direction reverses the sign of $\phi_B$ and of any induced EMF when the current changes, but the value of $L$ itself is completely unaffected. This is exactly analogous to how a resistor's resistance $R$ does not depend on which way current happens to be flowing through it.