Self and mutual inductance

Switch off a large electromagnet and, for an instant, a spark can jump across the switch contacts even though the supply voltage was modest. The coil is fighting the change in its own current, and that fight is called inductance. Every current-carrying coil sets up a magnetic flux linked with itself; when that current changes, the coil's own changing flux induces an EMF back into itself — self-inductance. Place a second coil nearby and the same changing current can induce an EMF in that neighbour too, with no wire connecting them — mutual inductance. Together these two ideas, building directly on Faraday's and Lenz's laws, explain why inductors resist sudden switching, how energy is stored in a magnetic field, and how a transformer can pass power from one coil to another across an air gap.

The core insight

A coil's self-inductance $L$ links its own flux to its own current via $\phi = LI$ , so a changing current induces a back-EMF $\varepsilon = -L\,dI/dt$ in the same coil, while a second nearby coil experiences a mutual EMF $\varepsilon_2 = -M\,dI_1/dt$ from the same changing current, purely through shared magnetic flux.

The core idea

Self-inductance. When current $I$ flows in a coil, it sets up a magnetic flux $\phi$ linked with the coil itself, proportional to $I$ :

\phi = LI

where: $\phi$ is the flux linked with the coil (weber, Wb), $I$ is the current (A), and $L$ is the coil's self-inductance, measured in henry (H). If the current changes, the linked flux changes, and by Faraday's law an EMF is induced in the same coil, opposing the change (Lenz's law applied to the coil's own current):

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

Self-inductance is a purely geometric property — it depends on the coil's shape, size, number of turns, and core material — never on the instantaneous current or its rate of change. For a long solenoid of $N$ turns, length $l$ , and cross-sectional area $A$ (air core), $L = \mu_0 N^2 A/l$ , derived below.

Energy stored in an inductor. Pushing a current $I$ through a coil against its own back-EMF requires work, and that work does not vanish — it is stored as energy in the coil's magnetic field:

U = \frac12 LI^2

where: $U$ is the stored energy (joule, J). This is the magnetic analogue of $\frac12 CV^2$ for a charged capacitor: a switched-off inductor releases this energy back into the circuit, which is exactly why breaking an inductive circuit abruptly can produce a spark.

Mutual inductance. When two coils sit near each other, a current $I_1$ in coil 1 produces a flux $\phi_2$ linked with coil 2:

\phi_2 = MI_1

where: $M$ is the mutual inductance of the pair, also measured in henry. $M$ is symmetric — the flux linked with coil 1 due to a current in coil 2 involves the same $M$ — so it is a property of the pair, not of either coil alone. If $I_1$ varies, an EMF is induced in coil 2 with no electrical connection between the coils, only a shared magnetic flux:

\varepsilon_2 = -M\frac{dI_1}{dt}

For two coaxial solenoids of the same length $l$ and area $A$ , wound with $N_1$ and $N_2$ turns (one inside the other, so their fluxes fully link), $M = \mu_0 N_1 N_2 A/l$ — note both turn counts appear, unlike the $N^2$ of self-inductance for a single coil. Mutual inductance depends on the geometry of both coils and how tightly their fluxes couple, and it is the physical basis of a transformer.

Figure

Two coaxial solenoids one inside the other, showing self flux looping back on the inner coil and mutual flux also linking the outer coil

Derivation ASelf-inductance of a long solenoid

The goal is to derive the self-inductance formula for a long solenoid from first principles, using only the solenoid's field formula and the definition of flux linkage, so that a bare geometric description of the coil — its turns, length, and area — directly yields a number in henry without any extra assumption about the current, since $L$ must come out independent of $I$ .

Step 1. Start from the magnetic field inside a long, tightly wound solenoid carrying current $I$ : $B = \mu_0 n I$ , where $n = N/l$ is the number of turns per unit length. This result follows from Ampere's law applied to the ideal solenoid and is taken as known here.

Step 2. Find the flux through one turn. Each turn encloses the full cross-sectional area $A$ , and the field is uniform inside, so the flux through a single turn is $\phi_{\text{turn}} = BA = \mu_0 n I A$ .

Step 3. Find the total flux linkage. All $N$ turns are threaded by essentially the same flux (the solenoid is long, so end effects are negligible), so the total flux linkage is $N\phi_{\text{turn}} = N\mu_0 n I A = \mu_0 N^2 I A/l$ , using $n = N/l$ .

Step 4. Match to the defining relation $\phi_{\text{total}} = LI$ and read off $L$ . Comparing $LI = \mu_0 N^2 I A/l$ , the current $I$ cancels from both sides, confirming $L$ depends only on geometry:

L = \frac{\mu_0 N^2 A}{l}

The identical logic, applied to the flux one solenoid's field produces through a second coaxial coil of $N_2$ turns occupying the same volume, gives $M = \mu_0 N_1 N_2 A/l$ — the only change is that the flux is now read off in the other coil's turns instead of the source coil's own turns.

Worked example

A solenoid has 800 turns, length 40 cm, and cross-sectional area $5\times10^{-4}\ \text{m}^2$ (air core). (a) Find its self-inductance. (b) If the current through it changes from 2 A to 5 A in 0.5 s, find the induced EMF. (c) Find the energy stored when the current is 5 A.

Step 1. Self-inductance: $L = \dfrac{\mu_0 N^2 A}{l} = \dfrac{(4\pi\times10^{-7})(800)^2(5\times10^{-4})}{0.40}$ . The numerator is $4\pi\times10^{-7}\times 6.4\times10^{5}\times5\times10^{-4} = 4\pi\times10^{-7}\times320 = 4.02\times10^{-4}$ , so $L = \dfrac{4.02\times10^{-4}}{0.40} = 1.0\times10^{-3}\ \text{H} = 1.0\ \text{mH}$ .

Step 2. Induced EMF from the average rate of change of current: $\varepsilon = L\dfrac{\Delta I}{\Delta t} = 1.0\times10^{-3}\times\dfrac{5-2}{0.5} = 1.0\times10^{-3}\times 6 = 6.0\ \text{mV}$ .

Step 3. Energy stored at $I = 5\ \text{A}$ : $U = \dfrac12 LI^2 = \dfrac12\times1.0\times10^{-3}\times(5)^2 = 1.25\times10^{-2}\ \text{J}$ . Answers: $L \approx 1.0\ \text{mH}$ , $\varepsilon = 6.0\ \text{mV}$ , $U = 1.25\times10^{-2}\ \text{J}$ .

Competitive stretch

If this same solenoid is wound as the primary of two coaxial coils, same length and area, with a 200-turn secondary wound over it, its mutual inductance is $M = \mu_0 N_1 N_2 A/l = (4\pi\times10^{-7})(800)(200)(5\times10^{-4})/0.40 \approx 2.5\times10^{-4}\ \text{H}$ ; a primary current changing at the same $6\ \text{A/s}$ as above would induce a secondary EMF $\varepsilon_2 = M\,dI_1/dt \approx 1.5\ \text{mV}$ — smaller than the self-induced EMF because the coupling is weaker with fewer secondary turns.

Common traps

Mixing up which coil's turns and current go where in $M = \mu_0 N_1 N_2 A/l$ — both $N_1$ and $N_2$ appear, unlike self-inductance which uses $N^2$ for a single coil.
Forgetting that $L$ and $M$ are geometric constants (turns, area, length, core material) that do NOT depend on the instantaneous current or its rate of change.
Dropping units or powers of ten: inductance is in henry, and typical solenoid values in exam problems land in the millihenry range — a bare answer like "1000" without checking the exponent is a red flag.
Thinking mutual inductance requires the coils to be wired together — the coupling is purely magnetic through shared flux, and the two coils remain electrically isolated circuits.

Ask the tutor: if you'd like to see how this mutual-inductance relation directly sets up the transformer turns-ratio formula covered next in this unit, just ask.

Why does an inductor resist a sudden change in current instead of a sudden change in voltage?

An inductor's defining behaviour comes from $\varepsilon = -L\,dI/dt$ : the induced EMF depends on how fast the current changes, not on the current's value or the voltage across it. If you try to change the current abruptly (open a switch suddenly), $dI/dt$ becomes enormous for an instant, producing a large back-EMF — which is exactly why breaking an inductive circuit can spark across the switch. The voltage across an inductor can jump instantly; the current through it cannot.

Is mutual inductance always smaller than the self-inductance of either coil?

Not necessarily, but in practice the coupling between two separate coils is rarely as complete as a coil's flux is linked with itself, so $M$ is often smaller than $\sqrt{L_1 L_2}$ would allow. For two coaxial solenoids of the same length and area, tightly wound one inside the other, $M = \mu_0 N_1 N_2 A/l$ can actually be comparable to or larger than one coil's self-inductance if that coil has fewer turns than the product $N_1N_2$ suggests — the comparison depends entirely on the specific turn counts, not on a general rule that self-inductance always wins.

How is mutual inductance related to a transformer?

A transformer is exactly two coils sharing mutual inductance $M$ , wound on a common (usually iron) core so their flux couples almost completely. An alternating current in the primary produces a changing flux that the secondary shares, inducing an EMF $\varepsilon_2 = -M\,dI_1/dt$ in the secondary with no direct electrical connection between the windings. The turns-ratio relation you meet later in this unit, $V_2/V_1 = N_2/N_1$ , is the practical, ratio-based way this same mutual-flux coupling is used to step voltage up or down.

Self and mutual inductance

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads