Mutual Inductance

Set your phone on a charging pad tonight and watch energy cross a plastic gap with no metal touching metal. Inside the pad, a coil carries a rapidly changing current; inside the phone, a second coil — connected to the first by nothing whatsoever — develops an emf and charges the battery. Two circuits, zero contact, energy flowing between them through empty space. The bridge is magnetic flux, and the number that prices the bridge is this module's subject. The idea is Self-Inductance turned outward. There, a coil's changing current induced an emf in itself, because its own flux threaded its own turns. But field lines do not stop at the coil that made them — they sprawl through space, and some fraction of them will thread any second coil in the neighbourhood. Changing current in coil 1 then means changing flux through coil 2, and Faraday's Law of Induction demands an emf in a circuit the first coil has never touched. The bookkeeping constant is the mutual inductance $M$ : how many webers coil 1's current threads through coil 2, per ampere. And $M$ hides a small miracle called reciprocity — the coupling works out identical in both directions, however lopsided the two coils are — which this article will make you see rather than memorize. For CBSE, ISC, JEE-Main, JEE-Advanced and NEET, mutual induction is a definition-plus-formula topic with one flagship derivation: the mutual inductance of two coaxial solenoids, which ISC names verbatim in its syllabus. It is also the doorway to the most consequential machine of the electrical age — the transformer, waiting in the Alternating Current chapter — so the effort invested here pays compound interest two chapters running.

Key takeaways

Two circuits couple through shared flux: current $I_1$ in coil 1 threads flux linkage $N_2\Phi_2 = M I_1$ through coil 2. The constant $M$ — the mutual inductance — is pure geometry: sizes, separation, orientation, core.
Change the current and the coupling goes live: $\varepsilon_2 = -M\,\dfrac{\Delta I_1}{\Delta t}$ — an emf in a circuit nothing touches. Unit: the henry, same as $L$ .
Reciprocity: $M_{12} = M_{21}$ , always — flux through 2 per ampere in 1 equals flux through 1 per ampere in 2, even for wildly mismatched coils. Compute the easy direction; the theorem hands you the hard one free.
The flagship geometry: two coaxial solenoids share $M = \mu_0 n_1 n_2 A \ell$ — with $A$ the inner (shared) area, the classic trap.
Coupling is an engineering dial: coaxial and close (or iron-linked) → strong; far apart or perpendicular → weak to zero. The transformer lives at the strong end; that story is the AC chapter's.
Key idea: mutual inductance $M = N_2\Phi_2/I_1$ prices the flux one circuit threads through another — a geometry-only constant, identical in both directions, that turns changing current here into emf there.

The core insight

Mutual inductance $M = N_2\Phi_2/I_1$ measures how much flux linkage one circuit's current establishes in another; it is fixed by geometry alone, obeys the exact reciprocity $M_{12} = M_{21}$ , and converts changing current in one circuit into emf in the other: $\varepsilon_2 = -M\,\Delta I_1/\Delta t$ .

Where this lives today

Mutual inductance is the grid's load-bearing constant. Between any power station and your desk lamp, the electricity has crossed four or five transformers — pairs of windings on shared cores, coupled by exactly equation (2), stepping voltage up for the long haul and down for the wall socket. That machine's own module, The Transformer, waits in the Alternating Current chapter with the turns-ratio machinery; what you have built here is its physical heart: changing current on one side, flux bridge in the middle, emf on the other, no contact anywhere. The same physics at deliberately weak coupling is the wireless-charging economy — pads and phones, electric toothbrushes sealed in plastic (no exposed contacts near a wet sink — the safety case for coupling without touching), and medical implants: a cochlear implant's processor powers the electronics inside the skull through intact skin, the body itself serving as the air gap. Engineering also spends effort driving $M$ down. Adjacent cables in a loom, neighbouring traces on a circuit board, parallel telephone pairs — every pair of nearby circuits is a mutual inductor, and unwanted coupling is "crosstalk": your signal leaking into my wire. Twisted-pair cable, the workhorse of every Ethernet run, defeats it with geometry — each twist flips the orientation, alternating the sign of $M$ segment by segment so the couplings cancel down the line. From gigawatt substations to the twist in a network cable, the design space is one constant, dialled from nearly-all-flux-shared to nearly-zero — and the dial is the geometry you can now price.

The physical picture

Lay two loops of wire on a bench, a hand-span apart, connected to nothing shared. Drive a current through the first. Its field sprawls outward — the looping patterns you learned to sketch in E4 — and some of those field lines happen to pass through the second loop's opening. What fraction? That depends on everything geometric: how big each loop is, how far apart they sit, whether they face each other squarely or edge-on. Freeze the geometry and the fraction is fixed; double the current and the threaded flux doubles with it, because every field formula is proportional to the current that sources it. So the flux through loop 2 is (a geometric constant) × $I_1$ — and that constant is $M$ .

Figure

Two circuits joined only by field — the fraction of flux caught by the second loop defines M

Static coupling is invisible — steady $I_1$ , steady flux, nothing induced. The bridge goes live only when the current changes: then the threaded flux changes, and Faraday's Law of Induction lights up loop 2 with an emf proportional to the rate. That conditional is the transformer's deepest design fact, two chapters early: mutual induction transfers nothing on DC. It is machinery built entirely out of change. How do you make $M$ big? Steal Faraday's own trick: make the two coils share their flux as completely as possible. Wind them on a common iron ring — the famous ring you met in Faraday's Law of Induction — or slide one solenoid snugly inside another, and nearly every line that threads one coil threads both. How do you make $M$ small? Distance and angle: pull the coils apart and the sprawling field dilutes fast; turn one perpendicular and lines that enter its opening on one side exit on the other, netting zero. Between the extremes lies an engineering dial that runs from the deliberately-strong coupling inside every charger to the deliberately-weak "crosstalk" spacing between adjacent cables.

The definition, the emf, and reciprocity

The defining relation prices the bridge. With coil 2 having $N_2$ turns, each threaded by flux $\Phi_2$ from coil 1's current:

N_2\Phi_2 = M\,I_1 \qquad (1)

where: $N_2\Phi_2$ is the flux linkage of coil 2 (the counting rule from Magnetic Flux), $I_1$ the source current, and $M$ the mutual inductance, in henries — the same unit as self-inductance, because it is the same species of quantity: webers of linkage per ampere of cause. Read $M$ as a question about hardware: "if I put one ampere in coil 1, how many webers land in coil 2?" For two loops on a bench, the answer is tiny — nanohenries. For solenoids nested snugly, microhenries to millihenries. For windings sharing an iron core, henries. The number is the geometry. Change the source current and Faraday's Law of Induction converts equation (1) into the working tool:

\varepsilon_2 = -\frac{N_2\,\Delta\Phi_2}{\Delta t} = -M\,\frac{\Delta I_1}{\Delta t} \qquad (2)

where: $\varepsilon_2$ is the emf induced around coil 2 (average over the interval; instantaneous reads the slope of the $I_1$ – $t$ tangent), and the minus sign is Lenz's Law doing its usual honest work — the induced current in coil 2, if the circuit is closed, flows so as to oppose the change in the flux arriving from coil 1. Note what drives $\varepsilon_2$ : the rate $\Delta I_1/\Delta t$ , not $I_1$ itself. A thousand steady amperes induce nothing; one ampere reversed in a microsecond induces plenty. And the roles commute: if instead coil 2's current changes, coil 1 receives $\varepsilon_1 = -M\,\Delta I_2/\Delta t$ with the same $M$ — which is the reciprocity claim, about to be paid for. Reciprocity deserves a moment of honest astonishment before it is used. Take a tiny 80-turn coil sitting inside a huge laboratory solenoid. Direction A: the solenoid drives, the little coil collects — easy, because the solenoid's interior field is uniform and known. Direction B: the little coil drives, the solenoid collects — a nightmare, because the little coil's field is a sprawling, non-uniform pattern that threads each of the solenoid's thousands of turns differently. The theorem $M_{12} = M_{21}$ says the two answers are equal before you start. It is a deep structural fact of magnetism (first written down by Franz Neumann in 1845, in a formula whose two-coil symmetry makes the equality visible at a glance), and at this level you should treat it as a licensed tool: always compute $M$ in the direction where the field is simple, then use it in whichever direction the question asks. The derivation below runs on exactly that licence.

Derivation AThe mutual inductance of two coaxial solenoids

The goal is the flagship formula $M = \mu_0 n_1 n_2 A \ell$ for two long solenoids wound one over the other — the geometry ISC names in its syllabus and every board's numericals prefer. The strategy is the definition run in the easy direction: let the outer solenoid drive (its interior field is the one formula we own), collect the flux linkage in the inner one, divide by the current. Watch two things: which solenoid's area enters (the shared flux lives only inside the inner bore — the classic trap), and how reciprocity then hands us the reverse direction, which would otherwise be a hard calculation about an escaping field.

Step 1. The setup: an inner solenoid of $n_2$ turns per metre and cross-sectional area $A$ , wound inside (or under) an outer solenoid of $n_1$ turns per metre, both of common length $\ell$ , sharing an axis. Drive the outer one with current $I_1$ . Deep inside, its field is uniform and axial: $B_1 = \mu_0 n_1 I_1$ — the flagship of Solenoids & Toroids, cited as a tool.

Step 2. Flux through one turn of the inner solenoid. The inner turn has area $A$ , squarely facing the field:

\Phi_2 = B_1 A = \mu_0 n_1 I_1 A.

Why the INNER area: the outer solenoid's field fills its whole bore, but a turn of the inner coil can only catch the flux passing through its own opening — area $A$ , the smaller one. If instead the small solenoid drove, its field would fill only its own bore, and the big coil's turns could still only catch flux where field exists — again area $A$ . Either way around, the shared area is the smaller one. Using the outer area is the single most common error in this calculation.

Step 3. Count the linkage. The inner solenoid has $N_2 = n_2\ell$ turns, each threaded by $\Phi_2$ :

N_2\Phi_2 = (n_2\ell)(\mu_0 n_1 I_1 A) = \mu_0 n_1 n_2 A \ell\; I_1.

Step 4. Divide by the driving current, per definition (1):

M = \frac{N_2\Phi_2}{I_1} = \mu_0\, n_1 n_2\, A\, \ell \qquad (3)

The current has cancelled — $M$ is geometry, as promised. And now collect reciprocity's dividend: driving the inner solenoid with 1 A would send its field leaking out of its open ends through the outer winding in some complicated pattern — a calculation beyond any exam — yet the total linkage it establishes in the outer solenoid is exactly $M$ webers, by $M_{12} = M_{21}$ . One easy calculation, both directions banked. Verified: units — $\mathrm{(T{\cdot}m/A)\,m^{-1}\,m^{-1}\,m^2\,m} = \mathrm{T{\cdot}m^2/A} = \mathrm{Wb/A} = \mathrm{H}$ ✓. Numbers — $n_1 = 1000$ /m, $n_2 = 2000$ /m, inner radius 1.0 cm ( $A = 3.14\times10^{-4}\ \mathrm{m^2}$ ), $\ell = 0.40$ m: $M = (4\pi\times10^{-7})(1000)(2000)(3.14\times10^{-4})(0.40) = 3.2\times10^{-4}$ H — recomputed in the competitive example below. The symmetry of formula (3) under swapping labels 1 ↔ 2 displays reciprocity on its face ✓.

The same assemble-from-parts logic prices any strongly-coupled pair; only the geometry line changes. And two practical dials follow immediately from the formula's anatomy. Core: fill the shared bore with soft iron and the flux — hence $M$ — multiplies by the core's $\mu_r$ (hundreds to thousands; the ledger behind $\mu_r$ belongs to Magnetisation, Intensity & Susceptibility). Geometry: anything that reduces flux sharing reduces $M$ — separation, misalignment, tilt. At perfect perpendicularity, the flux entering one side of the coil's opening exits the other, and $M = 0$ exactly: two circuits can sit centimetres apart, both live, and be magnetically invisible to each other.

Variant geometries at a glance

configuration	coupling	reading
coaxial solenoids, common length	$M = \mu_0 n_1 n_2 A_{\text{inner}} \ell$	the flagship; inner area only
small coil inside a big solenoid	$M = \mu_0 n_1 N_2 A_2$	uniform field × small coil's linkage
two coils on a shared iron ring	$M \to \mu_r$ -boosted, near-total sharing	Faraday's ring; the transformer limit
coplanar loops, far apart	$M$ tiny, falling fast with distance	stray coupling — "crosstalk"
axes perpendicular	$M = 0$	flux in = flux out; geometric decoupling
either coil's orientation flipped	$M$ changes sign	the induced polarity reverses
steady $I_1$ , any size	$\varepsilon_2 = 0$	coupling transfers change only

Common trapThe bigger coil's area goes in the formula

It does not — the shared area does, and that is always the smaller bore. In the coaxial-solenoid derivation, the flux that links both windings exists only where both can catch it: inside the inner solenoid's cross-section $A$ . Students who reach for the outer solenoid's area inflate $M$ by the area ratio — with a 2 cm outer radius over a 1 cm inner radius, a factor of 4, converting the correct 0.32 mH of the worked example into a confident 1.3 mH. The error feels plausible because the outer solenoid is "bigger, so it catches more"; but a turn cannot catch flux where there is none, and outside the inner bore (in the driven-inner case) or outside its own opening (in the driven-outer case) there is nothing to catch. Exam-safe rule: in any nested geometry, write $A = A_{\text{smaller}}$ before touching the calculator — and expect one distractor option built from the larger area.

Worked examples

Board level — a coil inside a solenoid (CBSE/ISC pattern, 3 marks)

A long solenoid has 1500 turns per metre. A compact coil of 80 turns and radius 1.2 cm is placed inside it, coaxially. (a) Find the mutual inductance of the pair. (b) The solenoid's current is increased steadily from 0 to 2.5 A in 50 ms; find the emf induced in the small coil. (c) What is the induced emf once the current is held steady at 2.5 A?

Step 1. Part (a): drive the solenoid (the easy direction — its interior field is uniform). Field per ampere: $B_1/I_1 = \mu_0 n_1 = (4\pi\times10^{-7})(1500)$ . The small coil's linkage per ampere is its $N_2$ turns times its own area $A_2 = \pi(0.012)^2 = 4.52\times10^{-4}\ \mathrm{m^2}$ :

M = \mu_0 n_1 N_2 A_2 = (4\pi\times10^{-7})(1500)(80)(4.52\times10^{-4}) = \boxed{68\ \mathrm{\mu H}}.

Step 2. Part (b): the ramp rate is $\dfrac{\Delta I_1}{\Delta t} = \dfrac{2.5}{0.050} = 50\ \mathrm{A/s}$ , so by equation (2): $\lvert\varepsilon_2\rvert = M\,\dfrac{\Delta I_1}{\Delta t} = (6.8\times10^{-5})(50) = \boxed{3.4\ \mathrm{mV}}$ .

Step 3. Part (c), the sentence worth its full mark: "Zero — the current, and hence the flux through the coil, is no longer changing; mutual induction responds only to changing current."

Step 4. Sanity habits. The small coil's area (not the solenoid's bore) entered, because the coil can only catch flux through its own opening — the shared-area rule. Units: H × A/s = V ✓. And note which quantities you never needed: the solenoid's radius and the coil's resistance — classic decoy data in this pattern.

Competitive level — nested solenoids and the reciprocity read (JEE-Main pattern, ~3 minutes)

Two coaxial solenoids have a common length of 40 cm. The inner one has 2000 turns per metre and radius 1.0 cm; the outer has 1000 turns per metre and radius 2.0 cm. (a) Find their mutual inductance. (b) Find the emf induced in the inner solenoid when the outer one's current rises uniformly from 0 to 5.0 A in 20 ms. (c) If instead the inner solenoid carried a current changing at the same rate, what emf would appear in the outer one?

Step 1. Part (a): equation (3) with the inner area, $A = \pi(0.010)^2 = 3.14\times10^{-4}\ \mathrm{m^2}$ :

M = \mu_0 n_1 n_2 A \ell = (4\pi\times10^{-7})(1000)(2000)(3.14\times10^{-4})(0.40) = \boxed{0.32\ \mathrm{mH}}.

The planted trap: using the outer radius gives $4\times$ the area and a wrong answer of 1.3 mH — expect it among the options.

Step 2. Part (b): rate $= 5.0/0.020 = 250\ \mathrm{A/s}$ , so $\lvert\varepsilon_2\rvert = (3.16\times10^{-4})(250) = \boxed{79\ \mathrm{mV}}$ (call it 0.08 V at two figures).

Step 3. Part (c): reciprocity, $M_{12} = M_{21}$ — the coupling constant is the same both ways, so the same rate of change induces the same emf: $\boxed{79\ \mathrm{mV}}$ in the outer solenoid. No new calculation, and crucially no attempt to work out the inner solenoid's leaking end-field — the theorem exists precisely so you never have to.

Step 4. Concept lock: which physical quantities did $M$ depend on? Turn densities, shared area, common length — geometry only. Neither current nor rate appears in $M$ itself; they enter only when the emf is asked for. Any option making $M$ "depend on the current" is auto-eliminable. Timing note: three minutes. Part (a) is a 60-second substitution once the inner-area decision is made — make it consciously, not by reflex. Part (c) should take ten seconds; if you find yourself computing the inner solenoid's field pattern, stop — that is the reciprocity read, and recognizing it is the mark. MCQ variant to expect: "M of a pair depends on which of the following?" with the current as bait.

Exam traps

The outer-area substitution. In nested geometries the shared (smaller) area carries the flux — see the warning box. One distractor option is nearly always built from the bigger area.

"M depends on the current." Never. $M$ is geometry (and core material); currents and their rates enter only through $\varepsilon = -M\,\Delta I/\Delta t$ . Options that scale $M$ with $I$ are elimination fodder.

Steady current, live coupling. A constant $I_1$ , however large, induces nothing. The flat segment of any current–time graph reads $\varepsilon_2 = 0$ .

Reciprocity forgotten — or overworked. $M_{12} = M_{21}$ means one constant per pair: use it to dodge impossible calculations (part c above), but do not "apply it" to claim the two coils' self-inductances are equal — $L_1$ and $L_2$ remain as different as the coils are.

Linkage versus flux per turn. $MI_1$ is coil 2's total linkage $N_2\Phi_2$ . Asked for the flux through each turn, divide by $N_2$ .

Unit slips. Henries with milliseconds and microhenries in the same line — convert once, at the start. $1\ \mathrm{H} = 1\ \mathrm{V\,s/A}$ runs every emf calculation.

Boards and constants. CBSE, ISC, JEE-Main, JEE-Adv and NEET all carry mutual induction; ISC names the coaxial-solenoids expression explicitly — learn Derivation A as a reproducible four-liner. A-Level (CIE 9702) has no inductance at all, and IB treats induction without $M$ — honesty for those boards' candidates: the concept, not the formula. Constants: $\mu_0 = 4\pi\times10^{-7}$ T·m/A, kept symbolic until the final line.

Origins · 1 minFranz Ernst Neumann and how this was discovered

Franz Ernst Neumann

1798–1895 · Königsberg

Neumann came to induction the long way: wounded as a teenage volunteer at the Battle of Ligny in 1815, he turned first to mineralogy — his law on the specific heats of compounds still bears his name — before physics claimed him. At Königsberg, where he co-founded the world's first true physics research seminar and trained a generation of German physicists (Kirchhoff among them), he read Faraday's qualitative discoveries and did what nobody had managed in fourteen years: turned them into equations. His 1845 memoir gave the induced emf a mathematical law and, within it, the first general formula for the mutual inductance of two circuits — a formula whose perfect symmetry between the two coils makes reciprocity, $M_{12} = M_{21}$ , visible at a glance rather than mysterious. The theorem this article licenses you to use ("compute the easy direction") is Neumann's gift. Faraday supplied the phenomenon; Neumann supplied the grammar — the quiet, professorial half of the induction story, and the one every calculation in this chapter silently leans on.

History

1831

— Faraday's iron ring: changing current in one winding induces current in another — mutual induction discovered

1832

— Henry, independently coupling coils in Albany, publishes the American thread of induction

1845

— Franz Neumann's memoir turns induction into equations; the mutual-inductance formula and its two-way symmetry appear

1885

— The ZBD engineers and William Stanley build practical closed-core transformers — near-perfect coupling goes commercial

2008

— Resonant wireless-power demonstrations revive weak-coupling energy transfer; phone-charging standards follow

Going further

Texts. NCERT Physics Class 12, Ch. 6 (§6.9) · HC Verma Vol. 2, Ch. 38 · SL Arora, ISC Physics XII, electromagnetic induction.

Provenance. All results standard. A verifier should recompute: board example $A_2 = 4.52\times10^{-4}\ \mathrm{m^2}$ , $M = \mu_0 n_1 N_2 A_2 = 68\ \mathrm{\mu H}$ , ramp 50 A/s → 3.4 mV; competitive example $A = 3.14\times10^{-4}\ \mathrm{m^2}$ , $M = 0.32$ mH (outer-area distractor 1.26 mH), rate 250 A/s → 79 mV both directions; dipole-falloff claim $1/r^3$ → ×8 per halving of distance.

Prerequisites. Self-Inductance & Inductors · Faraday's Law of Induction · Solenoids & Toroids. Leads to. LR Circuits: Growth & Decay (this chapter) · The Transformer (coming in the AC chapter) · Energy in an Inductor (coming up in this chapter) · Mutual Inductance (UG deep dive).

Reciprocity feels impossible — a tiny coil and a huge solenoid are so different. How can the coupling be equal both ways?

Because "the coupling" is a property of the pair, not of either coil. Here is a way to feel it: $M$ multiplies (field per ampere of the source) × (collecting power of the receiver). The big solenoid is a strong field-builder but the small coil is a weak collector; the small coil is a weak field-builder but the big solenoid is a strong collector. The two lopsidednesses are exact inverses of each other — strong × weak = weak × strong — and the product comes out identical. Neumann's 1845 formula makes this visible: $M$ is a single expression into which both coils' geometries enter completely symmetrically, so swapping the labels cannot change its value. At exam level, use it as a licensed shortcut; the formal proof is a UG pleasure waiting in the deep-dive article.

If flux from coil 1 threads coil 2, does coil 2's induced current then affect coil 1 back?

Yes — and you already own the rule that governs it. If coil 2's circuit is closed, its induced current creates its own flux, some of which threads coil 1 (with the same $M$ , by reciprocity), and by Lenz's Law that back-coupling opposes the original change. This is not exam-level detail for this module, but it is the secret of the loaded transformer: an unloaded secondary draws almost nothing from the primary, while a heavily loaded one forces the primary to draw more current, precisely through this back-and-forth flux negotiation. The full story — with the turns-ratio bookkeeping — is The Transformer, in the AC chapter.

Do the two coils need to be close together? How fast does M die with distance?

Very fast — faster than you would guess. Far from a current loop, its field weakens like $1/r^3$ (the dipole falloff you met with bar magnets and current loops), so the flux threading a distant coil — and with it $M$ — dies by roughly a factor of eight for every doubling of separation. That steep law is why wireless charging requires the phone to sit on the pad rather than near it, why crosstalk between cables is cured by simply spacing them, and why deliberate strong coupling always resorts to nesting the coils or joining them with an iron path. Distance is the cheapest decoupler in electrical engineering; iron is the strongest coupler.

What decides the direction of the induced current in the secondary coil?

Lenz's Law, applied to the flux arriving from the primary — the same four-step algorithm you use everywhere: find the direction of the flux through the secondary, decide whether it is growing or dying, point the induced field against the growth (or with the decay), and curl the right hand to read the current. One extra bookkeeping fact: physically flipping either coil (or reversing its winding sense) reverses the answer — which is why $M$ carries a sign in careful work, and why real transformer schematics mark winding polarity with dots. Board questions keep orientations fixed and ask only magnitudes plus "state the direction by Lenz's law"; give the algorithm's chain, not a memorized answer.

Is there energy stored in the coupling, the way an inductor stores energy?

Yes. Each coil stores its own $\tfrac{1}{2}LI^2$ (the account Energy in an Inductor opens later this chapter), and a coupled pair carrying simultaneous currents holds an additional cross-term proportional to $M I_1 I_2$ — energy that lives in the shared part of the field and can be positive or negative depending on whether the two fluxes aid or oppose. That cross-term is what makes coupled circuits able to hand energy back and forth, and its careful audit is what proves, at UG level, that $M$ can never exceed $\sqrt{L_1 L_2}$ — a perfect-coupling speed limit. For this module's exams you need only the per-coil stores; the cross-term is enrichment.

Why does an iron core help so dramatically?

Because iron is a flux amplifier and a flux guide at once. Amplifier: the coil's field magnetizes the core, and the core's own contribution multiplies the flux by $\mu_r$ — hundreds to thousands for soft iron (the ledger of Magnetisation, Intensity & Susceptibility). Guide: flux overwhelmingly prefers to travel inside high- $\mu_r$ material, so a closed iron loop — Faraday's ring — pipes nearly all of coil 1's flux through coil 2 instead of letting it sprawl uselessly through air. The two effects together turn a feeble air coupling into the near-perfect sharing a transformer needs. The fine print (iron saturates; $\mu_r$ is not a constant) is why real transformer design is an art — and why exam problems say "air-cored" unless they hand you a $\mu_r$ .

How is M different from L, really? The formulas look almost identical.

They are the same kind of constant — webers of linkage per ampere of cause, both in henries, both pure geometry — differing only in whose flux threads whom. $L$ is the diagonal case: a coil's own current linking its own turns. $M$ is the off-diagonal case: one circuit's current linking another's turns. The parallel is exact enough that the formulas rhyme ( $\mu_0 n^2 A\ell$ against $\mu_0 n_1 n_2 A\ell$ — the self-formula is the mutual formula with both roles played by the same coil), and it is why the pair share a unit and a chapter. What $M$ adds that $L$ cannot have is the two-body structure: reciprocity, orientation-dependence (including the sign), and the possibility of $M = 0$ between two live circuits — a coil can never be decoupled from itself.