Mutual inductance

Hold your phone over a wireless charging pad and nothing touches — no plug, no pins — yet the battery fills up. Underneath the pad is a coil of wire carrying a rapidly alternating current; inside the phone is another coil, a few millimetres away, wired to nothing but the phone's own circuit. There is no wire joining the two, and yet energy crosses the gap. The link is a changing magnetic field: the current in the pad's coil builds a field that threads through the phone's coil, and because that field keeps changing, it drags an EMF into existence in the second coil out of thin air. This coupling between two separate circuits — one coil's changing current writing an EMF into a completely different coil nearby — is called mutual induction, and the number that measures how strongly the two coils are coupled is the mutual inductance, $M$ . It is the natural sequel to self-inductance: where self-inductance asked how a coil's own changing current pushes back on itself, mutual inductance asks how one coil's changing current reaches across space and pushes an EMF into its neighbour. The same transformer sitting on an electricity pole outside your house, quietly stepping thousands of volts down to the 230 V in your wall socket, runs on nothing more exotic than this one idea, repeated at industrial scale. This article builds mutual inductance from the ground up: what $M$ means physically, why it is the same number no matter which coil you choose to drive, how to calculate it for a clean pair of coaxial solenoids, and how to turn a changing current in one coil into a predicted EMF, current, or even a definite pulse of charge in the other. Two worked problems anchor the arithmetic you will actually be asked to do — one board-style, one pushed a notch further for competitive exams — using the exact coil pairs and numbers this chapter is built around.

Key takeaways

Mutual inductance $M$ measures how much flux one coil links into a second coil per unit current in the first: $\phi_2 = MI_1$ , with $M$ in henry (H).
Faraday's law turns a changing $I_1$ into an EMF in coil 2: $\varepsilon_2 = -M\dfrac{dI_1}{dt}$ , the minus sign again encoding Lenz's law.
$M$ is geometric, not electrical — it depends on turns, shape, area, separation, orientation, and the medium (an iron core boosts it sharply), never on the current itself.
Reciprocity: the mutual inductance is exactly the same whichever coil you drive, $M_{12}=M_{21}=M$ — a fact you can quote and use, without needing its proof.
Two long coaxial solenoids give the one mutual-inductance formula worth memorising: $M=\mu_0 n_1 n_2 A\,l$ , built from the same $B=\mu_0 nI$ you already used for self-inductance.
A switched-off current delivers a fixed charge, $q = N_2\Delta\phi_2 / R$ , independent of how fast the switch-off happens — a favourite twist in numericals.
Key idea: a changing current in one coil links a changing flux into a neighbouring coil, inducing an EMF $\varepsilon_2=-M\,dI_1/dt$ where the coupling constant $M$ — the same in either direction — depends purely on the geometry of the two coils, not on the current flowing.

The core insight

Mutual inductance $M$ is the geometry-fixed constant linking the changing current in one coil to the EMF it induces in a separate, nearby coil through $\varepsilon_2=-M\,dI_1/dt$ , and it is exactly the same number whichever of the two coils is driven.

Where it fits in the syllabus

Mutual inductance closes the CBSE Class 12 chapter "Electromagnetic Induction" (Unit VI), sitting right after self-inductance. The board expects you to define $M$ through $\phi_2=MI_1$ , compute induced EMF from $\varepsilon_2=-M\,dI_1/dt$ , know its SI unit is the henry, and reason qualitatively about which factors (turns, area, separation, core material) make $M$ larger or smaller. A worked derivation of $M$ for a pair of long coaxial solenoids is fair game as an extension of the self-inductance solenoid work you have already done — but the detailed construction and quantitative treatment of a real transformer belongs to the next chapter, Alternating Current, and is only named here as the everyday application that makes mutual inductance worth knowing.

The physical picture

Picture two coils sharing the same neighbourhood — coil 1, the one you choose to drive with a current, and coil 2, the one you watch for an induced effect. When a current $I_1$ flows in coil 1, it sets up a magnetic field everywhere around it, including through the turns of coil 2. Because that field is directly proportional to $I_1$ (double the current, double the field, everywhere), the flux linked with coil 2 is also directly proportional to $I_1$ , for a fixed arrangement of the two coils:

\phi_2 = M I_1

where: $\phi_2$ is the magnetic flux linked with a single turn of coil 2 (in weber, Wb) due to the current $I_1$ (in ampere) flowing in coil 1, and $M$ is the mutual inductance of the pair, in henry (H). Read it physically: $M$ is simply "how many weber of flux land on coil 2 for every one ampere flowing in coil 1" — a conversion factor fixed entirely by the shapes, sizes, turns, spacing, and orientation of the two coils, and by whether a magnetic material threads through them. It does not change if you crank $I_1$ up or down; only $\phi_2$ changes, in step. The moment $I_1$ starts to change with time, $\phi_2$ changes with it, and Faraday's law does what it always does — turns a changing flux into an EMF, this time in coil 2, a circuit that carries no current of its own doing:

\varepsilon_2 = -\frac{d\phi_2}{dt} = -M\frac{dI_1}{dt}

where: $\varepsilon_2$ is the EMF induced in coil 2 (volt), and $dI_1/dt$ is the rate at which the driving current in coil 1 is changing (A/s). The negative sign is Lenz's law exactly as before: the induced EMF drives a current (if the secondary circuit is closed) whose own magnetic effect opposes the change in flux that created it. For most numerical work you only need the magnitude, $|\varepsilon_2| = M\,dI_1/dt$ , but always be ready to reason about the sign qualitatively when a question asks about direction.

SI unit. Mutual inductance, exactly like self-inductance, is measured in henry (H): a coil pair has $M = 1\ \text{H}$ if a current changing at $1\ \text{A/s}$ in coil 1 induces an EMF of exactly $1\ \text{V}$ in coil 2.

Reciprocity — $M_{12}=M_{21}$ . A genuinely surprising fact, which you should quote and use even though its formal proof (via a general energy argument) sits above this level, is that it makes no difference which coil you call the driver. If you instead pushed a changing current through coil 2 and measured the EMF induced in coil 1, you would get back the same number $M$ . Formally, the flux linked in coil 2 per unit current in coil 1 always equals the flux linked in coil 1 per unit current in coil 2: $M_{12}=M_{21}=M$ . So "primary" and "secondary" are just labels of convenience, not physically distinct roles.

What sets the size of $M$ . Four things, all purely geometric: (a) the number of turns on either coil — more turns on coil 1 means more flux produced per ampere, more turns on coil 2 means more of that flux gets linked and summed; (b) the common cross-sectional area shared by the two coils — more overlap, more linked flux; (c) the separation and relative orientation of the coils — closer and more directly facing means a stronger field reaches coil 2; (d) the medium threading both coils — slipping a soft-iron core through both coils multiplies $M$ by the core's relative permeability, often by a factor of hundreds to thousands, which is precisely why real transformers are wound on iron cores rather than left in air.

Figure

Two coaxial coils on a common axis, a longer outer solenoid of length l wound with n1 turns per metre carrying current I1, and a shorter inner secondary coil of n2 turns per metre and the same cross-section area A wound directly over the middle of it, with field lines from the outer solenoid threading straight through the inner coil

Derivation AMutual inductance of two long coaxial solenoids

The goal is to turn the abstract definition $\phi_2=MI_1$ into an actual number you can compute, for the one coil pair simple enough to handle with algebra alone — two long solenoids wound on the same axis, one inside the other, both sharing the same cross-sectional area $A$ . We already know, from the self-inductance work earlier in this chapter, that a long solenoid of $n$ turns per unit length carrying a current $I$ produces a (quoted, not re-derived) uniform field $B=\mu_0 n I$ inside itself; here we simply follow that field from coil 1 into coil 2 and count how much of it links coil 2's turns.

Step 1. Set up the two coils. Let coil 1 be the outer solenoid, length $l$ , with $n_1$ turns per unit length, carrying current $I_1$ . Let coil 2 be a second coil of $n_2$ turns per unit length wound directly over (or inside) the same length, sharing the same cross-sectional area $A$ — this keeps the field from coil 1 uniform across every turn of coil 2, which is what makes the algebra clean.

Step 2. Find the field coil 1 produces. Using the quoted solenoid result, the magnetic field inside coil 1 (and hence inside coil 2, since coil 2 sits inside the same region) is $B_1 = \mu_0 n_1 I_1$ , uniform and directed along the common axis.

Step 3. Find the flux through one turn of coil 2. Since $B_1$ is uniform over the shared cross-section $A$ and points straight along the axis, the flux linked with a single turn of coil 2 is simply the field times the area it passes through: $\phi_2 = B_1 A = \mu_0 n_1 I_1 A$ .

Step 4. Add up all the turns of coil 2. Coil 2 has $N_2 = n_2 l$ turns in total, and every one of them links the same flux $\phi_2$ , so the total flux linkage is $N_2\phi_2 = (n_2 l)(\mu_0 n_1 I_1 A) = \mu_0 n_1 n_2 A\, l\, I_1$ .

Step 5. Read off the mutual inductance. Comparing this total flux linkage with the defining relation $N_2\phi_2 = M I_1$ gives

M = \mu_0\, n_1 n_2\, A\, l

where: $n_1,n_2$ are the turns-per-unit-length of the two coils, $A$ is their common cross-sectional area (m²), $l$ is the shared length (m), and $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space. Notice the formula is symmetric in $n_1$ and $n_2$ — swap which coil you call "1" and which "2" and $M$ comes out identical, which is exactly the reciprocity $M_{12}=M_{21}$ stated earlier, now visible directly in the algebra rather than merely asserted. Verified: dimensional check — $[\mu_0][n_1][n_2][A][l] = (\text{T}\,\cdot\,\text{m/A})(\text{m}^{-1})(\text{m}^{-1})(\text{m}^2)(\text{m}) = \text{T}\,\cdot\,\text{m}^2/\text{A} = \text{Wb/A} = \text{H}$ , consistent with $M$ in henry.

Worked examples

Board level - EMF induced from a uniformly changing primary current

Two coils have a mutual inductance of $0.5\ \text{H}$ . The current in the primary coil changes uniformly from $2\ \text{A}$ to $8\ \text{A}$ in $0.3\ \text{s}$ . Find the EMF induced in the secondary coil.

Step 1. Find the rate of change of the primary current: $\dfrac{dI_1}{dt} = \dfrac{I_{\text{final}} - I_{\text{initial}}}{\Delta t} = \dfrac{8-2}{0.3} = \dfrac{6}{0.3} = 20\ \text{A/s}$ .

Step 2. Apply the mutual-EMF relation: $|\varepsilon_2| = M\dfrac{dI_1}{dt} = 0.5 \times 20 = 10\ \text{V}$ .

Answer: the induced EMF in the secondary coil is $10\ \text{V}$ , present throughout the $0.3\ \text{s}$ interval because the current changes at a constant rate.

Competitive level - mutual inductance, EMF, and total induced charge

In a pair of coils, a current of $4\ \text{A}$ in the primary coil produces a flux of $8\times10^{-3}\ \text{Wb}$ linked with each turn of the secondary coil (which has $100$ turns). Find (a) the mutual inductance, (b) the average EMF induced in the secondary if the primary current is switched off uniformly in $0.02\ \text{s}$ , and (c) the total charge that flows through the secondary coil during the switch-off, if the secondary circuit has a total resistance of $50\ \Omega$ .

Step 1. Total flux linkage in the secondary at $I_1=4\ \text{A}$ : $N_2\phi_2 = 100 \times 8\times10^{-3} = 0.8\ \text{Wb-turns}$ .

Step 2. Mutual inductance: $M = \dfrac{N_2\phi_2}{I_1} = \dfrac{0.8}{4} = 0.2\ \text{H}$ .

Step 3. Rate of change of primary current as it is switched off: $\dfrac{dI_1}{dt} = \dfrac{0 - 4}{0.02} = -200\ \text{A/s}$ , magnitude $200\ \text{A/s}$ ; so the average induced EMF is $|\varepsilon_2| = M\dfrac{dI_1}{dt} = 0.2 \times 200 = 40\ \text{V}$ .

Step 4. Total induced charge does not need the time at all — only the total change in flux linkage. Since $I_1$ falls from $4\ \text{A}$ to $0$ , the flux linkage falls from $0.8\ \text{Wb-turns}$ to zero, a change $\Delta(N_2\phi_2) = 0.8\ \text{Wb-turns}$ . Using $q = \dfrac{|\Delta(N_2\phi_2)|}{R} = \dfrac{0.8}{50} = 0.016\ \text{C} = 16\ \text{mC}$ .

Answer: $M = 0.2\ \text{H}$ , the average induced EMF while switching off is $40\ \text{V}$ , and the total charge through the secondary is $16\ \text{mC}$ , regardless of exactly how the $0.02\ \text{s}$ switch-off was timed. Timing note: this should take under two and a half minutes in an exam. Parts (a) and (b) are the same two-step chain as the board-level problem — find $M$ from flux linkage, then multiply by $dI_1/dt$ . Part (c) is the trap-and-shortcut combined: many students reach for $dI_1/dt$ again out of habit, but charge only cares about the total change in flux linkage divided by resistance, $q=N_2\Delta\phi_2/R$ , and is completely independent of how long the switch-off took. Spotting that a question asks for charge, not EMF, is the whole battle.

Common traps

Confusing $M$ with self-inductance $L$ — mutual inductance links the changing current in one coil to the induced EMF in a different coil, never the same one; $\varepsilon = -L\,dI/dt$ (self) and $\varepsilon_2=-M\,dI_1/dt$ (mutual) are answering different questions.
Forgetting $M_{12}=M_{21}$ and re-deriving $M$ from scratch when a question flips which coil is called "primary" — the number is unchanged, only the labels swap.
Treating $M$ as if it depended on the current $I_1$ flowing through the primary — it depends only on geometry, turns, area, separation, and the medium; the current only sets $\phi_2$ and $\varepsilon_2$ , not $M$ itself.
Dropping the secondary's turns $N_2$ when computing flux linkage — the defining relation uses total flux linkage $N_2\phi_2 = MI_1$ , not the flux through a single turn alone, and mixing the two silently multiplies (or divides) an answer by $N_2$ .
Using $dI_1/dt$ to find the charge $q$ through the secondary, when charge only needs the total change in flux linkage divided by resistance, $q = N_2\Delta\phi_2/R$ — the time taken cancels out entirely and is a distractor in the question.
Assuming an iron core always helps — it dramatically increases $M$ only while it stays unsaturated; at very large currents a real core saturates and the simple proportional picture, and hence a constant $M$ , breaks down (a detail flagged qualitatively, not tested quantitatively at this level).

Common trapMutual inductance is not "self-inductance shared between two coils"

The single most common board-exam slip is writing $\varepsilon_2 = -M\,dI_2/dt$ instead of $\varepsilon_2=-M\,dI_1/dt$ — differentiating the current in the wrong coil. Remember the logic: $M$ always couples the changing current in the driving coil to the induced EMF in the other coil. If coil 1 is driven, its own changing current never appears with an $M$ in front of it for its own EMF — that role belongs to $L$ (self-inductance). The two relations $\varepsilon_1=-L_1\,dI_1/dt$ (self, same coil) and $\varepsilon_2=-M\,dI_1/dt$ (mutual, other coil) can even happen simultaneously in the same real pair of coils, and full marks depend on keeping straight which effect, in which coil, is being asked for.

You should be able to…

Define mutual inductance through flux linkage per unit driving current, $N_2\phi_2 = M I_1$ , and state its SI unit, the henry.
Use $\varepsilon_2 = -M\,dI_1/dt$ to calculate the EMF induced in a secondary coil from a changing primary current.
Derive and use $M=\mu_0 n_1 n_2 A l$ for a pair of long coaxial solenoids sharing the same cross-section.
Quote and apply the reciprocity property $M_{12}=M_{21}$ without needing to reprove it each time.
List the geometric factors — turns, area, separation, orientation, core material — that make $M$ larger or smaller.
Compute the total induced charge $q=N_2\Delta\phi_2/R$ in a secondary circuit and recognise it does not depend on how fast the current changed.

Ask the tutor: want to work through a few more mutual-inductance numericals, or see how this exact idea scales up into the step-up and step-down transformers of the Alternating Current chapter?

How is mutual inductance different from self-inductance if they use such similar formulas?

Self-inductance $L$ describes a coil pushing back on itself — its own changing current creates a changing flux through its own turns, inducing an EMF in that same coil that opposes the change (this is why a coil resists sudden changes in its own current). Mutual inductance $M$ describes a completely different circuit reacting to a coil it is not electrically connected to at all — coil 2 can be sitting there with no current source of its own, and still get an EMF pushed into it purely because coil 1 nearby has a changing current. The formulas look alike ( $\varepsilon=-L\,dI/dt$ versus $\varepsilon_2=-M\,dI_1/dt$ ) because both come from Faraday's law applied to a changing flux, but one flux is self-generated and the other arrives from a neighbour.

Why should M really be the same in both directions? It feels like it should depend on which coil is bigger or has more turns.

It is a genuinely non-obvious result, and a full proof needs an energy argument beyond this level, so at Class 12 you are expected to know and use it, not prove it. But you can see why it is at least plausible from the coaxial-solenoid derivation above: the final formula $M=\mu_0 n_1 n_2 A l$ treats $n_1$ and $n_2$ completely symmetrically — swap the labels "1" and "2" and the formula, and therefore the number $M$ , does not change. That symmetry is not a coincidence of this particular geometry; it holds for any pair of coils, however oddly shaped, which is exactly the content of $M_{12}=M_{21}$ .

Does mutual inductance need the two coils to be touching, or even close together?

No — mutual induction works at any separation, as long as some of the magnetic flux from coil 1 manages to pass through coil 2. What changes with distance is only the size of $M$ : as the coils move apart, a smaller fraction of coil 1's field lines reach coil 2, so $M$ shrinks (the induced EMF for a given $dI_1/dt$ gets weaker), but it never strictly needs contact. This is exactly the principle behind wireless charging pads, which deliberately keep the coils close and well-aligned to keep $M$ , and hence the transferred power, as large as practically possible.

Is the iron core in a transformer just for support, or does it actually do something to M?

It does something very real. Air has almost no effect on a magnetic field, but a soft-iron core dramatically concentrates the field lines from the primary coil and guides almost all of them through the secondary coil as well, instead of letting them spread out and miss it. Since $M$ is essentially "how much of coil 1's flux actually reaches coil 2," channelling nearly all of it through an iron core can boost $M$ by a factor of hundreds or more compared with air-core coils of the same size and turns. That single fact — a core multiplying $M$ — is the whole engineering reason real transformers (studied in the next chapter) are wound around iron, not left as bare coils in air.