Self and Mutual Inductance, and the Coupling Between Two Coils

Q: Can the mutual inductance be bigger than either self-inductance?

Yes, as long as it stays below √(L1L2). With the defaults here, √(L1L2) is 181 µH while L2 is only 68 µH, so a well-coupled pair would have an M larger than coil 2's own inductance. What M can never exceed is the geometric mean of the two.

Two coils on one core have three inductances, not two: each coil’s own, and the mutual inductance M that measures how much of one coil’s flux the other one catches. This returns all three from the turns, the core area and the length, with the coupling coefficient k that ties them together — and shows why M can never exceed √(L₁L₂).

Last updated 11 Nov 2026

Solenoid self-inductance, mutual inductance and coupling

L = μ₀N²A/l · M = k√(L₁L₂)

Turns on coil 1Inductance goes as N², not as N. Double the turns and you get four times the inductance, because each of twice as many turns links twice as much flux. That square is the most useful single fact about an inductor and the most commonly dropped.

Turns on coil 2The second winding, wound over or beside the first. M depends on the product N₁N₂ and is the same measured either way round — drive coil 1 and read coil 2, or drive coil 2 and read coil 1, and you get the same M. That reciprocity is not obvious and is worth a mark.

Cross-sectional area of the coil (cm²)The area the flux goes through, in square centimetres. 6 cm² is a former about 28 mm across. Remember the squared conversion: 6 cm² = 6 × 10⁻⁴ m².

Length of the solenoid (cm)The winding length, not the wire length. The formula μ₀N²A/l assumes a long thin solenoid — strictly, that the length is much greater than the diameter — because it treats the field inside as uniform and the field outside as zero. For a short fat coil it overestimates, typically by ten or twenty per cent.

Coupling coefficient k (0 to 1)What fraction of coil 1’s flux reaches coil 2. k = 1 means every line that threads one threads the other, which needs a closed magnetic core and tight interleaved windings; k = 0 means the coils ignore each other entirely. Two air-cored coils side by side on a bench might manage 0.3 to 0.7, which is why 0.65 is the default here.

Two solenoids on one axis. The zigzags are the windings and the number of them drawn follows the turns you enter, so a coil with more turns looks like one. Coil 2 slides away down the axis as the coupling coefficient falls, and the flux arrows that link the two coils appear one at a time as the coupling rises: at k near zero the coils are far apart and nothing links them, at k near one they sit together and all three arrows run through both.

117.62µHExample

400 turns and 150 turns on a 6 cm², 25 cm solenoid, coupling k = 0.65

Three inductances and the number that links them

L = μ₀N²A/l · M = k√(L₁L₂) · perfect coupling: M = μ₀N₁N₂A/l · ε₁ = −L₁ dI₁/dt · ε₂ = −M dI₁/dt

L: self-inductance, in henries. OpenStax gives the solenoid case as L = μ₀N²A/l. A henry is a weber per ampere (SI Brochure, Table 4), so L is simply the flux linkage one ampere produces: NΦ = LI
M: mutual inductance. M = N₂Φ₂₁/I₁, the flux linkage in coil 2 per ampere in coil 1 — and it equals N₁Φ₁₂/I₂, the same quantity measured the other way round. That reciprocity is a theorem, not a coincidence
k: the coupling coefficient, between 0 and 1. It has no units and no formula of its own at this level — it is a measured property of the geometry. k = 1 needs every field line of one coil to pass through the other, which is the job a closed magnetic core does
why M ≤ √(L₁L₂): because k ≤ 1, and the physical reason k ≤ 1 is that coil 2 cannot link flux that coil 1 never produced. The two rows above make it concrete: √(L₁L₂) and μ₀N₁N₂A/l give the same number, and that number is the ceiling
the minus signs: Lenz’s law again. The EMF a changing current induces in its own coil opposes the change, which is why an inductor resists a sudden change of current and why breaking an inductive circuit makes a spark

Worked example

400 turns and 150 turns on a 6 cm², 25 cm solenoid, coupling k = 0.65
SI UNITS FIRST. A = 6 cm² = 6 × 10⁻⁴ m² and l = 25 cm = 0.25 m. The area conversion is squared and the length one is not; mixing those up is a factor of a hundred.
COIL 1's SELF-INDUCTANCE. L₁ = μ₀N₁²A/l = (1.2566 × 10⁻⁶)(400²)(6 × 10⁻⁴)/0.25. Take it in pieces: 400² = 160 000, and 1.2566 × 10⁻⁶ × 160 000 = 0.2011. Then × 6 × 10⁻⁴ = 1.2064 × 10⁻⁴, ÷ 0.25 = 4.825 × 10⁻⁴ H = 482.5 µH.
COIL 2's. Same sum with 150 turns: 150² = 22 500, which is 160 000/7.111, so L₂ = 482.5/7.111 = 67.86 µH. You can always get the second one from the first by the square of the turns ratio and save yourself the arithmetic.
THE CEILING. √(L₁L₂) = √(482.5 × 67.86) = √32 744 = 180.96 µH. That is the mutual inductance these coils would have if the coupling were perfect, and no arrangement of them can beat it.
APPLY THE COUPLING. M = k√(L₁L₂) = 0.65 × 180.96 = 117.62 µH. That is the headline.
CHECK IT INDEPENDENTLY. Perfect coupling can also be computed straight from the geometry: μ₀N₁N₂A/l = (1.2566 × 10⁻⁶)(400)(150)(6 × 10⁻⁴)/0.25 = 180.96 µH, the same figure to every digit. Two routes, one answer — which is what tells you that k = 1 really does mean “perfect”.

What each input does, starting from this page’s defaults

Row two is the one worth staring at. Doubling the turns on one coil *quadruples* that coil’s self-inductance but only *doubles* the mutual inductance, because L goes as N² for one coil while M goes as the product N₁N₂. Nothing in the table moves when you change k except the last column, because the self-inductances do not know the other coil exists.
Change	L₁	L₂	M at k = 0.65
Nothing (the defaults)	482.5 µH	67.86 µH	117.6 µH
Double the turns on coil 1 only	1930 µH	67.86 µH	235.2 µH
Double the turns on both coils	1930 µH	271.4 µH	470.5 µH
Double the area	965.1 µH	135.7 µH	235.2 µH
Double the length	241.3 µH	33.93 µH	58.81 µH

Why two coils have three inductances, and why M has a ceiling

Self-inductance is a coil arguing with itself. Change the current in a coil and its own flux changes, and that changing flux induces an EMF in the very coil that caused it — opposing the change, as Lenz’s law insists. L is the constant of proportionality: ε = −L dI/dt. Equivalently, and more usefully for calculation, NΦ = LI, so L is just the flux linkage one ampere buys you. A henry is literally a weber per ampere.

Mutual inductance is one coil talking to another. M = N₂Φ₂₁/I₁: the flux linkage set up in coil 2 by each ampere flowing in coil 1. The remarkable part is that the same number works backwards — drive coil 2 and measure coil 1 and you get the same M — which is not obvious from the geometry and is worth remembering as a check on an answer.

The coupling coefficient is where the honesty lives. M = k√(L₁L₂), and k is a measured property of how the coils sit, not something you can derive at this level. What you can show is that k cannot exceed 1, and this page shows it by computing the ceiling two independent ways: √(L₁L₂) from the two self-inductances, and μ₀N₁N₂A/l straight from the geometry assuming every line links both coils. Those two rows above always agree, and that agreement is the proof. If you ever calculate an M bigger than √(L₁L₂), you have made an arithmetic mistake.

Where this goes wrong is the solenoid assumption. L = μ₀N²A/l is derived for a long thin solenoid, where the field inside is uniform and the field outside is negligible. For a short fat coil the field bulges out of the ends, the flux linkage falls, and the real inductance is lower than this formula says — by ten or twenty per cent for a coil as long as it is wide, and much more for a single-layer pancake. The page warns you when your numbers go that way. The other classic slip is the N²: half the marks lost on inductance questions are someone doubling the turns and doubling the answer. For where the EMF comes from in the first place, see Faraday’s law, and for the direction of everything here, Lenz’s law.

Frequently asked questions

Why does inductance depend on N² and not on N?

Two factors of N, from two different places. More turns carrying the same current make a proportionally stronger field, so the flux per turn goes up as N; and then there are N turns to link that flux, so the flux linkage goes up as N again. Multiply the two and you have N². The same argument gives M its N₁N₂: one factor from the coil making the field, one from the coil catching it.

Can the mutual inductance be bigger than either self-inductance?

Yes, as long as it stays below √(L₁L₂). With the defaults here, √(L₁L₂) is 181 µH while L₂ is only 68 µH, so a well-coupled pair would have an M larger than coil 2’s own inductance. What M can never exceed is the geometric mean of the two.

What is a realistic value of k?

It depends entirely on the geometry and this page cannot work it out for you. Windings interleaved on a closed ferrite or laminated-iron core reach the high 0.99s, which is what makes a transformer efficient. Two separate air-cored coils pushed together end to end might manage 0.3 to 0.7. Coils a few diameters apart are down in the hundredths. Treat k as a measured number, and if a question gives you M and the two self-inductances, you can work k out from them.

Does the coupling change the self-inductances?

No, and the page demonstrates it: slide k from 0 to 1 and the two self-inductance rows do not move at all. A coil’s own inductance is set by its own turns, area and length. What coupling changes is what happens when both coils carry current — then the total inductance of the pair depends on k and on whether the windings aid or oppose.

Why does this use μ₀ and not μ for iron?

Because it is written for an air-cored or non-magnetic-cored solenoid, which is what school questions use. A ferromagnetic core multiplies the inductance by its relative permeability, which can be hundreds or thousands — and which is not a constant, because iron saturates. Putting a number on that properly needs a B-H curve, so this page does not pretend to.

Related calculators

References

Annual Syllabus, Class XII, Physics, Code 42, Session 2025-26, Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2025_26/12/12_Physics_EM.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside)”. Unit III Chapter 4 asks for “force on a moving charge in uniform magnetic and electric fields”, “torque experienced by a current loop in uniform magnetic field” and “Current loop as a magnetic dipole and its magnetic dipole moment”. Unit IV Chapter 6 is one line and covers three of this batch’s records: “Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction”. The Hall effect is not in this document.
University Physics Volume 2, chapters 11 and 14, OpenStax, Rice University (openstax.org/books/university-physics-volume-2, key-equations pages read 11 November 2026). An openly licensed university text, cited here because the coaching sites that dominate searches for this topic are not citable and mostly copy one another. Chapter 11 gives the radius of a charged particle’s path as r = mv/(qB), its period as T = 2πm/(qB), the Hall potential as V = IBl/(neA), the drift velocity in crossed fields as v_d = E/B, the magnetic dipole moment as μ = NIA and the torque as τ = μ × B. Chapter 14 gives L_solenoid = μ₀N²A/l, NΦ = LI, ε = −L dI/dt, M = N₂Φ₂₁/I₁ = N₁Φ₁₂/I₂ and U = ½LI². Neither key-equations page lists a cyclotron frequency or a field energy density, so those two are stated here from the definitions and not attributed to this source.
The International System of Units (SI), 9th edition 2019, version V4.01 of June 2026, Bureau International des Poids et Mesures, bipm.org/documents/d/guest/si-brochure-9-en-pdf (read 11 November 2026). Cited for the units, from Table 4 of coherent derived units with special names: magnetic flux, weber (Wb) = kg m² s⁻² A⁻¹ = V s; magnetic flux density, tesla (T) = kg s⁻² A⁻¹ = Wb/m²; inductance, henry (H) = kg m² s⁻² A⁻² = Wb/A. Those three equalities are worth more than they look: a weber is a volt-second, so a flux change of one weber in one second is one volt — which is Faraday’s law written as a unit. And a henry is a weber per ampere, so an inductance is just how much flux linkage one ampere produces.
Fundamental Physical Constants — Frequently used constants (2022 CODATA wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page comes from here, and the thing worth noticing is which numbers have a bracket after them. The elementary charge is printed 1.602 176 634 × 10⁻¹⁹ C and the speed of light 299 792 458 m s⁻¹, both with no bracket at all, because since 2019 they are exact by definition — the coulomb and the metre are now defined from them. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F m⁻¹ and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N A⁻²: both now carry an uncertainty of about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At three significant figures nothing changes; which constants are definitions changed completely.