In short

A coil of wire that carries a current I wraps its own magnetic flux around itself. Double the current and the flux doubles; the ratio is a property of the geometry alone — the self-inductance L:

\boxed{\;\Phi \;=\; L I\;}\qquad\text{SI unit: henry (H) } = \text{Wb/A} = \text{V·s/A}

If the current changes, the flux changes, and by Faraday's law an EMF appears across the coil that opposes the change:

\boxed{\;\varepsilon \;=\; -L\,\frac{dI}{dt}\;}

For a long solenoid of n turns per metre, cross-section A, and length \ell (total turns N = n\ell),

\boxed{\;L_\text{sol} \;=\; \mu_0 n^2 A\ell \;=\; \frac{\mu_0 N^2 A}{\ell}\;}

Mutual inductance. When a changing current I_1 in coil 1 wraps flux through coil 2, the EMF induced in coil 2 is

\boxed{\;\varepsilon_2 \;=\; -M\,\frac{dI_1}{dt}\;}\qquad \Phi_2 = M I_1

and M is symmetric (M_{12} = M_{21}, a deep reciprocity result from energy conservation). For a secondary of N_2 turns wound tightly over a primary solenoid (n_1 turns/m, area A, length \ell),

M \;=\; \mu_0 n_1 N_2 A

Why it matters. Self-inductance is why the points of a Maruti 800's ignition coil spark at 20 kV when a 12 V contact breaks — L\,dI/dt with dI/dt \sim 10^6 A/s. Mutual inductance is the physics of every transformer in the Indian grid, every wireless charger, every ISRO solenoid valve on the PSLV, and every BLDC fan motor humming in a Bengaluru apartment. The henry is the unit that governs how quickly you can change current — and at what voltage cost.

Flick the switch of a tubelight and you hear a faint tik from the choke. Unplug a running BLDC ceiling fan at the wall and watch the tiny spark between the plug's pin and the socket — the bright blue flash that lasts a millisecond. Stand near the ignition coil of a Maruti 800 with the bonnet open and the engine running and you can hear the rhythmic tik-tik-tik as 12 volts becomes 20,000 volts, over and over again, forty times a second. In each case, you are watching the same physics: a coil trying to preserve its own current against an outside world that is yanking it away.

A current-carrying coil does not just carry current. It wraps magnetic field around itself. That field stores energy. When the current changes, the field must change — and the flux through the coil's own loops must change. By Faraday's law, a changing flux induces an EMF. But the flux is being changed by the coil's own current — so the coil itself provides the EMF, and that EMF opposes whatever is trying to change the current. This is self-induction, and the proportionality between flux and current is the inductance. Any shape of wire — a straight segment, a single loop, a long solenoid — has one, though the number is tiny for a straight wire and huge for a tightly-wound coil around an iron core.

Extend the same idea to two coils. Coil 1 carries current I_1; its field passes through coil 2 and wraps a flux \Phi_2 there. If I_1 changes, the flux through coil 2 changes, and an EMF appears in coil 2 — even though coil 2 has no direct electrical connection to coil 1. This is mutual induction, and it is the entire working principle of transformers, wireless chargers, and the induction coil that Faraday built in 1831. This article derives both inductances from first principles, computes them for a solenoid, and shows where in Indian engineering they bite.

Self-inductance — the coil that resists itself

The physical picture

Take a coil with current I flowing through it. The coil produces a magnetic field \vec{B}; by Ampère's law or the Biot–Savart law, \vec{B} is proportional to I everywhere in space — doubling the current doubles the field at every point, because Maxwell's equations (in the magnetostatic regime) are linear in the sources. The flux through the coil's own loops is therefore also proportional to I:

\Phi \;=\; L\,I

where L is the self-inductance, a constant that depends only on the geometry of the coil and the magnetic permeability of any material filling the interior. No physics of the current, no time dependence — just the shape of the wire and whether there is iron inside.

Self-inductance — a coil wraps its own flux A solenoidal coil with current I flowing through it. Magnetic field lines thread through the interior of the coil and wrap around each loop. The flux linkage is the number of loops times the flux per loop, equal to L times I. B (from current I) I I (into page) flux linkage N·Φ = L·I L depends only on geometry (and μ of the core)
A coil carrying current $I$ wraps flux through each of its loops. The total flux linkage $N\Phi$ is proportional to $I$, and the constant of proportionality is the self-inductance $L$.

The self-induced EMF — change the current, pay the price

Now let the current vary with time: I = I(t). The flux varies with it: \Phi(t) = L\,I(t). By Faraday's law, a changing flux induces an EMF around the loop:

\varepsilon \;=\; -\,\frac{d\Phi}{dt} \;=\; -L\,\frac{dI}{dt}

Why: L is a geometric constant and comes out of the derivative. Faraday's law gives the minus sign — Lenz's law in disguise — and it means the EMF opposes the change in current, not the current itself.

This is the defining equation of an inductor:

\boxed{\;\varepsilon \;=\; -L\,\frac{dI}{dt}\;}

Read it carefully. If I is increasing (dI/dt > 0), the EMF is negative — which means the inductor acts like a battery whose + terminal points opposite to the current flow, fighting the supply that is trying to increase I. If I is decreasing (dI/dt < 0), \varepsilon > 0 — the inductor acts like a battery that tries to maintain the current, sourcing current to keep it going. The inductor is a momentum bank for current: it hates any dI/dt and always opposes it.

Henries, volts, amps, seconds. The SI unit of inductance is the henry (H): 1\text{ H} = 1\text{ Wb}/\text{A} = 1\text{ V·s}/\text{A}. An inductance of 1 H means that a current changing at 1 A/s induces an EMF of 1 V. Typical inductances:

Device Typical L
Straight 10 cm of wire \sim 100\text{ nH}
RF inductor on a circuit board 1–100\ \muH
Choke in a tubelight ballast 0.5–2 H
Maruti ignition coil (primary) 5–10 mH
Filter choke in a 10 kW power supply 10–100 mH
Transformer primary (mains) 10–1000 H (with iron core)

Iron cores boost inductance dramatically because ferromagnetic materials have relative permeability \mu_r in the hundreds or thousands — you will see the factor enter cleanly in the solenoid derivation below.

Animated: current ramp and back-EMF spike

Take an ideal inductor of L = 0.1 H. Ramp the current linearly from 0 to 2 A over 0.2 s, then break the circuit at t = 0.2 s — so the current must drop from 2 A to 0 in, say, 2 ms. Watch what the self-induced EMF does.

Animated current and self-induced EMF Two traces against time from 0 to 0.4 seconds. The current (red) ramps from 0 to 2 amps over 0.2 s, holds briefly, then collapses to 0 between 0.2 and 0.202 s. The self-induced EMF (dark) is a constant negative 1 volt during the ramp up and a huge positive spike (to about 100 V, clipped in the plot window) at the break. The mismatch between the gentle ramp and the violent spike is the physics of every ignition coil and every BLDC fan unplug-spark. time t (seconds, compressed near the break) I (A) and ε (V, scaled) +15 0 −15 0.1 0.2 0.3 0.4
Red: current $I(t)$, scaled so 2 A plots at +10 on the vertical axis. Dark: self-induced EMF $\varepsilon = -L\,dI/dt$ in volts. During the slow ramp ($dI/dt = 10$ A/s), the EMF is a gentle −1 V. At the break — where $dI/dt$ briefly reaches about −400 A/s — the EMF spikes to tens of volts (clipped to +12 V here for visibility). A real unplug-spark reaches kilovolts because $dI/dt$ is far more violent than this cartoon. Click **replay** to watch again.

The take-away is stark: a coil tolerates slow changes in current and punishes fast ones with enormous voltages. Pull the plug on a fan, and the current has nowhere to go; the fan's winding inductance demands dI/dt \approx 0, but the air gap forces dI/dt to be very large for the instant the contacts are separating. The resulting L\,dI/dt is what ionises the air and produces the little spark. The same physics, weaponised deliberately, is the ignition coil — driven by an electronic switch that breaks the primary current on a precise schedule to generate the high-voltage pulse that fires the spark plug.

Deriving the inductance of a solenoid

For a long tightly-wound solenoid with n turns per metre, cross-sectional area A, and total length \ell (so total turns N = n\ell), carrying current I, the magnetic field inside is uniform and axial:

B \;=\; \mu_0 n I

This follows from Ampère's law applied to a rectangular Ampèrian loop that runs along the axis inside the solenoid and returns outside (where B \approx 0 for an ideal long solenoid) — see Solenoid and Toroid for the full derivation.

Step 1. Flux through one turn. Each turn encloses cross-section A through which the uniform field B threads perpendicular to the plane of the turn:

\Phi_1 \;=\; B A \;=\; \mu_0 n I\,A

Step 2. Total flux linkage. There are N = n\ell turns, each threaded by the same flux:

N\Phi_1 \;=\; (n\ell)(\mu_0 n I\,A) \;=\; \mu_0 n^2 A\ell\,I

Why: for a coil, the "flux that counts" is the total flux linkage N\Phi — each turn contributes separately to the EMF, so every loop's flux appears once in the total.

Step 3. Read off L. Compare to the defining relation N\Phi = LI:

\boxed{\;L_\text{sol} \;=\; \mu_0 n^2 A\ell\;}

Equivalently, using N = n\ell so n = N/\ell:

L_\text{sol} \;=\; \mu_0 \frac{N^2 A}{\ell}

Step 4. With a ferromagnetic core (relative permeability \mu_r), replace \mu_0 with \mu = \mu_0 \mu_r:

L \;=\; \mu_0\mu_r\,n^2 A\ell

For silicon steel, \mu_r \sim 4000 at low fields — so wrapping iron through a 1 mH air-core solenoid turns it into a 4 H iron-core inductor. This is why every mains-frequency transformer has an iron core: a practical inductance at 50 Hz is impossible without one.

Sanity check the units. [\mu_0] = T·m/A; [n^2] = 1/m²; [A\ell] = m³. Multiply: T·m/A × 1/m² × m³ = T·m²/A = Wb/A = H. Clean.

What the formula tells you

Mutual inductance — two coils, one field

Two coils, coil 1 and coil 2, linked magnetically (either wound on the same core, or placed near each other). Coil 1 carries current I_1; its field \vec{B}_1 passes through coil 2 and wraps a total flux linkage N_2\Phi_{21} there:

N_2\Phi_{21} \;=\; M_{21}\,I_1

M_{21} is the mutual inductance from coil 1 to coil 2 — again, purely a geometric constant (plus permeability). Symmetrically, if coil 2 carries I_2 with coil 1 having no current, its field induces a flux linkage in coil 1:

N_1\Phi_{12} \;=\; M_{12}\,I_2

A deep theorem — provable either from the magnetic energy of two linked circuits, or from the symmetry of the Biot–Savart integral — states that

\boxed{\;M_{12} \;=\; M_{21} \;\equiv\; M\;}

This is not obvious: the two coils may have wildly different shapes and sizes. Coil 1 could be a one-turn loop of 1 m radius and coil 2 could be a 10,000-turn coil of 1 cm radius. The flux coil 1 wraps through coil 2 (when coil 1 carries 1 A) equals the flux coil 2 wraps through coil 1 (when coil 2 carries 1 A) — down to the last weber. The theorem is one of the small miracles of classical electromagnetism.

Induced EMF in the secondary

If coil 1's current varies, \Phi_{21}(t) = M\,I_1(t)/N_2 per turn, so the EMF induced around coil 2 is

\varepsilon_2 \;=\; -N_2\,\frac{d\Phi_{21}}{dt} \;=\; -M\,\frac{dI_1}{dt}

Symmetrically \varepsilon_1 = -M\,dI_2/dt. Transformers, wireless chargers, and ignition coils all run on this equation.

Mutual inductance of a coil wound over a solenoid

The cleanest derivable case is the coaxial pair: a long solenoid (coil 1) of n_1 turns per metre, cross-section A, length \ell, carrying current I_1; a secondary coil (coil 2) of N_2 turns wound tightly over the middle of the solenoid, so that every secondary turn encloses the same axial flux as each primary turn.

Step 1. Field inside the solenoid: B_1 = \mu_0 n_1 I_1 (derived above).

Step 2. Flux through one secondary turn: \Phi_{21} = B_1 A = \mu_0 n_1 I_1 A.

Step 3. Flux linkage in the full secondary: N_2\Phi_{21} = \mu_0 n_1 N_2 A\,I_1.

Step 4. Read off M from N_2\Phi_{21} = M\,I_1:

\boxed{\;M \;=\; \mu_0\, n_1 N_2 A\;}

Step 5. Sanity check against the reciprocity theorem. With n_1 = N_1/\ell where N_1 is the total primary turns,

M \;=\; \mu_0\,\frac{N_1 N_2 A}{\ell}

This is symmetric in N_1, N_2 — swapping which coil is primary and which is secondary gives the same M. Reciprocity holds, as it must.

Mutual inductance — secondary coil over a primary solenoid A long primary solenoid drawn as a row of loops. A shorter, tightly-wound secondary coil is wound around its middle. Field lines from the primary thread through both the primary and secondary loops. The secondary is connected to a voltmeter showing an induced EMF of minus M times dI1 over dt. Primary solenoid: n₁ turns/m, current I₁ Secondary coil: N₂ turns B₁ V ε₂ = −M dI₁/dt
A primary solenoid carries $I_1$; a secondary of $N_2$ turns is wound around its middle section. Every secondary turn sees the full axial flux $B_1 A = \mu_0 n_1 I_1 A$. When $I_1$ changes, the EMF induced in the secondary is $\varepsilon_2 = -M\,dI_1/dt$ with $M = \mu_0 n_1 N_2 A$.

Coupling coefficient and the bound M \le \sqrt{L_1 L_2}

The two self-inductances L_1, L_2 and the mutual inductance M between the same pair of coils are not independent. An energy argument — which you can follow in Energy Stored in an Inductor — shows that

M \;\le\; \sqrt{L_1 L_2}

with equality when every field line from coil 1 passes through coil 2 and vice versa (perfect flux linkage). Define the coupling coefficient

k \;=\; \frac{M}{\sqrt{L_1 L_2}},\qquad 0 \le k \le 1

Real transformers with laminated iron cores achieve k \approx 0.99; wireless chargers with a small air gap between coils have k \sim 0.5–0.8; two coils placed near each other with no shared core might have k \sim 0.01. The coupling coefficient is the single number that tells you how efficient a transformer, a wireless charger, or an ignition coil really is.

Worked examples

Example 1: The inductance of a BLDC fan stator coil

A stator coil in a Bajaj BLDC ceiling fan is a solenoid with N = 300 turns wound tightly around a soft-iron stator tooth. The tooth has cross-sectional area A = 2.0\text{ cm}^2 = 2.0\times 10^{-4}\text{ m}^2 and effective magnetic path length \ell = 4.0\text{ cm} = 0.04\text{ m}. The iron has \mu_r = 2000. Find the inductance of the coil and the self-induced EMF when the current changes at dI/dt = 500 A/s (typical for an unplug event).

Iron-core solenoid — stator coil schematic A schematic showing an iron-core solenoid. The iron core is drawn as a hatched rectangle in the centre. A coil of N turns is wound around it, drawn as a row of vertical loops. The length l and cross-section A are labelled, along with the 300 turns and mu_r equal to 2000. iron core (μ_r = 2000) ℓ = 4 cm A = 2 cm² N = 300 turns
Schematic of one stator coil — 300 turns wound on an iron tooth with cross-section $A = 2$ cm² and path length $\ell = 4$ cm.

Step 1. Use the iron-core solenoid formula, L = \mu_0\mu_r N^2 A/\ell.

Step 2. Plug in numbers.

L \;=\; \frac{(4\pi\times 10^{-7})(2000)(300)^2(2.0\times 10^{-4})}{0.04}

Why: \mu_0 = 4\pi\times 10^{-7} T·m/A; the iron multiplies this by \mu_r; N^2 = 90000; the area is in m²; the length is in metres. Plug-and-chug.

Step 3. Evaluate step by step.

\mu_0 \mu_r \;=\; 4\pi\times 10^{-7}\times 2000 \;=\; 8\pi\times 10^{-4} \;\approx\; 2.513\times 10^{-3}\text{ T·m/A}
N^2 A \;=\; 90000\times 2.0\times 10^{-4} \;=\; 18\text{ m}^2
L \;=\; \frac{2.513\times 10^{-3}\times 18}{0.04} \;=\; \frac{4.524\times 10^{-2}}{0.04} \;=\; 1.131\text{ H}

Why: breaking it into three factors keeps the powers of ten tractable.

Step 4. Self-induced EMF at dI/dt = 500 A/s.

|\varepsilon| \;=\; L\,|dI/dt| \;=\; 1.131\times 500 \;\approx\; 566\text{ V}

Result. L \approx 1.13 H, self-induced EMF at typical unplug dI/dt \approx 566 V.

What this shows. A modest ceiling-fan stator coil, with nothing fancy about its geometry, can generate several hundred volts of back-EMF during a sudden current break. This is why every fan, every washing-machine motor, and every BLDC controller in the country has a flyback diode or a snubber network across the winding — a one-way or energy-absorbing bypass that lets the inductor dump its stored energy somewhere harmless when the switch opens. Without snubbing, the first switching cycle would blow up the controller's transistor.

Example 2: Mutual inductance of a Maruti 800 ignition coil

A Maruti 800 ignition coil is a classic step-up transformer: primary of N_1 = 200 turns wound on a laminated iron bar; secondary of N_2 = 20{,}000 turns wound over it. Assume an effective coupling coefficient k = 0.95 and that the primary inductance is L_1 = 8 mH. The contact breaker chops the primary current from I_1 = 4 A to 0 in \Delta t = 40\ \mus. Estimate the peak voltage generated across the secondary (the voltage that fires the spark plug).

Step 1. First find L_2, then M.

The primary has L_1 = \mu_0 \mu_r N_1^2 A/\ell and the secondary L_2 = \mu_0 \mu_r N_2^2 A/\ell — same core, same path, just a different number of turns. So

\frac{L_2}{L_1} \;=\; \left(\frac{N_2}{N_1}\right)^2 \;=\; \left(\frac{20000}{200}\right)^2 \;=\; 10{,}000
L_2 \;=\; 10{,}000 \times 8\text{ mH} \;=\; 80\text{ H}

Why: for two coils sharing the same core, inductance scales as the square of turn count. The secondary has 100× the turns, so 10000× the inductance.

Step 2. Mutual inductance from the coupling coefficient.

M \;=\; k\sqrt{L_1 L_2} \;=\; 0.95\sqrt{(8\times 10^{-3})(80)} \;=\; 0.95\sqrt{0.64}
M \;=\; 0.95\times 0.8 \;=\; 0.76\text{ H}

Step 3. The EMF in the secondary when the primary current collapses.

|\varepsilon_2| \;=\; M\,\frac{|\Delta I_1|}{\Delta t} \;=\; 0.76 \times \frac{4}{40\times 10^{-6}} \;=\; 0.76 \times 10^5 \;=\; 76{,}000\text{ V}

Why: dI/dt during the break is approximately \Delta I_1/\Delta t = 4\text{ A}/40\ \mu\text{s} = 10^5 A/s. Multiply by M.

Step 4. Reality check against the actual spark voltage. Real spark plugs fire between 15 and 30 kV. Our estimate of 76 kV overshoots because the secondary's own self-inductance and capacitance, plus the finite spark breakdown voltage of the plug gap, clamp the voltage once the spark forms. The raw Faraday-law estimate is the voltage the coil would produce into an open circuit; the actual voltage is limited by the first load that breaks down.

Result. L_2 \approx 80 H; M \approx 0.76 H; peak secondary EMF \approx 76 kV open-circuit, clamped to 15–30 kV by the spark plug's breakdown. Enough to leap across the 0.7 mm plug gap and ignite the petrol–air mixture.

Ignition-coil schematic A schematic of an ignition coil. A battery labelled 12 V and a switch (the contact breaker) are in series with a primary coil of 200 turns. A secondary coil of 20000 turns is wound over the primary on the same iron core. One end of the secondary goes to a spark plug with a small gap. 12 V breaker iron core N₁ = 200 (primary) N₂ = 20000 (secondary) spark gap
The coil in series with the 12 V battery and a contact breaker forms the primary circuit. Over the iron core is wound the high-turn-count secondary, one end going to the spark plug. When the breaker opens, $dI_1/dt$ is huge; the resulting secondary EMF leaps the spark gap.

What this shows. The N_2/N_1 = 100 turns ratio gives the coil its step-up factor, and the sudden dI/dt at the breaker opening provides the needed rate of change. The same basic construction — two coils, one iron core, a turns ratio — is every transformer in India, from the tiny 230/12 V unit in your phone charger to the 400/220 kV transformers at a PGCIL substation.

Common confusions

You have the geometric formulas. What follows is JEE-advanced-level extensions: the LR circuit's time-constant, energy and reciprocity, the flux-linkage viewpoint versus the circuit viewpoint, and how inductance adds when coils are combined.

The LR circuit — switch-on and switch-off transients

Put an inductor L in series with a resistor R and a battery of EMF \varepsilon_0. Close the switch at t = 0. Kirchhoff's voltage law around the loop:

\varepsilon_0 \;=\; IR + L\,\frac{dI}{dt}

Why: the battery's EMF must equal the resistive drop plus the back-EMF across the inductor.

This is a first-order linear ODE. Try I(t) = I_\infty + Ce^{-t/\tau}. Substituting: I_\infty = \varepsilon_0/R (steady state) and \tau = L/R (time constant). With I(0) = 0, C = -I_\infty, so

I(t) \;=\; \frac{\varepsilon_0}{R}\left(1 - e^{-t/\tau}\right),\qquad \tau = L/R

The current rises exponentially toward its steady-state value. At t = \tau, the current is 63\% of final; at t = 5\tau, it is within 1\%. For the ceiling-fan stator coil of Example 1 (L = 1.13 H) connected through a winding resistance of R = 10 Ω, \tau = 0.113 s — the current takes almost a second to fully settle after switch-on. This "starting transient" is why a BLDC controller ramps the current gradually instead of slamming full voltage onto the winding — otherwise the inrush would trip the fuse.

Symmetrically, if the switch is opened (ideally, shorting the inductor through the resistor), I(t) = I_0 e^{-t/\tau}: exponential decay with the same time constant.

Energy stored in an inductor

Integrating P = \varepsilon I = LI\,dI/dt (the rate at which the source does work against the back-EMF) from 0 to I:

U \;=\; \int_0^I LI'\,dI' \;=\; \tfrac{1}{2}LI^2

The full derivation, including the reverse argument (U = \int B^2/2\mu_0\,dV integrated over all space), is in Energy Stored in an Inductor. The key point: an inductor storing current I is storing energy \tfrac{1}{2}LI^2 in the magnetic field around it. That energy has to come from somewhere (the battery during switch-on) and has to go somewhere (the resistor during switch-off). An unplug-spark is this energy being dumped, violently, into the air.

Reciprocity: why M_{12} = M_{21}

Write the total magnetic energy of two coupled coils carrying I_1 and I_2:

U \;=\; \tfrac{1}{2}L_1 I_1^2 + \tfrac{1}{2}L_2 I_2^2 + M_{12} I_1 I_2

Build this energy in two different orders: (A) ramp I_1 first, then I_2; (B) ramp I_2 first, then I_1. The final state is the same, so the final energy is the same. The work done against the mutual-coupling term during the second ramp is M_{21} I_1 I_2 in order A and M_{12} I_1 I_2 in order B. Equating, M_{12} = M_{21} \equiv M. No assumption about the geometry — it follows from energy conservation alone.

Writing the energy using M:

U \;=\; \tfrac{1}{2}L_1 I_1^2 + \tfrac{1}{2}L_2 I_2^2 + M I_1 I_2

Demanding U \ge 0 for every choice of I_1, I_2 is the quadratic form \tfrac{1}{2}L_1 x^2 + \tfrac{1}{2}L_2 y^2 + M xy \ge 0. Positivity requires M^2 \le L_1 L_2, i.e. |M| \le \sqrt{L_1 L_2}. This is where the coupling-coefficient bound comes from.

Flux-linkage (N\Phi) versus flux (\Phi) — the pedagogical trap

The defining equation you will see in some Indian textbooks is \Phi = LI, while others write N\Phi = LI. Both are correct, but they mean different things by \Phi. The clean way to avoid confusion:

  • \Phi = flux through one turn of the coil (just BA for a uniform field). This is a local quantity.
  • \Lambda \equiv N\Phi = total flux linkage of the coil. This is the quantity Faraday's law drives: \varepsilon = -d\Lambda/dt.
  • \Lambda = LI, always. This is the defining relation.

When a textbook writes \Phi = LI, they usually mean "flux linkage \Phi", not "flux per turn \Phi". Always check by dimensional sanity: inductance L has units Wb/A, so whatever goes on the left of LI must have units of total webers linking the coil — that is flux linkage, not single-turn flux.

Series and parallel combinations

In series, with no mutual coupling between them:

L_\text{eq} \;=\; L_1 + L_2

The same dI/dt flows through both, so the back-EMFs add.

In series, with mutual inductance M between them:

L_\text{eq} \;=\; L_1 + L_2 \pm 2M

The sign depends on whether the two coils' fluxes aid or oppose each other. If they aid (both windings wound the same way), +2M; if they oppose, -2M. This is why transformer designers care about "dot notation": the dots mark terminals of equal polarity so that series connections can be made with the correct sign.

In parallel (no mutual coupling):

\frac{1}{L_\text{eq}} \;=\; \frac{1}{L_1} + \frac{1}{L_2}

Same as resistors in parallel — the same voltage drops across both, currents divide.

Hysteresis and the effective inductance

In iron-core inductors, \mu_r is not a constant — it depends on H (the field strength), with a peak at some intermediate H and falling toward 1 at saturation. Moreover, because of hysteresis, \mu_r is multivalued: the same H on the way up and on the way down gives different B. See Hysteresis and Permanent Magnets for the full picture. The upshot: the inductance L of an iron-core coil is a slowly-varying function of the DC bias current, and textbook problems always assume you are in the linear low-field regime where \mu_r is approximately constant.

ISRO's solenoid valves — inductance at rocket-fuel pressures

Every liquid-propellant engine on the PSLV, GSLV, and LVM3 uses solenoid valves to switch propellant flow on a precise millisecond schedule. A typical valve has a stator coil with L \approx 10 mH; the control circuit must overcome this inductance to open the valve quickly. A 28 V bus driving into L = 10 mH through a 1 Ω resistance gives \tau = 10 ms — too slow for rapid valve switching. The trick is to drive the coil with a much higher voltage (say 100 V) for a few milliseconds during opening (hard switching), then drop to a holding voltage. The same trick — high-voltage pulse to open, low-voltage hold — is used in every injector solenoid in every Maruti fuel-injection system.

The inductance of a coaxial cable

A perhaps surprising example: a coaxial cable (inner conductor radius a, outer shield radius b, length \ell) has self-inductance per unit length

\frac{L}{\ell} \;=\; \frac{\mu_0}{2\pi}\ln(b/a)

derived by integrating B^2/(2\mu_0) between the conductors (where B = \mu_0 I/2\pi r by Ampère's law). A 1 m cable with b/a = 3 has L \approx 0.22 μH — small, but not negligible at GHz signalling rates. This is why long coaxial runs at high frequency need careful impedance matching.

Where this leads next