
Physics II course sequence · Interactive lesson
Self-inductance
Every circuit sits inside the magnetic field of its own current, and that field resists being changed. Here you turn that resistance into a single number computed from the winding alone — fixed long before any current flows — and learn to read its minus sign.
UPII-11.01 · Lesson 1 of 8 in this unit. Position is not a completion score.Build the model
A circuit's own field fights every change in its own current.
The key idea
Self-inductance is the flux linkage a circuit's own current produces per unit current, L = NΦB/I — a constant fixed by the winding's shape and the material inside it, not by the current.
Example. A 500-turn solenoid 25 cm long with a 4.0 cm² cross-section has L = μ₀N²A/ℓ = 0.50 mH, so a current changing at 200 A s⁻¹ induces 0.10 V across its ends.
The complete physical model and assumptions
Self-inductance is the one number that closes the loop between a circuit's current and the field that current makes. Every contribution to that field scales with I, so the flux linkage Λ = NΦB scales with I too, and the ratio Λ/I turns out to be a property of the hardware rather than of what the hardware is doing: turns, dimensions and the material filling the winding decide it, and nothing else does. Measure L at 10 mA and it reads the same at 10 A. That constant then does two jobs. Magnetically it says how much flux a coil wraps round itself per ampere. Electrically, through Faraday's law, it converts a rate of change of current into a terminal voltage, ε = −L dI/dt, which is why an inductor is invisible to a steady current and violent about a sudden one: the coil a 1.5 V cell drives quietly can throw hundreds of volts across an opening switch. This lesson computes L for a solenoid, a toroid and a coaxial line by one four-step recipe, and then marks the two places the linear model stops — a saturating iron core, where Λ(I) bends over and only dΛ/dI survives, and a moving geometry, where a second term I dL/dt reappears.
1A circuit sits inside its own field
Switch a current I into any closed loop and the loop finds itself threaded by the field that current is making. Biot–Savart gives every element of that field as a fixed geometric factor times I, so doubling the current doubles B everywhere at once, doubles the flux ΦB = ∫B⋅dA through one turn, and doubles the flux linkage Λ = NΦB — the flux counted once for each of the N turns it passes through. What survives the doubling is the ratio: L = Λ/I = NΦB/I is the same number at 0.10 A and at 10 A. That ratio is the self-inductance, and its unit is the henry, with 1 H = 1 Wb A⁻¹ = 1 V s A⁻¹ = 1 Ω s. Fix the sense of positive flux to match the sense of positive current by the right-hand rule and L comes out positive for every circuit, because a current's own flux always links it the same way round.
2Faraday's law, turned on the circuit itself
Faraday's law does not care whose current made the flux. Apply ε = −dΛ/dt to a coil's own linkage and substitute Λ = LI. If the coil is rigid, so that L is a constant, the product rule leaves one term: ε = −L dI/dt. Hold the current steady and the emf is exactly zero no matter how large that current is; change it and the emf is set by the rate alone. The minus sign is Lenz's law written for a single circuit: the induced emf acts against the change in the current, never against the current itself. In the passive sign convention, with the terminal voltage measured from the end the current enters, this appears as v = L dI/dt: positive while the current rises, so the coil draws energy from the source, and negative while it falls, so the coil returns it. If the geometry is allowed to move, L becomes a function of time and ε = −L dI/dt − I dL/dt; everything in this lesson assumes the first term alone.
3Four steps that compute any inductance
The recipe never changes. Assume a current I in the circuit; find B from Ampère's law or Biot–Savart; integrate B over the surface bounded by one turn to get ΦB; multiply by N and divide by I. The current cancels in the last step, and that cancellation is the proof that L is pure geometry. Run it on a long solenoid of N turns, length ℓ and cross-section A. Ampère's law on a rectangular loop straddling the winding gives a uniform axial interior field B = μ₀NI/ℓ, with almost nothing outside. The flux through one turn is ΦB = BA = μ₀NIA/ℓ, the linkage is N times that, and L = μ₀N²A/ℓ. For N = 500 turns over ℓ = 0.25 m with A = 4.0 cm², L = (4π × 10⁻⁷)(500)²(4.0 × 10⁻⁴)/0.25 = 5.0 × 10⁻⁴ H. The N² is the signature of self-inductance: N turns make N times the field, and that field then links N turns.
4When the field is not uniform, integrate across it
A toroid shows what to do when B varies across the cross-section. Ampère's law on a circle of radius r inside the windings gives B = μ₀NI/(2πr), so the field is strongest against the inner wall. Take the cross-section to be a rectangle of height h running from inner radius a to outer radius b, and slice it into strips of width dr at radius r. Each strip carries dΦB = B h dr, so the flux through one turn is ΦB = (μ₀NIh/2π) ∫dr/r from a to b = (μ₀NIh/2π) ln(b/a). Multiply by N and divide by I to get L = μ₀N²h ln(b/a)/(2π). This result is exact for an ideal winding, because a toroid's flux never leaves the ring and there are no ends to leak from. The logarithm is what a 1/r field always leaves behind, and the same integral returns for a coaxial cable, where the flux sits in the gap between the conductors and the circuit has only one turn.
5What the sign does when a switch opens
Numbers make the minus sign concrete. Take L = 0.50 mH carrying 3.0 A and let the current fall to zero in 2.0 ms: dI/dt = −1500 A s⁻¹, so ε = −L dI/dt = +0.75 V, an emf pointing the way the current was already flowing and trying to keep it alive. Now open a mechanical switch, so the same 3.0 A collapses in 2.0 μs. The rate is a thousand times larger and so is the emf: 750 V, across a coil that a 1.5 V cell was happily driving. That voltage appears across the separating contacts, and it is why they arc, why relay coils are fitted with a flyback diode, and why an inductor's terminals deserve respect after the supply is switched off. Nothing about the coil changed — only dI/dt did.
6Where the linear model stops
Two assumptions carry everything above: that B is proportional to I, and that the geometry holds still. Fill a 1.5 mH air solenoid with a ferromagnetic core and, while the material stays linear, the flux rises by a factor of its relative permeability — perhaps 2000, taking L to roughly 3 H for no extra copper. But iron's B against H curve bends. Once the domains are aligned, extra current adds almost no extra flux and Λ(I) flattens; beyond that knee L = Λ/I is no longer a constant, and designers quote the differential inductance dΛ/dI instead, which collapses towards the air-core value in saturation. Air and vacuum have no such limit, and their L holds up to the current that melts the wire. A second limit appears at high frequency: current crowds towards the conductor surfaces and the turn-to-turn capacitance of the winding starts carrying its own current, so a real coil stops behaving like a pure inductance well before Λ = LI itself fails.
Key equations
Read the units, assumptions and limits before substituting.
Self-inductance from flux linkage
L = Λ / I = N ΦB / I
One number for the whole winding, and ΦB ∝ I is what makes it independent of I.
Λ flux linkage in Wb, ΦB flux through one turn in Wb, I in A, L in HSelf-induced emf
ε = −L dI/dt
Converts a rate of change of current into a terminal voltage you can measure.
ε in V, L in H, dI/dt in A s⁻¹; rigid geometry, so L stays constantLong air-core solenoid
L = μ₀ N² A / ℓ
N² because N turns build the field and then N turns link it; needs ℓ ≫ diameter.
μ₀ = 4π × 10⁻⁷ T m A⁻¹, A cross-section in m², ℓ winding length in mToroid of rectangular cross-section
L = μ₀ N² h ln(b/a) / (2π)
Exact with no end correction, because a toroid's flux never leaves the ring.
a inner radius, b outer radius, h height, all in m; B falls as 1/r insideCoaxial line, external flux
L / ℓ = (μ₀ / 2π) ln(b/a)
Self-inductance with no coil at all, in H m⁻¹; add μ₀/8π per metre for flux inside the core wire.
a radius of the inner conductor, b inner radius of the outer, ℓ length in mInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Double the turns from 250 to 500 and watch the line steepen four times over, then sweep the current slider and see the marker travel along a line whose slope refuses to move — that slope is L.
Live results
- SELF-INDUCTANCE L
- 1.51 mH
- FLUX PER TURN ΦB
- 8.64 μWb
- LINKAGE Λ = N ΦB
- 3.02 mWb
- |ε| AT dI/dt = 100 A s⁻¹
- 151 mV
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Turns N: 350; Winding length ℓ: 20 cm; Current I: 2.00 A
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The minus sign in ε = −L dI/dt gets read as though the inductor opposed the current, so students expect a coil carrying a large steady current to hold a large back-emf, or to push current backwards. It does neither. With dI/dt = 0 the emf is exactly zero however big I is: an ideal inductor carrying 100 A steadily is a piece of wire. What the sign opposes is the change. While the current rises the induced emf fights the rise; while it falls, that same emf reverses and fights the fall, driving current forward through whatever path it can find. And L itself is never negative and never a function of I — it is decided by turns, dimensions and core the moment the coil is wound.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A single-layer air-core solenoid has 400 turns wound over a length of 20.0 cm on a form of radius 1.50 cm. Find its self-inductance, then find the magnitude of the emf induced in it while its own current rises steadily at 50.0 A s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Cross-section: A = πr² = π(1.50 × 10⁻² m)² = 7.069 × 10⁻⁴ m².
- The winding is long compared with its diameter (0.200 m against 0.0300 m), so the uniform interior field B = μ₀NI/ℓ applies and L = μ₀N²A/ℓ.
- L = (4π × 10⁻⁷ T m A⁻¹)(400)²(7.069 × 10⁻⁴ m²) / (0.200 m) = 7.11 × 10⁻⁴ H.
- |ε| = L |dI/dt| = (7.11 × 10⁻⁴ H)(50.0 A s⁻¹) = 3.55 × 10⁻² V. Note that the value of the current never entered either line.
Answer. L = 0.711 mH, and |ε| = 35.5 mV while the current ramps.
MediumExample 2
A toroid of rectangular cross-section carries 900 turns. Its inner radius is 40.0 mm, its outer radius 55.0 mm and its height 12.0 mm. Find its self-inductance exactly, then compare with the estimate obtained by treating the field as uniform at the mean radius.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Ampère's law on a circle of radius r inside the winding gives B = μ₀NI/(2πr). The field is not uniform across the cross-section, so the flux needs an integral.
- A strip at radius r, of width dr and height h, carries dΦB = B h dr, so ΦB = (μ₀NIh/2π) ∫dr/r from a to b = (μ₀NIh/2π) ln(b/a).
- Multiply by N and divide by I: L = μ₀N²h ln(b/a)/(2π). With μ₀/2π = 2.00 × 10⁻⁷ and ln(55.0/40.0) = 0.3185, L = (2.00 × 10⁻⁷)(900)²(0.0120 m)(0.3185) = 6.19 × 10⁻⁴ H.
- Uniform-field estimate: A = h(b − a) = 1.80 × 10⁻⁴ m² at mean radius r̄ = 47.5 mm gives L ≈ μ₀N²A/(2πr̄) = 6.14 × 10⁻⁴ H.
- The estimate falls 0.84% short because 1/r is convex, so its average across the cross-section exceeds 1/r̄. The shortfall grows as b/a grows and vanishes as the ring becomes thin.
Answer. L = 0.619 mH exactly; the uniform-field estimate gives 0.614 mH, low by 0.84%.
HardExample 3
You have 30.0 m of enamelled wire and a cylindrical form of radius 2.00 cm. Wound as a single layer, how many turns do you get, and what self-inductance results if the winding is spread over 30.0 cm rather than 15.0 cm? Then, with the shorter winding carrying I(t) = (4.00 A s⁻²)t², find the induced emf at t = 0.500 s and say how far you would trust it.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Each turn spends one circumference of wire: 2πr = 2π(0.0200 m) = 0.1257 m, so (30.0 m)/(0.1257 m) = 238.7 turns' worth of wire. Only 238 turns are complete — a 239th would need 30.03 m, which you do not have. The wire length fixes N; it does not fix ℓ.
- A = πr² = 1.2566 × 10⁻³ m². With ℓ = 0.300 m, L = μ₀N²A/ℓ = (4π × 10⁻⁷)(238)²(1.2566 × 10⁻³) / (0.300) = 2.98 × 10⁻⁴ H.
- Halving ℓ to 0.150 m doubles L to 5.96 × 10⁻⁴ H. With N fixed by the wire, packing the same turns into half the length doubles the turn density N/ℓ, hence the field and the flux.
- dI/dt = d[(4.00 A s⁻²)t²]/dt = (8.00 A s⁻²)t, which is 4.00 A s⁻¹ at t = 0.500 s.
- |ε| = L|dI/dt| = (5.963 × 10⁻⁴ H)(4.00 A s⁻¹) = 2.39 × 10⁻³ V.
- Trust check: the turns must physically fit, (0.150 m)/238 = 0.630 mm per turn, so the enamelled wire can be no thicker than that. And ℓ/2r = 3.75 is only moderately long, so end leakage makes μ₀N²A/ℓ an overestimate of order ten per cent.
Answer. N = 238 turns; L = 0.298 mH at 30.0 cm and 0.596 mH at 15.0 cm; |ε| = 2.39 mV at t = 0.500 s, good to roughly ten per cent because of end effects.
Quick check
Select an answer and explain it before checking.
Review & continue
Self-inductance is the flux linkage a circuit's own current produces per unit current, L = NΦB/I — a constant fixed by the winding's shape and the material inside it, not by the current.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.