Physics · Electrodynamics

Self-induction of a coil

Q: Why does switching off a coil produce a voltage surge?

Initial energy of a linear inductor is $W=\frac12LI^2$. At switch-off that energy must move into other stores or loads. Finite voltage cannot make ideal inductor current disappear discontinuously. At a real opening switch, parasitic capacitance, an arc or protection circuitry can provide the current path; the complete system determines peak voltage and its waveform. A freewheeling diode provides a different decay path and limits voltage according to the circuit, often slowing current decay. Inductance and initial current alone do not specify one turn-off time or voltage peak. An ignition system additionally uses magnetically coupled windings, rather than demonstrating a universal fixed multiplication from a single isolated coil.

Current change alters self-flux linkage and produces emf opposing that change. Ideal passive inductor voltage uses the opposite voltage reference.

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Formula

Calculate

U_{in d} = - L \cdot \frac{d I}{d t}

U_ind in V · L in H (henry) · dI/dt in A/s

Variables & units

Variables & units
Symbol	Meaning	Unit
$U_{\mathrm{ind}}$	Oriented self-emf in the chosen traversal; not ideal passive component voltage	V
$L$	Inductance of the coil	H
$dI/dt$	Rate of change of the current	A/s

LaTeX

U_{ind} = -L \cdot \frac{dI}{dt}

QUANTA STUDYquanta-study.de

Upper current trace rises 0→1 A, stays at 1 A and falls to 0.5 A. Lower passive voltage is +50,0 and−50 V over the open intervals. — Ideal constant L=0.100 H: current rises linearly during 0–2 ms, is constant during 2–5 ms and falls linearly during 5–6 ms. Passive voltage is +50 V,0 V,−50 V; emf has opposite sign. Derivative is not uniquely defined at corners. Final current remains 0.500 A.

Self-induction calculator

Find the induced emf, inductance or current change of an ideal coil when the current changes steadily, plus the stored energy.

Calculation model

Find

Inductance L

Inductance of the coil, e.g. 200 mH.

Initial current I_i

Current at the start of the change, with sign, e.g. 0 A when switching on.

Final current If

Current at the end of the change, with sign, e.g. 2 A.

Duration of the current change Δt

How long the current changes steadily, e.g. 100 ms.

Filled with example values. Type your own and the result updates straight away.

Result: Induced emf ℰ

ℰ = -4 V

With L = 200 mH, I_i = 0 A, If = 2 A and Δt = 100 ms, ℰ = -4 V.

Working · rearrange for ℰ, substitute, evaluate

1. Rearrange for ℰ
$\mathcal E=-L\frac{I_f-I_i}{\Delta t}$
2. Substitute values (in SI units)
- L = 200 mH = 0.2 H
- Δt = 100 ms = 0.1 s
$\mathcal E=(-1)\cdot 0.2\cdot\frac{2-0}{0.1}$
3. Result
ℰ = -4 V

Quick tips

Lenz's law: if the current rises, ℰ is negative and opposes the rise; if it falls, ℰ is positive and opposes the fall. The terminal voltage uL = −ℰ always has the opposite sign.
Unit check: 1 H · 1 A/s = 1 V and 1 H · (1 A)² = 1 J. Half the time for the current change means twice the voltage.

Assumptions and limits

Ideal inductor with constant L > 0 and fixed geometry. The specified current changes linearly from I_i to If over Δt > 0. This does not predict a real open-switch peak: winding resistance, protection paths, coupling and parasitic capacitance are excluded. Emf is positive along positive-current traversal: ℰ = −LΔI/Δt.

Derivation & background

Constant linear inductance gives flux linkage $\lambda=LI$ , hence $\mathcal E=-L\frac{dI}{dt}$ . Passive component voltage is $u_L=L\frac{dI}{dt}$ . A long air-core solenoid approximately has $L=\mu_0\frac{N^2A}{\ell}$ . Field energy is $W=\frac12LI^2$ . Switch-off peak voltage depends on the entire decay path; L and initial current alone do not determine it.

Exam blueprint

Validity range

Constant linear inductance and fixed geometry; emf and passive terminal voltage use different references. Real terminal voltage may include winding resistance. Variable geometry, nonlinear cores and parasitic capacitance need an extended model.

Derivation steps

Apply Faraday’s law to self-flux linkage while retaining voltage reference.

1Define flux linkage $\lambda=\sum_k\Phi_k$ ; equal turn flux gives $\lambda=N\Phi$ .
2Linear fixed geometry gives $\lambda=LI$ with positive constant L.
3Faraday’s law gives emf $\mathcal E=-\frac{d\lambda}{dt}$ .
4Constant L gives $\mathcal E=-L\frac{dI}{dt}$ .
5Ideal-component passive voltage instead is $u_L=+L\frac{dI}{dt}$ .
6A linear ramp permits constant finite-difference rate; other waveforms require interval average or actual derivative.

Rearrangements

Inductance

L=\frac{|\mathcal E|}{\left|\frac{\mathrm dI}{\mathrm dt}\right|}

Requires a nonzero measured current-change rate.

Rate of current change

\frac{dI}{dt}=-\frac{\mathcal E}{L}=\frac{u_L}{L}

Distinguish loop emf from ideal passive terminal voltage; a magnitude equation gives only magnitude of the rate.

Inductance of a long air-core solenoid

L\approx\mu_0\frac{N^2A}{\ell}

Tightly wound long air-core solenoid of fixed geometry; not automatically valid for a nonlinear iron core.

Task variant

Compute L of a coil with $N=1000$ , $A=20\,\mathrm{cm^2}$ , $\ell=0.4\,\mathrm m$ (no core).

$L\approx\mu_0\frac{N^2A}{\ell}=(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})\frac{1000^2(2.0\times10^{-3}\,\mathrm{m^2})}{0.40\,\mathrm m}\approx6.285\,\mathrm{mH}$ Tightly wound long air-core solenoid, fringe fields neglected.

A choke ( $L=0.3\,\mathrm H$ ) sees 12 V. How fast does the current rise?

$\frac{\mathrm dI}{\mathrm dt}=\frac{u_L}{L}=\frac{12\,\mathrm V}{0.30\,\mathrm H}=40\,\mathrm{A\,s^{-1}}$ The 12 V is positive passive voltage directly across the ideal inductance, with no winding resistance.

Common mistakes

Dropping the minus sign as mere convention.

It encodes Lenz rule: the induced voltage opposes the change in current.

Scaling L linearly with the number of turns.

At unchanged length, cross-sectional area and linear material model, a long solenoid has $L\propto N^2$ . Doubling turns then quadruples L; recheck altered winding geometry.

Expecting an induced voltage at constant current.

With constant self-current and fixed geometry, self-emf is zero. A real winding can still have resistive voltage drop $RI$ and heating loss; stored field energy does not vanish.

Substituting the area A in $cm^{2}$ .

Convert A to $m^{2}$ : $20 cm^{2} = 2\times{}10^{-3} m^{2}$ .

Exam context

Exams ask switch-on and switch-off transients (delayed current rise, voltage spike with a glow lamp), the energy $E=\frac12LI^2$ and deriving L for a long coil.

Formula cluster

Induction

Self-induction links the coil field, the induction law and the oscillating circuit.

ℰ = −N dΦB/dt

Faraday’s law of induction

B = μ₀·(N/l)·I

Magnetic Field of a Long Coil

T = 2π·√(L·C)

Thomson formula for an LC circuit

U₁/U₂ = n₁/n₂ = I₂/I₁

Transformer turns ratio

Worked example

For constant $L=0.5\,\mathrm H$ , current falls explicitly linearly from 2 A to zero over 10 ms. Approach: $\dot I=\frac{\Delta I}{\Delta t}=-200\,\mathrm{A/s}$ . Thus emf is $\mathcal E=+100\,\mathrm V$ in loop traversal and passive inductive voltage is $u_L=-100\,\mathrm V$ . Without a linear ramp these values are interval means rather than determined peaks.

Applications

Car ignition coil, freewheeling diodes in circuits, chokes and switch-mode power supplies, energy storage in coils, sparking at switches

How do you calculate with Self-Induction of a Coil?

Here is how to work through a typical Self-Induction of a Coil (U_ind = −L·(dI/dt)) task step by step:

1
Task
Compute L of a coil with $N=1000$ , $A=20\,\mathrm{cm^2}$ , $\ell=0.4\,\mathrm m$ (no core).
Solution path
$L\approx\mu_0\frac{N^2A}{\ell}=(1.257\times10^{-6}\,\mathrm{H\,m^{-1}})\frac{1000^2(2.0\times10^{-3}\,\mathrm{m^2})}{0.40\,\mathrm m}\approx6.285\,\mathrm{mH}$ Tightly wound long air-core solenoid, fringe fields neglected.
2
Task
A choke ( $L=0.3\,\mathrm H$ ) sees 12 V. How fast does the current rise?
Solution path
$\frac{\mathrm dI}{\mathrm dt}=\frac{u_L}{L}=\frac{12\,\mathrm V}{0.30\,\mathrm H}=40\,\mathrm{A\,s^{-1}}$ The 12 V is positive passive voltage directly across the ideal inductance, with no winding resistance.

Understand and apply the equation

Changing a coil current changes its own magnetic field. Induced emf opposes that change, whereas ideal-component terminal voltage uses the passive reference. A current waveform, an energy integral and a switch-off model clarify which voltage is meant and where the energy goes.

Why inductor current does not respond instantly

A source is switched on, but inductor current does not instantly reach its final value. The growing current builds a magnetic field whose change produces an emf opposing that growth. After switch-off, the same stored energy can keep current flowing. The component responds to current change, not merely current magnitude.[1][4]

Before calculating a sign, establish whether you mean the driving self-emf or the passive voltage drop across the ideal component. A real winding also has a resistive term. Our initial calculations use fixed geometry and constant linear L.[1][4]

Self-flux links every turn

\begin{aligned}\lambda&=\sum_{k=1}^{N}\Phi_k\\{}\lambda&=N\Phi\\{}\lambda&=LI\end{aligned}

λ is flux linkage in webers. Generally sum the oriented fluxes of all turns; λ=NΦ applies only when each turn links the same flux. Fixed geometry and a linear material response additionally give λ=LI.[1]

In the linear model, flux linkage produced by the same current increases in proportion to current. L combines geometry and suitable material response. Turn normals are chosen consistently with positive current direction, giving positive self-inductance. This does not claim that an arbitrary iron core has the same constant value at every current. Moving the winding can also change linkage.[1]

Distinguish induced emf and passive terminal voltage

\begin{aligned}\mathcal E_{\mathrm{self}}&=-\frac{d\lambda}{dt}=-L\frac{dI}{dt}\\{}u_L&=+L\frac{dI}{dt}\end{aligned}

Emf is positive in the chosen loop direction. uL is the passive voltage drop: positive current enters the positive voltage terminal. L is constant here.[1][2]

The emf minus sign describes Lenz’s opposition. It does not forbid a different correctly defined voltage reference from using a plus sign. Increasing positive current gives negative emf but positive passive terminal voltage, and the ideal component absorbs energy. Decreasing that current reverses both voltage signs. A real winding also has a resistive drop; total terminal voltage is not automatically only the inductive contribution.[1][4]

From Faraday to current-change rate

In the linear model L is initially the proportionality factor between current and self-flux linkage. Faraday’s law of induction acts on the time derivative of that linkage. Only the additional assumption of constant L allows it outside the derivative. This separates the field law from the component assumption.[1][2]

\begin{aligned}\mathcal E&=-\frac{d\lambda}{dt}\\{}\lambda&=LI\\{}\frac{d\lambda}{dt}&=L\frac{dI}{dt}\quad L=\mathrm{const}\\{}L&=\frac{|\mathcal E|}{\left|\frac{\mathrm dI}{\mathrm dt}\right|}\\{}\frac{dI}{dt}&=-\frac{\mathcal E}{L}=\frac{u_L}{L}\end{aligned}

Faraday’s law gives emf. The final rearrangement retains voltage reference; magnitudes alone cannot give a signed rate.[2][1]

The difference between final and initial current divided by interval length is an average rate. It replaces instantaneous derivative only for a linear ramp. At fixed L the resulting emf is otherwise the interval average, not necessarily the peak value. Area under the emf-time curve instead determines current change whether the curve is constant or curved.[1]

\Delta I=-\frac1L\int_{t_0}^{t_1}\mathcal E(t)\,dt

Assume the same sign convention and constant L.[1]

Explore further: inductance from coil geometry

\begin{aligned}B&\approx\mu_0\frac N\ell I\\{}\Phi&\approx BA\\{}\lambda&\approx NBA=\mu_0\frac{N^2A}{\ell}I\\{}L&\approx\mu_0\frac{N^2A}{\ell}\end{aligned}

Long tightly wound air-core solenoid; A is interior cross-sectional area and Φ approximates equal flux through each turn.[1]

More turns strengthen interior field at equal current, while the same flux links more turns: this produces quadratic dependence. This assumes unchanged length, area and material model. If winding also alters geometry, the simple factor-of-four rule needs further checking. A toroidal coil or a magnetic circuit with a gap needs its appropriate geometry expression.[1]

Voltage sign alone does not determine energy flow

\begin{aligned}p_L&=u_LI=LI\frac{dI}{dt}\\{}dW&=LI\,dI\\{}W(I)&=\frac12LI^2\\{}\Delta W&=\frac L2(I_f^2-I_0^2)\end{aligned}

Passive reference and lossless linear component. Positive power means energy absorption.[3]

Energy depends on current squared, not current sign. Even positive passive voltage can release energy if negative current decreases in magnitude: the deciding quantity is current times voltage. Henry equals volt-seconds per ampere; multiplying by amperes per second gives volts. Henry times amperes squared gives joules. A steady nonzero field can store energy while induced emf is zero at constant current.[3]

The Thomson LC oscillation relation leads to the ideal LC circuit, where an inductor and capacitor exchange energy. The inductor energy calculated here is only one contribution to that combined balance.[3][7]

Account fully for a ramp and switch-off energy

Current falls with explicit signs

Given: Constant $L=0.240\,\mathrm H$ ; positive current falls linearly from 1.80 A to 0.300 A in 12.0 ms. Passive voltage and emf use the references defined above.
Find: Find both voltages and change in stored energy.

Approach: the explicitly linear ramp makes the average rate equal to the constant instantaneous rate. Evaluate voltage signs and squared currents separately.
$\begin{aligned}\dot I&=\frac{I_f-I_0}{\Delta t}\\{}\mathcal E&=-L\dot I\\{}u_L&=L\dot I\\{}\Delta W&=\frac L2(I_f^2-I_0^2)\end{aligned}$
Convert milliseconds to seconds.
$\begin{aligned}\dot I&=\frac{0.300-1.80}{0.0120}\,\mathrm{A\,s^{-1}}=-125\,\mathrm{A\,s^{-1}}\\{}\mathcal E&=+30.0\,\mathrm V\\{}u_L&=-30.0\,\mathrm V\\{}\Delta W&=\frac{0.240}{2}(0.300^2-1.80^2)\,\mathrm J=-0.378\,\mathrm J\end{aligned}$

Result: Emf +30.0 V, passive voltage −30.0 V; stored energy falls by 0.378 J.

Check: Average current is 1.05 A; constant passive voltage times average current times 12.0 ms gives −0.378 J. Positive emf sustains positive current while it decays.

Energy of a long air-core solenoid

Given: $N=600$ , $A=3.00\,\mathrm{cm^2}$ , $\ell=0.300\,\mathrm m$ ; air-core model with $\mu_0=1.25663706\cdot10^{-6}\,\mathrm{H/m}$ . Current rises uniformly from 0 to 2.50 A over 20.0 ms.
Find: Calculate inductance, passive inductive voltage and final stored energy.

Approach: long-solenoid geometry gives L; the linear ramp gives voltage rate. Final energy must agree with supplied inductive work.
$\begin{aligned}L&=\mu_0\frac{N^2A}{\ell}\\{}u_L&=L\frac{I_f}{\Delta t}\\{}W_f&=\frac12LI_f^2\end{aligned}$
Convert area using the square of the length conversion.
$\begin{aligned}A&=3.00\cdot10^{-4}\,\mathrm{m^2}\\{}L&=(1.25663706\cdot10^{-6})\frac{600^2(3.00\cdot10^{-4})}{0.300}\,\mathrm H\\{}&=0.0004523893416\,\mathrm H\\{}u_L&=L\frac{2.50}{0.0200}\,\mathrm{A\,s^{-1}}\approx0.0565487\,\mathrm V\\{}W_f&=\frac12L(2.50)^2\approx0.00141372\,\mathrm J\end{aligned}$

Result: L≈0.452 mH, uL≈56.5 mV and Wf≈1.41 mJ.

Check: uL times average current 1.25 A times 20.0 ms gives the same energy. Copper loss would be an additional contribution, not part of this ideal inductive term.

Example: switch on a voltage

Resistance sets final current; L sets its rise

Given: At $t=0$ , a constant $U=6.00\,\mathrm V$ source is connected to ideal $L=0.200\,\mathrm H$ in series with total resistance $R=10.0\,\Omega$ . Initial current is zero; use passive inductor voltage and no other storage.
Find: Find current, inductor voltage and energy after one time constant.

Approach: source voltage divides between resistance and inductance. The differential equation determines the waveform.
$\begin{aligned}U&=RI+L\dot I\\\tau&=\frac LR\\I(t)&=\frac UR(1-e^{-t/\tau})\\u_L(t)&=Ue^{-t/\tau}\end{aligned}$
Substitution at one time constant: current has not reached its final value; inductor voltage is not yet zero.
$\begin{aligned}\tau&=20.0\,\mathrm{ms}\\I(\tau)&=(0.600\,\mathrm A)(1-e^{-1})\\&\approx0.379272\,\mathrm A\\u_L(\tau)&=(6.00\,\mathrm V)e^{-1}\approx2.20728\,\mathrm V\\W(\tau)&=\frac12LI(\tau)^2\approx14.3848\,\mathrm{mJ}\end{aligned}$

Result: After 20.0 ms: about 0.3793 A, +2.207 V passive inductor voltage and 14.38 mJ stored energy. Eventually they approach 0.600 A and 36.0 mJ.

Check: Resistor voltage $RI(\tau)\approx3.79272\,\mathrm V$ adds to the inductor voltage to give 6.00 V. Positive current and positive passive voltage still indicate energy absorption; $\mathcal E_{\mathrm{self}}=-u_L$ .

View diagram: switch-on response

Current rises while inductor voltage falls

QUANTA STUDYquanta-study.de

In the third example

\tau=20\,\mathrm{ms}

. The curves add to one because

RI+u_L=U

; the axis compares normalised quantities, not amperes with volts.[4]

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Five exercises with distinct voltage references

Calculate · Foundation

An ideal 80.0 mH inductor undergoes a linear current rise from 0.400 to 1.60 A in 6.00 ms. Find passive voltage and emf.

Show hint

First form the signed change rate.

Learning goal: Calculate terminal voltage and emf during a ramp.

Show solution and calculation

Approach: rate is +200 A/s.
Passive voltage +16.0 V; emf −16.0 V.

Check: positive current means absorption because passive voltage and current have the same sign.

\begin{aligned}\dot I&=\frac{1.60-0.400}{0.00600}=200\,\mathrm{A\,s^{-1}}\\{}u_L&=(0.0800)(200)\,\mathrm V=16.0\,\mathrm V\\{}\mathcal E&=-16.0\,\mathrm V\end{aligned}

[1]

Rearrange · Application

Current through constant linear L falls from 2.00 to 0.500 A. Measured emf-time integral is +0.0900 V·s in the chosen loop direction. Find L without assuming constant voltage.

Show hint

Integrate Faraday’s relation rather than using a peak value.

Learning goal: Determine L from an emf-time integral.

Show solution and calculation

Approach: the integral is $-L\Delta I$ .
With $\Delta I=-1.50\,\mathrm A$ , $L=0.0600\,\mathrm H$ .

Check: positive integral and negative current change give positive L; detailed waveform is unnecessary.

L=-\frac{\int\mathcal E\,dt}{\Delta I}=-\frac{0.0900\,\mathrm{V\,s}}{-1.50\,\mathrm A}=0.0600\,\mathrm H

[1]

Interpret · Application

An ideal $L=0.100\,\mathrm H$ has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

Show hint

At corners use one-sided values; derivative is not uniquely defined there.

Learning goal: Read voltage from curve slope rather than current height.

Show solution and calculation

Approach: multiply each slope by L.
Voltages are +50 V, 0 V, −50 V.

Check: the middle interval stores 0.050 J despite zero voltage; constant height does not mean loss of field.

\begin{aligned}u_1&=0.100\frac{1}{0.002}\,\mathrm V=50\,\mathrm V\\{}u_2&=0\\{}u_3&=0.100\frac{-0.500}{0.001}\,\mathrm V=-50\,\mathrm V\end{aligned}

[1]

Error analysis · Application

A real winding has $R=6.0\,\Omega$ and $L=0.200\,\mathrm H$ . At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.

Show hint

Total terminal voltage has two contributions.

Learning goal: Distinguish emf from resistive voltage drop.

Show solution and calculation

Approach: split terminal voltage into $RI$ and $L\dot I$ .
All 9.0 V is resistive drop, leaving zero inductive term and hence zero current rate.

Check: loss power is 13.5 W, not growing magnetic energy.

\begin{aligned}u_{\mathrm{coil}}&=RI+L\dot I\\{}\dot I&=\frac{9.0-(6.0)(1.50)}{0.200}\,\mathrm{A\,s^{-1}}=0\\{}P_R&=RI^2=13.5\,\mathrm W\end{aligned}

[4]

Transfer · Challenge

An ideal 0.120 H inductor initially carries 2.00 A. It is disconnected from the source but remains in a closed loop with 8.00 Ω total resistance. No capacitor or other loss path. Find time constant, initial passive inductor voltage and final heat.

Show hint

The loop is not open; resistance provides a determined current waveform.

Learning goal: Track switch-off energy through a closed resistive path.

Show solution and calculation

Approach: loop balance is $L\dot I+RI=0$ .
Time constant 15.0 ms; initial inductor voltage −16.0 V.
Initial energy 0.240 J becomes heat.

Check: integrating $RI^2$ over exponential decay gives the same energy; no arbitrary switch-off time is imposed.

\begin{aligned}\tau&=\frac LR=0.0150\,\mathrm s\\{}I(t)&=2.00e^{-t/\tau}\,\mathrm A\\{}u_L(0)&=-RI_0=-16.0\,\mathrm V\\{}Q_R&=\int_0^\infty RI_0^2e^{-2t/\tau}\,dt=\frac{RI_0^2\tau}{2}=0.240\,\mathrm J\end{aligned}

[4]

Voltage spikes require a complete switch-off path

An ideal L cannot change current discontinuously at finite voltage. Opening a real switch sends field energy through other paths: parasitic capacitance, an arc, protection circuitry or coupled components. Actual peak voltage follows from the entire system, not L and initial current alone. A specified turn-off interval permits an average voltage but says nothing certain about peak voltage without waveform information.[4]

Changing linear geometry adds $I\frac{\mathrm dL}{\mathrm dt}$ when differentiating LI. For a nonlinear fixed core, use the actual flux-current relation and its local slope. Hysteresis can make that relation history-dependent. The linear energy expression and a single L then need explicit validation. At high frequencies, winding capacitance and spatial field distribution may limit a lumped-component model.[1][6]

\begin{aligned}\frac{d(LI)}{dt}&=L\dot I+I\dot L\\{}\mathcal E&=-\frac{d\lambda}{dI}\dot I\end{aligned}

The second line requires fixed geometry and a single-valued flux-current relation; it does not apply to an arbitrary hysteretic or moving coil.[1]

In brief

Constant linear L connects self-flux linkage and current. Self-emf uses the opposition sign; passive voltage uses the voltage-drop reference. Current height sets energy; slope sets inductive voltage. Real switch-off requires the complete current and energy path.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Linked Faraday’s induction law and LC energy exchange in the relevant sections; preserved voltage references, exercises and switch-on diagram.
October 1, 2026Separated flux linkage, emf and passive voltage; developed linear energy, two original examples, five exercises and real switch-off limits.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

Cite this reference page

Quanta. Self-Induction of a Coil. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/selbstinduktion-spule

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Frequently asked questions about Self-Induction of a Coil

How do you calculate the self-induced voltage?+

For constant linear inductance, induced emf in the chosen loop direction is $\mathcal E=-L\frac{dI}{dt}$ . Ideal-component passive terminal voltage has the opposite sign: $u_L=L\frac{dI}{dt}$ . For an explicitly linear current ramp, replace the derivative with current change divided by interval length. Without that assumption the calculation gives an interval average, not automatically peak or instantaneous voltage. In the existing example, current falls from 2 A to zero in 10 ms at 0.5 H; emf is then +100 V and passive voltage −100 V. A real winding’s resistance creates an additional terminal-voltage contribution. Always state which voltage reference is being used before interpreting its sign.

Why does switching off a coil produce a voltage surge?+

Initial energy of a linear inductor is $W=\frac12LI^2$ . At switch-off that energy must move into other stores or loads. Finite voltage cannot make ideal inductor current disappear discontinuously. At a real opening switch, parasitic capacitance, an arc or protection circuitry can provide the current path; the complete system determines peak voltage and its waveform. A freewheeling diode provides a different decay path and limits voltage according to the circuit, often slowing current decay. Inductance and initial current alone do not specify one turn-off time or voltage peak. An ignition system additionally uses magnetically coupled windings, rather than demonstrating a universal fixed multiplication from a single isolated coil.

What does the inductance L mean intuitively?+

In a linear model, inductance describes self-flux linkage produced by current. One henry means one weber per ampere and also one volt at a current-change rate of one ampere per second. Larger inductance gives a smaller current-change rate for the same passive inductive voltage, resembling mechanical inertia. This analogy does not replace a complete circuit balance. Long air-core solenoid geometry gives $L=\mu_0\frac{N^2A}{\ell}$ . Turn count enters quadratically only when the remaining geometry is unchanged. A real iron core needs its material characteristic and possibly a model for air gaps, hysteresis and saturation; an arbitrary constant multiplier is not generally valid.

Why does the number of turns enter the inductance quadratically?+

Turn count acts twice in the long air-core solenoid model. First, field at fixed current and length grows proportionally to turn count: $B\approx\mu_0\frac N\ell I$ . Second, each turn’s flux links all equally oriented turns: $\lambda=N\Phi$ . Together these give $L\approx\mu_0\frac{N^2A}{\ell}$ . Doubling turns therefore quadruples inductance provided length, area and material model remain unchanged. For 1000 turns, area 20 square centimetres and length 0.4 metres, the rounded vacuum permeability gives about 6.3 millihenries; 2000 turns give about 25 millihenries. If additional winding appreciably changes radius or winding thickness, check the geometry again instead of applying the factor of four automatically.

What is the difference between induction and self-induction?+

Induction describes emf produced by changing oriented flux linkage: $\mathcal E=-\frac{\mathrm d\lambda}{\mathrm dt}$ . An external cause can be a moving magnet, another coil or a rotating loop. In self-induction, the flux under consideration comes from the same conductor’s current. A fixed linear coil has $\lambda=LI$ , giving $\mathcal E_{\mathrm{self}}=-L\frac{\mathrm dI}{\mathrm dt}$ . This is the same induction law applied to self-flux. Ideal passive component voltage uses the opposite reference. In a transformer, self- and mutual flux linkages occur together, requiring an appropriate balance of changes in both winding currents. Nonlinear cores, moving geometry or unequal turn fluxes need a more detailed model than one constant inductance.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §14.2 Self-Inductance and Inductors

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Retain Self-Induction of a Coil for exams

Create a curated FSRS exam set for U_ind = −L·(dI/dt): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Self-induction of a coil

Current change alters self-flux linkage and produces emf opposing that change. Ideal passive inductor voltage uses the opposite voltage reference.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

U_{\mathrm{ind}}

Oriented self-emf in the chosen traversal; not ideal passive component voltage

L

Inductance of the coil

dI/dt

Rate of change of the current

A/s

Derivation & background

Exam blueprint

Validity range

Derivation steps

Apply Faraday’s law to self-flux linkage while retaining voltage reference.

1Define flux linkage $\lambda=\sum_k\Phi_k$ ; equal turn flux gives $\lambda=N\Phi$ .
2Linear fixed geometry gives $\lambda=LI$ with positive constant L.
3Faraday’s law gives emf $\mathcal E=-\frac{d\lambda}{dt}$ .
4Constant L gives $\mathcal E=-L\frac{dI}{dt}$ .
5Ideal-component passive voltage instead is $u_L=+L\frac{dI}{dt}$ .
6A linear ramp permits constant finite-difference rate; other waveforms require interval average or actual derivative.

How do you calculate with Self-Induction of a Coil?

Here is how to work through a typical Self-Induction of a Coil (U_ind = −L·(dI/dt)) task step by step:

Task

Compute L of a coil with $N=1000$ , $A=20\,\mathrm{cm^2}$ , $\ell=0.4\,\mathrm m$ (no core).

Solution path

Task

A choke ( $L=0.3\,\mathrm H$ ) sees 12 V. How fast does the current rise?

Solution path

Why inductor current does not respond instantly

Self-flux links every turn

\begin{aligned}\lambda&=\sum_{k=1}^{N}\Phi_k\\{}\lambda&=N\Phi\\{}\lambda&=LI\end{aligned}

Distinguish induced emf and passive terminal voltage

\begin{aligned}\mathcal E_{\mathrm{self}}&=-\frac{d\lambda}{dt}=-L\frac{dI}{dt}\\{}u_L&=+L\frac{dI}{dt}\end{aligned}

Emf is positive in the chosen loop direction. uL is the passive voltage drop: positive current enters the positive voltage terminal. L is constant here.[1][2]

From Faraday to current-change rate

\begin{aligned}\mathcal E&=-\frac{d\lambda}{dt}\\{}\lambda&=LI\\{}\frac{d\lambda}{dt}&=L\frac{dI}{dt}\quad L=\mathrm{const}\\{}L&=\frac{|\mathcal E|}{\left|\frac{\mathrm dI}{\mathrm dt}\right|}\\{}\frac{dI}{dt}&=-\frac{\mathcal E}{L}=\frac{u_L}{L}\end{aligned}

Faraday’s law gives emf. The final rearrangement retains voltage reference; magnitudes alone cannot give a signed rate.[2][1]

\Delta I=-\frac1L\int_{t_0}^{t_1}\mathcal E(t)\,dt

Assume the same sign convention and constant L.[1]

Explore further: inductance from coil geometry

\begin{aligned}B&\approx\mu_0\frac N\ell I\\{}\Phi&\approx BA\\{}\lambda&\approx NBA=\mu_0\frac{N^2A}{\ell}I\\{}L&\approx\mu_0\frac{N^2A}{\ell}\end{aligned}

Long tightly wound air-core solenoid; A is interior cross-sectional area and Φ approximates equal flux through each turn.[1]

Voltage sign alone does not determine energy flow

\begin{aligned}p_L&=u_LI=LI\frac{dI}{dt}\\{}dW&=LI\,dI\\{}W(I)&=\frac12LI^2\\{}\Delta W&=\frac L2(I_f^2-I_0^2)\end{aligned}

Passive reference and lossless linear component. Positive power means energy absorption.[3]

Account fully for a ramp and switch-off energy

Current falls with explicit signs

Given: Constant $L=0.240\,\mathrm H$ ; positive current falls linearly from 1.80 A to 0.300 A in 12.0 ms. Passive voltage and emf use the references defined above.
Find: Find both voltages and change in stored energy.

Approach: the explicitly linear ramp makes the average rate equal to the constant instantaneous rate. Evaluate voltage signs and squared currents separately.
$\begin{aligned}\dot I&=\frac{I_f-I_0}{\Delta t}\\{}\mathcal E&=-L\dot I\\{}u_L&=L\dot I\\{}\Delta W&=\frac L2(I_f^2-I_0^2)\end{aligned}$
Convert milliseconds to seconds.
$\begin{aligned}\dot I&=\frac{0.300-1.80}{0.0120}\,\mathrm{A\,s^{-1}}=-125\,\mathrm{A\,s^{-1}}\\{}\mathcal E&=+30.0\,\mathrm V\\{}u_L&=-30.0\,\mathrm V\\{}\Delta W&=\frac{0.240}{2}(0.300^2-1.80^2)\,\mathrm J=-0.378\,\mathrm J\end{aligned}$

Result: Emf +30.0 V, passive voltage −30.0 V; stored energy falls by 0.378 J.

Check: Average current is 1.05 A; constant passive voltage times average current times 12.0 ms gives −0.378 J. Positive emf sustains positive current while it decays.

Energy of a long air-core solenoid

Given: $N=600$ , $A=3.00\,\mathrm{cm^2}$ , $\ell=0.300\,\mathrm m$ ; air-core model with $\mu_0=1.25663706\cdot10^{-6}\,\mathrm{H/m}$ . Current rises uniformly from 0 to 2.50 A over 20.0 ms.
Find: Calculate inductance, passive inductive voltage and final stored energy.

Approach: long-solenoid geometry gives L; the linear ramp gives voltage rate. Final energy must agree with supplied inductive work.
$\begin{aligned}L&=\mu_0\frac{N^2A}{\ell}\\{}u_L&=L\frac{I_f}{\Delta t}\\{}W_f&=\frac12LI_f^2\end{aligned}$
Convert area using the square of the length conversion.
$\begin{aligned}A&=3.00\cdot10^{-4}\,\mathrm{m^2}\\{}L&=(1.25663706\cdot10^{-6})\frac{600^2(3.00\cdot10^{-4})}{0.300}\,\mathrm H\\{}&=0.0004523893416\,\mathrm H\\{}u_L&=L\frac{2.50}{0.0200}\,\mathrm{A\,s^{-1}}\approx0.0565487\,\mathrm V\\{}W_f&=\frac12L(2.50)^2\approx0.00141372\,\mathrm J\end{aligned}$

Result: L≈0.452 mH, uL≈56.5 mV and Wf≈1.41 mJ.

Check: uL times average current 1.25 A times 20.0 ms gives the same energy. Copper loss would be an additional contribution, not part of this ideal inductive term.

Example: switch on a voltage

Resistance sets final current; L sets its rise

Given: At $t=0$ , a constant $U=6.00\,\mathrm V$ source is connected to ideal $L=0.200\,\mathrm H$ in series with total resistance $R=10.0\,\Omega$ . Initial current is zero; use passive inductor voltage and no other storage.
Find: Find current, inductor voltage and energy after one time constant.

Approach: source voltage divides between resistance and inductance. The differential equation determines the waveform.
$\begin{aligned}U&=RI+L\dot I\\\tau&=\frac LR\\I(t)&=\frac UR(1-e^{-t/\tau})\\u_L(t)&=Ue^{-t/\tau}\end{aligned}$
Substitution at one time constant: current has not reached its final value; inductor voltage is not yet zero.
$\begin{aligned}\tau&=20.0\,\mathrm{ms}\\I(\tau)&=(0.600\,\mathrm A)(1-e^{-1})\\&\approx0.379272\,\mathrm A\\u_L(\tau)&=(6.00\,\mathrm V)e^{-1}\approx2.20728\,\mathrm V\\W(\tau)&=\frac12LI(\tau)^2\approx14.3848\,\mathrm{mJ}\end{aligned}$

Result: After 20.0 ms: about 0.3793 A, +2.207 V passive inductor voltage and 14.38 mJ stored energy. Eventually they approach 0.600 A and 36.0 mJ.

View diagram: switch-on response

Current rises while inductor voltage falls

QUANTA STUDYquanta-study.de

In the third example

\tau=20\,\mathrm{ms}

. The curves add to one because

RI+u_L=U

; the axis compares normalised quantities, not amperes with volts.[4]

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Five exercises with distinct voltage references

Calculate · Foundation

An ideal 80.0 mH inductor undergoes a linear current rise from 0.400 to 1.60 A in 6.00 ms. Find passive voltage and emf.

Show hint

First form the signed change rate.

Learning goal: Calculate terminal voltage and emf during a ramp.

Show solution and calculation

Approach: rate is +200 A/s.
Passive voltage +16.0 V; emf −16.0 V.

Check: positive current means absorption because passive voltage and current have the same sign.

\begin{aligned}\dot I&=\frac{1.60-0.400}{0.00600}=200\,\mathrm{A\,s^{-1}}\\{}u_L&=(0.0800)(200)\,\mathrm V=16.0\,\mathrm V\\{}\mathcal E&=-16.0\,\mathrm V\end{aligned}

[1]

Rearrange · Application

Current through constant linear L falls from 2.00 to 0.500 A. Measured emf-time integral is +0.0900 V·s in the chosen loop direction. Find L without assuming constant voltage.

Show hint

Integrate Faraday’s relation rather than using a peak value.

Learning goal: Determine L from an emf-time integral.

Show solution and calculation

Approach: the integral is $-L\Delta I$ .
With $\Delta I=-1.50\,\mathrm A$ , $L=0.0600\,\mathrm H$ .

Check: positive integral and negative current change give positive L; detailed waveform is unnecessary.

L=-\frac{\int\mathcal E\,dt}{\Delta I}=-\frac{0.0900\,\mathrm{V\,s}}{-1.50\,\mathrm A}=0.0600\,\mathrm H

[1]

Interpret · Application

An ideal $L=0.100\,\mathrm H$ has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

Show hint

At corners use one-sided values; derivative is not uniquely defined there.

Learning goal: Read voltage from curve slope rather than current height.

Show solution and calculation

Approach: multiply each slope by L.
Voltages are +50 V, 0 V, −50 V.

Check: the middle interval stores 0.050 J despite zero voltage; constant height does not mean loss of field.

\begin{aligned}u_1&=0.100\frac{1}{0.002}\,\mathrm V=50\,\mathrm V\\{}u_2&=0\\{}u_3&=0.100\frac{-0.500}{0.001}\,\mathrm V=-50\,\mathrm V\end{aligned}

[1]

Error analysis · Application

A real winding has $R=6.0\,\Omega$ and $L=0.200\,\mathrm H$ . At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.

Show hint

Total terminal voltage has two contributions.

Learning goal: Distinguish emf from resistive voltage drop.

Show solution and calculation

Approach: split terminal voltage into $RI$ and $L\dot I$ .
All 9.0 V is resistive drop, leaving zero inductive term and hence zero current rate.

Check: loss power is 13.5 W, not growing magnetic energy.

\begin{aligned}u_{\mathrm{coil}}&=RI+L\dot I\\{}\dot I&=\frac{9.0-(6.0)(1.50)}{0.200}\,\mathrm{A\,s^{-1}}=0\\{}P_R&=RI^2=13.5\,\mathrm W\end{aligned}

[4]

Transfer · Challenge

An ideal 0.120 H inductor initially carries 2.00 A. It is disconnected from the source but remains in a closed loop with 8.00 Ω total resistance. No capacitor or other loss path. Find time constant, initial passive inductor voltage and final heat.

Show hint

The loop is not open; resistance provides a determined current waveform.

Learning goal: Track switch-off energy through a closed resistive path.

Show solution and calculation

Approach: loop balance is $L\dot I+RI=0$ .
Time constant 15.0 ms; initial inductor voltage −16.0 V.
Initial energy 0.240 J becomes heat.

Check: integrating $RI^2$ over exponential decay gives the same energy; no arbitrary switch-off time is imposed.

\begin{aligned}\tau&=\frac LR=0.0150\,\mathrm s\\{}I(t)&=2.00e^{-t/\tau}\,\mathrm A\\{}u_L(0)&=-RI_0=-16.0\,\mathrm V\\{}Q_R&=\int_0^\infty RI_0^2e^{-2t/\tau}\,dt=\frac{RI_0^2\tau}{2}=0.240\,\mathrm J\end{aligned}

[4]

Voltage spikes require a complete switch-off path

\begin{aligned}\frac{d(LI)}{dt}&=L\dot I+I\dot L\\{}\mathcal E&=-\frac{d\lambda}{dI}\dot I\end{aligned}

The second line requires fixed geometry and a single-valued flux-current relation; it does not apply to an arbitrary hysteretic or moving coil.[1]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Linked Faraday’s induction law and LC energy exchange in the relevant sections; preserved voltage references, exercises and switch-on diagram.
October 1, 2026Separated flux linkage, emf and passive voltage; developed linear energy, two original examples, five exercises and real switch-off limits.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

Cite this reference page

Quanta. Self-Induction of a Coil. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/selbstinduktion-spule

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Self-Induction of a Coil

How do you calculate the self-induced voltage?+

Why does switching off a coil produce a voltage surge?+

What does the inductance L mean intuitively?+

Why does the number of turns enter the inductance quadratically?+

What is the difference between induction and self-induction?+

Retain Self-Induction of a Coil for exams

Create a curated FSRS exam set for U_ind = −L·(dI/dt): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Self-induction of a coil

Variables & units

Self-induction calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction

Worked example

Applications

How do you calculate with Self-Induction of a Coil?

Why inductor current does not respond instantly

Self-flux links every turn

Distinguish induced emf and passive terminal voltage

From Faraday to current-change rate

Explore further: inductance from coil geometry

Voltage sign alone does not determine energy flow

Account fully for a ramp and switch-off energy

Current falls with explicit signs

Energy of a long air-core solenoid

Example: switch on a voltage

Resistance sets final current; L sets its rise

Current rises while inductor voltage falls

Five exercises with distinct voltage references

An ideal 80.0 mH inductor undergoes a linear current rise from 0.400 to 1.60 A in 6.00 ms. Find passive voltage and emf.

Current through constant linear L falls from 2.00 to 0.500 A. Measured emf-time integral is +0.0900 V·s in the chosen loop direction. Find L without assuming constant voltage.

An ideal L=0.100 HL=0.100\,\mathrm HL=0.100H has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

A real winding has R=6.0 ΩR=6.0\,\OmegaR=6.0Ω and L=0.200 HL=0.200\,\mathrm HL=0.200H. At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.

An ideal 0.120 H inductor initially carries 2.00 A. It is disconnected from the source but remains in a closed loop with 8.00 Ω total resistance. No capacitor or other loss path. Find time constant, initial passive inductor voltage and final heat.

Voltage spikes require a complete switch-off path

In brief

Sources for this explanation

Frequently asked questions about Self-Induction of a Coil

Subject references and further reading

More formulas: Electricity and magnetism

Retain Self-Induction of a Coil for exams

Self-induction of a coil

Variables & units

Self-induction calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction

Worked example

Applications

How do you calculate with Self-Induction of a Coil?

Why inductor current does not respond instantly

Self-flux links every turn

Distinguish induced emf and passive terminal voltage

From Faraday to current-change rate

Explore further: inductance from coil geometry

Voltage sign alone does not determine energy flow

Account fully for a ramp and switch-off energy

Current falls with explicit signs

Energy of a long air-core solenoid

Example: switch on a voltage

Resistance sets final current; L sets its rise

Current rises while inductor voltage falls

Five exercises with distinct voltage references

An ideal 80.0 mH inductor undergoes a linear current rise from 0.400 to 1.60 A in 6.00 ms. Find passive voltage and emf.

Current through constant linear L falls from 2.00 to 0.500 A. Measured emf-time integral is +0.0900 V·s in the chosen loop direction. Find L without assuming constant voltage.

An ideal L=0.100 HL=0.100\,\mathrm HL=0.100H has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

A real winding has R=6.0 ΩR=6.0\,\OmegaR=6.0Ω and L=0.200 HL=0.200\,\mathrm HL=0.200H. At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.

An ideal 0.120 H inductor initially carries 2.00 A. It is disconnected from the source but remains in a closed loop with 8.00 Ω total resistance. No capacitor or other loss path. Find time constant, initial passive inductor voltage and final heat.

Voltage spikes require a complete switch-off path

In brief

Sources for this explanation

Frequently asked questions about Self-Induction of a Coil

Subject references and further reading

An ideal $L=0.100\,\mathrm H$ has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

A real winding has $R=6.0\,\Omega$ and $L=0.200\,\mathrm H$ . At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.

An ideal $L=0.100\,\mathrm H$ has I(t): linear 0→1 A from 0 to 2 ms, constant 1 A from 2 to 5 ms, linear 1→0.500 A from 5 to 6 ms. Find passive voltage in the three open intervals.

A real winding has $R=6.0\,\Omega$ and $L=0.200\,\mathrm H$ . At steady 1.50 A a voltmeter reads 9.0 V. A solution infers current growth 45 A/s. Identify and correct the error.