Physics · Electrodynamics

Self-Induction of a Coil

Self-induction describes how a coil reacts to changes of its own current with an opposing voltage; the inductance L is its measure.

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Formula

LaTeX: U_{ind} = -L \cdot \frac{dI}{dt}
U_ind in V · L in H (henry) · dI/dt in A/s

Variables & units – Self-Induction of a Coil

SymbolMeaningUnit
U_indSelf-induced voltageV
LInductance of the coilH (Henry)
dI/dtRate of change of the currentA/s

Derivation & background – Self-Induction of a Coil

Joseph Henry and Michael Faraday discovered self-induction around 1831/32: when the current changes, the coil own magnetic flux changes, and by the law of induction a voltage arises that opposes the change (Lenz rule, hence the minus sign). For a long coil L = μ₀·N²·A/l; the number of turns enters squared. Consequences: delayed current rise at switch-on, high voltage spikes at switch-off and the magnetic field energy E = ½·L·I².

Exam blueprint

Validity range

Applies for constant inductance L, i.e. without magnetic saturation of the core. The voltage arises only while the current changes; at constant current U_ind = 0. The minus sign expresses Lenz rule.

Derivation steps

The coil induces a voltage in itself because its own current creates the flux.

  1. 1The total flux linkage is proportional to the current: N·Φ = L·I (definition of L).
  2. 2Induction law: U_ind = −N·dΦ/dt = −L·dI/dt.

Rearrangements

Inductance

Measurement rule: voltage per rate of current change.

Rate of current change

Limits how fast the current in a choke can rise.

Inductance of a long coil

The number of turns enters quadratically.

Task variant

Compute L of a coil with N = 1000, A = 20 cm², l = 0.4 m (no core).

L = μ₀·N²·A/l = 1.257×10⁻⁶ × 10⁶ × 2×10⁻³/0.4 ≈ 6.3×10⁻³ H = 6.3 mH.

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

dI/dt = U/L = 12/0.3 = 40 A/s, so the current initially grows by 40 amperes per second.

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.

L ∝ N²: doubling the turns quadruples the inductance.

Expecting an induced voltage at constant current.

Only the change dI/dt induces; on steady DC an ideal coil acts like a plain wire.

Substituting the area A in cm².

Convert A to m²: 20 cm² = 2×10⁻³ m².

Exam context

  • Exams ask switch-on and switch-off transients (delayed current rise, voltage spike with a glow lamp), the energy E = ½·L·I² and deriving L for a long coil.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

Coil with L = 0.5 H: at switch-off the current falls from 2 A to 0 within 10 ms, so dI/dt = −200 A/s: U_ind = −0.5 × (−200) = +100 V of self-induced voltage.

Applications

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

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Curated exam set for "Self-Induction of a Coil":

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Which formula describes Self-Induction of a Coil?

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How do you rearrange U_ind = −L·(dI/dt) for Inductance?

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Which common mistake happens with Self-Induction of a Coil?

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Scientific sources

Common notations & search queries

Uind=-L*dI/dtU = -L dI/dtSelbstinduktion FormelInduktivität Spuleself inductance formulaL=mu0 N^2 A/lHenry Einheit SpuleAbschaltspannung Spule

Related formulas

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Frequently asked questions about Self-Induction of a Coil

How do you calculate the self-induced voltage?+

Multiply the inductance by the rate of change of the current: U_ind = −L·(dI/dt). For a uniform change you may replace dI/dt by ΔI/Δt. Example: in a coil with L = 0.5 H the current collapses from 2 A to 0 within 10 ms at switch-off, so ΔI/Δt = −200 A/s and U_ind = −0.5 × (−200) = +100 V. The sign says: the voltage opposes the change in current (Lenz rule). For magnitudes in exams |U| = L·|ΔI|/Δt often suffices. The faster the change, the higher the voltage, which is why abrupt switch-offs in particular are critical.

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

The magnetic field of a current-carrying coil stores the energy E = ½·L·I², and it cannot vanish instantly. When the switch opens the circuit, the current tries to fall to zero within microseconds; the rate dI/dt becomes enormous, and with it the induced voltage U = −L·dI/dt. It can reach hundreds to thousands of volts although the operating voltage was only a few volts, arcing across the opening contact or destroying transistors. The car ignition coil uses exactly this effect constructively to create spark voltages above 20 kV from 12 V. To protect circuits, an antiparallel freewheeling diode lets the coil current decay gently after switch-off.

What does the inductance L mean intuitively?+

L is the electrical inertia of a coil: it states how strongly the coil resists changes of current. A coil of 1 henry produces 1 V of back voltage when its current changes by 1 A per second. Large inductance means a sluggish, slowly rising current, just as large mass means sluggish acceleration; in the mechanical analogy L corresponds to the mass m and the current to the velocity. Geometrically, for a long coil L = μ₀·N²·A/l: the number of turns enters squared, and an iron core multiplies by μ_r. Typical values range from microhenries (wire loops, RF coils) through millihenries (chokes) to several henries (mains transformer windings).

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

Because the number of turns acts twice. First, more turns at the same current create a stronger magnetic field, since B = μ₀·(N/l)·I grows linearly with N. Second, this flux also threads more turns: the total flux linkage is N·Φ, again proportional to N. Both effects together give L = N·Φ/I ∝ N². Concretely: doubling the turns of a coil at the same length and area quadruples L. Sample calculation: N = 1000, A = 20 cm², l = 0.4 m gives L = 1.257×10⁻⁶ × 10⁶ × 2×10⁻³/0.4 ≈ 6.3 mH; with N = 2000 it would be about 25 mH.

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

In ordinary induction an external cause produces the flux change: a moving magnet, another coil or a loop rotating in a field. The voltage follows from U = −N·dΦ/dt with the external flux Φ. In self-induction the coil is its own cause: its own current creates the flux, and any change of this current induces the back voltage U = −L·dI/dt in the same coil. Physically it is the same law of induction, only applied to the coil own flux; L bundles the geometry (N²·A/l). In a transformer both occur at once: self-induction in the primary coil and mutual induction across to the secondary.

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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. 1

    Task

    Compute L of a coil with N = 1000, A = 20 cm², l = 0.4 m (no core).

    Solution path

    L = μ₀·N²·A/l = 1.257×10⁻⁶ × 10⁶ × 2×10⁻³/0.4 ≈ 6.3×10⁻³ H = 6.3 mH.

  2. 2

    Task

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

    Solution path

    dI/dt = U/L = 12/0.3 = 40 A/s, so the current initially grows by 40 amperes per second.