Self-Inductance — Physica

Equivalent forms

L = \frac{N\Phi}{I}

U = \tfrac{1}{2}LI^2

One coefficient L captures a coil's entire reluctance to change current, independent of how fast you try.

Symbol	Name	SI unit	Dimension	Range
$\mathcal{E}$	Induced EMFoutput Back-EMF opposing the change	V	M·L²·T⁻³·I⁻¹	0 – 1000
$L$	Inductance Coil self-inductance	H	M·L²·T⁻²·I⁻²	0.001 – 1
$dIdt$	Current rate Rate of change of current dI/dt	A/s	I·T⁻¹	-5000 – 5000

Unit systems

Symbol	SI
$\mathcal{E}$	V
$L$	H
$dIdt$	A/s

Where it holds

Linear (constant L) for non-magnetic or unsaturated cores. With ferromagnetic cores L drops as the core saturates, and at high frequency L becomes complex due to core loss and skin effect.

Dimensional analysis

$[L][dI/dt] = H\cdot$ (A/ $s) = (V\cdot s$ /A $)\cdot$ (A/ $s) = V \checkmark$

Discovery

Joseph Henry · 1832

Henry discovered self-induction independently of Faraday, noticing violent sparks when he broke the circuit of a long coil. The unit of inductance, the henry, bears his name.

Try this

Switch off a coil and it bites back with a spark. Why does a coil fight any change to its own current?

A coil of 200 turns carries 2 A and links a flux of 5×10⁻⁴ Wb per turn. Find its inductance and the EMF when the current is switched off in 1 ms.

Research status: stable

Real-world applications

Flyback converters and ignition coils
EMI chokes and filters
Energy storage in SMPS
Relay snubber design

Common misconceptions

An inductor resists current — it resists CHANGES in current; in steady state it carries DC freely
The minus sign means energy is lost — it only encodes opposition (Lenz); energy is stored, not dissipated

Experimental verification

Drive a known L with a triangular current and observe the square-wave EMF whose height equals

L\cdot (dI/dt)

; measured slopes match the geometric inductance to a few percent.

Derivation

Flux linkage

\Lambda = N\Phi is

proportional to current:

\Lambda = L I

.

Faraday's law gives \mathcal

{E} = -d\Lambda /dt = -L dI/dt

for constant L.

Integrating power

-

\mathcal{E}I over a current ramp gives stored energy

U = \tfrac{1}{2}LI^{2}

.

Limiting cases

dI/dt \to 0

⟶ \mathcal

{E} \to 0

Steady DC current produces no back-EMF — an ideal inductor is a plain wire to DC.

dI/dt \to \infty

(hard switch-off)⟶ \mathcal

{E} \to

hugeInterrupting current almost instantly produces the large 'inductive kick' that sparks across switches.

What if…

What if you double the number of turns?

Inductance roughly quadruples — L scales as $N^{2} for$ a fixed geometry.

What if you switch the current off ten times faster?

The back-EMF is ten times larger — which is exactly how ignition coils make tens of kilovolts.

1

200-turn coil

Given ·

N:: 200
\Phi:: 0.0005
I:: 2
dIdt:: -2000

Find · L and switch-off EMF

Steps

$L = N\Phi /I = (200)(5\times 10^{-4})/(2) = 0.05\,\mathrm{H}$
Switching 2 A to 0 in 1 ms gives $dI/dt = -2000$ A/s
\mathcal ${E} = -L dI/dt = +100\,\mathrm{V}$ — the inductive kick

Result ·

L = 0.05\,\mathrm{H}

, and a 100 V spike appears at switch-off.

Solenoid Inductance→energy in inductorlenz lawmutual inductance