Self-Inductance of a Solenoid

Equivalent forms

L = \mu_0 n^2 A \ell

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

Inductance is a purely geometric quantity for linear materials — it doesn't depend on the current you put through it, only on shape and material.

Symbol	Name	SI unit	Dimension	Range
$L$	Self-inductanceoutput Inductance of the solenoid	H	M·L²·T⁻²·I⁻²	0 – 0.01
$N$	Total turns Total number of turns in the solenoid	—	dimensionless	10 – 5000
$A$	Cross-section area Area of one loop	m²	L²	0.0001 – 0.01
$ℓ$	Length Solenoid length along its axis	m	L	0.01 – 1

Unit systems

Symbol	SI	CGS	Imperial
$L$	H	abH	H
$N$	—	—	—
$A$	m²	cm²	ft²
$ℓ$	m	cm	ft

Where it holds

Valid for long, tightly wound solenoids with linear core (constant

\mu )

. Breaks down for ferromagnetic cores at saturation and for short solenoids where end effects matter.

Dimensional analysis

$[L] = [\mu _{0}][N^{2}][$ A]/[ℓ $] = (H/m)(1)(m^{2})/(m) = H \checkmark$

Discovery

Joseph Henry · 1832

Joseph Henry independently discovered electromagnetic induction (essentially simultaneously with Faraday) and emphasized the self-induction effect, observing huge sparks when interrupting coil currents. The SI unit of inductance bears his name.

Try this

Why does a hand-held drill kick back when you release the trigger — even with no load?

A solenoid 0.2 m long with 500 turns and cross-section 4 cm² is wound around an air core. What is its self-inductance?

Research status: stable

Real-world applications

Power-supply filter chokes smoothing rectified DC
Wireless charging pads using mutual inductance
Induction stoves coupling energy into ferromagnetic cookware
Tank circuits in radios setting the resonant frequency

Common misconceptions

Inductance is $N^{2}$ , not N — doubling the turns quadruples L
Inductance is a property of the coil's geometry, independent of the current
Inductors don't 'resist' current — they resist changes in current

Experimental verification

Drive a sinusoidal current through the coil and measure the voltage across it; the reactance

X_L = \omega L

extracts L to within a few percent. Bridge methods (Maxwell, Hay) reach much higher accuracy.

Derivation

Field inside solenoid:

B = \mu _{0} n I

with

n = N

/ℓ.

Flux per turn:

\Phi = BA = \mu _{0} (N

/ℓ) I A.

Total flux linkage:

N\Phi = \mu _{0} (N^{2}

/ℓ) A I.

By definition

L = N\Phi / I

, giving

L = \mu _{0} N^{2} A

/ ℓ.

Limiting cases

N \to 2N

⟶

L \to 4L

Doubling the number of turns quadruples inductance —

N^{2}

scaling.

ℓ

\to \infty

⟶

L \to 0

Spreading the same turns over a longer solenoid dilutes the flux, reducing L.

A \to 0

⟶

L \to 0

No cross-section means no flux can be enclosed.

What if…

What if a ferromagnetic core with

\mu

ᵣ

= 1000

is inserted?

Inductance grows $by \mu$ ᵣ — from 0.628 mH to 628 mH for the lab solenoid. This is how tiny ferrite chokes deliver large inductance.

What if the radius doubles?

Area quadruples, so L quadruples (turns and length unchanged).

What if you halve the length while keeping N fixed?

L doubles. The same turns concentrated in less space link more flux per turn.

1

Air-core lab solenoid

Given ·

N:: 500
A:: 0.0004
ℓ:: 0.2

Find · L

Steps

Compute $N^{2} = 250 000$
Apply $L = \mu _{0} N^{2} A$ / ℓ
$L = (4\pi \times 10^{-7})(2.5\times 10^{5})(4\times 10^{-4})/(0.2)$
$L = (1.257\times 10^{-6})(2.5\times 10^{5})(2\times 10^{-3})$
$L \approx 6.28\times 10^{-4} H$

Result ·

L \approx 6.28\times 10^{-4} H = 0.628\,\mathrm{mH}

2

Turns needed for 10 mH

Given ·

L target:: 0.01
A:: 0.0004
ℓ:: 0.2

Find · N

Steps

Rearrange: $N = \sqrt(L$ ℓ / $(\mu _{0} A))$
$N = \sqrt(0.01 \times 0.2 / (4\pi \times 10^{-7} \times 4\times 10^{-4}))$
$N = \sqrt{2\times 10^{-3} / 5.03\times 10^{-10}}$
$N = \sqrt{3.98\times 10^{6}} \approx 1995$

Result ·

N \approx 1995

turns

Magnetic Field Inside a Solenoid→Energy Stored in an Inductor→Faraday's Law of Induction→LC Oscillation Frequency→