Magnetic Field Inside a Solenoid

Equivalent forms

B = \mu_0 \frac{N}{L} I

B = \mu_0 n I \quad (\text{inside, far from ends})

An infinite solenoid's interior field depends only on turn density and current — not on radius or position. A rare case where geometry drops out entirely.

Symbol	Name	SI unit	Dimension	Range
$B$	Magnetic flux densityoutput Magnetic field along the solenoid axis	T	M·T⁻²·I⁻¹	0 – 0.1
$n$	Turn density Number of turns per unit length	1/m	L⁻¹	10 – 10000
$I$	Current Current through the solenoid winding	A	I	0 – 20

Unit systems

Symbol	SI	CGS	Imperial
$B$	T	G	T
$n$	1/m	1/cm	1/ft
$I$	A	statA	A

Where it holds

Valid deep inside a long, tightly wound solenoid where L ≫ radius. Edge effects reduce B near the ends; for finite solenoids B at the very end is exactly half the central value.

Dimensional analysis

$[B] = [\mu _{0}][n][I] = (T\cdot m$ /A)(1/m)(A $) = T \checkmark$

Discovery

André-Marie Ampère · 1820

Ampère wrapped wire around a glass tube and showed it behaved like a bar magnet — a discovery so important he proposed all magnetism arises from circulating microscopic currents.

Try this

How can a hollow tube of wire produce a magnetic field as uniform as a permanent magnet's?

A solenoid 0.4 m long with 1000 turns carries 2 A. Find the magnetic field strength along its axis.

Research status: stable

Real-world applications

MRI machines use superconducting solenoids producing 1.5–7 T fields
Relay coils, fuel injectors, and door locks in cars
Magnetic confinement coils in tokamak fusion reactors
Pickup coils in electric guitars

Common misconceptions

B does not depend on the solenoid's radius (in the ideal long limit)
The field is not zero outside — only approximately zero for long solenoids
Near the ends, B is roughly half the interior value (a useful sanity check)

Experimental verification

A Hall probe inserted along the axis of a calibrated solenoid reads B values that match

\mu _{0} n I

to within 1–2% well away from the ends. Confirmed daily in physics teaching labs since the 1950s.

Derivation

Apply Ampère's law to a rectangular path with one side of length ℓ inside the solenoid and the opposite side outside (where

B \approx 0)

:

\oint B\cdot dl = B

ℓ

= \mu _{0} I_enc = \mu _{0} (n

ℓ) I.

Solve:

B = \mu _{0} n I

.

Limiting cases

I \to 0

⟶

B \to 0

No current produces no field — solenoid is effectively absent.

n \to \infty

⟶

B \to \infty

Idealized limit: arbitrarily tight winding maximizes field, in practice limited by wire thickness and heating.

Outside the solenoid⟶

B \to 0

For an infinite solenoid, the external field vanishes exactly; for finite ones, only approximately.

What if…

What if you double the number of turns but also double the length?

n stays the same, so B is unchanged. Field strength depends on turn density, not absolute count.

What if you insert an iron core?

Replace $\mu _{0}$ with $\mu = \mu _{0} \mu$ ᵣ. For soft iron, $\mu$ ᵣ $\sim 5000$ , so B grows by orders of magnitude — the basis of electromagnets.

What if the current alternates at 50 Hz?

B alternates too, inducing eddy currents in nearby conductors — used in induction cooktops and transformers.

1

Lab-bench solenoid

Given ·

N:: 1000
L:: 0.4
I:: 2

Find · B

Steps

Compute $n = N/L = 1000/0.4 = 2500$ turns/m
Apply $B = \mu _{0} n I$
$B = (4\pi \times 10^{-7})(2500)(2)$
$B \approx 1.257\times 10^{-6} \times 5000 \approx 6.28\times 10^{-3} T = 6.28\,\mathrm{mT}$

Result ·

B \approx 6.28\,\mathrm{mT}

2

Current needed for 0.05 T

Given ·

n:: 5000
B target:: 0.05

Find · I

Steps

Rearrange: $I = B / (\mu _{0} n)$
$I = 0.05 / (4\pi \times 10^{-7} \times 5000)$
$I = 0.05 / (6.283\times 10^{-3})$
$I \approx 7.96$ A

Result ·

I \approx 7.96

A

Ampère's Law (with Maxwell's correction)→biot savartSelf-Inductance of a Solenoid→