Ampère's Law (with Maxwell's correction)

Equivalent forms

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \quad \text{(magnetostatics)}

\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}

B = \mu_0 n I \quad \text{(solenoid)}

Maxwell's displacement current addition was pure theoretical insight — it fixed a mathematical inconsistency and predicted radio waves before they were observed.

Symbol	Name	SI unit	Dimension	Range
$B$	Magnetic fieldoutput Magnetic field along the Amperian loop	T	M·T⁻²·I⁻¹	0 – 0.1
$I_enc$	Enclosed current Total current threading the Amperian loop	A	I	0 – 100
$n$	Turns per unit length Number of turns per meter (solenoid application)	1/m	L⁻¹	10 – 10000

Unit systems

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

Where it holds

Magnetostatic form valid for steady currents. The full Ampère–Maxwell form is universally valid and is one of Maxwell's four equations. Most useful when current distributions have cylindrical, planar, or toroidal symmetry.

Dimensional analysis

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

Discovery

André-Marie Ampère / James Clerk Maxwell · 1826

Ampère discovered the circuital law for steady currents in 1826. Maxwell added the displacement current term in 1861, completing the equation and predicting electromagnetic waves.

Try this

Why does a long solenoid create a nearly uniform field inside but almost nothing outside?

A solenoid has 500 turns per meter and carries 3 A. Find the magnetic field inside.

Research status: stable

Real-world applications

Solenoid and toroid magnetic field calculations for electromagnet design
Magnetic field inside coaxial cables for signal integrity analysis
MRI gradient coil design for medical imaging
Tokamak plasma confinement field calculations

Common misconceptions

Ampère's law is always true but only useful for calculating B when the current distribution has sufficient symmetry
The displacement current term is essential for capacitor charging — without it, the law gives contradictory results for different surfaces spanning the same loop
The enclosed current includes ALL currents passing through any surface bounded by the loop

Experimental verification

Ampère's original experiments with parallel current-carrying wires. Solenoid field uniformity verified by Gauss probes. Maxwell's displacement current confirmed by Hertz's 1887 detection of electromagnetic waves.

Derivation

From the Biot–Savart law, integrate

B\cdot dl

around a closed loop.

For a long straight wire:

B(2\pi r) = \mu _{0}I

, giving

B = \mu _{0}I/(2\pi r)

.

Maxwell noticed

\nabla \cdot (\nabla \times B)

must equal zero (identity),

but \mu _{0}\nabla \cdot J \neq 0

for time-varying charge distributions.

Adding

\mu _{0}\varepsilon _{0}\partial E/\partial t

(displacement current) fixes the continuity equation.

Limiting cases

I_enc = 0

⟶

\oint B\cdot dl = 0

No enclosed current means zero circulation — the field may still exist but has no net curl through the loop.

Infinite solenoid⟶

B = \mu _{0}nI

(uniform inside),

B = 0

outsideSymmetry and Ampère's law give a perfectly uniform interior field with no leakage.

What if…

What if the solenoid current doubles?

Field doubles: $B = 3.77\,\mathrm{mT}$ . Linear relationship — doubling current doubles enclosed current per loop.

What if you need

B = 1\,\mathrm{T}

inside the solenoid?

Need $nI = B/\mu _{0} = 1/(1.257\times 10^{-6}) = 795$ ,775 A-turns/m. With 10 A current, that' $s \sim 80$ ,000 turns/m — hence superconducting magnets for MRI.

What if the solenoid is finite (not ideal)?

Field near the ends drops $to \sim 50$ % of the center value. Outside the solenoid, B is weak but nonzero — it resembles a bar magnet's dipole field.

1

Solenoid interior field

Given ·

n:: 500
I:: 3

Find · B inside

Steps

Choose a rectangular Amperian loop: one side inside, one outside the solenoid
Outside $B \approx 0$ for an ideal solenoid; sides perpendicular to B contribute nothing
$\oint B\cdot dl = B \times L$ (inside segment only)
Enclosed current: $I_enc = nLI (n$ turns per meter, each carrying I)
Ampère's law: $BL = \mu _{0}nLI \to B = \mu _{0}nI = 1.885\times 10^{-3} T$

Result ·

B = \mu _{0}nI = 1.257\times 10^{-6} \times 500 \times 3 = 1.885\times 10^{-3} T = 1.885\,\mathrm{mT}

2

Field inside a toroid

Given ·

N:: 200
I:: 5
r:: 0.1

Find · B at the center of the toroid cross-section

Steps

Choose a circular Amperian loop of radius r inside the toroid
By symmetry, B is tangential and constant: $\oint B\cdot dl = B \times 2\pi r$
Enclosed current: $I_enc = NI (all N$ turns pass through the loop)
Ampère's law: $B \times 2\pi r = \mu _{0}NI$
$B = \mu _{0}NI/(2\pi r) = 2.0\times 10^{-3} T$

Result ·

B = \mu _{0}NI/(2\pi r) = 1.257\times 10^{-6} \times 200 \times 5 / (2\pi \times 0.1) = 2.0\times 10^{-3} T = 2.0\,\mathrm{mT}

Biot–Savart Law→Faraday's Law of Induction→Gauss's Law for Magnetism→