Ampère's law

Ampère’s law is the magnetic analog of Gauss’s law: the line integral of the magnetic field intensity $\mathbf{H}$ around a closed loop equals the total free current enclosed by the loop:

${\oint }_C\mathbf{H}\cdot d\mathbf{l}=I_{\text{enc}}\text{(integral form)},$

with the corresponding differential form

$\nabla \times \mathbf{H}=\mathbf{J}\text{(magnetostatic case)}.$

The two forms are equivalent via Stokes’ theorem: integrate $\nabla \times \mathbf{H}=\mathbf{J}$ over a surface $S$ bounded by $C$, and the surface integral of curl becomes the line integral around the boundary.

The “free” qualifier on the current matters: $\mathbf{H}$ is sourced only by free conduction currents, not by bound magnetization currents. In ferromagnetic materials this distinction is essential; in non-magnetic media (most of Electromagnetics) all current is “free” and the distinction is moot.

When Ampère beats Biot-Savart

When the current distribution has high symmetry, you can guess the shape of $\mathbf{H}$ and pull its magnitude outside the line integral. The recipe:

1. Identify the symmetry — typically axial (long wires), planar (current sheets), or solenoidal (long coils).
2. Choose an Amperian loop matched to the symmetry: one on which $|\mathbf{H}|$ is constant and $\mathbf{H}$ is parallel to $d\mathbf{l}$ everywhere on the loop.
3. The line integral collapses to $|\mathbf{H}|\cdot L$, with $L$ the loop length.
4. Set equal to enclosed current. Solve for $|\mathbf{H}|$.

Worked example: infinite straight wire

Wire on the $z$-axis carrying current $I$. By axial symmetry $\mathbf{H}=H_{\varphi }(r)\hat{\mathbf{\varphi }}$ — magnitude depends only on radial distance, direction is azimuthal.

Amperian loop: circle of radius $r$ in a plane perpendicular to the wire. On this loop $\mathbf{H}\cdot d\mathbf{l}=H_{\varphi }(r)\cdot rd\varphi$ (since $d\mathbf{l}=rd\varphi \hat{\mathbf{\varphi }}$):

$\oint H_{\varphi }(r)rd\varphi =H_{\varphi }(r)\cdot 2\pi r=I.$

Therefore

$\mathbf{H}=\hat{\mathbf{\varphi }}\frac{I}{2\pi r}.$

Field circles the wire, falling as $1/r$.

Worked example: long wire with uniformly distributed current

Same wire but now of finite radius $a$ with current $I$ uniformly distributed over its cross-section. The current density is $J=I/(\pi a^2)$ for $r<a$ and 0 for $r>a$.

Outside ($r>a$): Amperian loop encloses the full current $I$. Same result as the infinite-thin-wire case: $H_{\varphi }=I/(2\pi r)$.

Inside ($r<a$): Amperian loop of radius $r$ encloses only the fraction of current passing through its interior. The enclosed area is $\pi r^2$ and the current density is uniform, so $I_{\text{enc}}=J\cdot \pi r^2=I(r^2/a^2)$. Then

$H_{\varphi }\cdot 2\pi r=I\frac{r^2}{a^2}⟹H_{\varphi }=\frac{Ir}{2\pi a^2}.$

Linear growth from 0 at the center to $I/(2\pi a)$ at the surface ($r=a$), then $1/r$ decay outside. The two regions are continuous at $r=a$: both formulas give $I/(2\pi a)$ at the boundary.

Combined:

$H_{\varphi }(r)=\{\begin{array}{ll}\frac{Ir}{2\pi a^2},& r\le a\\ \frac{I}{2\pi r},& r\ge a\end{array}$

The plot of $H_{\varphi }$ vs $r$ has two regimes: a straight-line rise from origin to $(a,I/(2\pi a))$, then a hyperbolic decay $\propto 1/r$ for $r>a$. The peak value of $H$ is achieved exactly at the conductor surface.

Physical interpretation. Inside the wire, only the current “below” radius $r$ contributes; the outer shell of current cancels out by symmetry (analog of Gauss’s-law shell theorem for spherical mass distributions). At the surface, all the current is enclosed and the field reaches its maximum. Outside, the field drops as if all the current were concentrated on the axis — far away, a thick wire and a thin wire produce the same field.

This radial profile matters for proximity effect and internal inductance at high frequency, where current redistributes toward the conductor surface (skin effect) and the field pattern reshapes accordingly.

Worked example: long solenoid

A tightly wound solenoid of length $l$ with $N$ turns, current $I$. Use an Amperian loop that’s a rectangle with two sides parallel to the solenoid axis: one inside, one far outside (where $\mathbf{H}\approx 0$). Sides perpendicular to the axis contribute nothing (their $\mathbf{H}$ is perpendicular to $d\mathbf{l}$).

The inside-side contribution is $H\cdot l_{\text{loop}}$. Enclosed current is the number of turns inside the loop times $I$, equal to $(N/l)l_{\text{loop}}I$:

$H\cdot l_{\text{loop}}=(N/l)l_{\text{loop}}I⟹H=\frac{NI}{l},B=\frac{{\mu }_{0}NI}{l}.$

Uniform inside, zero outside. This is the textbook recipe for a uniform magnetic field.

In the time-varying case

In dynamic problems, Ampère’s law needs the Displacement current term:

$\nabla \times \mathbf{H}=\mathbf{J}+\frac{\partial \mathbf{D}}{\partial t}.$

The ${\partial }_t\mathbf{D}$ term is essential for closing the loop in problems like a capacitor being charged — without it, current conservation breaks at the capacitor gap. The corresponding integral form:

${\oint }_C\mathbf{H}\cdot d\mathbf{l}=I_{\text{free,enc}}+{\int }_S\frac{\partial \mathbf{D}}{\partial t}\cdot d\mathbf{s}.$

This extension was Maxwell’s crucial contribution — completing the symmetry between Faraday’s law and Ampère’s, and predicting electromagnetic waves.