Ampere's Circuital Law

Statement of the law. Ampere's circuital law states that, for any closed loop (called an

Amperian loop) drawn in space, the line integral of the magnetic field $\vec{B}$ around that loop

equals $\mu_0$ times the total (algebraic) current $I_{\text{enc}}$ passing through the surface

bounded by the loop:

The sign of each current threading the loop is fixed by a right-hand-rule convention: if the fingers

of the right hand curl in the direction the loop is traversed (the direction $d\vec{l}$ points),

the thumb gives the POSITIVE current direction; a current flowing the opposite way is counted as

negative in $I_{\text{enc}}$.

Analogy with Gauss's law. Ampere's circuital law plays, for magnetism, precisely the role

Gauss's law plays for electrostatics: both are EXACT, general laws relating a field to its source

(Gauss's law relates the electric flux through a closed SURFACE to the enclosed charge; Ampere's law

relates the magnetic circulation around a closed LOOP to the enclosed current), and both become

enormously useful CALCULATION tools only in situations with enough symmetry to argue, in advance,

that $\vec{B}$ (or $\vec{E}$) must be constant in magnitude and either parallel or perpendicular to

the chosen loop (or surface) at every point along it -- letting the integral be pulled out and solved

algebraically rather than genuinely integrated term by term. Choosing a well-suited Amperian loop is

exactly analogous to choosing a well-suited Gaussian surface: a poorly chosen loop (one without the

matching symmetry) makes the integral just as hard as attempting the corresponding Biot-Savart

integral directly, and gives no shortcut at all.

Re-deriving the straight-wire field. As a first application, and as a check against the direct

Biot-Savart integration of Section 4.4, take the Amperian loop to be a circle of radius $r$, centred

on and coplanar with (perpendicular to) a long straight wire carrying current $I$. By the symmetry

established from Oersted's experiment (Section 4.2), $\vec{B}$ has the SAME magnitude at every point

on this circle and is everywhere TANGENT to it (parallel to $d\vec{l}$ at every point), so

$\vec{B}\cdot d\vec{l} = B\,dl$ throughout, and $B$ itself can be pulled outside the integral: …