Ampere's law and enclosed current

Give the closed loop a direction, cut it into short vectors dl, and at each step take the component of B along dl. The sum once round is the circulation ∮B⋅dl, a scalar measured in T m. Try it on a field you already know. A long wire carrying 12 A makes a field μ₀I/(2πr) that is everywhere tangent to the circle of radius r about the wire, so on that circle B⋅dl is simply B dl and the sum is B(2πr). At r = 1.0 cm the field is 2.4 × 10⁻⁴ T and the path is 0.0628 m, giving 1.51 × 10⁻⁵ T m. At r = 2.0 cm the field halves and the path doubles, and the product is the same 1.51 × 10⁻⁵ T m, which is μ₀ × 12 A. The 1/r of the field and the 2πr of the path cancel, and that cancellation is the whole content of the law: the circulation has forgotten the radius and remembers only the current.

I_enc is a signed count, so the direction of travel round the loop must be fixed first. Curl the fingers of your right hand along the way you traverse the loop, and your thumb points along the positive normal of any surface capping it. Current crossing that surface along the thumb counts positive, current crossing the other way counts negative, and current that misses the surface does not count at all. Take a loop walked counterclockwise as the reader sees it, threaded by 5.0 A toward the reader, 8.0 A away and 1.0 A toward: I_enc = 5.0 − 8.0 + 1.0 = −2.0 A, so the circulation is μ₀ × (−2.0 A) = −2.5 × 10⁻⁶ T m. A negative answer is not a mistake; it reports that the field runs against the way you walked. Two consequences are worth stating. Reverse the traversal and both sides change sign together, so the law is untouched. And the capping surface may be a flat disc or a deep bag: for steady currents charge piles up nowhere, so every surface bounded by that loop is pierced by the same net current.

Ampere's law constrains one number, the circulation, not the field at a point. Turning it into B takes a further step the law does not licence: writing ∮B⋅dl as B times a length requires the path to fall into pieces on each of which either B runs along dl with one and the same |B|, or B⋅dl is zero because B lies across dl or is negligible there. That is a claim about the symmetry of the current distribution, and it must be argued before the law is used. Run the test in order. Does the geometry force |B| to depend on one coordinate only? Does it fix the direction of B? Can you draw a closed path made of pieces on which B⋅dl is either B dl or zero? A single circular coil fails at the first question: it has an axis of rotational symmetry, yet no closed path through a point on that axis keeps |B| constant, so the circulation is known while the field is not, and only Biot-Savart will deliver it.

Take a straight cylindrical conductor of radius R carrying a steady current I spread uniformly, so J = I/(πR²) everywhere in the metal. Rotational symmetry and the straightness of the wire together force B to be tangent to circles about the axis with |B| depending only on r, which is exactly the licence the symmetry test demanded. For a circle of radius r < R the circulation is B(2πr), and the enclosed current is the fraction of the cross-section inside that circle: I_enc = J πr² = I r²/R². Setting the two equal gives B = μ₀ I r/(2πR²), so the field climbs in a straight line from zero on the axis. For r > R the whole current is enclosed and B = μ₀I/(2πr). With R = 2.0 mm and I = 10 A the field is 0.50 mT halfway out, peaks at 1.0 mT on the surface, and is back to 0.50 mT at r = 4.0 mm. The maximum always sits at the surface, which is a quick check on any answer.

The two laws are one bargain struck in different integrals. Gauss's law integrates E over a closed surface and returns Q_enc/ε₀; Ampere's law integrates B around a closed loop and returns μ₀I_enc. Each is exact, each is blind to detail, and each pays out only when a symmetry argument lets the field come outside the integral — a sphere, cylinder or plane there, a circle, rectangle or toroid here. The contrast is just as instructive. The flux of B out of any closed surface is zero, ∮B⋅dA = 0, because no magnetic charge has ever been found for field lines to start on, so B lines close on themselves and can circulate. An electrostatic field is the mirror image: it has flux out of charges but no circulation at all, ∮E⋅dl = 0, which is exactly what allows a potential V to exist. There is no equally useful scalar magnetic potential, and the nonzero circulation of B is the reason.

Put an amperian loop around the wire feeding a charging capacitor. Cap it with a flat disc and the disc cuts the wire, so I_enc is the charging current, say 2.0 A. Cap the same loop with a bag that bulges through the gap between the plates, and no charge crosses it at all, so I_enc = 0. One loop, two answers: the steady-current law has been used outside its range, because charge is piling up on the plates and the current is not steady. Maxwell's repair is to notice what is changing inside the bag. The gap field is E = q/(ε₀A), so the electric flux through the bag is Φ_E = EA = q/ε₀ and ε₀ dΦ_E/dt = dq/dt = 2.0 A, exactly the missing amount. The mended law, ∮B⋅dl = μ₀I_enc + μ₀ε₀ dΦ_E/dt, returns the same answer for every capping surface, and that extra term is what later lets electromagnetic waves exist.