Ampere's circuital law

Why it matters

Ampère's circuital law is to magnetostatics what Gauss's law is to electrostatics: for symmetric current distributions it gives H in one line. Solenoids, toroidal cores, coaxial cables and busbars — the field sources inside current transformers, clamp meters, LVDTs and inductive sensors — are all analysed with it, and it is the basis of the magnetic-circuit method (MMF = NI).

Key ideas

The law. The line integral of H around any closed path equals the net steady current enclosed by that path: ∮H·dl = I_enc. In free space, ∮B·dl = μ₀I_enc. The "enclosed current" is the current crossing any surface bounded by the path, counted positive in the direction given by the right-hand rule (fingers along the path direction, thumb along positive current).

Using it. The law is always true, but it gives H directly only when you can choose an Amperian path on which H is either tangential with constant magnitude, or perpendicular to the path (contributing nothing). The usable symmetries are:

• infinite straight conductors and coaxial cables (circular paths);
• infinite current sheets (rectangular paths straddling the sheet);
• long solenoids (rectangular path, one side inside) and toroids (circular path inside the core).

Point form. Applying Stokes' theorem gives ∇ × H = J: the curl of H at a point equals the current density there. Where J = 0 (outside conductors) H is curl-free; and in every case ∇·B = 0 (there are no magnetic monopoles).

Inside conductors. For a uniform current in a round wire of radius a, only the fraction (r/a)² of the current is enclosed by a path of radius r < a, so H rises linearly from zero at the axis to a maximum at the surface, then falls as 1/r outside.

Coaxial cable. With +I on the inner conductor and −I returning on the outer, the field exists only between the conductors (and inside them); outside the outer conductor the enclosed current is zero, so H = 0. This is why coax does not radiate or pick up magnetic interference at low frequencies.

Limits. The form ∮H·dl = I_enc holds for steady currents. For time-varying fields the displacement current ∂D/∂t must be added (Maxwell's correction), otherwise the law gives contradictory answers for a charging capacitor.

Magnetic circuits. ∮H·dl = NI is the magnetomotive force (MMF) equation used to size coil current for a required core flux.

Formulas

• ∮ H·dl = I_enc — Ampère's circuital law (integral form); H in A/m, I in A.
• ∇ × H = J — point form; J in A/m².
• H = I / (2πr), B = μ₀I / (2πr) — outside a long straight conductor (r ≥ a).
• H = I r / (2πa²) — inside a long round conductor of radius a with uniform current (r ≤ a).
• H = I / (2πρ) (a < ρ < b) and H = 0 (ρ > c) — coaxial cable with inner radius a, outer conductor from b to c.
• H = nI, B = μ₀μr nI — inside a long solenoid; n = N/ℓ turns per metre; H ≈ 0 outside.
• H = N I / (2πr) — inside a toroid at radius r (between its inner and outer radii); zero outside.
• H = ½ K × a_n — infinite current sheet of surface current density K (A/m); a_n points from the sheet to the field point.
• μ₀ = 4π × 10⁻⁷ H/m.

Worked examples

Example 1 (standard). A long straight wire carries 5 A. Use Ampère's law to find B at 0.1 m.

1. Path: a circle of radius r = 0.1 m centred on the wire; H is tangential and constant on it.
2. H × 2πr = I → H = 5 / (2π × 0.1) = 7.96 A/m.
3. B = μ₀H = 4π × 10⁻⁷ × 7.96 = 10 μT.

Example 2 (field inside a conductor). A solid round conductor of radius a = 2 mm carries 10 A uniformly. Find H at r = 1 mm, at the surface and at r = 4 mm.

1. Inside: enclosed current = I(r/a)², so H = Ir/(2πa²) = 10 × 0.001 / (2π × 4 × 10⁻⁶) = 398 A/m.
2. Surface: H = I/(2πa) = 10 / (2π × 0.002) = 796 A/m (the maximum).
3. Outside at 4 mm: H = I/(2πr) = 10 / (2π × 0.004) = 398 A/m. Note that H at a/2 and at 2a are equal.

Example 3 (GATE level, toroid). A toroid with N = 500 turns and I = 2 A has an inner radius of 9 cm and an outer radius of 11 cm, with an air core. Find H at the inner edge, mean radius and outer edge, and the mean B.

1. Circular Amperian path of radius r inside the core encloses NI = 1000 A-turns: H = NI / (2πr).
2. r = 0.09 m: H = 1000 / (2π × 0.09) = 1768 A/m; r = 0.10 m: 1592 A/m; r = 0.11 m: 1447 A/m.
3. Mean B = μ₀ × 1592 = 2.00 mT.
4. The field is not uniform across the core (about ±11%); the "mean radius" result is a good approximation only when the core thickness is small compared with the radius. A long solenoid with the same 500 turns per metre and 2 A would give B = μ₀nI = 1.26 mT.

Common mistakes

• Using the total current instead of the enclosed current — for points inside a wire, or outside a coax where the net current is zero.
• Getting the sign of the enclosed current wrong (use the right-hand rule relative to the path direction).
• Claiming B is "constant in direction" along the circular path — it is constant in magnitude and always tangential.
• Using H = nI with n = total turns instead of turns per metre.
• Applying Ampère's law to a short wire or a single loop — the symmetry needed to take H outside the integral is missing; use Biot–Savart.
• Forgetting μr when the solenoid or toroid has a magnetic core.

For GATE IN

Typical items: H inside and outside a solid conductor (find where H is maximum, or the ratio at two radii); fields in the regions of a coaxial cable; solenoid and toroid fields with or without a core; current sheets and parallel sheets; finding J from a given H via ∇ × H = J. Practise the point form in cylindrical coordinates.

Quick check

1. What is H outside an ideal coaxial cable carrying equal and opposite currents?
2. Where is H maximum for a solid round wire carrying uniform current?
3. What is B inside a long air-core solenoid with 1000 turns/m carrying 1 A?
4. H = y a_x A/m. What is J?
5. Does Ampère's law hold for a short straight wire?

Answers: 1. Zero. 2. At its surface. 3. 1.26 mT. 4. −a_z A/m². 5. Yes, it always holds, but symmetry is missing so it cannot be used to find H directly.