Introduction to Ampere's Circuital Law
Gauss's law (Part 2) relates the electric field to its sources, the charges. Ampère's circuital law does the same job for the magnetic field and its sources, the currents. It states that the magnetic field circulating around any closed loop is fixed by the total current passing through that loop. Formulated by André-Marie Ampère in 1826, it is to magnetostatics what Gauss's law is to electrostatics: a fundamental law, and — for symmetric problems — a quick way to calculate fields.
For RF engineers, Ampère's law answers practical questions directly: what is the magnetic field around a cable, how much inductance does a coaxial line have per metre, how many turns does a toroidal inductor need? It also has a famous flaw. Applied to a charging capacitor, the original law gives two contradictory answers. Repairing it — the subject of Part 6 — was the step that led Maxwell to predict electromagnetic waves.
Relationship Between Electric Current and Magnetic Fields
Magnetic Field Around a Current-Carrying Wire
Ørsted's 1820 experiment showed that a compass needle placed near a wire swings when current flows. Mapping the field with iron filings reveals concentric circles around the wire. The field strength is proportional to the current and inversely proportional to the distance from the wire:
Right-Hand Rule
The direction of the field is given by the right-hand grip rule: grip the wire with your right hand with the thumb pointing in the direction of conventional current; your fingers curl in the direction of the magnetic field lines. The same rule, applied the other way round, gives the direction of the field inside a coil: curl the fingers along the current in the turns, and the thumb points along the field through the coil (toward its north pole).
The right-hand rule also fixes the sign convention in Ampère's law. If you traverse the Amperian loop with your fingers, currents passing through the loop in the direction of your thumb count as positive.
Magnetic Field Due to Different Current Configurations
| Configuration | Field | Notes |
|---|---|---|
| Long straight wire | B = μ₀I / (2πr) | Circles around the wire |
| Centre of a circular loop (radius R) | B = μ₀I / (2R) | Along the loop axis (from Biot–Savart) |
| Long solenoid (n turns per metre) | B = μ₀nI | Uniform inside, ≈ 0 outside |
| Toroid (N turns, radius r) | B = μ₀NI / (2πr) | Confined inside the core |
| Coaxial line (a < r < b) | B = μ₀I / (2πr) | Zero outside the outer conductor |
| Infinite current sheet (K A/m) | B = μ₀K / 2 | Uniform; model of a ground plane current |
Understanding Ampere's Circuital Law
Statement of Ampere's Circuital Law
The line integral of the magnetic field intensity 𝐇 around any closed path equals the total current enclosed by that path. Equivalently, the line integral of 𝐁 equals μ₀ times the enclosed current.
Mathematical Expression
We can verify the law for a straight wire. Take a circle of radius r around it: 𝐇 is tangent to the circle and constant in magnitude, so ∮𝐇·d𝐥 = H × 2πr. Setting this equal to I gives H = I/(2πr) — the result Ørsted and Biot–Savart found experimentally. As with Gauss's law, the 1/r fall-off of the field exactly balances the r growth of the path length, so any circle gives the same answer.
Understanding the Concept of Enclosed Current
"Enclosed current" means the net current that passes through any surface bounded by the loop C:
- Sign: currents in the direction of the right-hand-rule normal count positive, currents in the opposite direction count negative.
- Multiple conductors: only the algebraic sum matters. A loop around both conductors of a cable carrying equal and opposite currents encloses I − I = 0.
- Coils: a loop threading a coil of N turns, each carrying I, encloses NI ampere-turns — the "magnetomotive force" of magnetic circuit theory.
- Partial enclosure: if the loop lies inside a conductor, only the fraction of the current inside the loop counts: I_enc = ∫𝐉·d𝐀 over the enclosed area.
- Zero enclosed current does not mean zero field. It means the circulation is zero — the field may still exist, with contributions that cancel around the loop.
Integral Form of Ampere's Circuital Law
Writing the enclosed current as the flux of the current density 𝐉 (A/m²) through any surface S bounded by C gives the general integral form:
To use the law to find 𝐇, choose a loop on which 𝐇 is either tangent and constant in magnitude, or perpendicular to the path (contributing zero). The usual choices are circles around wires and coax, rectangles straddling solenoid walls and current sheets, and circles inside toroids.
Differential Form of Ampere's Circuital Law
Applying Stokes' theorem, ∮𝐇·d𝐥 = ∫(∇×𝐇)·d𝐀, and requiring the equality to hold for every surface gives the point form:
Applications of Ampere's Circuital Law
Magnetic Field Around a Straight Conductor
For a solid round conductor of radius a carrying current I uniformly distributed over its cross-section (true at DC; at RF the skin effect pushes current to the surface), a circular Amperian loop of radius r gives:
Magnetic Field Inside a Solenoid
For a long solenoid with n turns per metre, take a rectangular loop with one side of length ℓ inside the coil, parallel to the axis, and the opposite side outside. Outside the field is negligible, and the two short sides are perpendicular to the field, so only the inside side contributes: Hℓ = nℓI.
Magnetic Field Inside a Toroid
A toroid is a solenoid bent into a ring. A circular Amperian loop of radius r inside the core encloses NI ampere-turns, while a loop outside the ring encloses equal currents going in and coming out, giving zero:
Magnetic Fields in Transmission Lines
In a coaxial line the inner conductor carries current +I and the outer conductor carries the return current −I. A circular loop between the conductors (a < r < b) encloses only +I, so H = I/(2πr). A loop outside the cable encloses I − I = 0, so H = 0: a properly balanced coax has no external magnetic field. This is the field-theory reason coaxial cable does not radiate or pick up interference — as long as the outer conductor carries the full return current. Common-mode current on the outside of the shield breaks the balance, and that is exactly what baluns and ferrite chokes are used to suppress.
The magnetic energy stored between the conductors gives the cable its inductance per unit length:
Limitations of the Original Ampere's Circuital Law
Ampère's law in the form ∮𝐇·d𝐥 = I_enc is correct for steady currents but fails when currents change in time. Two arguments show the problem.
The capacitor paradox. Consider a capacitor being charged by a current I. Draw an Amperian loop around the wire leading to one plate. If we choose a flat surface S₁ bounded by the loop, the wire pierces it, so I_enc = I. But we may equally choose a bag-shaped surface S₂ bounded by the same loop that passes between the plates, where no conduction current flows — giving I_enc = 0. The same loop yields two different answers for the same field.
The mathematical contradiction. The divergence of a curl is always zero. Taking the divergence of ∇ × 𝐇 = 𝐉 therefore gives ∇·𝐉 = 0. But charge conservation requires ∇·𝐉 = −∂ρ_v/∂t. The two agree only when the charge density is constant in time — i.e., for steady currents. In a charging capacitor, charge is accumulating on the plates, ∂ρ_v/∂t ≠ 0, and the original law violates conservation of charge.
Maxwell resolved the contradiction in 1861–1865 by adding a second source of magnetic field — the rate of change of electric flux density, ∂𝐃/∂t, which he called the displacement current. The corrected law, ∇ × 𝐇 = 𝐉 + ∂𝐃/∂t, is consistent with charge conservation and predicts electromagnetic waves. That is the subject of Part 6: Maxwell's Displacement Current.
Numerical Examples
Use μ₀ = 4π × 10⁻⁷ H/m, so μ₀/(2π) = 2 × 10⁻⁷ H/m.
Field Around a Straight Wire
Problem: A long straight wire carries 10 A DC. Find H and B at 5 cm from the wire. At what distance does B fall to the Earth's field of about 50 μT?
A 10 A cable deflects a compass noticeably within a few centimetres. If the same 10 A has its return conductor right beside it (as in a power cord), the two fields almost cancel at a distance — the same principle as the coaxial line.
Field Inside a Thick Conductor
Problem: A copper rod of radius 2 mm carries 20 A DC, uniformly distributed. Find B at r = 1 mm (inside), r = 2 mm (surface) and r = 4 mm (outside).
The field is the same at half the radius and at twice the radius: linear growth inside, 1/r decay outside.
Field Inside a Solenoid
Problem: A solenoid 50 cm long has 1000 turns and carries 2 A. Find the field inside (a) with an air core and (b) if it is filled with a powdered-iron core of μ_r = 75. (c) If its cross-section is 4 cm², what is its air-core inductance?
H = nI = 4000 A/m in both cases — 𝐇 is set by the current alone. The core multiplies 𝐁 (and the inductance) by μ_r, until the core material approaches saturation.
Toroidal Inductor Field
Problem: A toroid with 500 turns and mean radius 4 cm carries 0.5 A. Find H and B at the mean radius for (a) an air core and (b) a powdered-iron core with μ_r = 100.
The 125 mT result is comfortably below the 0.5–1 T saturation level of typical powdered-iron materials. A high-permeability ferrite (μ_r ≈ 2000) with the same ampere-turns would try to reach 2.5 T and saturate, so ferrite-cored chokes are designed with fewer ampere-turns or with an air gap.
Coaxial Cable Inductance and Characteristic Impedance
Problem: Return to the RG-58-type cable of Part 2 (a = 0.45 mm, b = 1.47 mm, polyethylene ε_r = 2.25, C′ = 105.7 pF/m). It carries 1 A. Find (a) H at r = 1 mm, (b) H outside the cable, (c) the inductance per metre, (d) the characteristic impedance and (e) the propagation velocity.
Check: Z₀ = (60/√2.25) × 1.1837 = 47.3 Ω, and v = c/√ε_r = c/1.5. Commercial RG-58 is specified at 50 Ω; the small difference comes from the stranded inner conductor and dielectric tolerances. Two laws — Gauss and Ampère — applied to the cable's cross-section give its impedance and velocity factor from first principles.
Summary
| Concept | Expression | Key point |
|---|---|---|
| Ampère's law (integral) | ∮ 𝐇·d𝐥 = I_enc = ∫ 𝐉·d𝐀 | Circulation of 𝐇 = enclosed current |
| Ampère's law (differential) | ∇ × 𝐇 = 𝐉 | Currents are the "vortices" of 𝐇 |
| Straight wire | H = I/(2πr) | Right-hand rule for direction |
| Solenoid | H = nI; L ≈ μN²A/ℓ | Uniform field inside |
| Toroid | H = NI/(2πr) | Field confined to the core |
| Coaxial line | L′ = (μ/2π) ln(b/a); Z₀ = √(L′/C′) | No external field when balanced |
| Limitation | ∇·(∇ × 𝐇) = 0 ⇒ ∇·𝐉 = 0 | Fails for time-varying currents |
- Currents produce magnetic fields that circulate around them; Ampère's law relates the circulation of 𝐇 around any loop to the current through it.
- With a symmetric Amperian loop, the fields of wires, solenoids, toroids and coaxial lines follow in one line.
- Combined with Gauss's law, Ampère's law gives the inductance, capacitance and characteristic impedance of transmission lines.
- The original law contradicts conservation of charge when currents vary in time — a gap that Maxwell's displacement current fills.
Put It Into Practice
Use the NEXT ANTENNAS calculators to apply the ideas from this article to real designs.