Fields of loops, arcs and coils

Put the field point on the axis, a distance z from the centre of a ring of radius R, and three things go right at once. Every current element I dl runs along the ring while the vector to the field point leans out of the ring's plane, and the two are always perpendicular, so the cross product keeps its full magnitude I dl with no sine factor to carry. Every element lies the same distance r = √(R² + z²) away, so the 1/r² factor is a constant and leaves the integral. And every element has one diametrically opposite whose transverse contribution is equal and opposite, so only the axial share survives, cut down by cos α = R/√(R² + z²). What is left to integrate is ∫dl = 2πR, the circumference, and the Biot-Savart integral has become a multiplication: B = (μ₀I/4π) × (1/(R² + z²)) × (R/√(R² + z²)) × 2πR = μ₀IR²/2(R² + z²)³⁄².

The same geometry handles a piece of a circle. Bend a wire so that it runs radially inward, sweeps an arc of angle φ at radius R, then leaves radially outward. On the straight leads dl points at the centre, so dl × r̂ = 0 and those sections give nothing at the centre however long they are. On the arc dl is perpendicular to r̂ and r = R throughout, so the integral is again the arc length alone: B = (μ₀I/4πR²) × Rφ = μ₀Iφ/4πR, with φ in radians. A quarter circle of radius 5.0 cm carrying 3.0 A gives 10⁻⁷ × 3.0 × (π/2) / 0.050 = 9.4 μT at its centre, and the closed turn, φ = 2π, returns μ₀I/2R. Arcs superpose, so a hairpin of two concentric arcs at radii R₁ and R₂ carrying the same current in opposite senses gives the difference of their contributions, read off by angle alone.

Divide the on-axis field by its centre value and every coil collapses onto a single curve: B(z)/B₀ = R³/(R² + z²)³⁄² = [1 + (z/R)²]⁻³⁄². Only the ratio z/R appears, so a 2 cm coil and a 2 m coil have identical profiles once the axis is measured in radii. Three points are worth carrying. The field is half its centre value at z = 0.766R, the root of (R² + z²)³⁄² = 2R³. At one radius out it is 1/2³⁄² = 0.354 of the centre value, not a half. At two radii out it is 5⁻³⁄² = 0.089. The curve is also flat at the centre, because it is even in z and so has zero slope there, which is what makes a coil a usable source of near-uniform field over a small central region.

N tightly wound turns at the same radius are N copies of one loop, so every value above multiplies by N while the shape of the profile is untouched: a 45-turn coil of radius 8.0 cm carrying 2.4 A has B₀ = μ₀NI/2R = 8.5 × 10⁻⁴ T. Tightly wound means the winding's spread in radius is small next to R, and because B₀ goes as 1/R a winding spanning ±4% in radius moves the centre field by only about 0.05% — the error is second order. Geometry is the sharper lever. Wind a fixed length ℓ into N turns of radius R, so ℓ = 2πRN and B₀ = μ₀Iℓ/4πR². The same wire keeps the same resistance and so draws the same current from the same supply, yet at 2.0 A a 25 m length wound at R = 25 cm gives 80 μT while the same 25 m wound at R = 5.0 cm gives 2.0 mT — twenty-five times the field for nothing but a smaller former.

Take z well beyond R and factor the denominator: (R² + z²)³⁄² = z³(1 + R²/z²)³⁄² ≈ z³(1 + 3R²/2z²). The field therefore tends to B ≈ μ₀NIR²/2z³, which is μ₀m/2πz³ with m = NIπR², the coil's magnetic moment. Everything else has dropped out, so two coils sharing the same NIA — one wide and thin, one small and many-turned — are indistinguishable from far away. The correction term says how far away far is: the dipole form runs high by about 3R²/2z², that is 1.5(R/z)², which is 17% at three radii, 6% at five, 1.5% at ten, and does not fall to 1% until z = 12.3R. That is the everyday licence for treating a coil, a compass needle or a small magnet as a point dipole, and the reason a calibration done at three radii out is worth repeating further away.

Move off the axis and every simplification fails at once: the elements lie at different distances, the cross product keeps a varying sine, and nothing cancels in pairs. The exact answer needs elliptic integrals; the working answer is numerical. Replace the ring by n equal straight segments with their ends on the circle, apply the finite-straight-wire result to each, and add the vectors. At the centre that sum closes in form and gives B = μ₀I n tan(π/n)/2πR against the true μ₀I/2R. The polygon's sides run slightly inside the arc, so the estimate is high: 2.35% for n = 12, 0.575% for n = 24. The error falls as 1/n², so doubling the segment count quarters it and about a hundred segments reaches the third significant figure. The same loop over segments, evaluated off the axis, is how field-mapping software draws a coil's field lines.