Field of a long straight wire

Two symmetries survive on an infinite straight wire: slide along it and nothing changes, rotate about it and nothing changes. So the magnitude of B can depend on the perpendicular distance r and on nothing else. The direction comes from the cross product in the Biot-Savart law. Every current element points along the wire, and dl × r̂ is perpendicular both to the wire and to the line joining that element to the field point, so the contribution has no component along the wire and none along the radius. What is left is the azimuthal direction, tangent to the circle through the field point centred on the wire. The right hand fixes the sense: thumb along the conventional current, fingers curling the way B points. The field lines are therefore closed circles lying in planes perpendicular to the wire, with no source to start from and no sink to end on.

Put the wire on the x-axis and the field point P at perpendicular distance R from the origin. An element of length dx sits at position x, a distance √(x² + R²) from P, and the angle φ between the element and the line to P satisfies sin φ = R/√(x² + R²). Biot-Savart then gives dB = (μ₀ I/4π) dx sin φ/(x² + R²), and substituting for sin φ turns that into dB = (μ₀ I/4π) R dx/(x² + R²)³⁄². Because every element contributes in the same azimuthal direction, the vector integral collapses to this single scalar integral — the one feature that makes the straight wire solvable by hand. The integrand also shows why the answer converges: an element at x ≫ R contributes like R dx/x³, so the far reaches of the wire add almost nothing.

Set x = R tan θ, with θ measured at P from the perpendicular. Then dx = R sec²θ dθ and (x² + R²)³⁄² = R³ sec³θ, so the integrand simplifies to dB = (μ₀ I/4π R) cos θ dθ, which integrates in one line to B = (μ₀ I/4π R)(sin θ₂ − sin θ₁), where θ₁ and θ₂ locate the two ends. Nothing about being long has been assumed yet; this is the field of any straight segment. Now push the ends away. θ₂ → +90° and θ₁ → −90°, both sines saturate at ±1, their difference is 2, and B = μ₀ I/2π R. With μ₀ = 4π × 10⁻⁷ T m A⁻¹ the prefactor is μ₀/2π = 2 × 10⁻⁷ T m A⁻¹ exactly, so 30 A at 6.0 cm gives 1.0 × 10⁻⁴ T without touching π.

The 1/r law decays slowly. A 10 A wire gives 40 µT at 5.0 cm, 20 µT at 10 cm and still 2.0 µT at 1.0 m, where a point charge would have fallen by a factor of 400 over the same span and a dipole by 8000. That is why one power feeder can swing a magnetometer several metres away. The slow decay also makes the law easy to test: plot measured B against 1/r and the points lie on a straight line through the origin of slope μ₀I/2π, which is 2.0 × 10⁻⁶ T m for I = 10 A. Curvature in that plot means the wire is not long enough or a neighbour is contributing. Note what r is measured from: the axis, not the surface. Outside a round conductor the field does not know how thick it is, so a 1 mm wire and a 10 mm rod carrying the same current give the same field at 50 mm.

For a segment of half-length L with the field point on its perpendicular bisector the two sines are equal, and B = μ₀ I L/(2π R √(L² + R²)). The fraction L/√(L² + R²) is the price of finiteness. At L = 10R it is 0.995, so the infinite formula reads 0.5% high; at L = 5R it is 0.981 and the infinite formula reads 2.0% high; at L = 2R it is 0.894, 12% high; and at L = R only 0.707, 41% high. A usable rule: the wire should extend at least ten times the field distance either side of the foot of the perpendicular. Off the bisector the shortfall is worse — on the perpendicular through the end of a semi-infinite wire only one sine survives and B is exactly μ₀ I/4π R, half the full value. And an isolated finite current cannot exist alone: charge would pile up at its ends, so a real circuit closes, and near the corners the return path must be added.

Superposition holds exactly for magnetic fields, so with more than one wire you take each field from its own perpendicular distance and its own right-hand sense, then add the vectors. Two cases repay knowing. Parallel currents in the same direction give fields that oppose between the wires, so a null sits on the joining line wherever I₁/r₁ = I₂/r₂: for 4.0 A and 12.0 A separated by 0.16 m that is 0.040 m from the weaker wire, where each field is 2.0 × 10⁻⁵ T. Antiparallel currents reinforce between the wires and nearly cancel outside, and at a distance r much larger than their separation s the survivor is about μ₀ I s/2π r², an inverse-square tail. A pair carrying 8.0 A with s = 4.0 mm gives 3.2 µT at 0.50 m as a lone wire but only 26 nT as a pair, which is the whole reason a signal cable carries its return alongside.