Forces Between Parallel Currents

A wire carrying current is surrounded by its own magnetic field, and that field exerts no net force on the wire that made it. So nothing happens to a single isolated wire. Bring a second wire up and the situation changes, because wire 2 now sits in an external field — wire 1's field — and it is that external field alone that appears in the force law. The sequence is strictly two-step: first find B₁ at the position of wire 2 from the straight-wire result B₁ = μ₀I₁/(2πd), then feed that field into dF = I₂ dL × B₁. Put 50 A through wire 1 and stand wire 2 40 mm away: B₁ = (2 × 10⁻⁷ T m A⁻¹)(50 A)/(0.040 m) = 2.5 × 10⁻⁴ T, about five times Earth's field. Wire 2's own field, however large, never enters this calculation.

Set both wires along x̂, wire 1 at the origin and wire 2 at y = d. By the right-hand rule, wire 1's field at wire 2 points along ẑ, out of the plane the two wires define: B₁ = B₁ ẑ. Wire 2's current element is I₂ dL x̂. The force is dF = I₂ dL x̂ × B₁ ẑ = −I₂ dL B₁ ŷ, since x̂ × ẑ = −ŷ. Two things follow. The angle between dL and B₁ is 90°, so sin θ = 1 and the magnitude is simply I₂ L B₁ — the cross product costs nothing here, which is exactly why this geometry is the one worth memorising. And the direction is −ŷ, straight back toward wire 1: parallel currents attract. Reverse I₂ and every sign in that line flips, giving repulsion.

Substituting B₁ gives the working relation F/L = μ₀I₁I₂/(2πd), and the product I₁I₂ is symmetric: swap the labels and nothing changes. That is Newton's third law, and it holds in a way worth watching. Take I₁ = 100 A and I₂ = 10 A, 50 mm apart. Wire 1's field at wire 2 is 4.0 × 10⁻⁴ T, so wire 2 feels (10 A)(4.0 × 10⁻⁴ T) = 4.0 × 10⁻³ N m⁻¹. Wire 2's field at wire 1 is only 4.0 × 10⁻⁵ T, ten times weaker — but wire 1 carries ten times the current, so it feels (100 A)(4.0 × 10⁻⁵ T) = 4.0 × 10⁻³ N m⁻¹, the same number. The weaker field and the larger current cancel exactly, because both forces are built from the same product I₁I₂.

From 1948 until 2019 this equation was a definition rather than a result. The ampere was the constant current which, in two infinite parallel wires of negligible cross-section 1 m apart in vacuum, produces a force of exactly 2 × 10⁻⁷ N per metre. Reading that backwards, μ₀ was fixed at exactly 4π × 10⁻⁷ T m A⁻¹, and current balances — instruments that weigh the force between coils — realised the unit in the laboratory. Since 20 May 2019 the ampere is fixed instead by the elementary charge, e = 1.602176634 × 10⁻¹⁹ C exactly, and μ₀ has become a measured quantity: 1.25663706 × 10⁻⁶ T m A⁻¹, differing from 4π × 10⁻⁷ by well under a part in 10⁹. For every calculation in this course, 4π × 10⁻⁷ is exact enough.

Two laboratory wires pull on each other with micronewtons; two busbars in a switchboard do not. Run 15 kA down one bar and back along its partner 80 mm away and the force per unit length is (2 × 10⁻⁷)(15 000)²/(0.080) = 5.6 × 10² N m⁻¹ — about 57 kg of thrust on every metre, pushing the bars apart because the currents oppose. The bars themselves are strong; the insulators carrying them are what fails, so the design number is the reaction at each support, F/2 for a span simply supported at both ends. Two further facts make it worse. The force goes as the square of the current, and the first peak of an asymmetric fault can reach roughly 2.5 times the rms value, so the peak thrust is about six times the steady figure.

Three assumptions carry the result, and they fail at different rates. Infinite length is the gentlest: a pair ten separations long delivers about 90% of the infinite-wire force, and twenty separations about 95%, the shortfall coming from the weaker field near the ends. Negligible thickness is gentler still — by Ampere's law, a round conductor with uniform current density produces outside itself exactly the field of a filament on its axis, so the relation is exact for round wires at any centre-to-centre distance greater than the sum of the radii. What does bite is uniformity: the force redistributes the current across each conductor, and for bars closer than a few widths a geometry factor is needed. Nearby iron changes B outright, and the formula no longer applies at all.