How this instrument works
The force between two straight, parallel wires comes from each wire's own magnetic field acting on the current flowing in the other. A wire carrying current I produces a magnetic field that circles it, falling off as 1 ⁄ r with distance; a second wire sitting in that field, carrying its own current, feels a sideways push or pull along its length. Ampère's force law packages this into F ⁄ L = μ₀I₁I₂ ⁄ (2πr): multiply the two currents, multiply by the permeability of free space, divide by twice pi times the separation. Run the currents the same direction and the wires attract; run them opposite ways and they push apart.
The 2π in the denominator is not decoration — it falls straight out of the geometry of a long, straight conductor. Ampère's law gives the field circling such a wire at radius r as B = μ₀I ⁄ (2πr); the force law is just that field acting, through F = IL × B, on the second wire's current. This is also why the result is exact only for wires long compared with their separation. Real finite conductors, coils, or bent busbars need the fuller Biot–Savart treatment, and this straight-line case is the limit they are checked against.
Because μ₀ is a small number — about 1.26×10⁻⁶ T·m/A — the force at household currents is tiny, micronewtons per metre, nowhere near enough to feel. It stops being trivial during a short-circuit fault, when currents in adjacent busbars or transformer windings can reach tens of kiloamps for a fraction of a second. Because the force scales with the product of the two currents, a fault current ten times larger than normal produces roughly a hundred times the mechanical stress on the mounting hardware, which is exactly the number switchgear engineers check before specifying busbar supports.
- Enter the current in wire 1, in amps — this is I₁ in the formula.
- Enter the current in wire 2, in amps — I₂. The fields are independent; the wires need not carry equal current.
- Set the distance between the wires, the centre-to-centre separation r, choosing cm, mm, or m to suit closely spaced busbars or widely spaced feeders.
- Read the force per unit length, N/m — the sideways pull each metre of wire feels, an attraction when both currents run the same way.
Worked example — two 5 A wires, 10 cm apart
Take two straight wires, each carrying 5 A, running parallel and 10 cm (0.1 m) apart — a fair description of two current-carrying conductors bundled in a household sub-panel. Multiply the currents: 5 × 5 = 25. Multiply by μ₀ ⁄ (2π), which works out to exactly 2×10⁻⁷ T·m/A, then divide by r: F ⁄ L = 2×10⁻⁷ × 25 ⁄ 0.1 = 5×10⁻⁵ N/m — fifty micronewtons of attraction, assuming the currents flow the same direction, for every metre the wires run side by side.
This isn't an arbitrary demonstration figure. Before the SI was redefined in 2019, the ampere itself was fixed by this exact setup: the current that, flowing through two infinite parallel wires exactly 1 metre apart, produces a force of precisely 2×10⁻⁷ N per metre of length was defined to be one ampere. Scale that historical reference down to two 5 A wires 0.1 m apart and the geometry moves the number to the 5×10⁻⁵ N/m this instrument returns — 250 times larger than the defining case, because the currents are five times bigger (25× the product) and the wires are ten times closer (10× the force).
Questions
Why do the wires attract instead of repel?
Because in this example both currents flow the same direction. Ampère's force law gives an attraction when I₁ and I₂ point the same way and a repulsion when they run opposite; flip the direction of either current and the same 5×10⁻⁵ N/m becomes a push apart instead of a pull together. This is the same rule that keeps twisted-pair cable resistant to self-induced movement — the two conductors carry opposite currents and so repel, holding the pair pulled taut rather than clumped.
Is the force always this small?
At everyday currents, yes — a few amps a few centimetres apart gives micronewtons, far below what you would feel. The force scales with the product of the two currents and inversely with distance, so it grows fast: two 1000 A busbars 5 cm apart feel roughly 4 N per metre, and short-circuit fault currents of tens of kiloamps can generate hundreds of newtons per metre, enough to bend or tear loose busbar supports not designed for it.
Why does the formula assume infinite wires?
Because that is the case Ampère's law solves exactly: a straight conductor with no ends, so its magnetic field depends only on distance from the wire. Real finite wires have end effects this formula ignores, but the approximation holds well whenever the wires are much longer than the gap between them — a typical busbar run or parallel cable trunk easily qualifies, which is why engineers reach for this straight-line result rather than a full Biot–Savart integral for routine checks.
What if the two wires carry different currents?
The formula still applies directly — enter I₁ and I₂ as given, since the force depends on their product, not their average or sum. A 2 A wire beside a 20 A wire 5 cm apart feels the same force per metre as any other pairing multiplying to 40: F ⁄ L = μ₀ × 2 × 20 ⁄ (2π × 0.05) ≈ 1.6×10⁻⁴ N/m, regardless of which field holds which value.
How is this related to the definition of the ampere?
This exact relation used to define it. From 1948 to 2019, one ampere was the constant current that, maintained in two straight infinite parallel wires one metre apart in a vacuum, would produce a force of exactly 2×10⁻⁷ newtons per metre between them. The 2019 SI redefinition replaced that with a fixed value for the elementary charge instead, but the force law and the μ₀ figure used here carry on unchanged — only the reference point moved, not the physics.
Does the formula work for alternating current?
Only as an instantaneous or RMS approximation. Ampère's force law as written assumes steady current; with AC the relation still holds at each instant, but because it depends on the product of two currents, feeding in RMS values gives the average force over a cycle, not the peak. That is why switchgear designers separately check peak asymmetrical fault current rather than relying on RMS alone.