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Force Between Parallel Wires Calculator

Calculate the magnetic force between two parallel current-carrying wires — the same relationship that historically defined the ampere.

Force per unit length
400 µN/m
Total force over the given length
400 µN
Interaction
Attractive (currents flow the same direction)

How it works

Each current-carrying wire sits in the magnetic field the other one creates. Combining the field of an infinite straight wire, B = μ₀I ÷ (2πd), with the force on a current-carrying conductor, F = ILB, gives the force per unit length between two parallel wires: F ÷ L = μ₀ × I₁ × I₂ ÷ (2πd), where d is the separation between them. Verified via web search against HyperPhysics's and OpenStax University Physics's 'Magnetic Force between Two Parallel Conductors' derivations.

Separation and wire length get this page's metric/imperial toggle (m or ft), converted to meters before the formula runs; currents are left as plain, un-toggled amps, since the ampere is already the same SI unit worldwide. This relationship is also the historical basis for how the ampere itself used to be defined: two infinitely long, infinitely thin parallel wires each carrying exactly 1A, 1m apart in a vacuum, attract with a force of exactly 2×10⁻⁷ newtons per meter of length — this calculator reproduces that figure when you plug in those exact numbers.

Whether the wires pull together or push apart depends only on whether the currents flow the same way or opposite ways, not on the currents' magnitude or the separation — currents in the same direction attract, and currents in opposite directions repel, with identical force magnitude either way.

FAQ

Does this force depend on the wires' thickness or material?

No — the formula treats each wire as an infinitely thin filament carrying a given current, and the force depends only on the two currents, their separation, and the wire length, not on wire gauge, resistance, or material. Real wires are finite and carry current distributed across their cross-section, but for typical separations much larger than the wire diameter, this idealization is an excellent approximation.

Why is the force so small for household currents?

Because μ₀ ÷ (2π) is an extremely small number (about 2×10⁻⁷ in SI units), so unless the currents are very large or the wires are extremely close together, the resulting force is tiny — two wires each carrying 10A a few centimeters apart pull with only a few tens of micronewtons, far too small to feel by hand. This force becomes practically significant mainly in high-current contexts like busbars, power distribution, or MRI magnet windings.

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