On this page
The force between two current-carrying wires is a short derivation, but students lose marks on its direction, on which wire feels the bigger force, and on what the ampere actually means today. This post covers the formula, the old and new definitions of the ampere, and three worked problems.
Why do parallel currents attract?
Wire a carries current Iₐ and produces a magnetic field around it. At wire b, a distance d away, that field is Bₐ = μ₀Iₐ/(2πd). Wire b carries current I_b through this field, so it feels a force F = I_b L Bₐ. Putting the two together gives the result in the NCERT Class 12 chapter Moving Charges and Magnetism:
F/L = μ₀IₐI_b / (2πd)
The direction comes from the right-hand rule applied twice: once for the field of a at b, and once for the force on b in that field. The NCERT chapter states the result plainly: "Parallel currents attract, and antiparallel currents repel." It also notes this is the opposite of electrostatics, where like charges repel.
The JEE (Main) 2026 syllabus lists "the force between two parallel currents carrying conductors-definition of ampere" in Unit 13. If the right-hand rules themselves are shaky, start with our guide to magnetic force and right-hand-rule mistakes.
Does the wire with the bigger current feel a bigger force?
No. NCERT shows that the force on b due to a is equal and opposite to the force on a due to b, in line with Newton's third law. The formula is symmetric in Iₐ and I_b, so a 10 A wire and a 5 A wire pull on each other equally.
Worked example 1: force between two wires
Two long parallel wires are 2.0 cm apart. One carries 10 A and the other 5.0 A in the same direction. Find the force on a 50 cm length of either wire.
Use μ₀/(2π) = 2 × 10⁻⁷ T m/A.
F/L = 2 × 10⁻⁷ × 10 × 5.0 / 0.020 = 5.0 × 10⁻⁴ N/m.
On 0.50 m: F = 2.5 × 10⁻⁴ N, attractive, the same on both wires.
If one current is reversed, the size stays the same and the force becomes repulsive.
Worked example 2: where is the net force zero?
A third long wire is placed between the two wires above, in the same plane and parallel to them. Where must it go so that the net force on it is zero?
At a distance x (in cm) from the 10 A wire, the forces from the two wires balance when 10/x = 5/(2 − x). So 10(2 − x) = 5x, which gives x = 4/3 cm, or about 1.33 cm from the 10 A wire. The third wire's own current cancels out, so its size and direction do not matter. A common slip is placing the point nearer the larger current; it must be nearer the smaller one.
How did this force define the ampere?
NCERT gives the classic definition: the ampere is the steady current which, in two very long, straight, parallel conductors of negligible cross-section placed one metre apart in vacuum, would produce a force of 2 × 10⁻⁷ newtons per metre of length. Check it with the formula: with 1 A in each wire and d = 1 m, F/L = 2 × 10⁻⁷ × 1 × 1 / 1 = 2 × 10⁻⁷ N/m.
NIST explains the trouble with this definition: infinitely long wires in a vacuum cannot be built, so the ampere could only be approximated, and it depended on the newton and therefore on the kilogram artefact. Starting on 20 May 2019, the ampere is based on the elementary charge.
What is the definition of the ampere now?
The BIPM defines the ampere by fixing the elementary charge e at exactly 1.602176634 × 10⁻¹⁹ C, with C equal to A s. One ampere is then the current carried by 1/(1.602176634 × 10⁻¹⁹) elementary charges per second.
A side effect matters for problem solving. μ₀ used to be exactly 4π × 10⁻⁷ N/A². It is now measured: the CODATA value on NIST's site is 1.25663706127(20) × 10⁻⁶ N A⁻², which agrees with 4π × 10⁻⁷ to within about two parts in ten billion. Keep using μ₀/(2π) = 2 × 10⁻⁷ in exams.
Common mistakes and how to avoid them
Using μ₀/(4π) instead of μ₀/(2π). For a long straight wire, B = μ₀I/(2πd), so the factor is 2 × 10⁻⁷, not 10⁻⁷.
Forgetting to convert cm to m. In example 1, d = 0.020 m.
Getting the direction backwards by carrying the electrostatic rule over. Like currents attract.
Answering the force instead of force per length, or the other way round. Read what the question asks.
For the field formulas behind this result, see our post on Biot-Savart and Ampere's law.
A practice routine
Derive F/L from B = μ₀I/(2πd) and F = ILB until you can do it in under two minutes.
Solve three problems with currents in the same and opposite directions, drawing the field direction at each wire.
Solve two zero-net-force problems with a third wire, both between and outside the pair.
If the idea of the field of one wire acting on another is hard to picture, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise the field of a straight wire before returning to this topic.
Next, rework example 2 with the currents in opposite directions and find where the zero-force point moves.
