Explain attraction and repulsion between currents

Use the other wire’s field

Two parallel current-carrying wires can attract or repel. The field and force arrows are not interchangeable: the field near a wire circulates around it, while the interaction force in this geometry points toward or away from the other wire.

Start at 2 cm with both currents +2 A. Wire 1’s field at wire 2 is upward, while wire 2’s current points toward you. Predict the force on wire 2 with the magnetic-force direction rule. Then reverse wire 2 to −2 A and predict whether the interaction becomes attraction or repulsion.

From moving charges to a force on a conductor

For a straight segment carrying current in an externally produced uniform field, the force is F = I (ℓ × B). The vector ℓ points along the chosen positive current direction and its length is the segment length in metres. I is signed current in amperes. Use conventional current; this relation already accounts for the charge-carrier direction. Reversing electron drift a second time would introduce a sign error.

Here the wires run along z and each other’s field runs along y. The relevant cross product is +z × +y = −x. Thus positive current in wire 2 feels force toward wire 1, which is to its left. The interaction is attractive for currents in the same direction. Opposite current directions give repulsion.

Because the ideal wires are infinitely long, the browser reports force per unit length, not a total force on an infinite object. Define f₂ = F₂x/ℓ, the signed x force per length on wire 2. In this geometry f₂ = −I₂ B₁y = −μ₀ I₁ I₂/(2πr), where B₁y is the field from wire 1 at wire 2. Force per length is measured in newtons per metre (N/m); multiplying it by a relevant length gives force only where the assumed uniform-per-length approximation applies.

Work through the default pair

At r = 0.020 m, I₁ = +2 A and I₂ = +2 A, B₁y = +20 µT. Therefore f₂ = −2 × 20 × 10⁻⁶ = −40 × 10⁻⁶ N/m = −40 µN/m. Wire 2 is forced left. Wire 1 feels +40 µN/m, to the right. Each wire is pulled toward the other.

For a 0.25 m portion sufficiently far from the ends of much longer wires, the predicted magnetic force magnitude is 40 µN/m × 0.25 m = 10 µN on that portion. This does not establish the total force between two isolated 0.25 m wires: their ends and return paths would matter. State the length and geometry whenever you convert force per length into force.

Capture +2 and −2 A at the same separation. The sign of each force reverses, while the magnitude stays 40 µN/m. At I₂ = 0, both interaction forces vanish even though B₁y remains 20 µT. A magnetic field is not itself a force; the receiving current also matters.

Unequal fields can give equal forces

Set I₂ = +4 A at 2 cm. Wire 1 still supplies +20 µT at wire 2. Wire 2 now supplies −40 µT at wire 1. Nevertheless the force magnitudes are equal:

• On wire 2: 4 A × 20 µT = 80 µN/m toward wire 1.
• On wire 1: 2 A × 40 µT = 80 µN/m toward wire 2.

Each current is a source in one calculation and the receiver in the other. The product I₁I₂ is the same in both, producing equal-and-opposite forces in this ideal steady pair. Unlike the previous lesson’s separate hypothetical particle trials, these rows describe one mutual interaction. The forces act on different wires; they do not cancel the force on either individual wire.

The wires are held at fixed positions. Supports would supply balancing forces, but those are not drawn. No acceleration is inferred from the magnetic-force rows alone. Currents are prescribed, and the model omits electrical power, resistive heating and the response of the source circuits. Comparing separations does not calculate the work required to move actual wires while maintaining current.

Compare one change at a time

Hold I₂ = +2 A and compare separations 1, 2 and 4 cm. Force per length magnitudes are 80, 40 and 20 µN/m. Doubling separation halves the interaction. Hold separation fixed and double I₂: the force doubles. If both currents doubled, the force would quadruple; the browser fixes I₁, so that last comparison is a symbolic transfer rather than another slider setting.

A useful check combines all three dependencies: |f| is proportional to |I₁I₂|/r. For signs, return to the vector rule or compare current directions. Do not use a negative force magnitude; a negative component instead means leftward force in the chosen axis.

Transfer and diagnose a proposed explanation

Two long parallel wires carry +3 A and −5 A at 0.040 m separation. Find the interaction type and force per length magnitude. Estimate the force on a 0.20 m interior portion. A learner says the stronger-current wire must feel a stronger force. Use source fields to evaluate that explanation.

Check the parallel-current transfer

The opposite currents repel. The force magnitude per length is (2 × 10⁻⁷) × 3 × 5 / 0.040 = 75 × 10⁻⁶ N/m = 75 µN/m. A 0.20 m interior portion has approximately 15 µN force under the long-wire assumption. The 3 A wire supplies 15 µT at the 5 A wire, while the 5 A wire supplies 25 µT at the 3 A wire. Products 5 × 15 and 3 × 25 are both 75 µN/m. Unequal source fields do not imply unequal interaction-force magnitudes.

Use two captures to justify attraction versus repulsion. Identify the receiving current and external field in each force calculation, give units, and state why fixed positions do not imply zero magnetic force. The current values are simulation inputs, not a physical wiring procedure.

Reference for the interaction law: OpenStax: force between parallel currents. This lesson uses the force relationship with a rounded magnetic constant; it does not adopt the source’s historical discussion as the present SI definition of the ampere. Worked examples and transfer tasks are original.