Force Between Parallel Wires

F/L = µ0 · I1 · I2 / (2π · d)

Quick answer

This calculator solves F/L = µ₀ × I1 × I2 / (2π × d) for Force per Length. With Current 1 2 A, Current 2 4.4 A, Separation 10 mm, the result is 0.176 mN/m. It solves in 4 directions, so you can enter the values you have and get the one you need.

Calculator

Force per Length (N/m)
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Formula

F/L = µ₀ × I1 × I2 / (2π × d)

Description

Two parallel current-carrying conductors exert a force per unit length on each other through their magnetic fields. Currents in the same direction attract; opposite currents repel. This effect once defined the ampere itself and is the source of the mechanical stress that high-current busbars and fault currents place on conductors.

Variables

  • F/L — Force per unit length (N/m)
  • I1, I2 — Currents in the two wires (A)
  • d — Separation between wires (m)
  • µ₀ = 4π×10⁻⁷ H/m —

Practical Notes

During a short-circuit fault, busbars can see thousands of amps and large attractive/repulsive forces—mechanical bracing must be rated for it. Twisting or transposing conductors reduces the net force.

Symbols and units

Symbol Meaning Default unit Accepted units
FL Force per Length N/m µN/m, mN/m, N/m
I1 Current 1 A pA, nA, µA, mA, A, kA
I2 Current 2 A pA, nA, µA, mA, A, kA
d Separation mm mm, cm, m, km

Step by step

  1. 1. What you know: Current 1 2 A, Current 2 4.4 A, Separation 10 mm
  2. 2. The relation: F/L = µ₀ × I1 × I2 / (2π × d)
  3. 3. Substitute: I1 = 2 A, I2 = 4.4 A, d = 10 mm
  4. 4. Result: Force per Length = 0.176 mN/m

Worked examples

Minimum range

Given: Current 1 2 mA, Current 2 4.4 mA, Separation 0.1 mm

Result: Force per Length = 0.0176 µN/m

Low range

Given: Current 1 20 mA, Current 2 44 mA, Separation 1 mm

Result: Force per Length = 0.176 µN/m

Typical

Given: Current 1 2 A, Current 2 4.4 A, Separation 10 mm

Result: Force per Length = 0.176 mN/m

High range

Given: Current 1 100 A, Current 2 0.22 kA, Separation 10 cm

Result: Force per Length = 44 mN/m

Maximum range

Given: Current 1 1 kA, Current 2 2.2 kA, Separation 1 m

Result: Force per Length = 0.44 N/m

Solving for each variable

Solve for Force per Length

Current 1 2 mA, Current 2 4.4 mA, Separation 0.1 mm → 0.0176 µN/m

Solve for Current 1

Force per Length 100 mN/m, Current 2 4.4 mA, Separation 0.1 mm → 11.36 kA

Solve for Current 2

Force per Length 100 mN/m, Current 1 2 mA, Separation 0.1 mm → 25 kA

Solve for Separation

Force per Length 100 mN/m, Current 1 2 mA, Current 2 4.4 mA → 1.76 × 10⁻⁸ mm

How Force per Length changes with Current 1

Current 1 (input) Force per Length (result)
0.2 A 17.6 µN/m
0.5 A 44 µN/m
1 A 88 µN/m
1.5 A 132 µN/m
2 A 0.176 mN/m
3 A 0.264 mN/m
4 A 0.352 mN/m
8 A 0.704 mN/m
14 A 1.232 mN/m
20 A 1.76 mN/m

The same result in other units

Unit Value
µN/m 176
mN/m 0.176
N/m 0.000176

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Frequently asked questions

How do you calculate Force Between Parallel Wires?

Use F/L = µ₀ × I1 × I2 / (2π × d). With Current 1 2 A, Current 2 4.4 A, Separation 10 mm, Force per Length works out to 0.176 mN/m.

What is the formula for Force Between Parallel Wires?

The formula is F/L = µ₀ × I1 × I2 / (2π × d). Two parallel current-carrying conductors exert a force per unit length on each other through their magnetic fields.

How do you find Current 1 when you know the other values?

Rearrange for Current 1: with Force per Length 100 mN/m, Current 2 4.4 mA, Separation 0.1 mm, Current 1 is 11.36 kA.

How do you find Current 2 when you know the other values?

Rearrange for Current 2: with Force per Length 100 mN/m, Current 1 2 mA, Separation 0.1 mm, Current 2 is 25 kA.

What units does Force Between Parallel Wires use?

Force per Length is shown in µN/m, mN/m, N/m. Values are held in SI base units internally, so you can mix prefixes without converting by hand.