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
Formula
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. What you know: Current 1 2 A, Current 2 4.4 A, Separation 10 mm
- 2. The relation: F/L = µ₀ × I1 × I2 / (2π × d)
- 3. Substitute: I1 = 2 A, I2 = 4.4 A, d = 10 mm
- 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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<p><a href="https://vbatpower.com/formulas/force-between-wires/">Force Between Parallel Wires</a> Calculator by vbatpower.com</p> 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.
Related Concepts
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