Ampere's force law

This verification example compares the force of attraction between two thin parallel current-currying wires given by Ampere's law and calculated in QuickField using three formulations: magnetostatics, time harmonics and transient magnetics.

Geometry:

Given:
I = 1 A - current of each wire;
f = 50 Hz - frequency in time harmonics and transient magnetics problems;
r = 1 m - distance between the wires.

Task:
Calculate interaction force (per meter of length) between two wires and compare with the value given by the Ampere's force law.

Solution:
Currents are set as linear ones. This way the model completely corresponds to Ampere's formulation.
In the time harmonics problem peak amplitude value of current is set √2·I.
In the transient magnetics current is set via formula I(t) = √2·I · sin(2·180·50·t).

According to the Ampere's law^* the force of the interaction between two parallel wires is determined as:
F = 2·(μ₀/4π) · I·I / r [N/m]

Results:

Ampere's law:
F = 2·(μ₀/4π) · 1·1/ 1 = 2·10^-7 [N/m]

Magnetostatics:

Time harmonics:

Transient magnetics:

Time

Current

Force

0.01 s

2 A

4.0092·10^-7 N/m

0.015 s

1 A

2.0046·10^-7 N/m

0.02 s

0 A

6.255·10^-22 N/m

F, *10^-7 N/m

Error

Ampere's formula

2.000

Magnetostatics

2.0042

0.2%

Time harmonics

2.0042

0.2%

Transient magnetics (at t=0.015 s)

2.0046

0.2%