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# 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.

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%
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