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Main >> Applications >> Sample problems >> Ampere's force law
This verification example compares the force of attraction between two thin parallel currentcurrying 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·(μ_{0}/4π) · I·I / r [N/m]
Results:
Ampere's law:
F = 2·(μ_{0}/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% 
*Wikipedia: Ampere's force law.