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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:

two conductors carrying current for Ampere force calculation

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:

Ampere force in DC magnetics

Time harmonics:

Ampere force in 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

Ampere force in transient magnetics

 

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.

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