Industrial Educational Scientific Sample problems Examples gallery Step-by-step tutorials Verification examples Programming examples Distributive examples Success stories Customers Main >> Applications >> Sample problems >> Ampere's force law Ampere's force law Ampere force law, magnetic force between two wires 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% *Wikipedia: Ampere's force law. Watch on YouTube. View simulation report in PDF. Download simulation files (files may be viewed using any QuickField Edition).

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