Inductance of a pair of concentric cylinders - QuickField simulation example
A coaxial cable has an inner core of radius 1.0 mm and an outer sheath of internal radius of 4.0 mm. Determine the inductance of the cable per meter length.
How to find inductance per unit length of concentric cylindrical conductors?
Answer Typical applications Geometry
Inner radius a = 1 mm, outer radius b = 4 mm.
Given
Analytical solution*:
QuickField simulation results:
Note: QuickField calculates the total current and the total flux. To get RMS we should divide the corresponding amplitude value by root 2:
1% / 1.55%
0.17% / 1.03%
0.08% / 0.26%
0.05% / 0.14%
0.02% / 0.05%
0.02% / 0.02%
*References: John Bird, "Electrical circuit theory and technology", p.520. ISBN-13: 978 0 7506 8139 1.
Engineering question
Set up a plane-parallel QuickField AC Magnetics problem for concentric cylindrical conductors and evaluate inductance per unit length from computed field results.
coaxial conductors, shielded cables, cylindrical transmission lines
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Simulation problem
Problem Type
Plane-parallel problem of AC magnetics.
Current I = 0.001 A;
The total inductance per meter at low frequency is given by L = μ/2π · (1/4 + ln(b/a)) H/m
L = 4π·10-7 / 2π · (1/4 + ln(4)) = 3.27259·10-7 H/m
L = Flux / I = 3.272·10-10/1.4142 / 0.001/1.4142 = 3.272·10-7 H/m
L = 2·W / I² = 2·8.181·10-14 / (0.001/1.4142)² = 3.272·10-7 H/m
Mesh size
QuickField
Discrepancy with analytical solution*
L = Flux / I
L = 2·W / I²
251 (QuickField Student Edition)
3.23·10-7 H/m
3.22·10-7 H/m
1301 (automatic refinement in Professional Edition)
3.267·10-7 H/m
3.239·10-7 H/m
4003 (automatic refinement 2)
3.27·10-7 H/m
3.263·10-7 H/m
8814 (automatic refinement 3)
3.271·10-7 H/m
3.268·10-7 H/m
26474 (automatic refinement 5)
3.272·10-7 H/m
3.271·10-7 H/m
78585 (automatic refinement 10)
3.272·10-7 H/m
3.272·10-7 H/m
Video
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