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 calculate the inductance per unit length of concentric cylindrical conductors and verify?
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
Calculate inductance from either flux linkage or magnetic energy and compare the converged mesh results with the analytical coaxial-cable formula.
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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