Long solenoid inductance - QuickField simulation example
Engineering question
How to find long solenoid inductance calculation?
Answer
Set up an axisymmetric QuickField DC Magnetics problem for a long solenoid and evaluate inductance from computed field results.
Typical applications
long solenoid coils, laboratory solenoids, axisymmetric coil windings
Download
- Download simulation files (files may be viewed using any QuickField Edition).
Simulation problem
Problem TypeAxisymmetric problem of DC magnetics.
Geometry
R1 = 30 mm, R2 = 35 mm.
Given
Current in the conductor i = 10 A;
Number of turns N = 100,
Length L = 0.5 m;
Task
Calculate inductance of the long solenoid, side effects are neglected.
Solution
Inductance of the long solenoid can be obtained from the equation:
L = N ² · μ0 A / L,
where A is a cross-section area of the core (m²).
To calculate the inductance in QuickField you should divide magnetic flux by the source current.
L = N · Φ / i
Results
Cross-section area of the core
A = π * ( (R1 + R2)/2 )² = 3.142 * ( (0.03+0.035)/2 )² = 0.00332 m².
Inductance of the long solenoid
L = 100² · 4e-7 · 3.142 * 0.00332 / 0.5 = 83.45 μH.
| QuickField | Theory | Δ, % | |
|---|---|---|---|
| Inductance L, μH | 83.55 | 83.45 | 0.12% |
* Reference: John Bird, Electrical circuit theory and technology, p.77. ISBN-13: 978 0 7506 8139 1.
Video
- Long solenoid inductance
Watch on YouTube. - View simulation report in PDF