QuickField

A new approach to field modelling

Main >> Applications >> Sample problems

Long solenoid inductance

long solenoid inductance, magnetic FEA

Problem Type
Axisymmetric problem of DC magnetics.

Geometry
Long solenoid inductance Calculate inductance of the long solenoid, side effects are neglected. Winding R1 R2

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.

Long solenoid magnetic field

  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 online on YouTube.
  • View simulation report in PDF
  • Download simulation files (files may be viewed using any QuickField Edition).
  • Eddy currents simulation with QuickField

    Product
    Order
    Evaluation request
    Editions
    Version history
    Packages
    Components
    Programming
    Consulting

    Applications
    Industrial
    Educational
    Scientific
    Sample problems
    Success stories
    Customers

    Support
    Webinars
    Virtual Classroom
    Online courses
    Customer login
    Glossary
    QuickField help
    FAQ

    Downloads
    QuickField Student Edition
    User manual
    Data Libraries
    Video
    Free tools

    News
    Product News
    Events
    Publications
    Subscription

    Contacts
    Tera Analysis info
    Resellers
    Contact us
    Online chat
    About us