Created By : Vaibhavi Kumari

Reviewed By : Rajashekhar Valipishetty

Last Updated : May 14, 2023

The free Solenoid Inductance Calculator tool makes it simple and quick to calculate a solenoid's self-inductance. To check the inductance as output, enter the length, cross-section area, radius, and the number of turns data, then press the calculate button.

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Number of turns(N):

Inductance of a Solenoid

The RLC Circuit uses a solenoid, which is a coil. The internal element is the characteristic inductance. The self-induced potential difference describes how current flows through a coil.

The voltage in a solenoid is equal to the ratio of potential to current change rate. V = -L x dI/dt

A solenoid's inductance can be expressed as L = µ₀ x N^2 x A/l

  • Where, L = self-inductance of a solenoid
  • N = number of turns
  • r = coil radius
  • A = cross-sectional area of the solenoid
  • l = length
  • µ₀ = vacuum permeability (µ₀ = 1.25664 x 10^-6 T*m/A)

How do you find the Inductance of a Solenoid?

The steps for calculating the self-inductance of a solenoid are listed below

  • Step 1: Check the solenoid's number of turns, radius, and length.
  • Step 2: Calculate the cross-sectional area using the solenoid's radius.
  • Step 3: Multiply the cross-section area and vacuum permeability by the square of the number of turns.
  • Step 4: To calculate the inductance of a solenoid, multiply the product by the length of the solenoid.

How to Use the Solenoid Inductance Calculator?

The following is the procedure how to use the solenoid inductance calculator

  • Step 1: Input the unknown value's length, cross-section area, radius and number of turns, and x in the appropriate input fields.
  • Step 2: To acquire the result, click the "Calculate the Unknown" button.
  • Step 3: Finally, the output field will show the self-inductance of a solenoid.

Examples on Calculating Inductance of a Solenoid

Question 1: A solenoid has 43 turns and can carry 4 amps of current. Find the inductance of a circle with a radius of 7 cm and a length of 8 cm?


Given: Number of turns of solenoid N = 43

Radius r = 7 cm

Length l = 9 cm = 0.09 m

Cross section area A = πr²

A = π x 7 x 7

= 153.93 cm^2 = 0.001539 m^2

Inductance of the solenoid L = µ₀ x N^2 x A/l

= 1.25664 x 10^-6 x 43² x 0.001539/0.09

= 39.73 x 10^-6 µH

Therefore, the inductance of a solenoid is 39.73 x 10^-6 µH.

For more concepts check out to get quick answers by using this free tool.

FAQs on Solenoid Inductance

1. Is a solenoid an inductor?

An inductor is a solenoid. Not all solenoids are inductors. A solenoid is a moving coil with a ferrous core. A wire coil with a fixed core, which may or may not be present, is known as an inductor.

2. What is the coil's inductance?

Even a straight piece of wire has some inductance, which is the name given to the property of a component that opposes the change of current passing through it.

3. How does the number of turns in a solenoid's coil affect its self-inductance?

The coil's self-inductance is determined by the cross-sectional area, the number of turns per unit length, the length of the solenoid, and the permeability of the core material.

4. Why is an electromagnet stronger than a solenoid?

Inside the solenoid, a magnetised ferromagnetic substance adds to the strength of the magnetic field created by the current alone.

5. How does the solenoid inductance calculator work?

The inductance of a solenoid calculator quickly calculates the inductance of a coil by entering the required information, such as the number of turns, radius, cross-sectional area, and length.

6. What is the significance of self-inductance?

The electrical energy is stored in the form of a magnetic field by the inductor. Filters, sensors, induction motors, chokes, turning circuits, transformers, ferrite beads, and relays all use self-inductance.