Free Solenoid Inductance Calculator tool is helpful to calculate the self-inductance of a solenoid effortlessly & fastly. You have to enter the length, cross-section area, radius and number of turns details and hit the calculate button to check the inductance as output within no time.

**Solenoid Inductance Calculator: **Calculate the
self-inductance of a solenoid easily by using this handy & user-friendly
calculator. This calculator tool gives the accurate result instantly and
it shows the detailed calculations to solve it. Check out the step by
step process on how to find the inductance and solenoid inductance
formula here. Also, find the example questions and useful information
about solenoid.

Below provided are the steps for calculating the self-inductance of a solenoid easily.

- Check the number of turns, radius, length of the solenoid.
- Find the area of cross-section from the radius of the solenoid.
- Multiply the square of number of tuns with the cross-section area and vaccum permeability.
- Divide the product by the solenoid length to get the inductance of a solenoid.

A solenoid is a coil that is used in RLC Circuit. The characteristic inductance act as an interial element. The change in current flows through a coil is the self-induced potential difference.

The voltage in the solenoid is the ratio between the potential and the rate of change of current.

V = -L x dI/dt

The inductance of a solenoid can be stated as

L = µ₀ * N² * A/l

Where,

L is the self-inductance of a solenoid

N is the number of turns

r is the coil radius

A is the area of cross-section of the solenoid

l is the length

µ₀ is the vacuum permeability and its value is 1.25664 * 10^{-6}
T*m/A

**Example**

**Question: A solenoid has 33 turns and carries a current of 4 amps. It
has a radius of 5 cm, length of 8 cm find its inductance?**

**Solution:**

Given that

Number of turns of solenoid N = 33

Radius r = 5 cm

Length l = 8 cm = 0.08 m

Coss section area A = πr²

A = π x 5 x 5

= 78.54 cm² = 0.007854 m²

Inductance of the solenoid L = µ₀ * N² * A/l

L = 1.25664 x 10^{-6} x 33² * 0.007854/0.08

= 134.35 µH

Therefore, the inductance of a solenoid is 134.35 µH.

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** 1. What is the self-inductance of a solenoid?**

A solenoid is a long, tightly wound coil. Self-inductance is a property of a current-carrying coil that resist the change of current flowing through it. In simple terms, the self-inductance of a solenoid is a phenomenon where the induction of a voltage in a current-carrying wire.

**2. What is the inductance of a solenoid formula?**

The inductance of a solenoid formula is L = µ₀ * N² * A/l. The parameters included are the number of turns, length of coil, area of the cross-section.

**3. How does the solenoid inductance calculator works?**

The inductance of a solenoid calculator gets the inductance of a coil easily by taking the required details such as the number of turns, radius, area of cross-section and length.

**4. What are the uses of self-inductance?**

The inductor stores the electrical energy in the form of a magnetic field. Self-inductance is used in filters, sensors, induction motors, chokes, turning circuits, transformers, ferrite beads and as relays.