# Solenoid Inductance Calculator

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.

### 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?

**Solution:**

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 physicscalculatorpro.com 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.