### Solenoid Inductance Calculator:

Solenoid Inductance Calculator Finds the self-inductance of a solenoid.

By reading the text below, you will find out how a solenoid works in electrical circuits and what its inductance is.

If you want to know the effective resistance of a solenoid, we recommend checking out our inductance calculator.

Easily calculate the self-inductance of a solenoid using this simple and user-friendly calculator. This calculator tool gives accurate result instantly and shows detailed calculations to solve it.

Here is a step by step procedure on how to find inductance and solenoid inductance formula.

Below are the steps to easily calculate the self-inductance of a solenoid.

Check number of turns, radius, length of solenoid.

Find the cross-sectional area from the radius of the solenoid.

Multiply the square of the number of tons by the cross-sectional area and the void permeability.

To get the inductance of a solenoid, divide the product by the length of the solenoid.

A solenoid is a coil used in an RLC circuit. Characteristic induction acts as an internal element. A change in current through a coil is a self-induced potential difference.

Enter the Reference Values to calculate the Inductance value of the Solenoid coil.

### Solenoid Inductance Formula:

Solenoid inductance L_{(μH)} in micro Henries is equal to the vacuum permeability μ_{0(} _{T *m /A)} in Telsa per meter per Ampere and multiply square of the number of turns N^{2} and multiply the area of cross-section of the solenoid A_{(m2)} in meter^{2} and divided by the the length I_{(m)} in meter.Hence the solenoid inductance formula can be written as

L_{(μH)} = μ_{0(} _{T *m /A) }* N^{2} * A_{(m2)} /I_{(m)}

### Derivation:

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= μ_{} * N^{2} * A /I

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

I=is the length

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

### Example:1

A Solenoid has 45 turns and carries a current of 5 amps.It has a radius of 9cm ,length of 7cm find its inductance?

### Answer:

Given that

Number of turns of solenoid N =45

Radius r = 9cm

Length I = 7cm = 0.07m

A =π * r^{2}

A =π *9 *9

=254.46 cm^{2}=0.0025446 m^{2}

Inductance of the solenoid L = μ_{} * N^{2} * A /I

L= 1.25664 *10^{-6}*45^{2} *0.0025446 / 0.07

L= 0.00009250 μH

### Example:2

Calculate the number of turns 250 and radius of cross section 300m and length 200m.Find the inductance value?

### Answer:

N=250

R=300 m

I =200 m

By using formula

L = μ_{} * N^{2} * A /I

L= 1.25664 *10^{-6} *250^{2}* π * 300^{2} /200

=110.977020000H