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?
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.