**Inductance** is the name given to the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance.

Inductors do this by generating a self-induced emf within itself as a result of their changing magnetic field. In an electrical circuit, when the emf is induced in the same circuit in which the current is changing this effect is called **Self-induction**, ( L ) but it is sometimes commonly called back-emf as its polarity is in the opposite direction to the applied voltage.

When the emf is induced into an adjacent component situated within the same magnetic field, the emf is said to be induced by **Mutual-induction**, ( M ) and mutual induction is the basic operating principal of transformers, motors, relays etc. Self inductance is a special case of mutual inductance, and because it is produced within a single isolated circuit we generally call self-inductance simply, **Inductance**.

The basic unit of measurement for inductance is called the **Henry**, ( H ) after Joseph Henry, but it also has the units of **Webers per Ampere** ( 1 H = 1 Wb/A ).

Lenz’s Law tells us that an induced emf generates a current in a direction which opposes the change in flux which caused the emf in the first place, the principal of action and reaction. Then we can accurately define **Inductance** as being: “a coil will have an inductance value of one Henry when an emf of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second”.

In other words, a coil has an inductance, ( L ) of one Henry, ( 1H ) when the current flowing through it changes at a rate of one ampere/second, ( A/s ) inducing a voltage of one volt, ( V_{L} ) in it. This mathematical representation of the rate of change in current through a coil per unit time is given as:

Where: *di* is the change in the current in Amperes and *dt* is the time taken for this current to change in seconds. Then the voltage induced in a coil, ( V_{L} ) with an inductance of L Henries as a result of this change in current is expressed as:

Note that the negative sign indicates that voltage induced opposes the change in current through the coil per unit time ( di/dt ).

From the above equation, the inductance of a coil can therefore be presented as:

Where: L is the inductance in Henries, V_{L} is the voltage across the coil and di/dt is the rate of change of current in Amperes per second, A/s.

Inductance, L is actually a measure of an inductors “resistance” to the change of the current flowing through the circuit and the larger is its value in Henries, the lower will be the rate of current change.

We know from the previous tutorial about the Inductor, that inductors are devices that can store their energy in the form of a magnetic field. Inductors are made from individual loops of wire combined to produce a coil and if the number of loops within the coil are increased, then for the same amount of current flowing through the coil, the magnetic flux will also increase.

So by increasing the number of loops or turns within a coil, increases the coils inductance. Then the relationship between self-inductance, ( L ) and the number of turns, ( N ) and for a simple single layered coil can be given as:

- Where:
- L is in Henries
- N is the Number of Turns
- Φ is the Magnetic Flux Linkage
- Ι is in Amperes

This expression can also be defined as the flux linkage divided by the current flowing through each turn. This equation only applies to linear magnetic materials.

A hollow air cored inductor coil consists of 500 turns of copper wire which produces a magnetic flux of 10mWb when passing a DC current of 10 amps. Calculate the self-inductance of the coil in milli-Henries.

Calculate the value of the self-induced emf produced in the same coil after a time of 10mS.

The self-inductance of a coil or to be more precise, the coefficient of self-inductance also depends upon the characteristics of its construction. For example, size, length, number of turns etc. It is therefore possible to have inductors with very high coefficients of self induction by using cores of a high permeability and a large number of coil turns. Then for a coil, the magnetic flux that is produced in its inner core is equal to:

Where: Φ is the magnetic flux linkage, B is the flux density, and A is the area.

If the inner core of a long solenoid coil with N number of turns per metre length is hollow, “air cored”, then the magnetic induction within its core will be given as:

Then by substituting these expressions in the first equation above for Inductance will give us:

By cancelling out and grouping together like terms, then the final equation for the coefficient of self-inductance for an air cored coil (solenoid) is given as:

- Where:
- L is in Henries
- μ
_{ο}is the Permeability of Free Space (4.π.10^{-7}) - N is the Number of turns
- A is the Inner Core Area (π.r
^{ 2}) in m^{2} - l is the length of the Coil in metres

As the inductance of a coil is due to the magnetic flux around it, the stronger the magnetic flux for a given value of current the greater will be the inductance. So a coil of many turns will have a higher inductance value than one of only a few turns and therefore, the equation above will give inductance L as being proportional to the number of turns squared N^{2}.

As well as increasing the number of coil turns, we can also increase inductance by increasing the coils diameter or making the core longer. In both cases more wire is required to construct the coil and therefore, more lines of force exists to produce the required back emf. The inductance of a coil can be increased further still if the coil is wound onto a ferromagnetic core, that is one made of a soft iron material, than one wound onto a non-ferromagnetic or hollow air core.

Ferrite Core

If the inner core is made of some ferromagnetic material such as soft iron, cobalt or nickel, the inductance of the coil would greatly increase because for the same amount of current flow the magnetic flux generated would be much stronger. This is because the material concentrates the lines of force more strongly through the the softer ferromagnetic core material as we saw in the Electromagnets tutorial.

So for example, if the core material has a relative permeability 1000 times greater than free space, 1000μ_{ο} such as soft iron or steel, then the inductance of the coil would be 1000 times greater so we can say that the inductance of a coil increases proportionally as the permeability of the core increases. Then for a coil wound around a former or core the inductance equation above would need to be modified to include the relative permeability μ_{r} of the new former material.

If the coil is wound onto a ferromagnetic core a greater inductance will result as the cores permeability will change with the flux density. However, depending upon the ferromagnetic material the inner cores magnetic flux may quickly reach saturation producing a non-linear inductance value and since the flux density around the coil depends upon the current flowing through it, inductance, L also becomes a function of current flow, i.

In the next tutorial about inductors, we will see that the magnetic field generated by a coil can cause a current to flow in a second coil that is placed next to it. This effect is called Mutual Inductance, and is the basic operating principle of transformers, motors and generators.

Related product: Common Mode Chokes

Error! Please fill all fields.

I want data for drum core inductor.

specificatons:

5mh

12ohm

0.16amp

Nice tutorial….hi supriya how is tutorial?

i like but if it is more brief,its so good4

A current of 4 A is passed through a solenoid (l = 5 m) which has 1800 coils. Assuming that there is a core (relative permeability = 130) in the center of the solenoid, calculate the total magnetic field, B at a point in the center of the solenoid.

An inductor supplied with 50 V ac with a frequency of 10kHz passes a current of 7.96 mA. The value of inductor is

(A) 1 mH

(B) 10 mH

(C) 100 mH

(D) 1 H

which formula solves it?

XL = V/I and XL = 2pi.f.L

therefore: XL = 50/0.00796 = 6281.4 ohms

and L = XL/(2pi.f) = 6281.4/(2pi x 10000) = 100mH

Here is the simple way to measure the Inductive Reactance of a Coil in AC circuit

1 –just measure the pure resistance if the given coil – in ohms (#…)

2 – apply AC Voltage to the coil and measure the Current through this coil

3- now calculate the total Resistance based on applied Voltage and measured Current, (per Ohm’s law)

4 – the Total calculated Resistance(actually the Impedance) will be the SUM of:

Pure resistance – plus – Inductive Reactance (of the coil) and the result will come straight

In Ohms

5 – then, the Inductive Reactance will be a result of difference between:

(Total Resistance of the Circuit – minus-Pure Resistance of the Coil) – in Ohms!!!

All done!

Rethink No4, impedance, Z is the Vector Sum.

A great tutorial it is

Is the coil inductance equation you gave for one layer of coil or does it apply to multiple layers as well? If not, what would the equation be for multiple layers of coils?

How do I determine the rate of flux in a circuit in which 15 V is induced in a 30-turn coil?

You need one of Faraday’s Law’s as the voltage (v) induced by the magnetic flux cutting the turns of a coil will depend on the number of turns, (N) and the speed by which the flux cuts the coil.

Giving v = N(dΦ/dt), therefore the rate of flux will be dΦ/dt = v/N = 15/30 = 0.5W/s

Then the flux is changing at a rate of: 500 milli-webers per second (500mW/s).

Can induction occur in cable that is sheaved that is 1.5meters in length coiled around a plastic core.

All the best

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