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Self Inductance

Last updated date: 18th May 2024
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Let’s observe the magic of the concept of self-inductance.

Let’s consider a coil and pass current through it, not only current but the changing current.

Now, because of the changing current, an additional current induces in it, i.e., the induced current. 

So, do you know what this induced current means?

Well, the induced current is intangible, and it’s a property of the coil to generate it because of the changing current supplied by us through the battery. 

Isn’t it so simple? 

However, our focus is on self-inductance, so let’s get back to observe this magical concept.

So, what did you observe in this phenomenon, and why is it magical?

So, the magic is: On passing the current to the coil, it induces a current within itself known as self-induced. That’s why this phenomenon is known as self-inductance.

Self Inductance of Coil

Let’s consider a coil and supply current through it. The current which is supplied is the primary current and we will consider two cases here, which are:

  1. The supplied that goes on increasing, and

  2. The supplied current is decreasing.

Case a: Consider a coil, in which the primary (supplied) current goes on increasing in the direction, as shown below in the diagram.

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As you know that the increasing (changing) current generates an induced current by itself, i.e., the self-induced current and within itself, but that flows in a direction opposite to that of the current supplied. The direction of this self-induced current is described through the diagram below.

(image will be uploaded soon)

This induced current opposes any change (or increasing current) in that current because of which it is produced.

Now, taking another case:

Case b: Consider a coil, in which the current is decreasing, what happens here is, this induced current favors the change (or decrease) in the applied current.

This means the induced current flows in the direction of the current applied and favors its increase.

(image will be uploaded soon)

We understood that induced current opposes the increase in the current and supports the diminishing current.

(image will be uploaded soon)

What is Self-Inductance?

Let’s consider a circular in which changing current produces a magnetic field (B).

The direction of this field can be determined by curling the fingers of our right-hand and we get the direction of B, which is pointing inwards, that can be seen as cross marks in the diagram below:

(image will be uploaded soon)

Now, on increasing the current, the magnetic field lines also increase. This means B α i.

Due to the increase in B, flux (ΦB) also increases. 

As soon as the flux increases, then according to Faraday’s law of induction, an induced emf generates in this coil.

By Lenz’s law, 

This induced emf is the potential difference between the two points in this coil because of which an induced current is generated. This induced current will decrease the primary current. Its direction is outward, i.e., opposite to the direction of B.

This current produces its flux that opposes the flux (ΦB) because of which it was generated.

So, this is the phenomenon of self-inductance.

Self-Inductance Formula

Let’s consider a current-carrying coil having N number of turns, as shown below:

(image will be uploaded soon)

If the flux through one coil is Φ, then for N number of coils, it will be:

                       ΦT (total flux) = NΦ, and  

                         ΦT α i

On removing the proportionality sign, we get 

                        ΦT = Li ⇒ L = \[\frac{N\Phi T}{i}\]

Where L is the coefficient of self-inductance.

Here, L is a constant that doesn’t depend upon Φ and i. Rather it depends upon the following:

  1. Geometry,

  2. Shape, and

  3. Size of the inductor (coil).

This means the inductance doesn’t change with the increase or decrease with the change in Φ and i.

Unit of Self-Inductance

The unit of self-inductance is Weber/Ampere or Henry ‘H’.

The dimension of self-inductance is [ML2T-2A-2].

Coefficient of Self-Inductance Definition

By Faraday’s law, the emf, \[e = -\frac{\Phi T}{dt}\]

So, \[e = |- L \frac{di}{dt}| \Rightarrow L = \frac{e}{|\frac{di}{dt}|}\]

If the value of a change in current or di/dt is 1 Ampere/second, then L = e. This is the definition of the coefficient of self-inductance.

We know that inductance is the property of an electric conductor by which the change in current produces an EMF.

Self-Induction and Mutual Induction


Self - Induction

Mutual Induction

1. Definition

If the rate of current produces an emf or voltage in the same coil, then this type of induction is the self-induction.

If the rate of change of current produces an induced emf in the neighboring coil, then it is mutual induction.

2. Formula

\[e = - L \frac{di}{dt}\], and \[L = \frac{e}{|di/dt|}\]

For N number of turns in a coil, \[L = \frac{N \Phi T}{i}\]

If the current is flowing in the primary coil, then the coefficient of mutual induction,  \[M = \frac{N_{2} \Phi_{12}}{i_{1}}\]

If the current is flowing in the secondary coil, then \[M = \frac{N_{1} \Phi_{21}}{i_{2}}\]

FAQs on Self Inductance

Q1: If the coefficient of self-inductance of a coil is 4 H, and the amount of current flowing through it per unit time is 5 Ampere/Sec, then find the emf induced in the coil.

Ans: Here, L = 4 H, di/dt = 5 Ampere/second, e = ?

We know that e = |-L di/dt|= L di/dt = 4 x 5 = 20 Volt

Q2: What is the Value of Self-Inductance?

Ans: We know that e = |-L di/dt|. When the rate of change of current is 1 Ampere per second, and the emf induced in the coil is 1 V, then the value of inductance, i.e., L = 1.

Q3: Does Current Affect Inductance?

Ans: No. Inductance is a property that opposes the change in current. This property can be increased by forming the conductor into a loop.

In a loop, the magnetic lines of force affect more of the conductor at a time. This increases self-induced emf.

Q4: Find the change in current in an inductor of 12 H in which the emf induced is 400 V in 10-2 sec. Also, find the change in magnetic flux.

Ans: Here, L = 12 H,  dt = 10-2 sec, e = 400 V, and di = ?, dΦ = ?

Using the formula, e = |-L di/dt|

⇒ 400 = 12 x di/10-2 ⇒ di = 0.3 A

Now, using  the formula, 

e = dΦ/dt

⇒ dΦ = e * dt = 400 x 10-2 sec = 4 Wb