What is Self Induction

ExpertsMind.com > Physics HelpMagnetism 

Self induction is the property of a coil by virtue of which the coil opposes any change in the strength of current flowing through it by inducing an in itself. For this reason, self induction is also called the inertia of electricity.

If current in a coil L is changed by varying the contact position on a variable resistor, a self induced appears in the coil – while the current is changing.

When current (I) is increasing the self induced appears across the coil in a direction such that it opposes the increase.

When current (I) is decreasing the self induced (e) appears across the coil in a direction such that it opposes the decrease. Therefore it would be in the direction of (I)

Coefficient of self induction

Suppose I= strength of current flowing through a coil at any time.

∅ = amount of magnetic flux linked with all the turns of the coil at that time.

It is found that ∅ ∝I

∅ = LI

Where L is a constant of proportionality and is called coefficient of self induction or self inductance of the coil. The value of L depends on number of turn’s area of cross-section and nature of material of the core on which it he coil is wound.

If I = 1, ∅ = L X 1 or L = ∅

Therefore, coefficient of self induction of a coil is numerically equal to the amount of magnetic flux linked with the coil when unit current flows through the coil.

Now, the induced in the coil is given by

E = d ∅ / dt = – d / dt (LI)

E = – L  dI / dt

If dI / dt = 1, then e = – L X1

Or L = – e 

Hence coefficient of self induction of a coil is equal to the induced in the coil when rate of change of current through the coil is unity.

Inductance is a scalar quantity.

The SI unit of L is Henry. Self inductance of a coil is said to be one Henry (H) when a current change at the rate of 1 ampere/sex through the coil induced an of 1 volt in the coil.

From (12) L = ∅/I

∴ 1 Henry = 1 Weber/ampere

From (13) L = – e / dI / dt

∴ 1 hennery = 1 volt / 1 ampere/ ec 

Thus 1 Henry = 1 Weber/ ampere = 1 volt – sec / ampere 

Smaller units of L are

1 millinery (1 mh) = 10 -3 Henry,

1 micrometry (1μ H) = 10-6 Henry,

Dimensions of inductance L

From (13) L = e dt / dI

As e = work (W) / charge (q) ∴ L = W / q dt / dI

L = [M L2 T-2 ] [ T] / {AT} {AT} = M L T A-2

A wire cannot act as an inductor as magnetic flux linked with wire of negligible area of cross section is zero. The wire has to be in the form of a coil to serve as an inductor. Further, self induced appears only during the time the current is changing.

What will be induced in a 10 H inductor in which current changes from 10 A to 7 A in 9 X 10 -2 s?

Here e = ? L = 10 H, I = 10 A 

I2 + 7 A, dt = 9 X 10 -2s

As e = – L dI / dt

= – L (I2 – I1) / dt = – 10 (7- 10) / 9 X 10-2

∴ e = 333.3 volt.

A coil of inductance 0.5 H is connected to a 13 V battery. Calculate the rate of growth of current?

Here L = 0.5 He = 18 V, dt/dt =?

From e = L dI / dt (leaving neg. sign)

dI / dt = e / L = 18 / 0.5 = 36 As -1

Related Tags: Electromagnetic Induction, Electric Current

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s