Inductance

Inductance of a circuit, mutual inductance, and the energy stored in terms of inductance

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Contents

Overview
Calculation Of Self Inductance
Mutual Inductance
The Reciprocal Property Of Inductance
The Induced Voltage Of Two Coils In Series Using Inductance
Energy Stored In Terms Of Inductance
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

$\inline I$ = Moment of Inertia.
$\inline \omega$ = Angular velocity
$\inline \Omega$ = Angular velocity of precession.

Overview

Key facts

The inductance of a circuit characterized by the magnetic flux $\inline \Phi$ , and through which passes a current of $\inline i$ amperes through a coil of $\inline N$ turns, is:

$L = N \frac{d\Phi}{di}$

For a circuit passed by a coil of N turns, and characterized by the length $\inline l$ , cross-sectional area $\inline A$ , and relative magnetic permeability $\inline \mu_r$ , the inductance can also be calculated as:

$L = A \frac{\mu_0 \mu_r N^2}{l}$

where $\inline \mu_0$ is the magnetic permeability of free space.

For a coupled coil circuit, the emf induced in coil 2 due to changes in coil 1 can be expressed as:

$emf_2 = M \frac{di_1}{dt}$

where $\inline M$ , the mutual inductance, is given by:

$M = N_2 \frac{d\Phi_{1.2}}{di_1}$

The induced voltage of two coils in series with additive turns is given by:

$e = (L_1 + L_2 + 2M) \frac{di}{dt}$

or, if the turns would have been opposite:

$e = (L_1 + L_2 - 2M) \frac{di}{dt}$

The energy stored in a magnetic field can be expressed in terms of inductance as:

$E_{stored} = \frac{1}{2} L I^2$

where $\inline I$ is the current through the inductor

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Constants

$\mu_0 = 4\pi \cdot 10^{-7} \; \frac{N}{A^2}$

In order to define the inductance, consider a coil of wire as the one diagramed in Figure 1.

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The induced voltage $\inline e$ at any instant is:

$e = N \frac{d\Phi}{dt}$

(2)

where $\inline N$ is the number of wire turns, and $\inline \Phi$ the magnetic flux. As we can also write that:

$\frac{d\Phi}{dt} = \frac{d\Phi}{di} \frac{di}{dt}$

(3)

equation (2) becomes:

$e = N \frac{d\Phi}{di} \frac{di}{dt}$

(4)

The term $\inline N \frac{d\Phi}{di}$ is denoted by $\inline L$ and is called (self-) inductance. Thus, the inductance is:

$L = N \frac{d\Phi}{di}$

(5)

and equation (4) becomes:

$e = L \frac{di}{dt}$

(6)

Inductance can be illustrated by the behavior of a coil of wire which resists any change of electric current that passes through it. The unit of inductance is the Henry ( $\inline H$ ). Thus, a coil has an inductance of $\inline 1\;H$ if an induced voltage of $\inline 1\;V$ flows through it with a rate of change of current of $\inline 1\; A/s$ .

Calculation Of Self Inductance

In order to calculate the self inductance, consider a circuit of length $\inline l$ and cross-sectional area $\inline A$ , which is passed by a coil of $\inline N$ turns (see Figure 2).

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Also, assume that the area of the core is small, and that the magnetic field strength $\inline H$ is constant across the area. We can then write that:

$H l = N I$

(7)

or, furthermore, that:

$H = \frac{N I}{l}$

(8)

where $\inline I$ is the current which passes through the coil (for a more detailed discussion on the magnetic field strength see Field Strength ). We also know that the magnetic field strength $\inline H$ is related to the magnetic flux density $\inline B$ by the equation:

$B = \mu_0 \mu_r H$

(9)

where $\inline \mu_0$ is the magnetic permeability of free space, and $\inline \mu_r$ the relative magnetic permeability. Taking into account (8), equation (9) becomes:

$B = \frac{\mu_0 \mu_r N I}{l}$

(10)

As the magnetic flux density $\inline B$ is defined as:

$B = \frac{\Phi}{A}$

(11)

where $\inline \Phi$ is the magnetic flux, equation (10) becomes:

$\frac{\Phi}{A} = \frac{\mu_0 \mu_r N I}{l}$

(12)

from which we obtain the magnetic flux as:

$\Phi = A \frac{\mu_0 \mu_r N I}{l}$

(13)

From the definition of the inductance $\inline L$ (see equation 5), we can also write that:

$L = \frac{N \Phi}{I}$

(14)

Using (13) in (14) gives the following expression of the self inductance:

$L = A \frac{\mu_0 \mu_r N^2}{l}$

(15)

Mutual Inductance

Mutual inductance represents the generation of an electromotive force ( $\inline emf$ ) in a coil as a result of a change in current in a coupled coil as the one diagramed in Figure 3.

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The $\inline emf$ induced in coil 2 due to changes in coil 1 can be expressed as:

$emf_2 = N_2 \frac{d\Phi_{1.2}}{dt}$

(16)

or, furthermore, as:

$emf_2 = N_2 \frac{d\Phi_{1.2}}{di_1} \frac{di_1}{dt}$

(17)

We defined the mutual inductance between coil 1 and coil 2, $\inline M$ , as:

$M = N_2 \frac{d\Phi_{1.2}}{di_1}$

(18)

In this definition, $\inline M$ represents the ratio between the $\inline emf$ generated in coil 2 and the change in current in coil 1 responsible for generating this $\inline emf$ .

In the case of no magnetic saturation, the mutual inductance can be written as:

$M = N_2 \frac{\Phi_{1.2}}{I_1}$

(19)

Using (18), equation (17) becomes:

$emf_2 = M \frac{di_1}{dt}$

(20)

The best known application of mutual inductance is the transformer.

The Reciprocal Property Of Inductance

Consider the coupled coil circuit diagramed in Figure 4.

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From our previous discussion on mutual inductance, and also considering that there is no magnetic saturation, we can write that:

$M = N_2 \frac{\Phi_{1.2}}{I_1}$

(21)

and also that:

$M = N_1 \frac{\Phi_{2.1}}{I_2}$

(22)

Taking into account (21) and (22), we can write that:

$\frac{\Phi_{1.2}}{\Phi_{2.1}} = \frac{N_1 I_1}{N_2 I_2}$

(23)

As the two coils are wound on the same core, we have that $\inline \Phi_{1.2} = N_1$ and $\inline \Phi_{2.1}=N_2$ . However, for more complicated cases, such relations are harder to see.

The Induced Voltage Of Two Coils In Series Using Inductance

Consider two coils arranged in series, as diagramed in Figure 5.

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There are four induced electromotive forces due to self and mutual inductance. As such, the induced voltage at any instant can be written as:

$e = L_1 \frac{di}{dt} + L_2 \frac{di}{dt} + M \frac{di}{dt} + M \frac{di}{dt}$

(24)

or, furthermore, as:

$e = (L_1 + L_2 + 2M) \frac{di}{dt}$

(25)

where $\inline L_1$ is the inductance of the first coil, $\inline L_2$ the inductance of the second coil, and $\inline M$ the mutual inductance.

Equation (25) applies only when the turns are additive. If the turns are opposite, equation (25) becomes:

$e = (L_1 + L_2 - 2M) \frac{di}{dt}$

(26)

where the term $\inline (L_1 + L_2 - 2M)$ is called equivalent inductance.

Energy Stored In Terms Of Inductance

The energy stored in a magnetic field can also be expressed using the (self-) inductance (for a more detailed discussion on the energy stored in a magnetic field see Stored Energy ). For example, equation (6) can also be written as:

$e\cdot i = L\cdot i \cdot \frac{di}{dt}$

(27)

or, furthermore, as:

$e\cdot i\cdot dt = L\cdot i\cdot di$

(28)

As the energy released from the coil in a time $\inline dt$ is:

$E = e\cdot i \cdot dt$

(29)

and also taking into account (28), we can write the energy stored in terms of inductance as:

$E_{stored} = \int Lidi = \frac{1}{2} L I^2$

(30)

where $\inline I$ is the current through the inductor.

Reference

http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html