Inductance

Inductance

•Definition and Calculation of Self-Inductance

• To obtain an expression for the Energy Stored by an Inductor

•Definition and Calculation of Mutual-Inductance

Learning Objectives

The phenomenon of self-inductance was discovered by Joseph Henry in

1832 (Princeton University).

Joseph Henry 1797-1878

Let’s start with:

SELF-INDUCTANCE

When current in the circuit changes, the flux changes also, and a self-induced voltage appears in the circuit

Self Inductance

I constant, = 0

L is the self-inductance of the coil.

I increasing or decreasing

, = Vab>0ab

€

∝LdI

dt

(a) Definition used to find LSuppose a current I in a coil of N turns causes a flux B to thread each turn

€

NφB ∝ B∝ IThe self-inductance L is defined by the equation

€

NφB = LI

€

NφB = LI

€

L =NφB

I

(b) Definition that describes the behaviour of an inductor in a circuit

From Faraday’s Law of Induction

€

=−d

dtNΦB

€

=−d

dtNΦB = −

dLI

dt= −L

dI

dt

€

NφB = LI

dt

dIL−=

Two equivalent definitions of L

SI unit for inductanceV s A-1

This is called the henry (H)

dt

dIL−=

If a current changing by 1A/s is to generate 1V, the inductance is 1H.

Calculation of Self-Inductance

€

L =NΦB

I

The Self-Inductance of a Solenoid

n turns per unit length, radius Rand the length of the solenoid is l

Set up a current I, and we have a B field

nIB 0μ=Total number of turns is N=nl

Flux through each turn

€

φB = πR2B

€

NΦB = (nl)πR2B

= (nl)πR2(μ0nI) = πR2μ0n2Il

€

L =NΦB

I= μ0n

2πR2l

€

L =NΦB

I= μ0n

2πR2l

The inductance does not dependon current or voltage, it is a property

of the coil. (length, width, and number of turns per unit length)

Find the self-inductance of a solenoid of length 10 cm, area 5 cm2, and 100 turns.

n = 100/0.1 = 1000 turns/m

€

L = μ0n2πR2l

= 4π ×10-7( ) ×106 × 5 ×10−4

( ) × 0.1

= 6.27 ×10−5 HAt what rate must the current in the solenoid change to induce a voltage of 20 V?

Answer: 3.18 105 A/s dt

dIL−=

The Self-Inductance of a Toroid

ab

a

b

The Self-Inductance of a Toroida

b

Consider an elementary strip of area hdr

r

dr

h

€

B =μ0IN

2πr

( )∫=∫=b

aB hdrBAdB. ∫=

b

a r

drINh

π

μ

20

a

bINhln

20

πμ

=

€

dφB =μ0IN

2πrhdr

The Self-Inductance of a Toroid

⎟⎠

⎞⎜⎝

⎛=

=a

bhN

I

NL B ln

2

20

π

μ

Inductance – like capacitance – depends only on geometric factors

Unit of μ0 is H m-1

μ0 = 4π 10-7 H m-1

From the worked examples it can be seen that:

μ0 = 4π 10-7 wb/Am

€

L =NΦB

I= μ0n

2πR2l

The Energy Stored by an Inductor

I increasing

dt

dIL=

dt

dILIIP ==

The energy dU supplied to the inductor during an infinitesimal time interval dt is:

€

dU = Pdt = LI × dI

a b

(Faraday’s law in disguise)

The Energy Stored by an Inductor

The total energy U supplied while the current increases from zero to a final value I is

22

1

0

LIIdILUI

∫ ==

This energy is stored in the magnetic field

The energy stored in the magnetic field of an inductor is analogous to that in the electric field of a capacitor

22

1LIU =

C

QU

2

2

1=

Example: the energy stored in a solenoid

2

2

1LIU =

Energy per unit volume (magnetic energy density)

2202 2

1In

lR

UuB μ

π==

nIB 0μ=0

222

0 μμ B

In =

22202

1lIRn πμ=

MAGNETIC ENERGY DENSITY IN A VACUUM

0

2

2μB

uB =

The equation is true for all magnetic field configurations

Compare with the energy density in an electric field

202

1EuE =

The self-inductance L is defined by the equations

€

NφB = LI

dt

dIL−=

Review and Summary

Review and Summary• An inductor with inductance L carrying

current I has potential energy

22

1LIU =

•This potential energy is associated with the magnetic field of the inductor. In a vacuum, the magnetic energy per unit volume is

0

2

2μB

uB =

How would the self-inductance of a solenoid be changed if

(a)the same length of wire were wound onto a cylinder of the same diameter but twice the length?

(b)twice as much wire were wound onto the same cylinder?

(c)the same length of wire were wound onto a cylinder of the same length but twice the diameter?

(a) Since the diameter does not change, the number of turns and the area A remain constant. However, n2 is diminished by a factor of 4 and l is increased by a factor of 2. Thus L is reduced by a factor of 2.

(b) Using twice as much wire and making no other change, n2 and L are increased by a factor of 4.

(c) With twice the diameter, n2 is reduced by a factor of 4, but A is increased by the same factor; L is unchanged.

Mutual Inductance

A changing current in loop 1 causes a changing flux in loop 2 inducing a voltage

dt

dN B2

22

−=

122 iN B ∝

Mutual Inductance

(1) The mutual inductance M21 is defined by the equation:

12122 iMN B =(2) The mutual inductance M21 may also be

defined by the equation:

dt

diM 1

212 −=

Mutual InductanceIt can be proved that the same value is obtained for M if one considers the flux threading the first loop when a current flows through the second loop

2

11

1

22

i

N

i

NM BB

=

= (mutual inductance)

dt

diM

dt

diM 2

11

2 & −=−=

(mutually induced voltages)

A Metal DetectorSinusoidally varying current

Parallel to the magnetic field of Ct

Review and Summary•If two coils are near each other, a changing current in either coil can induce a voltage in the other. This mutual induction phenomenon is described by

dt

diM

dt

diM 2

11

2 & −=−=

where M (measured in henries) is the mutual inductance for the coil arrangement

Revision – Ampere’s Circuital Law

enclosed0 Ild.B∫ =μ

∫ ldB.

Where• is the line integral round a closed loopand• Ienclosed is the current enclosed by the loop

∫ ldB.

The B Field Due to a Long Straight Wire

( ) IrB 02 μπ =×

r2

IB 0

πμ

=

Magnetic Materials

Next Installment