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Chapter 32 Inductance and Magnetic Materials

The appearance of an induced emf in a circuit associated with changes in its own magnet field is called self-induction. The corresponding property is called self-inductance.

A circuit element, such as a coil, that is designed specificallyto have self-inductance is called an inductor.

The appearance of an induced emf in one circuit due to changes in the magnetic field produced by a nearby circuit is called mutual induction. The response of the circuit is characterized by their mutual inductance.

Can we find an induced emf due to its own magnetic field changes? Yes!

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32.1 Inductance

Open: When the switch is opened, the flux rapidly decreases. This time the self-induced emf tries to maintain the flux.

Close: As the flux through the coil changes, there is an induced emf that opposites this change. The self induced emf try to prevent the rise in the current. As a result, the current does not reach its final value instantly, but instead rises gradually as in right figure.

When the current in the windings of an electromagnet is shut off, the self-induced emf can be large enough to produce a spark across the switch contacts.

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32.1 Inductance (II)

)( 1211111 Φ+Φ=Φ NNThe net emf induced in coil 1 due to changes in I1 and I2 is

Since the self-induction and mutual induction occur simultaneously, both contribute to the flux and to the induced emf in each coil.

)( 12111emf Φ+Φ−=dtdNV

The flux through coil 1 is the sum of two terms:

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Self-Inductance

dtdILV 1

111emf_ −=

It is convenient to express the induced emf in terms of a current rather than the magnetic flux through it.

The magnetic flux is directly proportional to the current flowing through it.

where L1 is a constant of proportionality called the self-inductance of coil 1. The SI unit of self-inductance is the henry (H). The self-inductance of a circuit depends on its size and its shape.

The self-induced emf in coil 1 due to changes in I1 takes the form

11111 ILN =Φ

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Mutual Inductance

dtdIMV 2

12emf_ −=

The magnetic flux produced by coil 2 is proportional to I2. Thus the flux contributed from coil 2 may be written as

where the constant of proportionality is M is called the mutual inductance of the two coils. The SI unit of mutual inductance is the henry (H). The mutual inductance of the two coils depends on its separation and orientation.

The emf induced in coil 1 due to changes in I2 takes the form

2121 MIN =Φ

Are the mutual inductances M12 and M21 the same? Why?

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Example 32.1

ALnLLIN

IALNBA

20

0

µ

µ

=

=Φ

==Φ

A long solenoid of length L and cross-sectional area A has Nturns. Find its self-inductance. Assume that the field is uniform throughout the solenoid.

Solution:

Notice that the self-inductance is proportional to the square of the number density.

This result may be compared to the capacitance of two parallel plates (C=ε0A/d), which depends on the geometry of the capacitor.

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abL

LIabIdx

xI

dxxIBdAd

xIB

b

a

ln2

ln22

2 ,

2

0

00

00

πµ

πµ

πµ

πµ

πµ

l

ll

l

=

===Φ

==Φ=

∫

Example 32.2A coaxial cable consists of an inner wire of radius a that carries a current I upward, and an outer cylindrical conductor of radius b that carries the same current downward. Find the self-inductance of a coaxial cable of length L. Ignore the magnetic flux within the inner wire.

Solution:

Hint1: The direction of the magnetic field.

Hint2: What happens when considers the inner flux?

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Example 32.3

F 54.7

)0004.0)(10)(1500)(104( 7

11202

121

12201212

u

ANnI

NM

AInAB

=×=

=Φ

=

==Φ

−π

µ

µ

A circular coil with a cross-sectional area of 4 cm2 has 10 turns. It is placed at the center of a long solenoid that has 15turns/cm and a cross-sectional area of 10 cm2, as shown below. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?Solution:

Notice that although M12=M21, it would have been much difficult to find Φ21 because the field due to the coil is quite nonuniform.

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32.2 LR Circuits

)1(

:0

0 : 0

:

Let

0

emf

emf0

emfemf

00

emf

tLR

t

tt

eRVI

RVIt

RVRV

LRe

eIdtdIeII

dtdILIRV

−

−

−−

−=∴

−=−==

=⇒=−

=

−=⇒+=

=−−

β

ββ

α

αβ

α

αα

How does the current rise and fall as a function of time in a circuit containing an inductor and a resistor in series?

Rise

The quantity τ=L/R is called the time constant.

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32.2 LR Circuits

τ

α

αα

α

α

ttLR

t

tt

eRVe

RVI

RVIt

LRe

eIdtdIeII

dtdILIR

−−

−

−−

==∴

==

=

−=⇒=

=−−

emfemf

emf0

00

:0

:

Let

0

Decay

The quantity τ=L/R is called the time constant.

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31.3 Energy Stored in an InductorThe battery that establishes the current in an inductor has to do work against the opposing induced emf. The energy supplied by the battery is stored in the inductor.

In Kirchhoff’s loop rule, we obtain

22emf

2emf

emf

2

1 where, LIU

dtdURiiV

dtdiLiRiiV

dtdiLiRV

LL =+=

+=

+=

power supplied by the battery

power dissipatedin the resistor

energy change ratein the inductor

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Energy Density of the Magnetic FieldWe have expressed the total energy stored in the inductor in terms of the current and we know the magnetic field is proportional to the current. Can we express the total magnetic energy in terms of the B-field? Yes.

Let’s consider the case of solenoid.

space) freein field magnetic a ofdensity energy (The2

2)(

2

1

2

1

0

2

0

22

00

2

20

µ

µµ

µ

µ

Bu

ABAnILIU

AnL

B

L

=

===

=

ll

l

Although this relation has been obtained from a special case, the expression is valid for any magnetic field.

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Example 31.4

)21( d)(

)(

)1( , c)(

ms 5.111.0ln)1(90 (b)

ms 5 isconstant timeThe (a)

/2/2

emf2

/2/2

emf

/emf

/00

ττ

ττ

τ

τ τ

ttR

ttL

tL

t

eeRVRRIP

eeRVR

dtdU

eRVI

dtdILI

dtdU

teII.L/Rτ

−−

−−

−

−

+−

==

−

=

−==

=−=⇒−=

==

A 50-mH inductor is in series with a 10-Ω resistor and a battery with an emf of 25 V. At t=0 the switch is closed. Find: (a) the time constant of the circuit; (b) how long it takes the current to rise to 90% of its final value; (c) the rate at whichenergy is stored in the inductor; (d) the power dissipated in the resistor.Solution:

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Example 31.5

38

7-

22

0

3

2612-20

J/m 102.3

1042

20

2

1

J/m 40

)10)(3100.5)(8.85(2

1

×=

××==

=

××==

πµ

ε

BU

EU

B

E

The breakdown electric field strength of air is 3x106 V/m. A very large magnetic field strength is 20 T. compare the energy densities of the field.

Solution:

Magnetic fields are an effective means of storing energy without breakdown of the air. However, it is difficult to produce such large fields over large regions.

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Example 31.6

)ln(2

2

1)ln(

44

)(

4

)()2(

22

2

20

222

02

0

20

0

2

0

2

0

abhNL

LIabIhNdr

rNIhU

drrNIhrdrh

µBdV

µBdU

rNIB

b

aB

B

πµ

πµ

πµ

πµπ

πµ

=

===

===

=

∫

Use the expression for the energy density of the magnetic field to calculate the self-inductance of a toroid with a rectangular cross section.Solution:

Can we use the concept of magnetic flux to derive the self-inductance?

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31.4 LC OscillationsThe ability of an inductor and a capacitor to store energy leads to the important phenomena of electrical oscillations.

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31.4 LC Oscillations (II)Consider the situation shown in Fig. 32.11. A circuit consists of a inductor and a capacitor in series. According to Kirchhoff’sloop rule,

condition. initial from detremined becan and

,frequency)angular (natural 1

where

)sin(

0

sign) minus a is there(note

0

0

0

00

2

2

φ

ω

φω

QLC

tQQdtQdL

CQ

dtdQI

dtdIL

CQ

=

+=

=+

−=

=−

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31.4 LC Oscillations (III)

)cos(

)sin(

000

00

φωωφω+−=

+=tII

tQQThe current and charge relation

The energy stored in the system is

(constant) 2

1

)(cos2

1

)(cos2

1

2

1

)(sin2

1

2

1

20

022

0

022

020

2

022

02

QC

UUU

tQC

tLQLIU

tQC

QC

U

BE

B

E

=+=

+=

+==

+==

φω

φωω

φω

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The Analogies Between LC Oscillations and the Mechanical Oscillations

0001

2

2

2

2

2

2

=+=+=+ θθl

gdtdx

mk

dtxdQ

LCdtQd

LC Osc. Spring Pendulum.

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Example 32.7In an LC circuit, as in Fig. 32.10, L=40 mH, C=20 uF, and the maximum potential difference across the capacitor is 80 V. Find: (a) the maximum charge on C; (b) the angular frequency of the oscillation; (c) the maximum current; (d) the total energy.

Solution:

J 104.62

1 (d)

A 79.1 c)(

rad/s 112010201040

11 (b)

mC 4.2)80)(1020( (a)

220

000

630

60

−

−−

−

×==

==

=×××

==

=×==

QC

U

QILC

CVQ

ω

ω

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31.5 Damped LC Oscillations

.detremined be toparameters are and

,frequency)angular (damped )2

(1

where

)cos(

01

0

since 0

0

2

/0

2

2

φ

ω

φω

QLR

LC

teQQ

QLC

QLRQ

LCQ

dtdQ

LR

dtQd

dtdQI

dtdILRI

CQ

LRt

−=′

+′=⇒

=+′+′′∴=++

−==−−

−

Consider the situation shown in Fig. 32.14. A circuit consists of a inductor, a capacitor, and a resistor in series. According to Kirchhoff’s loop rule,

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31.5 Damped LC Oscillations (II)The behavior of the system depends on the relative value of ω0 and R/2L.

under damping critical and over damping

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32.6 Magnetic Properties of Matter

lity)susceptibi magnetic:( m0mM χχ BB =

When a material is placed in an external magnetic field B0, the resultant field within the material is different from B0. The field due to the material itself, BM, is directly proportional to B0:

The total field within the material is

ty)permeabili relative:(

)1(

m0m

0mM0

κκχ

BBBBB

=+=+=

The magnetic properties of matter can be classified according to their magnetic susceptibility.

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32.6 Magnetic Properties of Matter (II)

Ferromagnetic (χm=103~105): Fe, Ni, Co, Gd, and Dy.

Depends on temperature

Paramagnetic (χm=10-5): Al, Cr, K, Mg, Mn, and Na.

Depends on temperature

Diamagnetic (χm=-10-5): Cu, Bi, C, Ag, Au, Pb, and Zn.

Not depends on temperature

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32.6 Magnetic Properties of Matter (III)

When the surrounding temperature is greater than the Curie temperature, the ferromagnetic material becomesparamagnetic.

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32.6 Magnetic Properties of Matter (IV)

When the induced magnetic field “lags” behind the change in applied field B0, this lagging effect is called magnetic hysteresis.

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Exercises and Problems

Ch.32:

Ex. 13,16, 24, 29

Prob. 3, 4, 5, 9, 10