National Open University of Nigeria: School of Science and Technology
National Open University of Nigeria: School of Science and Technology
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PHY 407                                 SOLID STATE PHYSICS II
    COURSE
     GUIDE
PHY 407
SOLID STATE PHYSICS II
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PHY 407                               MODULE 1
Abuja Office
5 Dar es Salaam Street
Off Aminu Kano Crescent
Wuse II, Abuja
E-mail: centralinfo@nou.edu.ng
URL: www.nou.edu.ng
Published by
National Open University of Nigeria
Printed 2014
ISBN: 978-058-981-4
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PHY 407                            SOLID STATE PHYSICS II
CONTENTS PAGE
Introduction …………………………………………….              iv
The Course ……………………………………………..               iv
Course Aim …………………………………………….                iv
Course Objectives ……………………………………...          v
Working through the Course …………………………..      v
Course Material ………………………………………...           v
Study Units ……………………………………………..              v
Textbooks ………………………………………………                 vi
Assessment ……………………………………………..               vi
Summary ……………………………………………….                  vi
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PHY 407                                                           MODULE 1
INTRODUCTION
What is the importance of solid state physics? Solid state physics applies
everywhere around us. It involves how the macro-scale properties of
solid materials result from their micro-scale properties. Most of the
solids are arranged in such a way that their atoms are in an orderly
repetitive arrays so the solids we see around are the examples of solid
state physics e.g. sugar molecules, salt (Nacl), glass, etc. So, it forms the
theoretical basis of material science. Its applications are in the
technology of transistors and semiconductors. Quantum mechanics,
metallurgy, electromagnetism, crystallography, etc. are all part and
parcel of solid state physics.
THE COURSE
COURSE AIMS
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PHY 407                                              SOLID STATE PHYSICS II
COURSE OBJECTIVES
You will be provided with a course guide and study material containing
the study units.
At the end of the course, you will find a list of recommended textbooks
which are necessary as supplements to the course material. However,
note that it is not compulsory for you to acquire or indeed read all of
them.
STUDY UNITS
Module 1
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PHY 407                                                         MODULE 1
TEXTBOOK
ASSESSMENT
SUMMARY
This course is designed to lay a foundation for you for further studies in
solid state physics. At the end of this course, you should be able to:
We wish you all the best as you work through the course.
       MAIN
      COURSE
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PHY 407                                    SOLID STATE PHYSICS II
CONTENTS PAGE
Module 1 ………………………………………….. 1
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PHY 407                                                          MODULE 1
MODULE 1
CONTENTS
1.0      Introduction
2.0      Objectives
3.0      Main Content
         3.1    Dielectric Properties
         3.2    Local Electric Field
         3.3    Dielectric Constant and Polarisability
         3.4    Dipole Relaxation and Dielectric Losses
4.0      Conclusion
5.0      Summary
6.0      Tutor-Marked Assignment
7.0      Reference/Further Reading
1.0 INTRODUCTION
In the previous course on solid state physics, you learnt the crystal
structure of solids, thermal and elastic properties of the crystal lattice.
You will recall that a solid has been defined as a rigid substance in
which the average positions of the constituent atoms form a lattice
structure. In this course, our discussion is on the properties of solids. We
will distinguish between the properties of different materials. Glass
flows like liquids at high temperatures and copper exhibits
characteristics of a crystal structure.
Solid materials can be divided into two distinct groups on the basis of
their behaviour under the influence of an external electric field.
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PHY 407                                                SOLID STATE PHYSICS II
2.0 OBJECTIVES
C = KCo (1.1)
              H
          K                                                (1.2)
              Ho
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PHY 407                                                            MODULE 1
                  Vo
          V                                               (1.4)
                  K
where Vo is the potential difference when the dielectric is not inserted.
Equation (1.4) states that the potential difference V between the plates
decreases by a factor of K.
Q = KQo (1.5)
iii.   f the electric field between the parallel plate capacitor is Eo and it
       is E after the material with dielectric constant K is inserted into
       the space between the plates, then
                  Eo
          E                                               (1.6)
                  K
from equation (1.6), we see that the electric field within the dielectric is
reduced by a factor equal to the dielectric constant
where Eo is the original field and Eind is the electric field induced
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PHY 407                                                               SOLID STATE PHYSICS II
                           §        1·
                  V ind   V ¨1      ¸                  (1.9)
                           ©        K¹
          and
                            §   1·
                  Qind    Q ¨1  ¸                      (1.10)
                            © K¹
Suppose the charges -q and +q are placed, respectively, at d/2 and –d/2
from the origin, as shown in Figure 1. The magnitude of the potential
due to this system at a point X is given by
                       q §1 1·
          )r              ¨  ¸                                               1.11
                     4SH o ¨© r1 r2 ¸¹
where r1, r2 are the distances of X from + q and – q respectively.
                               r2                                r1
                                                  r
                                              θ
      -q                                                              +q
                                          d
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PHY 407                                                         MODULE 1
If θ is the angle between the axis of the charges and the position vector
of X, we can write
          1 1 d cosT
            | 
          r1 r  2r 2
and
          1 1 d cosT
            | 
          r2 r  2r 2
Substituting in (11)
                       d cosT
          )r 
                    q
                                                                1.12
                  4SH o r 2
The dipole moment, p, is the vector along the axis of the dipole pointing
in the direction – q to + q and having magnitude qd.
In terms of the dipole moment, you can see that Eq. 1.12 becomes
                  p cosT
          )r 
                                     p.r
                                                                1.13
                  4SH o r 2        4SH o r
                                             3
                                  §1·
          )r 
                           p
                               ¨ ¸             p  ) o .   1.14
                       4SH o r 2
                                  ©r¹
where ) o is the potential of a unit charge.
                                    § pr ·
                                  ¨ 3 ¸
                               1
          E   )          
                            4SH o ¨ r ¸
                                    ©     ¹
                1 ª3
                    «
                            
                           p r r  r pº
                                      2
                                        »                       1.15a
              4SH o ¬          r5       ¼
          E
                   
              3 pr r  r2 p
                                                                1.15b
                   r5
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PHY 407                                                SOLID STATE PHYSICS II
          P   ¦q    r
                   n n                                            1.16
Let us imagine that a small sphere is cut out of the specimen around the
reference point; then E1 is field of the polarisation charges on the inside
of the cavity left by the sphere, and E2 is the field of the atoms within
the cavity. The polarisation charges on the outer surface of the specimen
produce the depolarisation field, E3. The depolarisation field is opposite
to P, that is, the field E3 is called depolarisation field, because within the
body it tends to oppose the applied field Eo as in Figure 1.2. The
dielectric acquires a polarisation due to the applied electric field Eo. This
polarisation is the consequence of redistribution of charges inside the
dielectric.
The local electric field Eloc at any atom of the dielectric may be written
as
Eloc = Eo + E1 + E2 + E3 1.17
and in SI unit
          E1  E 2  E3
                            1
                                «
                                    
                                ª3 p  r r  r 2 p º
                                                   »              1.18b
                          4SH o ¬       r5         ¼
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PHY 407                                                          MODULE 1
Lorentz showed that the field E1 due to the polarisation charges on the
surface formed a continuous distribution. The density of the surface
charge is equal to Pcosθ, where θ is the polar angle along the direction
of polarisation in Figure 3.
θ a
Fig. 3: Calculation of E1
The field E2 due to the dipoles within the spherical cavity is the only
term that depends on the crystal structure. It can be shown that for a
reference site with cubic surroundings in a sphere that E2 = 0 if all the
atoms may be replaced by point dipoles parallel to each other.
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PHY 407                                                                   SOLID STATE PHYSICS II
                     NP
          E                                                                         1.20a
               Ho
The constant N is known as the depolarisation factor.
If Px, Py, Pz are the components of the polarisation P referred to the
principal axes of an ellipsoid, then the components of the depolarisation
field are written
                         N x Px             N y Py             N z Pz
          E3 x                   ; E3 y              ; E3 z              ;         1.20b
                          Ho                 Ho                  Ho
SELF-ASSESSMENT EXERCISE
                 D              P
          H              1           1 F                                          1.21a
                 E             HoE
where χ is the electric susceptibility.
                  P
          F                 H 1                                                    1.21b
                 HoE
If the field is not too large, the strength of the induced dipole moment in
an atom i is proportional to the local electric field acting on the
dielectric. That is
         pi D Eloc
         pi D i Eloc
                   i
                                                                  1.22
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PHY 407                                                                                  MODULE 1
P   ¦N
     i
              i   pi       ¦E
                           i
                                     i
                                    loc   N iD i                                        1.23
                   P
                           ¦E         i
                                     loc   N iD i
          F
                   E                       4S
                           Eloc              P
                                            3
Using Eqs. (1.23) and the Lorentz relation, you will have:
P
                  ¦E   i
                            i
                           loc     N iD i              ¦N D
                                                        i
                                                                 i   i
                                                                                 H 1
                                                                                          1.24
                    4S                                  4S                        4S
                           ¦E                                ¦N D
E         i
      E loc                           i
                                      loc   N iD i   1                  i   i
                     3         i                         3   i
You can calculate the dielectric constant of various materials from the
polarisability by use of the Clausius-Mossotti equation. This equation
holds fairly accurately for non-polar substances but it fails completely in
pure polar liquids or solids.
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PHY 407                                             SOLID STATE PHYSICS II
SELF-ASSESSMENT EXERCISE
                Do
          D                                                       1.27
              1  iZ W
              4SK a 3
          W                                                        1.28
                kT
4.0 CONCLUSION
5.0 SUMMARY
                  Eloc = Eo +
                                  
                                3 pr r  r2 p
                                     r5
and in SI unit
                  Eloc = Eo + 1 « 3 p  r r5  r p »
                                 ª                2
                                                     º
                             4SH   o   ¬   r        ¼
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PHY 407                                               SOLID STATE PHYSICS II
      of atom i.
                                 3 § H 1 ·
x     In   CGS,    ¦N D
                    i
                        i   i      ¨       ¸ known
                                4S © H  2 ¹
                                                     as   Clausius-Mossotti
      equation, expresses the relation between the dielectric constant
      and the atomic polarisabilities.
1.    Two parallel plates have equal and opposite charges. When the
      space between the plates is evacuated, the electric field is 1.2 x
      105 V/m, the electric field within the dielectric is 2 x 105 V/m.
      Calculate the induced charge density on the surface of the
      dielectric.
2.    A parallel plate air capacitor is made of 0.2 m square tin plates
      and 1 cm apart. It is connected to a 50 V battery. What is the
      charge on each plate?
3.    The plates of a parallel-plate capacitor are 2 mm apart and 5 m2
      in area. The plates are in vacuum. A potential difference of 2000
      volts is applied across the capacitor. Calculate the magnitude of
      the electric field between the plates.
4.    Find the frequency dependence of the electronic polarisability of
      an electron having the resonance frequency ωo, treating the
      system as a simple harmonic oscillator.
5.    Calculate the individual dipole moment p of a molecule of carbon
      tetrachloride given the following data.
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PHY 407                                                       MODULE 1
UNIT 2 MAGNETISM
CONTENTS
1.0   Introduction
2.0   Objectives
3.0   Main Content
      3.1    Magnetisation
      3.2    Diamagnetism
      3.3    Paramagnetism
      3.4    Ferromagnetism
             3.4.1 Curie-Weiss Law
             3.4.2 Spontaneous Magnetisation
             3.4.3 Domain Theory of Ferromagnetism
             3.4.4 The Bloch Wall
      3.5    Domain Dimensions
      3.6    Antiferromagnetism
      3.7    Ferrites
      3.8    Magnetic Resonance
             3.8.1 Nuclear Magnetic Resonance
             3.8.2 Ferromagnetic Resonance
             3.8.3 Antiferromagnetic Resonance
4.0   Conclusion
5.0   Summary
6.0   Tutor-Marked Assignment
7.0   Reference/Further Reading
1.0 INTRODUCTION
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PHY 407                                             SOLID STATE PHYSICS II
2.0 OBJECTIVES
3.1 Magnetisation
                 ^
          Qm l   en IdS                                     2.1
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PHY 407                                                     MODULE 1
M Fm H 2.2
where χ
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PHY 407                                              SOLID STATE PHYSICS II
                    eH
          ZL                                                 2.4
                    2mc
          I
               ZeeH   2mc
                                                              2.5
                     2S c
          P       § Ze 2     · 2
                ¨¨      2
                             ¸¸ U                                  2.6
          H       © 4mc       ¹
                3 2
          r2      U                                                2.7
                2
for a distribution of charge which on the average is spherically
symmetrical, so that x 2 y 2 z 2 . Then the diamagnetic susceptibility
per unit volume is, if N is the number of atoms per unit volume,
                   Ze 2 N 2
          F              r                                        2.8
                   6m c 2
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PHY 407                                                          MODULE 1
SELF-ASSESSMENT EXERCISE
3.3 Paramagnetism
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PHY 407                                                SOLID STATE PHYSICS II
V P  H 2.9
M NPLa 2.10
          La  ctnh a 
                           1
                           a
M # NP 2 H / 3kT . 2.11
F M H NP 2 / 3kT C T, 2.12
SELF-ASSESSMENT EXERCISE
3.4 Ferromagnetism
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PHY 407                                                          MODULE 1
We now derive the Curie-Weiss law. We discuss the physical origin and
properties of the saturation magnetisation in ferromagnetic. The
properties of ferromagnetic materials of interest in technical applications
are closely related to the domain structure. Iron, nickel and cobalt are
known as transition elements; they and their alloys react very strongly in
a magnetic field. They are ferromagnetic and have very high values of
magnetic susceptibility. In ferromagnetic materials, the mutual coupling
forces between neighboring molecular dipole moments are sufficiently
stronger than the randomising effect of thermal agitation and the dipoles
are all aligned parallel within a small region known as domain.
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PHY 407                                                  SOLID STATE PHYSICS II
BE OM 2.13
              M           C
                                                                    2.14
          B E  OM        T
Note that we have taken the magnetic field as the sum of the applied
field B and the Weiss field BE.
               M        C
          F                                                         2.15
               H      T  CO
This gives a non-zero magnetisation for zero applied field at the Curies
point expressed by
          Tc   CO ,                                                 2.16
Therefore,
                 C
          F           .                                             2.17
               T  Tc
Eq. (2.17) is known as the Curie-Weiss law. This law describes the
observed susceptibility variation in the paramagnetic region above the
Curie point.
Let us now determine the value of the mean field constant λ. An atom
with angular momentum quantum number J has 2J + 1 equally spaced
energy levels in a magnetic field and the magnetisation is given by
M NgJP B BJ x , 2.18
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PHY 407                                                         MODULE 1
                 2J  1     § 2 J  1x · 1      § x ·
          BJ            ctnh¨            ¸   ctnh¨    ¸          2.19
                  2J        © 2J         ¹ 2J     © 2J ¹
                g>J J  1@ 2 .
                             1
          p                                                       2.21
                        NS S  1g 2 P B2 / 3kTc ,
                 C
          O1                                                     2.22
                 Tc
In this section you will see how we can determine the spontaneous024 54] TJETBT1 0 0 1 72.0
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PHY 407                                               SOLID STATE PHYSICS II
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PHY 407                                                         MODULE 1
Alloys used for the strongest permanent magnets are "hard" alloys made
with many defects in their crystal structure where the domain walls
"catch" and stabilise. The net magnetisation can be destroyed by heating
and then cooling (annealing) the material without an external field,
however. The thermal motion allows the domain boundaries to move,
releasing them from any defects, to return to their low-energy unaligned
state.
The term Bloch wall denotes the transition layer which separates
adjacent domains magnetised in different directions. The essential idea
of the Bloch wall is that the entire change in spin direction between
domains magnetised in different directions occurs in a gradual way over
many atomic planes as shown in Figure 2.2.
This gradual change in direction is due to the fact that for a given total
change of spin direction the exchange energy is lower when the change
is distributed over many spins than when the change occurs suddenly.
Let us attempt to explain this behaviour. The exchange energy between
two spins making a small angle φ with each other is given as:
wex JS 2I 2 2.25
                JS 2 Io N  .
                            2
          wex                                                  2.26
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PHY 407                                               SOLID STATE PHYSICS II
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PHY 407                                                             MODULE 1
          Vw       S      2
                                            
                               J S 2 Na 2  KNa                     2.31
          N       S   2
                           J S 2 Ka 3   
                                        12
                                                .                   2.32
                                                    D
                                        Fig.2.3: Domain Width
The wall energy per unit area of the crystal surface is approximately
wwall Vw L D 2.34
wanis KD 2 . 2.35
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PHY 407                                             SOLID STATE PHYSICS II
w V w L D KD 2 2.36
The condition for the minimum of w with respect to the domain width D
is
D 2V w L K 1 2 , 2.37
w     2V w LK 1 2 .                                          2.38
The energy per unit volume is
f domain 2V w K L1 2 2.39
3.6 Antiferromagnetism
Eex 2 J S i  S j 2.40
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PHY 407                                                    MODULE 1
SELF-ASSESSMENT EXERCISE
3.7 Ferrites
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PHY 407                                            SOLID STATE PHYSICS II
although nuclei from isotopes of many other elements (e.g. 2H, 6Li, 10B,
11
   B
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PHY 407                                                          MODULE 1
          U       P  B a                                      2.42
              ^
if B a    Bo z, then
          U P z Bo          JBo I z                          2.43
ωo = γBo 2.44
The Lorentz field and the exchange field do not contribute to the torque.
The components of the spin equation of motion M = γ(M x Bi) for an
                               ^
applied static field Bo z,
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PHY 407                                                                 SOLID STATE PHYSICS II
        J M y Bzi  M z B yi  J >Bo  N y  N z M @M y
dM x
 dt
dM y
        J >M  N x M x   M x Bo  N z M @ J >Bo  N x  N z M @M x .         2.45
 dt
          iZ                                      J >Bo  N y  N z M @
                                                                               0,
           J >Bo  N x  N z M @ iZ
Zo J Bo P o M 2.47
       J >Bo Bo  P o M @
                              1
Zo                                2                                                 2.48
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PHY 407                                                          SOLID STATE PHYSICS II
We have set M 1z M ; M 2z M .
If we can define M 1 M 1x  iM 1y ; M 2        M 2x  iM 2y , then (2.50) and (2.51)
become
           iZM 1          >
                      iJ M 1 B A  OM   M 2 OM  ;   @
           iZM 2      >
                      iJ M 2 B A  OM   M 1 OM  .@
These equations have a solution if, with exchange field BE ≡ λM,
          J B A  B E   Z       JBE
                                                            0.
          JBE                      J B A  B E   Z
SELF-ASSESSMENT EXERCISE
4.0 CONCLUSION
You have seen the basic idea of magnetisation of materials in this unit.
The differences between diamagnetism and paramagnetism have been
highlighted to you and you should be able to differentiate between
ferromagnetism and antiferromagnetism. We concluded this unit with
explanation on magnetic resonance.
5.0 SUMMARY
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PHY 407                                                                MODULE 1
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PHY 407                                       SOLID STATE PHYSICS II
42
PHY 407                                                         MODULE 1
CONTENTS
1.0    Introduction
2.0    Objective
3.0    Main Content
       3.1    Crystalline Defects
              3.1.1 Point Defects
              3.1.2 Linear Defects (Dislocations)
              3.1.3 Planar (Interfacial) Defects
              3.1.4 Volume (Bulk) Defects
4.0    Conclusion
5.0    Summary
6.0    Tutor-Marked Assignment
7.0    References/Further Reading
1.0 INTRODUCTION
2.0 OBJECTIVE
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PHY 407                                               SOLID STATE PHYSICS II
SELF-ASSESSMENT EXERCISE
There are two main types of geometrical defects in a crystal. There are
those which are very localised and are of atomic dimensions. These are
called point defects. An example of this is an impurity atom which can
be either a substitutional or an interstitial impurity.
(a)    When some atoms are not exactly in their right place, the lattice
       is said to contain imperfections or defects.
(b)    Many properties of solids e.g. the electrical resistance and
       mechanical strength are governed by the presence of certain types
       of defects in the lattice.
x Vacancy
x Self-interstitial or interstitialcy
x Impurities
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PHY 407                                                           MODULE 1
Unlike point defects, these are types of disorder which extend beyond
the volume of one or two atoms. It is a line defect which can extend
right through a crystal or it can form closed loops.
x Edge dislocation
x Screw dislocation
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PHY 407                                               SOLID STATE PHYSICS II
x External surfaces
x Grain boundaries
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PHY 407                                                        MODULE 1
x     turbine blades
x     silicon wafer chips.
x     Twin boundaries – a special type of grain boundary across
      which there exists a mirror image of the crystal lattice. It is
      produced by mechanical shear stresses and/or annealing some
      materials.
x     Stacking faults – the interruption of the stacking sequence of
      close packed planes.
x     Phase boundaries - the surface area between the grains in
      multiphase materials.
x     Ferromagnetic domain walls – the boundary between regions
      that have a different orientation of magnetic dipoles.
x     Pores
x     Cracks
x     Foreign inclusions.
4.0 CONCLUSION
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PHY 407                                              SOLID STATE PHYSICS II
5.0 SUMMARY
http://fog.ccsf.cc.ca.us
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