3.magnetic materials 1dkr

MAGNETIC MATERIALS
SD-JIITN-PH611-MAT-SCI-2013

FUNDAMENTAL RELATIONS
1. RELATION BETWEEN B, H and M
A magnetic field can be expressed in terms of Magnetic field
intensity (H) and Magnetic flux density. In free space, these
quantities are related as
HB 0µ= (1.1)
In a magnetic material, above relation is written as
HB µ= (1.2)
Here µ0 = absolute permeability of free space,
µ = absolute permeability of the medium and
µ/ µ0 = µr = relative permeability of the magnetic
material.

MAGNETIZATION (M)
Magnetization is defined as magnetic moment per unit
volume and expressed in ampere/ meter. It is proportional to
the applied magnetic field intensity (H).
HM χ= (1.3)
Here, χ = µr – 1 = Magnetic susceptibility (cm-3
).
HB µ=
HHHHB rr 0000 µµµµµµ +−==⇒
HHB r 00)1( µµµ +−=⇒
HHB 00 µχµ +=⇒
HMB 00 µµ +=⇒ (1.4)
Let us consider
)(0 HMB +=⇒ µ

CLASSIFICATION OF MAGNETIC MATERIALS
Diamagnetic:
χ (Mn)= 98 cm-3
Ferromagnetic: Magnetic materials with +ve and very large
magnetic susceptibility χ.
Examples:
χ (Hg) = - 3.2 cm-3
χ (H2O)) = - 0.2X10-8
cm-3
)
Paramagnetic: Magnetic materials with +ve and small
magnetic susceptibility χ.
Examples:
Examples:
Materials with –ve magnetic susceptibility χ.
χ (Au) = - 3.6 cm-3
,
χ (Al) = 2.2X10-5
cm-3
Normally of the order of 105
cm-3
.

2. A MICROSCOPIC LOOK
In an atom, magnetic effect may arise due to:
1. Effective current loop of electrons in atomic orbit (orbital
Motion of electrons);
2. Electron spin;
3. Motion of the nuclei.

MAGNETIC MOMENTS AND ANGULAR MOMENTUM
Consider a charged particle moving in a circular orbit (e.g. an
electron around a nucleus),
IA=µ
2
2
rqω
µ =⇒
1. ORBITAL MOTION
But prL

×=
Therefore, L
m
q 
2
=µ (2.1.1)
The magnetic moment µ may be
given as
2
rq πν×=
vmr

×=
ω2
mrmrvL ==⇒


For an electron orbiting around the nucleus, magnetic
moment would be given as
L
m
e
L

2
−=µ
In the equation (2.1.2)
m
e
L
L
2
=

µ
orbital gyro-magnetic
ratio, γ.
(2.1.2)
(2.1.3)
24
1027.9
2
−
== x
m
e
B

µ = Bohr Magneton
)1(
2
+= ll
m
e )1( += llBµ
Where,

S
m
e
S

−=µ
2. ELECTRON SPIN
(2.2.1)
But for reasons that are purely quantum mechanical, the ratio
between µ to S for electron spin is twice as large as it is for a
orbital motion of the spinning electron:
Electrons also have spin rotation about their own axis. As a
result they have both an angular momentum and magnetic
moment.

3. NUCLEAR MOTION
p
n
m
e
2

=µ
(2.3.1)
Nuclear magnetic moment is expressed in terms of nuclear
magneton
mp is mass of a proton.

What happens in a real atom?
J
m
e
gJ







−=
2
µ
In any atom, there are several electrons and some
combination of spin and orbital rotations builds up the total
magnetic moment.
The direction of the angular momentum is opposite to that of
magnetic moment.
Due to the mixture of the contribution from the orbits and spins
the ratio of µ to angular momentum is neither -e/m nor –e/2m.
Where g is known as Lande’s g-factor. It is given as
)1(2
)1()1()1(
1
+
+−+++
+=
JJ
LLSSJJ
gJ

DIAMAGNETISM
rmF 2
ω=
Diamagnetism is inherent in all substances and arises out of
the effect of a magnetic field on the motion of electrons in an
atom.
Suppose an electron is revolving around the nucleus in atom,
the force, F, between electron and the nucleus is
When this atom is subjected to a magnetic field, B, electron
also experiences an additional force called Lorentz force
rBeFL ω−=
Thus when field is switched on, electron revolves with the new
frequency, ω, given by
rmrBeF 2
'ωω =− rmrBerm 22
'ωωω =−⇒

ωµ ∆−=∆
2
2
er
m
Beω
ωω =×∆ 2
For small B,
Thus change in magnetic moment is
m
eB
2
=∆⇒ ω
]
2
[
2
reω
µ −=
m
eBer
22
2
−=
m
Bre
4
22
−=∆⇒ µ
rmrBerm 22
'ωωω =− Bemm ωωω =−⇒ 22
'
m
Beω
ωω =−⇒ )'( 22
m
Beω
ωωωω =+−⇒ )')('(
ωωω ∆=− ' ωωω 2'≈+and

Suppose atomic number be Z, then equation (2.9) may be
written as,
B
m
re
zi
i
i
4
1
22
∑
=
=
−=∆µ
Where, summation extends over all electrons. Since core
electrons have different radii, therefore
If the orbit lies in x-y plane then,
><+>>=<< 222
yxr
If R represents average radius, then for spherical atom
><+><+>>=<< 2222
zyxR
B
m
rZe
4
22
><
−=∆µ

3
2
222 ><
>=>=<>=<<
R
zyx
For spherical symmetry,
><+>>=<< 222
yxr
Therefore,
3
2 2
2 ><
>=⇒<
R
r
Therefore, equation (2.10) may be written as
)
3
2
(
4
2
2
><−=∆ R
m
BZe
µ

If there are N atoms per unit volume, the magnetization
produced would be
µ∆= NM 2
2
6
R
m
BNZe
−= 2
2
0
6
R
m
HNZeµ
−=
Susceptibility, χ, would be
B
M0µ
χ = 2
2
0
6
R
mB
BNZeµ
−=
2
2
0
6
R
m
eNZµ
χ −=⇒
Thus,
3
2
4
22 R
m
BZe
−=∆⇒ µ
2
2
6
R
m
BZe
−=∆⇒ µ
This is the Langevin’s formula
for volume susceptibility of
diamagnetism of core electrons.

m
RNZe
H
M
6
22
0 ><
−==
µ
χ
Conclusions:
1. Since χ α Z, bigger atoms would have larger susceptibility.
Example: R = 0.1 nm, N = 5x1028
/ m3
36
31
29219287
103
101.96
)101.0()106.1(105104 −−
−
−−−
×−=
××
×××××××
−= cm
π
χ
2. χdia depends on internal structure of the atoms which is
temperature independent and hence the χdia.
3. All electrons contribute to the diamagnetism even s
electrons.
4. All materials have diamagnetism although it may be
masked by other magnetic effects.
Diamagnetic susceptibility

PARAMAGNETISM
Paramagnetism occurs in those substances where the
individual atoms, ions or molecules posses a permanent
magnetic dipole moments.
The permanent magnetic moment results from the following
contributions:
- The spin or intrinsic moments of the electrons.
- The orbital motion of the electrons.
- The spin magnetic moment of the nucleus.

- Free atoms or ions with a partly filled inner shell: Transition
elements, rare earth and actinide elements. Mn2+
, Gd3+
, U4+
etc.
Examples of paramagnetic materials:
- Atoms, and molecules possessing an odd number of
electrons, viz., free Na atoms, gaseous nitric oxide (NO) etc.
- Metals.
- A few compounds with an even number of electrons
including molecular oxygen.

CLASSICAL THEORY OF PARAMAGNETISM
In presence of magnetic field,
potential energy of magnetic
dipole
θµµ cos. BBV −=−=

Where, θ is angle between
magnetic moment and the
field.
0when(minimum) =−= θµBV
It shows that dipoles tend to line up with the field. The effect
of temperature, however, is to randomize the directions of
dipoles. The effect of these two competing processes is that
some magnetization is produced.
B =0, M=0 B ≠0, M≠0
Let us consider a medium containing N magnetic dipoles per
unit volume each with moment µ.

Suppose field B is applied along z-axis, then θ is angle made
by dipole with z-axis. The probability of finding the dipole
along the θ direction is
kT
θμB
eef(θ kT
V cos
) ==
−
f(θ) is the Boltzmann factor which indicates that dipole is more
likely to lie along the field than in any other direction.
The average value of µz is given as
∫
∫
Ω
Ω
=
df
dfz
z
)(
)(
θ
θµ
µ
Where, integration is carried out over the solid angle, whose
element is dΩ. The integration thus takes into account all the
possible orientations of the dipoles.

Substituting µz = µ cosθ and dΩ = 2π sinθ dθ
∫
∫
= π θµ
π θµ
θθπ
θθπθµ
µ
0
cos
0
cos
sin2
sin2cos
de
de
kT
B
kT
B
z
a
kT
B
=
µ
Let
∫
∫
= π
θ
π
θ
θθ
θθθµ
µ
0
cos
0
cos
sin
sincos
de
de
a
a
z
∫
∫
= π θµ
π θµ
θθ
θθθµ
0
cos
0
cos
sin
sincos
de
de
kT
B
kT
B
Let cosθ = x, then sinθ dθ = - dx and Limits -1 to +1
∫
∫
+
−
+
−
= 1
1
1
1
dxe
dxxe
ax
ax
z
µ
µ 





−
−
+
=⇒ −
−
aee
ee
aa
aa
z
1
µµ







−=
a
a
1
)coth(µ
Langevin
function, L(a)
⋅⋅⋅⋅−+−=
945
2
453
)(
53
aaa
aL
L(a)µµ =⇒ z







−
−
+
= −
−
aee
ee
aa
aa
z
1
µµ
)coth(a
][
kT
B
a
µ
=
Variation of L(a) with a.
In most practical situations a<<1,
therefore,
33
a 2
kT
B
z
µ
µµ ==⇒
3
)(
a
aL ≈
The magnetization is given as
3
2
kT
BN
NM z
µ
µ ==
( N = Number of dipoles per unit volume)

3
0
2
kT
HN µµ
=
3
2
0
kT
N
H
M µµ
χ ==⇒
This equation is known as CURIE LAW. The susceptibility is
referred as Langevin paramagnetic susceptibility. Further,
contrary to the diamagnetism, paramagnetic susceptibility is
inversely proportional to T
Above equation is written
in a simplified form as:
T
C
=χ
Curie constant
3
2
kT
BN
NM z
µ
µ ==
3
where,
2
0
k
N
C
µµ
=

Self study:
1. Volume susceptibility (χ)
2. Mass susceptibility (χm)
3. Molecular susceptibility (χM)
Reference: Solid State Physics by S. O. Pillai

QUANTUM THEORY OF PARAMAGNETISM
J
2m
e
-g

=Jµ
Where g is the Lande’ splitting factor given as,
)1(2
)1()1()1(
1
+
+−+++
+=
JJ
LLSSJJ
g
Recall the equation of magnetic moment of an atom, i. e.
L
2m
e
-

=LµS
m
e
-

=Sµ
Consider only spin,
SJ =⇒0=L
)1(2
)1()1(
1
+
+++
+=⇒
SS
SSSS
g 2=
S
2m
e
-2

== SJ µµ S
m
e
-

=⇒ Sµ
SLJ

+=here,

Let N be the number of atoms or ions/ m3
of a paramagnetic
material. The magnetic moment of each atom is given as,
In presence of magnetic field,
according to space quantization.
Consider only orbital motion, LJ =⇒0=S
)1(2
)1()1(
1
+
+−+
+=⇒
LL
LLLL
g 1=
L
2m
e
-1

== LJ µµ L
2m
e
-

=⇒ Lµ
)1(2
)1()1()1(
1
+
+−+++
+=
JJ
LLSSJJ
g
J
2m
e
-g

=Jµ
Jz MJ =
Where MJ = –J, -(J-1),…,0,…(J-1), J i.e. MJ will have (2J+1)
values.

The magnetic moment of an atom along the magnetic field
corresponding to a given value of MJ is thus,
BJJz gM µµ =⇒
If dipole is kept in a magnetic field B then potential energy of the
dipole would be
BV Jz

•−= µ
Therefore, Boltzmann factor would be,
kT
BgM BJ
ef
µ
=
Thus, average magnetic moment of
atoms of the paramagnetic material
would be ∑
∑
+
−
+
−
=
J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egM
µ
µ
µ
µ
zJ
2m
e
g

−=Jzµ JM
2m
e
g−= 2m
e
BJzµ−= BgMV BJ µ−=⇒
Represents fraction of dipoles with
energy MjgµBB.
The magnetic moment of such atoms
would be
kT
BgM
BJ
BJ
egM
µ
µ

Therefore, magnetization would be
∑
∑
+
−
+
−
==⇒ J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egM
NNM µ
µ
µ
µ
Let, x
kT
Bg B
=
µ
Case 1:
∑
∑
+
−
+
−
= J
J
xM
J
J
xM
BJ
J
J
e
egM
NM
µ
][ln ∑
+
−
=⇒
J
J
xM
B
J
e
dx
d
NgM µ
∑
∑
+
−
+
−
=
J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egM
µ
µ
µ
µ
∑
∑
+
−
+
−
= J
J
xM
J
J
xM
J
B
J
J
e
eM
Ngµ
Average magnetic moment
1>>⇒
kT
BgM BJ µ

Since Mj = -J, -(J-1),….,0,….,(J-1), J, therefore,
)].....[ln( )1( JxxJJx
B eee
dx
d
NgM −−
+++= µ
)].....1([ln 2JxxJx
B eee
dx
d
NgM −−
+++=⇒ µ
]
2
sinh
2
)12(
sinh
[ln]
2
sinh
)
2
1
sinh(
[ln
x
x
J
dx
d
Ng
x
xJ
dx
d
NgM BB
+
=
+
= µµ
Simplifying this equation, we get (consult Solid State
Physics by S.O. Pillai, see next two slides),
]
2
coth
2
1
2
)12(
coth
2
12
[
x
x
JJ
NgM B −
++
= µ
][ln ∑
+
−
=
J
J
xM
B
J
e
dx
d
NgM µ

)].....1([ln 2JxxJx
B eee
dx
d
NgM −−
+++=⇒ µ
x
Jxn
x
x
xx
Jxx
e
e
ra
ra
sumSo
Jvaluesntotaland
e
e
ee
rtermfirstawhere
GPSeriesee
−
+−
−
−
−−
−−
−
−
=
−
−
=
+=
====
=+++
1
1
12
1
,,1
).....1(
)12(
2
2

]
2
sinh
)
2
1
sinh(
[ln
}][ln{
}][ln{
log
}]
1
[ln{
)]
1
1
([ln
)].....1([ln
2/2/
)2/1()2/1(
22
22
2
)12(
2
x
xJ
dx
d
Ng
ee
ee
dx
d
Ng
ee
eeee
dx
d
Ng
ebytermdividingandgMultiplyin
e
eee
dx
d
Ng
e
e
e
dx
d
Ng
eee
dx
d
NgM
B
xx
xJxJ
B
xx
x
Jx
x
Jx
B
x
x
xJxJx
B
x
Jx
Jx
B
JxxJx
B
+
=
−
−
=
−
−
=
−
−
=
−
−
=
+++=⇒
−
+−+
−
−
−
−
−−
−
+−
−−
µ
µ
µ
µ
µ
µ

Let a = xJ, above equation may be written as,
]
2
coth
2
1
2
)12(
coth
2
12
[
x
x
JJ
NgM B −
++
= µ
]
2
coth
2
1
2
)12(
coth
2
12
[
J
a
J
a
J
J
J
J
JNgM B −
++
= µ
)(aJBNgM JBµ=⇒
Here, BJ(a) = Brillouin function.
J
a
J
a
J
J
J
J
JB
2
coth
2
1
2
)12(
coth
2
12
)( −
++
=
]
2
coth
2
1
2
)12(
coth
2
12
[
J
xJ
J
xJ
J
J
J
J
JNgM B −
++
=⇒ µ
kT
Bg
x Bµ
=
kT
BgJ
a Bµ
=

The maximum value of
magnetization would be
JNgM Bs µ=
Thus,
]
2
coth
2
1
2
)12(
coth
2
12
[
J
a
J
a
J
J
J
J
MM s −
++
=
)(aB
M
M
J
s
=⇒
For J = 1/2
....
3
tanh
2
+−==
a
aa
M
M
s
For J = ∞
)(
1
coth aL
a
a
M
M
s
=−=
)(aJBNgM JBµ=
)(aBMM Js=⇒

Case 2:
∑
∑
+
−
+
−
= J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egM
NM µ
µ
µ
∑
∑
+
−
+
−
+
+
=⇒ J
J
BJ
J
J
BJ
BJ
kT
BgM
kT
BgM
gM
NM
)1(
222
µ
µ
µ
1Let <<
kT
BgM BJ µ
∑
+
−
=
J
J
JM and0But
Thus above equation becomes,
12
]
3
)12)(1(
[
22
+
++
=
J
JJJ
kT
Bg
N
M
Bµ
)1(
3
22
+=⇒ JJ
kT
Bg
NM Bµ
∑
∑
+
−
+
−
+
+
= J
J
BJ
J
J
BJ
BJ
kT
BgM
kT
BgM
gM
NM
)1(
)1(
µ
µ
µ
∑ ∑
∑ ∑
+
−
+
−
+
−
+
−
+
+
= J
J
J
J
J
B
J
J
J
J
J
B
JB
M
kT
Bg
M
kT
Bg
Mg
N
µ
µ
µ
1
2
22
)
3
)12)(1(
(
2 ++
=∑
+
−
JJJ
M
J
J
J

)1(
3
0
22
+== JJ
kT
g
N
H
M B µµ
χ
Thus
kT
Np Beff
3
0
22
µµ
χ =⇒
2
1
)]1([ += JJgpeff
T
C
=⇒ χ
k
Np
C Beff
3
0
22
µµ
=
Thus Peff is effective number of Bohr Magnetons. C is Curie
Constant. Obtained equation is similar to the relation obtained by
classical treatment.
where,
where,
Further, JBeffp µµ =
This is curie law.
kT
N J
3
2
0µµ
=
T
C
=⇒ χ
B
J
effp
µ
µ
=⇒
)1(
222
+= JJg Bj µµ
3
2
0
kT
N
H
M µµ
χ ==⇒

J
a
J
a
J
J
J
J
JB
2
coth
2
1
2
)12(
coth
2
12
)( −
++
=
1:Case <<
kT
BJg Bµ
3//1~)coth( aaa +
BJ(J)= [gμB B (J+1)]/(3kBT)
)1(
3
22
+=⇒ JJ
kT
Bg
NM Bµ
)(aJBNgM JBµ=
]
2
coth
2
1
2
)12(
coth
2
12
[
J
a
J
a
J
J
J
J
JNgM B −
++
= µ
kT
BgJ
a Bµ
=

Calculation of peff:
2. Find orbital quantum number (l) for partially filled sub-shell.
222
221 pss
3. Obtain magnetic quantum number. In the given example:
1. Write electronic configuration.
Partially filled
sub-shell
1=l
1,0,1 −=lm
4. Accommodate electrons in d sub shell according to
Pauli’s exclusion principle.
ml 1 0 -1
ms
Say for 6
C
In the given case:
In the given case:
For the given case:

5. Apply following three Hund’s rules to obtain ground state:
(i) Choose maximum value of S consistent with Pauli’s
exclusion principle.
(ii) Choose maximum value of L consistent with the Pauli’s
exclusion principle and rule 1.
In the given example:
1
2
1
2 =×==∑ smS
In the given example: 1=L
ml 1 0 -1
ms

5. Obtain J. Since, shell is less than half filled therefore,
011 =−=−= SLJ
6. Obtain g. In the given example:
)1(2
)1()1()1(
1
+
+−+++
+=
JJ
LLSSJJ
gJ
Now calculate peff using 2
1
)]1([ += JJgpeff
(iii) If the shell is less than half full, J = L – S and if it is more
than half full the J = L + S.
For 6
C, peff = 0, hence it does not show paramagnetism.
ml 1 0 -1
ms

WEISS THEORY OF PARAMAGNETISM
Langevin theory failed to explain some complicated
temperature dependence of few compressed and cooled
gases, solid salts, crystals etc. Further it does not throw light
on relationship between para and ferro magnetism.
Weiss introduced concept of internal molecular field in order to
explain observed discrepancies. According to Weiss, internal
molecular field is given as
MHi λ= Where λ is molecular field coefficient.
MHHe λ+=⇒
But, we know from classical treatment of paramagnetism that
)
3
(
a
MM s=
(For a << 1) )
3
(
kT
B
a
µ
=
Therefore the net effective field should be

3
)
3
( 0
2
kT
HNa
MM e
s
µµ
== )(
3
0
2
MH
kT
N
M λ
µµ
+=⇒
)
3
1(3 0
2
0
2
kT
N
kT
HN
M
µµ
λ
µµ
−
=⇒H
kT
N
kT
N
M
3
)
3
1( 0
2
0
2
µµ
λ
µµ
=−⇒
H
M
=⇒ χ
cT
C
θ
χ
−
=⇒ where
Paramagnetic
curie point
)
3
1(3 0
2
0
2
kT
N
kT
N
µµ
λ
µµ
−
=
3
0
2
k
N
C
µµ
=
λ
µµ
θ
k
N
c
3
0
2
=
and
)
3
(3 0
2
0
2
k
N
Tk
N
µµ
λ
µµ
−
=
Curie-Weiss Law.
Curie constant

FERROMAGNETIC
MATERIALS

FERROMAGNETIC MATERIALS
MHB 00 µµ +=
Certain metallic materials posses permanent magnetic
moment in the absence of an external field, and manifest
very large and permanent magnetization which is termed as
spontaneous magnetization.
χ ~ 106
are possible for ferromagnetic materials.
MB 0µ≈⇒
Spontaneous magnetization decreases as the temperature
rises and is stable only below a certain temperature known as
Curie temperature.
Example: Fe, Co, Ni and some rare earth metals such as Gd.
Therefore, H<<M and

Atomic magnetic moments of un-cancelled electron spin.
Coupling interaction causes net spin magnetic moments of
adjacent atoms to align with one another even in absence of
external field. This mutual spin alignment exists over a
relatively larger volume of the crystal called domain.
The maximum possible
magnetization is called saturation
magnetization.
µNMs =
ORIGIN:
Orbital motion also contributes but its contribution is very small.
There is also a corresponding
saturation flux density Bs (=µ0Ms).
Dimension ~ 10-2
cm, No. of atom/ domain 1015
to 1017

)(aBNgJM JBµ=
Where
and kT
HgJ
a Bµµ0
=
We know that, Bs NgJM µ=)0(
For spontaneous
magnetization H = 0,
)()0()( aBMTM JsS =
)(
)0(
)(
aB
M
TM
J
s
S
=⇒
Where,
kT
MHgJ
a B )(0 λµµ +
=
J
a
J
a
J
J
J
J
JB
2
coth
2
1
2
)12(
coth
2
12
)( −
++
=
WEISS THEORY OF SPONTANEOUS
MAGNETIZATION
kT
MHgJ B )(0 λµµ +
=
kT
MgJ sBλµµ0
=

kT
MgJ
a sBλµµ0
=
λµµ B
s
gJ
kTa
TM
0
)( =⇒
BBs
s
NgJgJ
kTa
M
TM
µλµµ
1
)0(
)(
0
=⇒
λµµ
2
0
22
)0(
)(
Bs
s
JNg
kTa
M
TM
=⇒
)(
)0(
)(
aB
M
TM
J
s
S
=

Let us consider Brillouin function (BJ) again,
J
a
JJ
aJ
J
J
aBJ
2
coth
2
1
2
)12(
coth
2
12
)( −
++
=
For a<<1












+−












+
+
+
+
=
J
a
J
aJJ
aJ
J
aJJ
J
aBJ
23
1
2
1
2
1
2
)12(
3
1
2
)12(
1
2
12
)(
(Since for x<<1
3
1
coth
x
x
x += Neglecting higher terms.)




+−




 +
+
+
+
=
J
a
a
J
JJ
aJ
aJ
J
J
J
aBJ
6
2
2
1
6
)12(
)12(
2
2
12
)(





 +
−





+
+++
=⇒
Ja
aJ
JaJJ
aJJ
J
J
aBJ
6
12
2
1
)12(6
])12[(12
2
12
)(
2222





 +
−




 +++
=⇒
aJ
aJ
aJ
aJJJ
aBJ 2
22
2
222
12
12
12
)414(12
)(






 +
−




 +++
=
aJ
aJ
aJ
aJJJ
aBJ 2
22
2
222
12
12
12
)414(12
)(





 −−+++
=⇒
aJ
aJJaaaJJ
aBJ 2
2222222
12
124412
)(





 +
=⇒
aJ
JaaJ
aBJ 2
222
12
44
)(





 +
=⇒
aJ
JJa
aBJ 2
2
12
)1(4
)(





 +
=⇒
J
Ja
aBJ
3
)1(
)(
Therefore, from the equation of magnetization
)()0()( aBMTM Jss =
J
J
aMTM ss
3
1
)0()(
+
=⇒
Thus, slope of the curve for a<<1, J
J
aM
TM
s
s
3
1
)0(
)( +
= (A)

λµµ
2
0
22
)0(
)(
Bs
s
JNg
kTa
M
TM
=We also know that,
λµµ
2
0
22
)0(
)(
Bs
s
JNg
kT
aM
TM
=
Thus,
(B)
Comparing (A) and (B) at T = θ,
J
J
JNg
k
B
3
1
2
0
22
+
=
λµµ
θ
λ
µµ
θ
k
JJNg B
3
)1(
2
0
2
+
=⇒ (C)
Equation gives relation between curie temperature, θ and
molecular field constant, λ.
J
J
aM
TM
s
s
3
1
)0(
)( +
= (A)

Now let us consider the case of T>θ, a<<1, then
Magnetization can be written as,
a
J
JJ
Ng
J
Ja
NgJaBNgJM BBJB
3
)1(
3
)1(
)(
+
=
+
== µµµ
Since,
kT
MHgJ
a B )(0 λµµ +
=
Therefore,
kT
MHgJ
J
JJ
NgM B
B
)(
3
)1( 0 λµµ
µ
++
=
)(
3
)1(2
0
2
MH
kT
JJ
NgM B λµµ +
+
=
H
kT
JJ
Ng
kT
JJ
NgM BB
3
)1(
)
3
)1(
1(
2
0
22
0
2 +
=
+
− µµλµµ
)
3
)1(
1(
3
)1(
2
0
2
2
0
2
λµµ
µµ
χ
kT
JJ
Ng
kT
JJ
Ng
H
M
B
B
+
−
+
==⇒

)
3
)1(
1(
3
)1(
2
0
2
2
0
2
λµµ
µµ
χ
kT
JJ
Ng
kT
JJ
Ng
H
M
B
B
+
−
+
==⇒
)
3
)1(
(
)1(
2
0
2
2
0
2
λµµ
µµ
χ
k
JJ
NgTk
JJNg
H
M
B
B
+
−
+
==⇒
)( θ
χ
−
=⇒
T
C (C)
Where,
k
JJNg
C B )1(
2
0
2
+
=
µµ
and λµµθ
k
JJ
Ng B
3
)1(2
0
2 +
=
(D)
(E)

Variation of magnetization with temperature below Curie
temperature.

WEISS THEORY OF SPONTANEOUS
MAGNETIZATION - CLASSICAL
MHm λ=
Where λ is constant independent of temperature, called
molecular field constant or Weiss constant.
A molecular field tends to produce a parallel alignment of
the atomic dipoles despite effect of thermal energy. This
internal magnetic field is, say, Hm is proportional to the
magnetization M of a domain i.e.
Effective field experienced by each dipole would be then,
MHHe λ+=

Let us consider a ferromagnetic solid containing N number of
atoms/ m3
, then magnetization due to spins (J=1/2) can be
given as
]
)(
tanh[ 0
kT
MH
NM B
B
λµµ
µ
+
=⇒
At sufficiently high temperature,
1
)(0
<<
+
kT
MHB λµµ
Then
kT
MH
kT
MH BB )(
]
)(
tanh[ 00 λµµλµµ +
≈
+
]tanh[ 0
kT
H
NM B
B
µµ
µ=

Where,
kT
MHN B )(
2
0 λµµ +
=
kT
MN
kT
HN
M BB λµµµµ 2
0
2
0
+=⇒
kT
HN
kT
N
M BB
2
0
2
0
)1(
µµλµµ
=−⇒
H
M
=⇒ χ
)(
2
0
2
0
k
N
Tk
N
B
B
λµµ
µµ
−
=
)( θ
χ
−
=⇒
T
C
C
k
N
k
N
C BB
λ
λµµ
θ
µµ
===
2
0
2
0
and
Therefore,
]
)(
tanh[ 0
kT
MH
NM B
B
λµµ
µ
+
=
)1(
2
0
2
0
kT
N
kT
N
B
B
λµµ
µµ
−
=

Now when H = 0, i.e. for spontaneous magnetization
]
)(
tanh[ 0
kT
MH
NM B
B
λµµ
µ
+
=
]tanh[ 0
kT
M
M
M
N
M B
sB
λµµ
µ
==⇒
α
µ
tanh==⇒
sB M
M
N
M
Where,
kT
MBλµµ
α 0
=
]tanh[ 0
kT
M
N B
B
λµµ
µ=

Now let us consider
kT
MBλµµ
α 0
=
kTN
MN
B
B
µ
λµµ
α
2
0
=It can be written as ,
λµµ
α
µ 2
0 BB N
kT
N
M
=⇒
θ
α
λ
α T
C
T
M
TM
s
==⇒
)0(
)(
where,
k
N
C B
2
0µµ
=
C
k
N B
λ
λµµ
θ ==
2
0
αtanhalso =
sM
M

DOMAINS AND HYSTERESIS
What happens when magnetic field is applied to the ferromagnetic
crystal?

According to Becker, there are two independent processes
which take place and lead to magnetization when magnetic
field is applied.
1. Domain growth:
2. Domain rotation:
Volume of domains oriented favourably w. r. t to the field
at the expense of less favourably oriented domains.
Rotation of the directions of magnetization towards the
direction of the field.

ORIGIN OF DOMAINS
According to Neel, origin of domains in the ferromagnetic materials
may be understood in terms of thermodynamic principle that
Total energy:
1. Exchange energy;
2. Magnetic energy;
3. Anisotropy energy and
4. Domain wall or Bloch wall energy.
IN EQUILIBRIUM, THE TOTAL ENERGY OF THE SYSTEM IS MINIMUM.

1. Exchange Energy
This arises because the magnetized specimen has free
poles at the ends and thus produce external field H.
Magnitude of this energy is
∫ dvH 2
8
1
π
Value of this energy is very high and can be reduced if the
volume in which external field exists is reduced and can be
eliminated if free poles at the ends of the specimen are
absent.
It is lowered when spins are parallel. Thus, it favours an
infinitely large domain or a single domain in the specimen.
∑ •−= jiee SSJE

2
2. Magnetic Energy

3. Anisotropy energy
For bcc Fe
[100] easy direction
[110] medium direction
[111] hard direction
The excess energy required to magnetize a specimen in a
particular direction over that required to magnetize in the
easy direction is called crystalline anisotropy energy.
4. Domain wall or Bloch wall energy
Domain wall creation involves energy which is known as domain
wall energy of Bloch energy.
For Ni
[111] easy direction
[110] medium direction
[100] hard direction

EXCHANGE INTERACTION IN MAGENTIC MATERIALS
212 sJsEexch −=
Heisenberg (1928) gave theoretical explanation for large
Weiss field in ferromagnetic materials.
Parallel arrangements of spins in ferromagnetic materials
arises due to exchange interaction in which two neighboring
spins. The exchange energy of such coupling is
Here J = exchange integral. Its value depends upon separation
between atoms as well as overlap of electron charge cloud.
J > 0, favors parallel configuration of spins, while for J < 0,
spins favors anti-parallel.

2
2ZS
K
Jij
θ
≈⇒
This energy must be equal to Kθ as at θ, ferromagnetic
order is destroyed. Thus,
θKSZJij ≈2
2
Thus criteria for
ferromagnetism (due to Slater)
becomes – atoms must have
unbalanced spins and the
exchange integral J must be
positive. Alloys like Mn-As, Cu-
Mn and Mn-Sb show
ferromagnetism.
2
1
2.2 ZJSSSJE
j
jiijexch −≈−= ∑=

If there are Z nearest neighbors to a central ith
spin, the
exchange energy for this spin is

jzizee SSZJU 2−=
B
ize
Ng
M
SZJ
µ
2−=
Now if exchange field be BE, the energy of the dipole
would be
Ee BU µ−=
Thus,
B
izeEizB
Ng
M
SZJBSg
µ
µ 2−=−
EizB BSgµ−=
M
Ng
ZJ
B
B
e
E 22
2
µ
=⇒ Mλ=
Where, 22
2
B
e
Ng
ZJ
µ
λ =
Expression for Exchange field (BE) (Stoner):

Expression for Exchange field (BE) (Stoner):
Assuming exchange integral, Je, to be constant over all
neighboring pairs, Exchange energy is,
∑=
−=
1
.2
j
jiee SSJU

∑=
−=
1
.2
j
jie SSJ

Suppose there are Z nearest neighbors and exchange
energy is contributed by nearest neighbors only,
jiee SSZJU

.2−=
Suppose magnetization is along z-axis,
jzizee SSZJU

.2−=
Also magnetization is given as
jzBSNgM µ=
BNg
M
S jz
µ
=⇒
jzize SSZJ2−=

Based on the definition of these energies a scheme is drawn
below which helps in minimization of energy of the system:
Domain closure
Single domain
Magnetic energy
high
Domain halved
magnetic energy
reduced
Elimination of magnetic
energy by domain closure
Idea of magnetic energy due to domain:

BLOCH WALL
Bloch Wall
The entire change in spin
direction between domains does
not occur in one sudden jump
across a single atomic plane
rather takes place in a gradual
way extending over many atomic
planes.
Because for a given total
change in spin direction, the
exchange energy is lower
when change is distributed
over many spins than when
the change occurs abruptly.
(Due to Bloch)
Bloch Wall HD.mp4

From the Heisenberg model, exchange energy is
jieexch SSJE

.2−=
0
2
cos2 φSJE eexch −=⇒
(Where ϕ0 is the angle
between two spins.)
Substituting ...
2
1cos
2
0
0 +−=
φ
φ
...)
2
1(2
2
02
+−−=
φ
SJE eexch
For small angle ϕ, the change in exchange energy when
angle between spins change from 0 to ϕ is
)0()( exchexchexch EEE −=∆ φ
)2(...)
2
1(2 2
2
02
SJSJE eeexch −−+−−=∆⇒
φ

2
0
2
φSJE eexch =∆⇒
Thus exchange energy increases when two spins are rotated
by an angle from exact parallel arrangement between them.
Now suppose the total change of angle between two domains
occurs in N equal steps.
2
2
02
N
SJE eexch
φ
=∆
)2(...)
2
1(2 2
2
02
SJSJE eeexch −−+−−=∆
φ
2
2
022
2...
2
22 SJSJSJ eee +++−=
φ
Thus the change of angle between
two neighbouring spins = N
0φ

Thus total energy change decreases when N increases.
Q. Why does not the wall becomes infinitely thick.
Ans. Because of increase of the anisotropy energy. Therefore
competing claims between exchange energy and anisotropy
energy leads to an equilibrium thickness.
2
2
02
N
SJE eexch
φ
=∆
Thus total energy change in N equal steps
N
N
SJE eTexch 2
2
02
)(
φ
=∆
N
SJE eTexch
2
02
)(
φ
=∆⇒
Exchange energy per unit area (refer to the
figure),
2
2
02
Na
SJE eexch
φ
=∆ (Where ϕ0 = π)

Anisotropy energy is KNaEanis =
Where K = anisotropy constant, a = lattice constant
Thus total wall energy would be
KNa
Na
SJE ew += 2
2
02 φ
Ka
aN
SJ
dN
dE
e +−=⇒ 22
2
02 φ
For minimum E wrt N
Ka
aN
SJe =⇒ 22
2
02 φ
3
2
022
Ka
SJN e
φ
=⇒ 2
1
3
2
02
)(
Ka
SJN e
φ
=⇒
Thus
Ka
aN
SJ
dN
dE
e +−== 22
2
02
0
φ
KNa
Na
SJE ew += 2
2
02 φ
a
Ka
SJK
a
Ka
SJ
SJ e
e
e
2
1
3
2
02
22
1
3
2
02
2
02
)(
)(
φ
φ
φ
+=

Thus 2
1
0 )(2
a
KJ
SE e
w φ=
This is equation for Block wall energy.
a
Ka
SJK
a
Ka
SJ
SJE e
e
ew
2
1
3
2
02
22
1
3
2
02
2
02
)(
)(
φ
φ
φ
+=
a
aK
SKJ
a
aK
SJ
SJE e
e
ew
2
1
2
3
2
1
0
2
1
2
2
3
2
1
0
2
1
2
02 φ
φ
φ
+=⇒
2
1
0
2
1
2
1
2
1
0
2
1
2
1
a
SJK
a
SKJ
E e
e
w
φφ
+=⇒

Domain growth reversible
boundary displacements.
Domain growth irreversible
boundary displacements.
Magnetization by
domain rotation

Soft and Hard magnetic materials
a. The area within the Hysteresis loop represents magnetic energy
loss per unit volume of the material per magnetization and
demagnetization cycle.
b. Both Ferri- and Ferro-magnetic materials are classified as soft or
hard on the basis of their Hysteresis characteristic.
Examples:
Soft magnetic materials: Commercial Iron
ingot (99.95Fe), Silicon-Iron (97Fe, 3Si), 45
Permalloy (55Fe, 45Ni), Ferroxcube A
(48MnFe2O4, 52ZnFe2O4) etc.
Hard magnetic materials: Tungsten steel
(92.8 Fe, 6 W, 0.5 Cr, 0.7 C), Sintered
Ferrite 3 (BaO-6Fe2O3), Cobalt rare earth 1
(SmCo5) etc.

1. High initial permeability.
2. Low coercivity.
3. Reaches to saturation
magnetization with a relatively low
applied magnetic field.
4. It can be easily magnetized and
demagnetized.
5. Low Hysteresis loss.
6. Applications involve, generators,
motors, dynamos and switching
circuits.
Characteristics of soft magnetic materials:
Important: Saturation magnetization can be altered by altering
composition of the materials. For example substitution of Ni2+
in place of
Fe2+
changes saturation magnetization of ferrous-Ferrite. However,
susceptibility and coercivity which also influence the shape of the
Hysteresis curve are sensitive to the structural variables rather than
composition. Low value of coercivity corresponds to the easy movement of
domain walls as magnetic field changes magnitude and/ or direction.

Characteristics of Hard magnetic materials:
1. Low initial permeability.
2. High coercivity.
3. High remanence.
4. High saturation flux density.
5. Reaches to saturation
magnetization with a high applied
magnetic field.
6. It can not be easily magnetized and
demagnetized.
7. High Hysteresis loss.
8. Used in permanent magnets.
Important: Two important characteristics related to applications of these
materials are (i) Coercivity and (ii) energy product expressed as (BH)max with
units in kJ/m3
. This corresponds to the area of largest B-H rectangle that
can be constructed within the second quadrant of the Hysteresis curve.
Larger the value of energy product harder is the material in terms of its
magnetic characteristics.

3.magnetic materials 1dkr

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Similaire à 3.magnetic materials 1dkr

Similaire à 3.magnetic materials 1dkr (20)

3.magnetic materials 1dkr