3.magnetic materials 1dkr
-
Upload
devyani-gera -
Category
Documents
-
view
43 -
download
2
Transcript of 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 0 (1.1)
In a magnetic material, above relation is written as
HB (1.2)
Here 0 = absolute permeability of free space,
SD-JIITN-PH611-MAT-SCI-2013
= absolute permeability of the medium and
/ 0 = r = 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 (1.3)
Here, = r – 1 = Magnetic susceptibility (cm-3).
HB
HHHHB rr 0000
HHB r 00)1(
HHB 00
HMB 00 (1.4)
SD-JIITN-PH611-MAT-SCI-2013
Let us consider
)(0 HMB
CLASSIFICATION OF MAGNETIC MATERIALS
Diamagnetic:
(Mn)= 98 cm-3
Ferromagnetic: Magnetic materials with +ve and very large magnetic susceptibility .
SD-JIITN-PH611-MAT-SCI-2013
Examples:
(Hg) = - 3.2 cm-3
(H2O)) = - 0.2X10-8 cm-3)
Paramagnetic: Magnetic materials with +ve and small magnetic susceptibility .
Examples:
Examples:
Materials with –ve magnetic susceptibility .
(Au) = - 3.6 cm-3,
(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:
SD-JIITN-PH611-MAT-SCI-2013
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
2
2rq
1. ORBITAL MOTION
But prL
Therefore, Lm
q 2
(2.1.1)
SD-JIITN-PH611-MAT-SCI-2013
The magnetic moment may be given as
2rq
vmr
2mrmrvL
For an electron orbiting around the nucleus, magnetic moment would be given as
Lm
eL
2
In the equation (2.1.2)m
e
LL
2
orbital gyro-magnetic ratio, .
(2.1.2)
(2.1.3)
241027.92
xm
eB
= Bohr Magneton
SD-JIITN-PH611-MAT-SCI-2013
)1(2
llm
e )1( llB
Where,
Sm
eS
2. ELECTRON SPIN
(2.2.1)
But for reasons that are purely quantum mechanical, the ratio between 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
pn m
e
2
(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?
Jm
eg J
2
SD-JIITN-PH611-MAT-SCI-2013
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 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
LLSSJJg J
DIAMAGNETISM
rmF 2
SD-JIITN-PH611-MAT-SCI-2013
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
Thus when field is switched on, electron revolves with the new frequency, , given by
rmrBeF 2' rmrBerm 22 '
2
2er
m
Be 2
SD-JIITN-PH611-MAT-SCI-2013
For small B,
Thus change in magnetic moment is
m
eB
2
]2
[2re
m
eBer
22
2
m
Bre
4
22
rmrBerm 22 ' Bemm 22 '
m
Be )'( 22
m
Be )')('(
' 2'and
Suppose atomic number be Z, then equation (2.9) may be written as,
Bm
rezi
ii
41
22
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
Bm
rZe
4
22
3
2222
R
zyx
For spherical symmetry,
222 yxr
Therefore,
3
2 22
R
r
Therefore, equation (2.10) may be written as
)3
2(
42
2
Rm
BZe
If there are N atoms per unit volume, the magnetization produced would be
NM 22
6R
m
BNZe 2
20
6R
m
HNZe
Susceptibility, , would be
B
M0 22
0
6R
mB
BNZe
22
0
6R
m
eNZ
Thus, 3
2
4
22 R
m
BZe 2
2
6R
m
BZe
This is the Langevin’s formula for volume susceptibility of diamagnetism of core electrons.
m
RNZe
H
M
6
220
Conclusions:
1. Since Z, bigger atoms would have larger susceptibility.
Example: R = 0.1 nm, N = 5x1028/ m3
3631
29219287
103101.96
)101.0()106.1(105104
cm
SD-JIITN-PH611-MAT-SCI-2013
2. dia depends on internal structure of the atoms which is temperature independent and hence the dia.
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
SD-JIITN-PH611-MAT-SCI-2013
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.
SD-JIITN-PH611-MAT-SCI-2013
CLASSICAL THEORY OF PARAMAGNETISM
In presence of magnetic field, potential energy of magnetic dipole
cos. BBV
Where, is angle between magnetic moment and the field.
0 when (minimum) BV
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 .
Suppose field B is applied along z-axis, then is angle made by dipole with z-axis. The probability of finding the dipole along the direction is
kT
θμB
eef(θ kT
V cos
)
f() 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 z is given as
df
dfz
z)(
)(
Where, integration is carried out over the solid angle, whose element is d. The integration thus takes into account all the possible orientations of the dipoles.
SD-JIITN-PH611-MAT-SCI-2013
Substituting z = cos and d = 2 sin d
0
cos0
cos
sin2
sin2cos
de
de
kT
B
kT
B
z
SD-JIITN-PH611-MAT-SCI-2013
akT
B
Let
0
cos
0
cos
sin
sincos
de
de
a
a
z
0
cos0
cos
sin
sincos
de
de
kT
B
kT
B
Let cos = x, then sin d = - dx and Limits -1 to +1
1
1
1
1
dxe
dxxe
ax
ax
z
aee
eeaa
aa
z
1
aa
1)coth(
Langevin function, L(a)
945
2
453)(
53 aaaaL
L(a) z
aee
eeaa
aa
z
1
)coth(a
][kT
Ba
Variation of L(a) with a.
In most practical situations a<<1, therefore,
3
3
a 2
kT
Bz
3
)(a
aL
The magnetization is given as
3
2
kT
BNNM z
( N = Number of dipoles per unit volume)
SD-JIITN-PH611-MAT-SCI-2013
3
02
kT
HN
3
20
kT
N
H
M
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
Curie constant
SD-JIITN-PH611-MAT-SCI-2013
3
2
kT
BNNM z
3
where,2
0
k
NC
Self study:
1. Volume susceptibility ()
2. Mass susceptibility (m)
3. Molecular susceptibility (M)
SD-JIITN-PH611-MAT-SCI-2013
Reference: Solid State Physics by S. O. Pillai
QUANTUM THEORY OF PARAMAGNETISM
J2m
e-g
J
Where g is the Lande’ splitting factor given as,
)1(2
)1()1()1(1
JJ
LLSSJJg
SD-JIITN-PH611-MAT-SCI-2013
Recall the equation of magnetic moment of an atom, i. e.
L2m
e-
LS
m
e-
S
Consider only spin,
SJ 0L)1(2
)1()1(1
SS
SSSSg 2
S2m
e-2
SJ S
m
e-
S
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 0S
)1(2
)1()1(1
LL
LLLLg 1
L2m
e-1
LJ L
2m
e-
L
)1(2
)1()1()1(1
JJ
LLSSJJg
J2m
e-g
J
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
If dipole is kept in a magnetic field B then potential energy of the dipole would be
BV Jz
Therefore, Boltzmann factor would be,
kT
BgM BJ
ef
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
SD-JIITN-PH611-MAT-SCI-2013
zJ2m
eg
Jz JM
2m
eg 2m
e
BJz BgMV BJ
Represents fraction of dipoles with energy MjgBB.
The magnetic moment of such atoms would be
kT
BgM
BJ
BJ
egM
Therefore, magnetization would be
J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egMNNM
Let, xkT
Bg B
Case 1:
J
J
xM
J
J
xMBJ
J
J
e
egMNM
][ln
J
J
xMB
Jedx
dNgM
SD-JIITN-PH611-MAT-SCI-2013
J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egM
J
J
xM
J
J
xMJ
BJ
J
e
eMNg
Average magnetic moment
1kT
BgM BJ
Since Mj = -J, -(J-1),….,0,….,(J-1), J, therefore,
)].....[ln( )1( JxxJJxB eee
dx
dNgM
)].....1([ln 2JxxJxB eee
dx
dNgM
]
2sinh
2)12(
sinh[ln]
2sinh
)21
sinh([ln
x
xJ
dx
dNg
x
xJ
dx
dNgM BB
Simplifying this equation, we get (consult Solid State Physics by S.O. Pillai, see next two slides),
]2
coth2
1
2
)12(coth
2
12[
xx
JJNgM B
SD-JIITN-PH611-MAT-SCI-2013
][ln
J
J
xMB
Jedx
dNgM
)].....1([ln 2JxxJxB eee
dx
dNgM
x
Jxn
xx
xx
Jxx
e
e
ra
rasumSo
Jvaluesntotaland
ee
eertermfirstawhere
GPSeriesee
1
1
121
,,1
).....1(
)12(
2
2
]
2sinh
)21
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
dNg
ee
ee
dx
dNg
ee
eeee
dx
dNg
ebytermdividingandgMultiplyin
e
eee
dx
dNg
e
ee
dx
dNg
eeedx
dNgM
B
xx
xJxJ
B
xx
xJx
xJx
B
x
x
xJxJx
B
x
JxJx
B
JxxJxB
Let a = xJ, above equation may be written as,
]2
coth2
1
2
)12(coth
2
12[
xx
JJNgM B
]2
coth2
1
2
)12(coth
2
12[
J
a
Ja
J
J
J
JJNgM B
)(aJBNgM JB
Here, BJ(a) = Brillouin function.
J
a
Ja
J
J
J
JJB
2coth
2
1
2
)12(coth
2
12)(
]2
coth2
1
2
)12(coth
2
12[
J
xJ
JxJ
J
J
J
JJNgM B
SD-JIITN-PH611-MAT-SCI-2013
kT
Bgx B
kT
BgJa B
The maximum value of magnetization would be
JNgM Bs
Thus,
]2
coth2
1
2
)12(coth
2
12[
J
a
Ja
J
J
J
JMM s
)(aBM
MJ
s
For J = 1/2
....3
tanh2
a
aaM
M
sFor J =
)(1
coth aLa
aM
M
s
)(aJBNgM JB
)(aBMM Js
SD-JIITN-PH611-MAT-SCI-2013
Case 2:
J
J
kT
BgM
J
J
kT
BgM
BJ
BJ
BJ
e
egMNM
J
J
BJ
J
J
BJBJ
kTBgM
kTBgM
gMNM
)1(
222
1Let kT
BgM BJ
J
JJM and 0 But
Thus above equation becomes,
12
]3
)12)(1([
22
J
JJJ
kT
BgN
M
B)1(
3
22
JJkT
BgNM B
SD-JIITN-PH611-MAT-SCI-2013
J
J
BJ
J
J
BJBJ
kT
BgMkT
BgMgM
NM)1(
)1(
J
J
J
JJ
B
J
J
J
JJ
BJB
MkT
Bg
MkT
BgMg
N
1
222
)3
)12)(1(( 2
JJJM
J
JJ
)1(3
022
JJkT
gN
H
M B
Thus
kT
Np Beff
30
22 2
1
)]1([ JJgpeff
T
C
k
NpC Beff
30
22
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
This is curie law.
SD-JIITN-PH611-MAT-SCI-2013
kT
N J
3
20
T
C
B
Jeffp
)1(222 JJg Bj
3
20
kT
N
H
M
J
a
Ja
J
J
J
JJB
2coth
2
1
2
)12(coth
2
12)(
1:Case kT
BJg B
3//1~)coth( aaa
BJ(J)= [gμB B (J+1)]/(3kBT)
)1(3
22
JJkT
BgNM B
)(aJBNgM JB
]2
coth2
1
2
)12(coth
2
12[
J
a
Ja
J
J
J
JJNgM B
kT
BgJa B
)1(3
022
JJkT
gN
H
M B
Thus
kT
Np Beff
30
22 2
1
)]1([ JJgpeff
T
C
k
NpC Beff
30
22
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
This is curie law.
SD-JIITN-PH611-MAT-SCI-2013
kT
N J
3
20
T
C
B
Jeffp
)1(222 JJg Bj
3
20
kT
N
H
M
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:
SD-JIITN-PH611-MAT-SCI-2013
1. Write electronic configuration. Partially filled sub-shell
1l
1,0,1 lm
4. Accommodate electrons in d sub shell according to Pauli’s exclusion principle.
ml 1 0 -1
ms
Say for 6C
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:
12
12 smS
In the given example: 1L
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
LLSSJJg J
Now calculate peff using 2
1
)]1([ JJgpeff
SD-JIITN-PH611-MAT-SCI-2013
(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 6C, 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
MH i Where is molecular field coefficient.
MHH e But, we know from classical treatment of paramagnetism that
)3
(a
MM s
SD-JIITN-PH611-MAT-SCI-2013
(For a << 1) )3
(kT
Ba
Therefore the net effective field should be
3
)3
( 02
kT
HNaMM e
s
)(
30
2
MHkT
NM
)3
1(3
02
02
kT
NkT
HNM
H
kT
N
kT
NM
3)
31( 0
20
2
H
M
cT
C
where
Paramagnetic curie point
SD-JIITN-PH611-MAT-SCI-2013
)3
1(3
02
02
kTN
kT
N
3
02
k
NC
k
Nc 3
02
and
)3
(3
0
20
2
kN
Tk
N
Curie-Weiss Law. Curie constant
FERROMAGNETIC MATERIALS
SD-JIITN-PH611-MAT-SCI-2013
SD-JIITN-PH611-MAT-SCI-2013
FERROMAGNETIC MATERIALS
MHB 00
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.
~ 106 are possible for ferromagnetic materials.
MB 0
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.
NM s
SD-JIITN-PH611-MAT-SCI-2013
ORIGIN:
Orbital motion also contributes but its contribution is very small.
There is also a corresponding saturation flux density Bs (=0Ms).
Dimension ~ 10-2 cm, No. of atom/ domain 1015 to 1017
)(aBNgJM JB
Where
and kT
HgJa B0
We know that,Bs NgJM )0(
For spontaneous magnetization H = 0,
)()0()( aBMTM JsS
)()0(
)(aB
M
TMJ
s
S
Where,kT
MHgJa B )(0
J
a
Ja
J
J
J
JJB
2coth
2
1
2
)12(coth
2
12)(
WEISS THEORY OF SPONTANEOUS MAGNETIZATION
kT
MHgJ B )(0
kT
MgJ sB0
kT
MgJa sB0
Bs gJ
kTaTM
0
)(
BBs
s
NgJgJ
kTa
M
TM
1
)0(
)(
0
20
22)0(
)(
Bs
s
JNg
kTa
M
TM
)()0(
)(aB
M
TMJ
s
S
Let us consider Brillouin function (BJ) again,
J
a
JJ
aJ
J
JaBJ 2
coth2
1
2
)12(coth
2
12)(
For a<<1
J
a
J
aJJ
aJ
J
aJJ
JaBJ 23
1
2
1
2
1
2
)12(
3
1
2
)12(1
2
12)(
(Since for x<<13
1coth
x
xx Neglecting higher terms.)
J
a
a
J
JJ
aJ
aJ
J
J
JaBJ 6
2
2
1
6
)12(
)12(
2
2
12)(
Ja
aJ
JaJJ
aJJ
J
JaBJ 6
12
2
1
)12(6
])12[(12
2
12)(
2222
aJ
aJ
aJ
aJJJaBJ 2
22
2
222
12
12
12
)414(12)(
aJ
aJ
aJ
aJJJaBJ 2
22
2
222
12
12
12
)414(12)(
aJ
aJJaaaJJaBJ 2
2222222
12
124412)(
aJ
JaaJaBJ 2
222
12
44)(
aJ
JJaaBJ 2
2
12
)1(4)(
J
JaaBJ 3
)1()(
Therefore, from the equation of magnetization
)()0()( aBMTM Jss J
JaMTM ss 3
1)0()(
Thus, slope of the curve for a<<1, J
J
aM
TM
s
s
3
1
)0(
)( (A)
20
22)0(
)(
Bs
s
JNg
kTa
M
TMWe also know that,
20
22)0(
)(
Bs
s
JNg
kT
aM
TMThus,
(B)
Comparing (A) and (B) at T = ,
J
J
JNg
k
B3
12
022
k
JJNg B
3
)1(20
2 (C)
Equation gives relation between curie temperature, and molecular field constant, .
J
J
aM
TM
s
s
3
1
)0(
)( (A)
Now let us consider the case of T>, a<<1, then Magnetization can be written as,
aJ
JJNg
J
JaNgJaBNgJM BBJB 3
)1(
3
)1()(
Since, kT
MHgJa B )(0
Therefore, kT
MHgJ
J
JJNgM B
B
)(
3
)1( 0
)(3
)1(20
2 MHkT
JJNgM B
HkT
JJNg
kT
JJNgM BB 3
)1()
3
)1(1( 2
022
02
)3
)1(1(
3
)1(
20
2
20
2
kT
JJNg
kT
JJNg
H
M
B
B
)3
)1(1(
3
)1(
20
2
20
2
kT
JJNg
kT
JJNg
H
M
B
B
)3
)1((
)1(
20
2
20
2
k
JJNgTk
JJNg
H
M
B
B
)(
T
C (C)
Where, k
JJNgC B )1(2
02
and k
JJNg B 3
)1(20
2
(D)
(E)
Variation of magnetization with temperature below Curie temperature.
WEISS THEORY OF SPONTANEOUS MAGNETIZATION - CLASSICAL
MHm
Where 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,
MHH e
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
MHNM B
B
At sufficiently high temperature,
1)(0
kT
MHB
ThenkT
MH
kT
MH BB )(]
)(tanh[ 00
]tanh[ 0
kT
HNM B
B
Where,
kT
MHN B )(20
kT
MN
kT
HNM BB 2
02
0 kT
HN
kT
NM BB
20
20 )1(
H
M
)(2
0
20
kN
Tk
N
B
B
)(
T
C
Ck
N
k
NC BB
2
02
0 and
Therefore,
])(
tanh[ 0
kT
MHNM B
B
)1(2
0
20
kTN
kT
N
B
B
Now when H = 0, i.e. for spontaneous magnetization
])(
tanh[ 0
kT
MHNM B
B
]tanh[ 0
kT
M
M
M
N
M B
sB
tanhsB M
M
N
M
Where, kT
MB 0
]tanh[ 0
kT
MN B
B
Now let us considerkT
MB 0
kTN
MN
B
B
20It can be written as ,
20 BB N
kT
N
M
T
C
T
M
TM
s
)0(
)(
where,
k
NC B
20
Ck
N B 2
0
tanh also 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
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.
22ZS
KJ ij
This energy must be equal to K as at , ferromagnetic order is destroyed. Thus,
KSZJ ij 22
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 ZJSSSJEj
jiijexch
If there are Z nearest neighbors to a central ith spin, the exchange energy for this spin is
jzizee SSZJU 2B
ize Ng
MSZJ
2
Now if exchange field be BE, the energy of the dipole would be
Ee BU
Thus, B
izeEizB Ng
MSZJBSg
2
EizB BSg
MNg
ZJB
B
eE 22
2
M
Where, 22
2
B
e
Ng
ZJ
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
.2j
jiee SSJU
1
.2j
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 BNg
MS jz
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
02 cos2 SJE eexch
(Where 0 is the angle between two spins.)
Substituting ...2
1cos2
00
...)2
1(22
02
SJE eexch
For small angle , the change in exchange energy when angle between spins change from 0 to is
)0()( exchexchexch EEE
)2(...)2
1(2 22
02 SJSJE eeexch
20
2SJE 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
202
NSJE eexch
)2(...)2
1(2 22
02 SJSJE eeexch
22
022 2...2
22 SJSJSJ eee
Thus the change of angle between two neighbouring spins = N
0
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
202
NSJE eexch
Thus total energy change in N equal steps
NN
SJE eTexch 2
202)(
NSJE eTexch
202)(
Exchange energy per unit area (refer to the figure),
2
202
NaSJE eexch
(Where 0 = )
Anisotropy energy is KNaEanis
Where K = anisotropy constant, a = lattice constant
Thus total wall energy would be
KNaNa
SJE ew 2
202
KaaN
SJdN
dEe
22
202
For minimum E wrt N
KaaN
SJ e 22
202
3
2022
KaSJN e
2
1
3
202 )(
KaSJN e
Thus
KaaN
SJdN
dEe
22
2020
KNaNa
SJE ew 2
202
aKa
SJK
aKa
SJ
SJ e
e
e2
1
3
202
22
1
3
202
202 )(
)(
Thus 2
1
0 )(2a
KJSE e
w
This is equation for Block wall energy.
aKa
SJK
aKa
SJ
SJE e
e
ew2
1
3
202
22
1
3
202
202 )(
)(
a
aK
SKJa
aK
SJSJE e
e
ew
2
1
2
3
2
102
1
2
2
3
2
102
1
202
2
102
12
1
2
10
2
1
2
1
a
SJK
a
SKJE e
ew
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.