Magnetic Materials 1

MAGNETIC MATERIALS

DKR-JIITN-PH611-MAT-SCI-2011

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,


= 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)


Let us consider

)(0 HMB

CLASSIFICATION OF MAGNETIC MATERIALS

Diamagnetic:

(Mn)= 98 cm-3

Ferromagnetic: Magnetic materials with +ve and very large magnetic susceptibility .


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:


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)


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


)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


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


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


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


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


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

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.


Substituting z = cos and d = 2 sin d

0

cos0

cos

sin2

sin2cos

de

de

kT

B

kT

B

z


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)


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


3

2

kT

BNNM z

3

where,2

0

k

NC

Self study:

1. Volume susceptibility ()

2. Mass susceptibility (m)

3. Molecular susceptibility (M)


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


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


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


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


][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


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


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


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.


kT

N J

3

20

T

C

B

Jeffp

)1(222 JJg Bj

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

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


(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


(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


)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



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


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

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

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

Thus exchange energy becomes

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

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.

Domain growth reversible boundary displacements.

Domain growth irreversible boundary displacements.

Magnetization by domain rotation

ORIGIN OF DOMAINSAccording 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.

Q. If Exchange energy, then why not all spins of the piece of material become parallel and form only one single domain

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

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

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)

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

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.

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Magnetic Materials 1