dia para magnetic materials

8/6/2019 Dia Para Magnetic Materials

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MAGNETIC MATERIALS

DKR-JIITN-PH611-MAT-SCI-2010




Review

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 freespace, these quantities are related as

H B 0μ = (1.1)

In a magnetic material, above relation is written as

H B μ = (1.2)

Here μ0 and μ are known as absolute permeability of free space and absolute permeability of the medium

respectively. The ratio μ/ μ0 is known as relative

permeability of the magnetic material and isrepresented as μr .




Magnetization (M)

Magnetization (expressed in ampere/ meter) is defined

as magnetic moment per unit volume. It is proportional

to the applied magnetic field intensity (H).

H M χ = (1.3)

Here χ is known as magnetic susceptibility. Further, χ =

μr – 1. χ is expressed in cm-3 . Let us consider equation

(2), H B μ =

H H H H B r r 0000 μ μ μ μ μ μ +−==⇒

H H B r 00)1( μ μ μ +−=⇒

H H B 00 μ χμ +=⇒

H M B 00 μ μ +=⇒ (1.4)


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Classification of magnetic materials

Diamagnetic: Magnetic susceptibility χ is –ve for diamagnetic materials.

Examples: χ= -3.6 cm-3

for Au, -3.2cm-3

for Hg and -0.2X10-8cm-3 for water)

Paramagnetic: Magnetic susceptibility χ is +ve for

paramagnetic materials but is very small.

Examples: χ = 2.2X10-5 cm-3for Al, 98 cm-3 for Mn

Ferromagnetic: Magnetic susceptibility χ is +ve andvery large for ferromagnetic materials.

Examples: 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 charge particle moving in a circular orbit

(e.g. an electron around a nucleus), magnetic

moment μ may be given as

r v

J

μ

m,q

22

2 r x

q

r xq IA π π

ω

π υ μ ===

2

2r qω μ =⇒

1.Orbital Motion

But ω 2mr mrvv xmr p xr L ====rrrrr

Therefore, Lmq

rr

2=μ (2.1)



For an electron orbiting around the nucleus, magnetic

moment would be given as

)1(22

+=−= Ll m

e L

m

e L

hrrμ

Here negative sign shows that μL and L are in oppositedirection. Here,

In the equation (2.2)m

e

L

L

2=r

rμ

is known is orbital gyro-magnetic ratio (γ).

Thus, m

e

2mometumAngular

momentMagnetic

==γ

(2.2)

(2.3)

(2.4)

24

1027.92

−

== xm

e B

h

μ is Bohr Magneton



Equation (2.2) may also be written as

)1(22

+=−= L Lm

e L

m

e L

hrrμ



Magnetic moments and angular momentum

Electrons also have spin rotation about their own axis.As a result they have both an angular momentum andmagnetic moment.

S m

e rr−=μ

2. Electron spin

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:

(2.5)



3. Nuclear motion

In nuclei there are protons and neutrons which may

move around in some kind of orbit and at the same

time, like an electron, have an intrinsic spin. Again themagnetic moment is parallel to the angular momentum

and we have

pm

e

2

hr=μ (2.6)



What happens in a real atom?

In any atom, several electrons and some combination of spin and orbital rotations builds up the total magneticmoment.

The direction of the angular momentum is opposite tothat of magnetic moment.

Due to the mixture of the contribution from the orbits andspins the ratio of μ to angular momentum is neither -e/m nor –e/2m.

J m

e g J ⎟

⎠ ⎞⎜

⎝ ⎛ −=

2μ (2.6)

Where g is known as Lande’s g-factor. gL = 1, gS =2and 1< gJ <2.



Lande’s g can be obtained from the equation

)1(2

)1()1()1(

1 +

+−+++

+= J J

L LS S J J

g J

Example: If, S = 1, L = 1, J = 0,1, 2

g= 1, 1.5, 1.5

(2.7)



Diamagnetism

Suppose we slowly turn on a magnetic field in the vicinityof atom. As the magnetic field changes an electric field isgenerated by magnetic induction (Faraday’s law, Maxwell’s

3rd equation).• From Faraday’s law, the line integral of E around any

closed path is the rate of change of atomic flux through thepath.

B

F

e

r

Closed Path Γ

dt

dBr E

ds Bdt

d dl E

2

..

−=⇒

−= ∫ ∫

∫ds=π r 2 and ∫dl=2 π r where

(2.8)



B

Fe

r

Closed Path Γ

dt

dBer

dt

dL

2

2

=

This is the extra angular

momentum from the twist given toelectrons as the field is turned on.

B

m

r e L

m

e

42

22

−=Δ−=Δμ

This added angular momentum gives

an extra magnetic moment

Here ∆B = B – 0 = B Ber Ber L22

22

=Δ=Δ

dt

dLeEr Fxr =−==τ

(2.9)



Suppose atomic number be Z, then equation (2.9) may

be written as,

Bm

r e

4

22∑−=Δμ

Where, summation extends over all electrons. Sincecore electrons have different radii, therefore

If the orbit lies in x-y plane then,

><+>>=<<222

y xr

If r 0 represents average radius then

><+><+>>=<<2222

0 z y xr

Bm

r Z e

4

22 ><−=Δμ (2.10)



or,

m

r BZ e

6

2

0

2 ><

−=Δμ

If there are N number of atoms per unit volume

then,

m

r B NZe N M

6

2

0

2 ><−=Δ= μ (2.11)

Equation (2.11) is the Langevin’s formula for volume

susceptibility of diamagnetism of core electrons.

and,

m

r NZe

H

M

6

2

0

2

0 ><−==μ

χ (2.12)



m

r NZe

H

M

6

20

20 ><−==

μ χ

Conclusions

1. Susceptibility of diamagnetic material is proportional tothe atomic number. Thus bigger the atom larger would

be the susceptibility.

2. Diamagnetic susceptibility depends on internal structure

of the atoms which is temperature independent and

hence the diamagnetic susceptibility.

3. All electrons contribute to the diamagnetism even s

electrons.

Example: r 0 = 0.1 nm, N = 5x1028/m3

6

31

29219287

103101.96

)101.0()106.1(105104 −−

−−−

−=−= x x x

x x x x x x xπ χ



Paramagnetism

• B>B0 , χ is positive, μr >1, .• The permanent magnetic moment results from the

following contributions:

1. The spin or intrinsic moments of the electrons.2. The orbital motion of the electrons.

3. The spin magnetic moment of the nucleus.



• Paramagnetism is observed in:

1. Metals

2. Atoms and molecules possessing an odd number of electrons, that is free sodium atoms, gaseous nitric

oxide etc.3. A few compounds having an even number of electrons

(example Oxygen molecule)

4. Free atoms or ions having a partially filled inner shelle.g. rare earth and actinide elements, ions of sometransition elements such as Mn2+

Manganese, platinum, tungsten, some members of rare earth group and ions formed by removing andadding electrons to basic atoms there by creatingunpaired spins.



Paramagnetism has net magnetic moments

No field : M=0 Field is applied, low Temp.

B

Field is applied, and High Temp.

B

T

χ



from http://www.geo.umn.edu



Classical theory of paramagnetism

In presence of magnetic field, potential energy of

magnetic dipole may be given as

θ μ μ cos. B BV −=−= rr

Where, θ is angle between magnetic moment and the

field. From equation it is obvious that

0when(minimum) =−= θ μ 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.

Bμ

θ

θ cos= z



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

θ μ B

ee f( θ kT

V cos

) ==−

f(θ) is the Boltzmann factor which indicates that dipole ismore likely to lie along the field than in any other

direction.The average value of μz is given as

∫ ∫

Ω

Ω

= d f

d f z

z )(

)(

θ

θ μ

μ

Where, integration is carried out over the solid angle,

whose element is dΩ. The integration thus takes intoaccount all the possible orientations of the dipoles.



Substituting μz = μ cosθ and dΩ = 2π sinθ dθ

∫ ∫ =

θ θ π

θ θ π θ μ μ θ μ

θ μ

d e

d e

kT

B

kT

B

z cos

cos

sin2

sin2cos

kT Ba

va z

μ

μ μ μ

=

=−=⇒

where

L(a)1

)coth(

Where L(a) is known as Langevin function, expanded

as

∫

∫ =⇒

θ θ

θ θ θ μ μ

θ μ

θ μ

d e

d e

kT

B

kT

B

z cos

cos

sin

cossin

⋅⋅⋅⋅−+−=945

2

453)(

53aaa

a L

Bμ

θ

θ cos= z



kT

Ba

μ =

)L( a

Variation of L(a) with a.

From the variation of L(a), it is

clear that at low a and henceat low field dipole moment is

proportional to B and at high

a i.e. B, it saturates.

In most practical situations

a<<1, therefore,

3

3

a 2

kT

B z

μ μ μ ==

The magnetization is given as

3

2

kT

B N N M z

μ μ ==

Where N is the number of

dipoles per unit volume



3

0

2

kT

H N M

μ μ =⇒

3

2

0

kT

N

H

M μ μ 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:

3

re,whe2

0

k

N C

T

C μ μ χ ==

Here C is known as curie constant.

χ

T



f



The magnetic moment of an atom corresponding to a given

value of mJ is thus,

B J J g M μ μ =

If dipole is kept in a magnetic field B then potential energy

of the dipole would be B g M B BV B J J J μ μ μ −=−=−=

rr

Therefore, Boltzmann factor would be,

kT

B g M B J

e f

μ

=

Thus, average magnetic moment of atoms of the

paramagnetic material would be

∑

∑+

−

+

−= J

J

kT B g M

J

J

kT

B g M

B J

B J

B J

e

e g M

μ

μ

μ

μ



Therefore, magnetization would be

∑

∑+

−

+

−== J

J

kT

B g M

J

J

kT

B g M

B J

B J

B J

e

e g M

N N M μ

μ

μ

μ

Let,kT

B g x Bμ = Therefore,

∑

∑

∑

∑+

−

+

−+

−

+

− == J

J

xM

J

J

xM

J

B J

J

xM

J

J

xM

B J

J

J

J

J

e

eM

Ng

e

e g M

N M μ

μ

][ln∑+

−

=⇒ J

J

xM

B J e

dx

d Ng M μ



Since M j = -J, -(J-1),….,0,….,(J-1), J, therefore,

)].....[ln( )1( Jx x J Jx

B eeedx

d Ng M −− +++= μ

)].....1([ln 2 Jx x Jx B eeedxd Ng M −− +++=⇒ μ

]

2sinh

)2

1sinh(

[ln

x

x J

dx

d Ng M B

+= μ

Simplifying this equation, we get (consult Solid State

Physics by S.O. Pillai),

]

2

coth

2

1

2

)12(coth

2

12[

x x

J J Ng M B −

++= μ



Let a = xJ, above equation may be written as,

]2

coth2

1

2

)12(coth

2

12[

x x

J J Ng M B −

++= μ

]2

coth2

1

2

)12(coth

2

12[

J

aa

J

J J Ng M B −

++= μ

]2

coth2

1

2

)12(coth

2

12[

J

a

J a

J

J

J

J J Ng M B −

++= μ

)(a JB Ng M J B

μ =⇒

Here, BJ(a) is known as Brillouin function defined as,

J

a

J a J

J

J

J

J B 2coth2

1

2

)12(

coth2

12

)( −

++

=

]2

coth2

1

2

)12(coth

2

12[

J

xJ xJ

J

J J Ng M B −

++=⇒ μ



The maximum value of magnetizationwould be

J Ng M B s =

Thus,

]2

coth2

1

2

)12(coth

2

12[

J

a

J a

J

J

J

J M M s −

++=

)(a BM M

J

s

=

For J = 1/2

....3tanh

2

+−==a

aaM

M

s

For J = ∞

)(1

coth a LaaM

M

s=−=

)(a JB Ng M J Bμ =

)(a BM M J s=⇒



Thus



T

C

kT

N J J

kT

g N

H

M J B ==+==⇒3

)1(3

2

00

22 μ μ μ μ χ

Thus

kT

Np Beff

3

0

22μ μ

χ =⇒2

1

)]1([ += J J g peff

T

C =⇒ χ

k

NpC

Beff

3

0

22μ μ

=

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, B

J eff J Beff p p

μ

μ μ μ =⇒=

This is curie law.

How to obtain p ?



How to obtain peff ?

1. Find orbital quantum number (l) for partially filled shell. For

example, for iron electronic configuration is6262622 3433221 d s p s p s s

Thus the only orbital outside the closed shell are 3d6

. For d orbital, l =2.

2. Corresponding to l value, obtain magnetic quantum number ml (l,

(l-1),…..0,….,-(l-1), -l). For example for l = 2, ml = 2, 1, 0, -1, -2.

3. Apply following three Hund’s rules to obtain ground state:

(i)Choose maximum value of S consistent with Pauli exclusion

principle.

(ii)Choose maximum value of L consistent with the Pauli exclusionprinciple and rule 1.

(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 the taken example



For the taken example,

ml

2 1 0 ‐1 ‐2

ms

2== ∑ l m L 2== ∑ smS

4. Obtain J. Since, shell is more than half filled therefore,

422 =+=+= S L J

Thus g can be obtained by using formula

)1(2

)1()1()1(1+

+−++++= J J

L LS S J J g J

Then peff can be obtained by using formula2

1

)]1([ += J J g peff

Weiss Theory of Paramagnetism



Weiss Theory of Paramagnetism

Langevin theory failed to explain some complicated temperaturedependence 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

M H i λ =

Where λ is molecular field coefficient. Therefore the net effective field

should be

M H H e λ +=

But, we know from classical treatment of paramagnetism that

)3

( aM M s=

2 HN2N μμ



Or, 3

0

2

kT

H N M eμ μ

= )(3

0 M H kT

N M λ

μ μ +=⇒

)

3

1(3

0

2

0

2

kT

N kT

H N M

μ μ λ

μ μ

−

=⇒ H kT

N

kT

N M

3)

31( 0

2

0

2 μ μ λ

μ μ =−⇒

)3

(3

)3

1(3

0

2

0

2

0

2

0

2

k

N T k

N

kT

N kT

N

H

M

μ μ λ

μ μ

μ μ λ

μ μ χ

−

=

−

==⇒

)(3 0

2

cT k

N

θ

μ μ χ

−=⇒

k

N c

3 0

2 μ μ θ =⇒

θc is known as paramagnetic curie point. Above equation is known

as Curie-Weiss Law.

dia para magnetic materials

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