Magnetism – HT10 - RJ Nicholas 1
Magnetic Properties of Materials1. Magnetisation of materials due to a set of isolated atoms
(or ions)a) Diamagnetism - magnetic moment of filled shells of
atoms. Induced moment opposes applied fieldb) Paramagnetism - unfilled shells have a finite magnetic
moment (orbital angular momentum) which aligns along the magnetic field direction.
2. Collective magnetisation - magnetic moments of adjacent atoms interact with each other to create a spontaneous alignment - Ferromagnetism, Ferrimagnetism, Antiferromagnetism
Some useful backgroundDefinition of the fields:B is the magnetic flux density (units Tesla)H is the magnetic field strength (units Am-1)M is the magnetisation (the magnetic dipole moment per
unit volume, units Am-1)B = μ0 (H + M) = μ0 μr H = μ0 H (1 + χ)
All materialsRelative
permeabilitysusceptibility
linear materials only
0fieldinlinearnonwhen
→∂∂=−
HHMχ
Magnetism – HT10 - RJ Nicholas 2
Magnetic Energy
Energy in Magnetic Field = ½B.H = ½μ0 (H + M).H
= ½μ0 H2 + ½M.H
Energy of a magnetic moment m in magnetic flux
energy to align one dipole = - m.B = -mzBz
Energy density due to magnetisation
of a material: E = M.B
Magnetic moment from a current loop:
Magnetic Flux density B is:
M is magnetic dipole moment/unit volume
im IA= =∫I dS
0 0 0 ( )NV
μ μ μ= + = +imB H H M
dm= IdS
Magnetism – HT10 - RJ Nicholas 3
Langevin DiamagnetismElectrons in an atom precess in a magnetic field at the Larmor frequency:
Act as a current loop which shields the applied field
I = charge/revolutions per unit time =
Area of loop = π<ρ2> = π(<x2> + <y2>) = 2/3 π<r2>
Hence magnetic moment induced/atom mi
meB2
=ω
⎟⎠⎞
⎜⎝⎛−
meBZe22
1)(π
22
6r
mBZe−=
22
00
6r
mNZe
BNmi μμχ −==For N atoms susceptibility
(per unit vol or per mole)
Magnetic Levitation - An example of magnetic energy density
Energy due to magnetisation = -m.BMagnetic moment = VχB/ μ0 so
force = -mdB/dx = - Vχ(B dB/dx)/ μ0 = - ½∇B2 Vχ/ μ0
force = gradient of Field energy density ½ B2 Vχ
Levitation occurs when force balances gravitationalforce = Vρg, therefore:-½∇B2 = ρg μ0 / χTypical values of χ are of order 10-5 - 10-6
Magnetism – HT10 - RJ Nicholas 4
ParamagnetismUnfilled shells.
Magnetic moment of an atom or ion is:μ = -gμBJ, where J is the total angular momentum
J = L + S and μB is the Bohr magneton (e /2m)
A magnetic field along z axis splits energy levels so:
U = - μ.B = mJ gμBB
mJ runs from J to -J
Spin ½ systemN ions with mJ = ±½ , g = 2, then U = ± μBB,ratio of populations is given by Boltzman factor so we can write:
Upper state N2 has moment -μB , so writing x = μBB/kT
)exp()exp(
)exp(
)exp()exp(
)exp(
2
1
kTB
kTB
kTB
NN
kTB
kTB
kTB
NN
BB
B
BB
B
μμ
μ
μμ
μ
−+
−=
−+=
xNeeeeNNNM Bxx
xx
BB tanh)( 21 μμμ =+−=−= −
−
Magnetism – HT10 - RJ Nicholas 5
Magnetisation
Magnetisation saturates at high fields and low temperatures
Low field, higher temperature limit tanh x → x
Curie’s Law:
Compare with Pauli paramagnetism
TkBNM
B
BB
μμ tanh=
TkN
HM
B
B2
0μμχ ==
FB
B
TkN2
3 20μμ=
General Result for mJ … J → -J
2 1 2 1 1 1( ) coth( ) ( ) coth( )2 22 2BJ JM NgJ y yJ JJ J
μ + +⎡ ⎤= −⎢ ⎥⎣ ⎦
( ) 2 2
0
10,3
13
as ,
BB
B
sat B
gJ BJas y M NgJJ kT
J J gM CNH kT T
y M Ng J
μμ
μχ μ
μ
+→ =
+= = =
→ ∞ =
Curie’s Law
Magnetism – HT10 - RJ Nicholas 6
Saturation behaviour
• Curie Law at low field
• Saturation at high field
Apply a magnetic field to split up energy levels and observe transitions between them
For a simple spin system ω = gμBB - selection rule is ΔmJ
= 1 (conservation of angular momentum)When ions interact with the crystal environment or each
other then extra splittings can occur.
Magnetic Resonance
Magnetic Field
Ene
rgy
Magnetism – HT10 - RJ Nicholas 7
Magnetic Resonance ExperimentExperiment uses microwaves (10 -
100 GHz), with waveguides and resonant cavities
Resonance Spectrum
Relate toCrystal fieldsplittings
Magnetism – HT10 - RJ Nicholas 8
Nuclear Magnetic Resonance
Similar to spin resonance of electrons, but from spin of nuclei. Energy is smaller due to the much smaller value of μBN
Resonance condition is ω = gμN(B + ΔB)ΔB is the chemical shift due to magnetic flux from orbital
motion of electrons in the atom.
Diamagnetic magnetic moment induced in a single atom
302
2
30 .
6 rr
mBZe
rmB μμ −==Δ
N.M.R. and M.R.I.
Chemical sensitivity is used in monitoring many biochemical and biological processes
NMR
Scan magnetic field with a field gradient in order to achieve chemically sensitive imaging
MRI
Magnetism – HT10 - RJ Nicholas 9
Magnetic Cooling (adiabatic demagnetisation)
Cool a solid containing alot of magnetic ions in a magnetic field. Energy levels are split by U = ± μBBi
Population ratio is
Remove magnetic field quickly while keeping population of spins the same - Adiabatically
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−
+
i
iB
kTB
NN μ2exp
iB
i
fB
f
BkT
BkT
μμ=
Tf is limited by small interactions between ions which split energy levels at B = 0
Interactions between magnetic ionsDipole - dipole interactionsNeighbouring atoms exert a force on each other which tries to
align dipoles m. Interaction energy is U = -m.B.Magnetic flux B is μ0(m/4πr3 - 3(m.r)r/4πr5) so
U = -m.B ~ - μ0 m1.m2/4πr3 = - μ0 μB2/4πr3
r is separation of atoms (approx. 0.3nm) giving U = 4 x 10-25 J = 0.025 K
If all atomic dipoles are aligned the magnetisation is:M = NμB giving a total magnetic flux B = μ0 NμB ~ 1Tso: U = 0.9 x 10-23 J ~ 0.7 K
Magnetism – HT10 - RJ Nicholas 10
Collective Magnetism -Ferro-, Antiferro- and Ferrimagnetism
Oldest piece of Condensed Matter physics known about and exploited is ferromagnetism.
Why do spins align at temperatures as high as 1000 K ??It is definitely not due to magnetic dipole-dipole
interactions.
The strong interaction is due to the action of the Pauli exclusion principle which produces:
Exchange Interactions
Consider two adjacent atoms and two electrons with a total wavefunction ψ (r1,r2;s1,s2)
Pauli exclusion principle: ψ (r1,r2;s1,s2) = - ψ (r2,r1;s2,s1)If r1 = r2 and s1 = s2 then ψ = 0
Electrons with same spin ‘repel’ each other and form symmetric and antisymmetric wavefunctions [φa(r1) = φ3d(r1-ra)]
ψs (1,2) = [φa(r1) φb(r2) + φa(r2) φb(r1)]χs(↑↓)
ψt (1,2) = [φa(r1) φb(r2) - φa(r2) φb(r1)]χt(↑↑)
Energy difference between singlet and triplet is dependent on alignment of the spins U = -2Js1.s2
Exchange Interaction
Magnetism – HT10 - RJ Nicholas 11
energy difference given by the ‘exchange integral’
Comes from electrostatic (Coulomb) interactions
23
13*
10
2
120
2* )2()2(
42
4)1()1(2 rdrd
re
reJ baba φφ
πεπεφφ ⎥
⎦
⎤⎢⎣
⎡−= ∫
Positive termelectron-electron
coupling
Negative termelectron-ion
coupling
J > 0Electron spins alignFerromagnetic coupling
J < 0Electron spins antiparallel
Anti-ferromagnetsCovalent bonding
Molecular Field ModelEnergy of one spin in an external magnetic flux B0
which can be considered as due to an effective flux Beff
Spins are aligned by the combination of the external flux B and the internal magnetisation.
. .
. .
ij B
ij B
U J g
J g
μ
μ
= − +
= − +
∑i j 0 i
i j 0 i
S S B S
S S B S
eff 0 0
2 20
B B M ,2
where M ij
j iB
B
JNg S and
N g
μ λ
μ λμ μ
≠
= +
= − =∑
Magnetism – HT10 - RJ Nicholas 12
Curie - Weiss LawAssuming Curie’s Law to hold for the total fields
Curie - Weiss Law shows that at around Tc there will be a spontaneous magnetisation
λμχ
λμ
λμμ
CTTT
CBM
CTCBM
TMBC
TCB
M
cc
eff
=−
==
−=
+==
,
)(
0
0
00
000
Saturation Magnetisation
Around Tc the spins begin to align.
No simple analytical solution, but graph shows numerical result is reasonable description of experiment
TkMNM
yBNgJM
B
BB
JB
00
00
tanh
1/2spin For
,)(
μμμμ
μμμ
=
=
Magnetism – HT10 - RJ Nicholas 13
Typical values of Tc are 1000K, so if z = 8 we find
J = ~ 0.02 eV,
At saturation (all spins aligned) M = NμB giving:
Beff = μ0λM = μ0NμBTc/C = kBTc/ μB ~ 1500 T
This is not a real Magnetic Field
Bferromagnet = μ0 NμB ~ 1T
neighboursnearestofNo.isz,2
so
2and,2
1spinfor 220
20
B
ijc
B
ijij
B
B
kzJ
CT
gN
J
kNC
==
==∑
≠
λ
μμλμμ
Result is a non-integer average magnetic moment per electron e.g.
Metal Ms (μB / atom) gJIron 2.2 Fe3+ 5 Tc = 1043Cobalt 1.7 Co2+ 6 Tc = 1388Nickel 0.6 Ni2+ 6 Tc = 627Gd 6.8 Gd3+ 7 Tc = 292Dy 10.2 Dy3+ 10 Tc = 88
Rare-earths have very narrow bands so result is given by free ion result
Magnetism – HT10 - RJ Nicholas 14
Exchange interaction shifts energy levels of electrons by less than typical band widths in metals, so we have to remember that just as in Pauli paramagnetism not all of the spins can be aligned.
Band model of Ferromagnetism
Domains
One thing is missing from our explanation:Real magnetic materials need an external magnetic field to
be applied in order to produce strong or permanent magnetisation - depends on crystal orientation
Magnetism – HT10 - RJ Nicholas 15
Why do Domains form?Magnetisation of a single crystal costs a large amount of
energy of magnetisation =Balanced by anisotropy energy - spins only like to align
along particular crystal directions - energy cost is K per electron spin
dM.B0∫
Domains in real crystalsSingle ‘whisker’
of ironLarger crystal of
NickelDomains can be ‘pinned’ by the presence of impurities, which make it easy for the domain boundary to sit in one place
Causes the difference between ‘soft’ and ‘hard’magnetic materials
Magnetism – HT10 - RJ Nicholas 16
Domain boundaries - Bloch Wall
Energy is minimised by changing the spin slowly in N steps by a small angle θ = π/N.
Energy cost calculated from exchange energy
U = -Σ2Js1.s2
ΔU =2NJS2(1-cos(θ))= 2NJS2(1-(1 - θ2/2)) = JS2 π2/N
Anisotropy energy cost = KN/2
Total = JS2 π2/N + KN/2,
Minimum when N = (2JS2 π2/K)½, e.g. 300 in Fe
Antiferromagnetism
What if exchange integral J < 0?
Adjacent spins line up antiparallel.
This causes antiferromagnetism
An example is MnO
Can model this with a molecular field model with twomolecular fields corresponding to spin up and spin down
Result gives: χ = C/(T + θ)
Magnetism – HT10 - RJ Nicholas 17
A A A AB B B
Aeff 0 0 B
Beff 0 0 A
2 20
B B M ,
B B M
4 ijj i
B
J
N g
μ λμ λ
λμ μ
≠
= −
= −
=∑
Sum these two equations
( )
0
0 0
0 0
0 0
,2
2 2
2 2
22
AA eff A B
A B
B A
CM B M M MT
C CM B MT TC CM B MT T
C C C BM B MCT T T
μ
λμ μ
λμ μ
λμ μλ
= = +
= −
= −
= − =+
Magnetism – HT10 - RJ Nicholas 18
Summary of Collective Magnetism
Some crystals can have ions with both J > 0 and J < 0This causes Ferrimagnetism. Best known example is ferrite
Fe3O4 (Fe2+O,Fe3+2O3).
Moment = (2 x 5/2 - 2) μB /formula unit
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