Magnetic Properties of Materials
1. 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 field
b) 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 background
Definition 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 materials
Relative
permeability
susceptibility
linear materials only
∂M
when non − linear in field χ =
∂H H → 0
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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:
mi = I v∫ dS = IA
dm
= IdS
Magnetic Flux density B is:
B = μ0 H + μ0
Nm i
= μ0 ( H + M )
V
M is magnetic dipole moment/unit volume
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Langevin Diamagnetism
Electrons in an atom precess in a
magnetic field at the Larmor frequency:
ω =
eB
2m
Act as a current loop which shields the applied field
⎛ 1 eB ⎞
⎟
⎝ 2π 2m ⎠
I = charge/revolutions per unit time = (− Ze) ⎜
Area of loop = π<ρ2> = π(<x2> + <y2>) = 2/3 π<r2>
Ze 2 B 2
r
6m
μ Nm
μ NZe 2 2
r
χ = 0 i = − 0
B
6m
Hence magnetic moment induced/atom mi = −
For N atoms susceptibility
(per unit vol or per mole)
Magnetic Levitation - An example of
magnetic energy density
Energy due to magnetisation = -m.B
Magnetic 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 gravitational
force = Vρg, therefore:
-½∇B2 = ρg μ0 / χ
Typical values of χ are of order 10-5 - 10-6
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Paramagnetism
Unfilled 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 system
N ions with mJ = ±½ , g = 2, then U = ± μBB,
ratio of populations is given by Boltzman factor so we can write:
exp( μ B B )
N1
kT
=
B
μ
N
B
exp(
) + exp(− μ B B )
kT
kT
B
−μ
exp( B
)
N2
kT
=
μ B ) + exp(− μ B B )
N
exp( B
kT
kT
Upper state N2 has moment -μB , so writing x = μBB/kT
M = ( N1 − N 2 ) μ B
Magnetism – HT10 - RJ Nicholas
e x − e− x
= Nμ B x
= Nμ B tanh x
e + e−x
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Magnetisation
Magnetisation saturates at high fields and low
temperatures
μ B
M = Nμ B tanh B
k BT
Low field, higher temperature limit tanh x → x
Nμ 0 μ B2
M
=
χ =
H
k BT
Curie’s Law:
Compare with
Pauli paramagnetism
3Nμ 0 μ B2
=
2k BTF
General Result for mJ … J → -J
2J + 1
⎡ 2J + 1
⎤
M = NgJ μ B ⎢(
) coth(
) y − ( 1 ) coth( 1 ) y ⎥
2J
2J ⎦
2J
⎣ 2J
Curie’s Law
J + 1 gJ μB B
kT
3J
J ( J + 1) g 2 μB 2 C
M
χ =
= μ0 N
=
H
T
3kT
as y → 0, M = NgJ μB
as y → ∞,
Magnetism – HT10 - RJ Nicholas
M sat = Ng μB J
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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
Energy
Magnetic Resonance
Magnetic Field
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.
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Magnetic Resonance Experiment
Experiment uses microwaves (10 100 GHz), with waveguides and
resonant cavities
Resonance Spectrum
Relate to
Crystal field
splittings
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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.
ΔB =
μ0m
r3
=
−
Ze 2 B 2 μ 0
r . 3
6m
r
Diamagnetic magnetic moment induced in a single atom
N.M.R. and M.R.I.
Chemical sensitivity is used in
monitoring many
biochemical and biological
processes
NMR
Magnetism – HT10 - RJ Nicholas
Scan magnetic field with a
field gradient in order to
achieve chemically
sensitive imaging
MRI
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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
N+
⎛ 2μ B ⎞
= exp⎜⎜ − B i ⎟⎟
N−
kTi ⎠
⎝
Remove magnetic field quickly while keeping population
of spins the same - Adiabatically
kT f
μB B f
=
kTi
μ B Bi
Tf is limited by small interactions
between ions which split energy
levels at B = 0
Interactions between magnetic ions
Dipole - dipole interactions
Neighbouring 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 ~ 1T
so:
U = 0.9 x 10-23 J ~ 0.7 K
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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
Exchange Interaction
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
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energy difference given by the ‘exchange integral’
⎡ e2
2e 2 ⎤ *
3
3
−
J = 2 ∫ φ a (1)φb (1) ⎢
⎥φ a (2)φb (2)d r1d r2
⎣ 4πε 0 r12 4πε 0 r1 ⎦
*
Positive term
electron-electron
coupling
Negative term
electron-ion
coupling
Comes from electrostatic (Coulomb) interactions
J>0
Electron spins align
Ferromagnetic coupling
J<0
Electron spins antiparallel
Anti-ferromagnets
Covalent bonding
Molecular Field Model
Energy of one spin in an external magnetic flux B0
U = −S i .∑ J ijS j + g μ B B0 .S i
= − J ij S i . S j
+ g μ B B0 .S i
which can be considered as due to an effective flux Beff
Beff = B0 + μ0λ M ,
where M = − Ng μ B S and λ =
2∑ J ij
j ≠i
N μ0 g 2 μ B2
Spins are aligned by the combination of the external flux
B and the internal magnetisation.
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Curie - Weiss Law
Assuming Curie’s Law to hold for the total fields
μ0 M =
CBeff
=
C ( B0 + λμ 0 M )
T
T
CB0
μ0 M =
T − Cλ
μM
C
χ = 0
=
,
B0
T − Tc
Tc = Cλ
Curie - Weiss Law shows that at around Tc there will be a
spontaneous magnetisation
Saturation Magnetisation
Around Tc the spins begin to
align.
μ 0 M = μ 0 NgJμ B BJ ( y ) ,
For spin 1/2
μ 0 M = Nμ B tanh
μ B μ0 M
k BT
No simple analytical solution, but graph shows numerical
result is reasonable description of experiment
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2∑ J ij
2
μ
μ
N
j ≠i
0 B
for spin 1 , C =
and λ =
2
kB
Nμ 0 g 2 μ B2
so Tc = Cλ =
zJ ij
2k B
, z is No. of nearest neighbours
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
Result is a non-integer average magnetic moment per electron
e.g.
Metal
Iron
Cobalt
Nickel
Gd
Dy
Ms (μB / atom)
2.2
1.7
0.6
6.8
10.2
Fe3+
Co2+
Ni2+
Gd3+
Dy3+
gJ
5
6
6
7
10
Tc =
Tc =
Tc =
Tc =
Tc =
1043
1388
627
292
88
Rare-earths have very narrow bands so result is given by
free ion result
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Band model of Ferromagnetism
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.
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
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Why do Domains form?
Magnetisation of a single crystal costs a large amount of
energy of magnetisation = ∫ B0 .dM
Balanced by anisotropy energy - spins only like to align
along particular crystal directions - energy cost is K
per electron spin
Domains in real crystals
Single ‘whisker’
of iron
Domains can be ‘pinned’ by
the presence of impurities,
which make it easy for the
domain boundary to sit in
one place
Larger crystal of
Nickel
Causes the difference
between ‘soft’ and ‘hard’
magnetic materials
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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 two
molecular fields corresponding to spin up and spin down
Result gives: χ = C/(T + θ)
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A
B
A
B
A
B
A
BA eff = B0 − μ0 λ M B ,
BBeff = B0 − μ0 λ M A
λ =
μ0 M A =
4∑ J ij
j ≠i
N μ0 g 2 μ B2
C A
Beff ,
2T
M = MA + MB
λC
C
μ0 M B
B−
2T
2T
λC
C
μ0 M B =
μ0 M A
B−
2T
2T
μ0 M A =
μ0 M =
λC
C
μ0 M =
B−
2T
T
Magnetism – HT10 - RJ Nicholas
Sum these two
equations
(
CB
T + λC
2
)
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Summary of Collective Magnetism
Some crystals can have ions with both J > 0 and J < 0
This 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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