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Home , Bohr magneton, Magnetic dipole, Magnetic moment, Magnetism, Saturation (magnetic)

Magnetic Properties of Materials

1. Magnetisation of materials due to a set of isolated (or ions) a) - magnetic of filled shells of atoms. Induced moment opposes applied b) - unfilled shells have a finite (orbital ) which aligns along the direction. 2. Collective magnetisation - magnetic moments of adjacent atoms interact with each other to create a spontaneous alignment - , ,

Some useful background Definition of the fields: B is the (units ) H is the magnetic field strength (units Am-1) M is the magnetisation (the magnetic moment per unit volume, units Am-1) μ μ μ μ χ B = 0 (H + M) = 0 r H = 0 H (1 + )

Relative susceptibility All materials permeability linear materials only ∂M when non − linear in field χ = ∂H H → 0

Magnetism – HT10 - RJ Nicholas 1 Magnetic

μ Energy in Magnetic Field = ½B.H =½ 0 (H + M).H μ 2 = ½ 0 H +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: dm mIA==IdS = IdS i v∫

Magnetic Flux density B is: Nm BH=+μμi = μ() HM + 00V 0 M is moment/unit volume

Magnetism – HT10 - RJ Nicholas 2 Langevin Diamagnetism in an precess in a eB magnetic field at the Larmor frequency: ω = 2m Act as a current loop which shields the applied field ⎛ 1 eB ⎞ I = charge/revolutions per unit time = ( −Ze) ⎜ ⎟ ⎝ 2π 2m ⎠ π ρ2 π 2 2 2 π 2 Area of loop = < > = ( + ) = /3 2 Ze B 2 Hence magnetic moment induced/atom mi = − r 6m μ Nm μ NZe2 For N atoms susceptibility χ = 0 i = − 0 r 2 (per unit vol or per mole) B 6m

Magnetic Levitation - An example of density

Energy due to magnetisation = -m.B χ μ Magnetic moment = V B/ 0 so χ dB μ ∇ 2 χ μ = -mdB/dx = - V (B /dx)/ 0 = - ½ B V / 0 force = of Field energy density ½ B2 Vχ

Levitation occurs when force balances gravitational force = Vρg, therefore: ∇ 2 ρ μ χ -½ B = g 0 / Typical values of χ are of order 10-5 -10-6

Magnetism – HT10 - RJ Nicholas 3 Paramagnetism Unfilled shells. Magnetic moment of an atom or ion is: μ μ = -g BJ, where =J is the total angular momentum μ J = L + S and B is the (e =/2m)

A magnetic field along z axis splits energy levels so: μ μ U = - .B = mJ g BB

mJ runs from J to -J

Spin ½ system ± ±μ N ions with mJ = ½ , g = 2, then U = BB, ratio of populations is given by Boltzman factor so we can write: μ B exp( B ) N1 = kT μ B − μ B N exp( B ) + exp( B ) kT kT − μ B exp( B ) N2 = kT μ B − μ B N exp( B ) + exp( B ) kT kT μ μ Upper state N2 has moment - B , so writing x = BB/kT ex − e− x M = (N − N )μ = Nμ = Nμ tanh x 1 2 B B ex + e−x B

Magnetism – HT10 - RJ Nicholas 4 Magnetisation

Magnetisation saturates at high fields and low temperatures μ = μ B B M N B tanh kBT

Low field, higher temperature limit tanh x → x M Nμ μ 2 Curie’s Law: χ = = 0 B H kBT Compare with μ μ 2 = 3N 0 B Pauli paramagnetism 2kBTF

→ General Result for mJ …J -J

++ =−μ ⎡⎤21JJ 21 11 M NgJB ()coth()()coth() y y ⎣⎦⎢⎥22JJ22JJ Curie’s Law J + 1 gJμ B as y→=0, M NgJ μ B B 3J kT M JJ()+ 1 g22μ C χμ==N B = H 0 3kT T

→∞ = μ asyMNgJ , sat B

Magnetism – HT10 - RJ Nicholas 5 behaviour

• Curie Law at low field • Saturation at high field

Magnetic

y

g

Apply a magnetic field r

e

to split up energy n

E levels and observe transitions between them Magnetic Field ω μ Δ For a simple system = = g BB - selection rule is mJ = 1 (conservation of angular momentum) When ions interact with the crystal environment or each other then extra splittings can occur.

Magnetism – HT10 - RJ Nicholas 6 Magnetic Resonance Experiment

Experiment uses microwaves (10 - 100 GHz), with waveguides and resonant cavities

Resonance Spectrum

Relate to Crystal field splittings

Magnetism – HT10 - RJ Nicholas 7 Nuclear Magnetic Resonance

Similar to spin resonance of electrons, but from spin of nuclei. Energy is smaller due to the much smaller value of μ BN ω μ Δ Resonance condition is = = g N(B + B) ΔB is the due to magnetic flux from orbital of electrons in the atom. μ m Ze2 B μ ΔB = 0 = − r 2 . 0 r 3 6m r 3 Diamagnetic magnetic moment induced in a single atom

N.M.R. and M.R.I.

Chemical sensitivity is used in Scan magnetic field with a monitoring many field gradient in order to biochemical and biological achieve chemically processes sensitive imaging NMR MRI

Magnetism – HT10 - RJ Nicholas 8 Magnetic Cooling (adiabatic demagnetisation) Cool a solid containing alot of magnetic ions in a magnetic ±μ field. Energy levels are split by U = BBi Population ratio is N+ ⎛ μ ⎞ = − 2 B Bi exp⎜ ⎟ N− ⎝ kTi ⎠

Remove magnetic field quickly while keeping population of spins the same - Adiabatically

kT T is limited by small interactions f = kTi f μ B μ B between ions which split energy B f B i levels at B = 0

Interactions between magnetic ions Dipole - dipole interactions Neighbouring atoms exert a force on each other which tries to align m. Interaction energy is U = -m.B. μ π 3 π 5 Magnetic flux B is 0(m/4 r -3(m.r)r/4 r ) so μ π 3 μ μ 2 π 3 U = -m.B ~ - 0 m1.m2/4 r = - 0 B /4 r 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: μ μ μ M = N B giving a total magnetic flux B = 0 N B ~ 1T so: U = 0.9 x 10-23 J ~ 0.7 K

Magnetism – HT10 - RJ Nicholas 9 Collective Magnetism - Ferro-, Antiferro- and Ferrimagnetism

Oldest piece of 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 ψ wavefunction (r1,r2;s1,s2) ψ ψ Pauli exclusion principle: (r1,r2;s1,s2) = - (r2,r1;s2,s1) ψ If r1 = r2 and s1 = s2 then = 0 Electrons with same spin ‘repel’ each other and form symmetric φ φ and antisymmetric wavefunctions [ a(r1) = 3d(r1-ra)] ψ φ φ φ φ χ ↑↓ s (1,2) = [ a(r1) b(r2) + a(r2) b(r1)] s( ) ψ φ φ φ φ χ ↑↑ t (1,2) = [ a(r1) b(r2) - a(r2) b(r1)] t( ) Energy difference between singlet and triplet is dependent on

alignment of the spins U = -2Js1.s2

Magnetism – HT10 - RJ Nicholas 10 energy difference given by the ‘exchange integral’

2 2 * ⎡ e 2e ⎤ * 3 3 J = 2 φ a (1)φ (1) − φ a (2)φ (2)d r d r ∫ b ⎢ πε πε ⎥ b 1 2 ⎣4 0r12 4 0r1 ⎦

Positive term Negative term -electron electron-ion coupling coupling Comes from electrostatic () interactions J > 0 J < 0 Electron spins align Electron spins antiparallel Ferromagnetic coupling Anti-ferromagnets Covalent bonding

Molecular Field Model

Energy of one spin in an external magnetic flux B0 =− + μ UJgSSij..∑ ij B BS 0i =− + μ JgijSSij.. B BS 0i

which can be considered as due to an effective flux Beff =+μ λ BBeff 0 0 M,

2∑ Jij where M =−Ngμλ S and = ji≠ B μμ22 Ng0 B Spins are aligned by the combination of the external flux B and the internal magnetisation.

Magnetism – HT10 - RJ Nicholas 11 Curie - Weiss Law

Assuming Curie’s Law to hold for the total fields

CB C(B + λμ M ) μ M = eff = 0 0 0 T T CB μ M = 0 0 T − Cλ μ M C χ = 0 = , T = Cλ − c B0 T Tc

Curie - Weiss Law shows that at around Tc there will be a spontaneous magnetisation

Saturation Magnetisation

Around Tc the spins begin to align. μ = μ μ 0M 0 NgJ B BJ (y) ,

For spin 1/2 μ μ μ = μ B 0M 0M N B tanh kBT

No simple analytical solution, but graph shows numerical result is reasonable description of experiment

Magnetism – HT10 - RJ Nicholas 12 2 J 2 ∑ ij Nμ μ ≠ for spin 1 , C = 0 B and λ = j i 2 μ 2μ 2 kB N 0 g B zJ = λ = ij so Tc C , z is No. of nearest neighbours 2kB

Typical values of Tc are 1000K, so if z = 8 we find J = ~ 0.02 eV, μ At saturation (all spins aligned) M = N B giving: μ λ μ μ μ Beff = 0 M = 0N BTc/C = kBTc/ B ~ 1500 T This is not a real Magnetic Field μ μ Bferromagnet = 0 N B ~ 1T

Result is a non-integer average magnetic moment per electron e.g.

μ Ms ( B / atom) gJ 3+ 2.2 Fe 5Tc = 1043 2+ 1.7 Co 6Tc = 1388 2+ 0.6 Ni 6Tc =627 3+ Gd 6.8 Gd 7Tc = 292 3+ Dy 10.2 Dy 10 Tc =88

Rare- have very narrow bands so result is given by free ion result

Magnetism – HT10 - RJ Nicholas 13 Band model of Ferromagnetism

Exchange interaction shifts energy levels of electrons by less than typical band widths in , 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

Magnetism – HT10 - RJ Nicholas 14 Why do Domains form?

Magnetisation of a single crystal costs a large amount of energy of magnetisation = B .dM ∫ 0 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’ Larger crystal of of iron Nickel Domains 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 15 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 Σ U = - 2Js1.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)

Magnetism – HT10 - RJ Nicholas 16 AAAABBB

A =−μλ BBeff 0 0 M, B B =−μλ BBeff 0 0 M A

4∑ Jij λ = ji≠ μμ22 Ng0 B

C μ MB==+A , MMM 0 A 2T eff A B

CCλ Sum these two μμMB=− M 00AB22TT equations CCλ μμMB=− M 00BA22TT

λ μμ=−CC = CB 00MB M TT2 ()T + λC 2

Magnetism – HT10 - RJ Nicholas 17 Summary of Collective Magnetism

Some crystals can have ions with both J > 0 and J < 0 This causes Ferrimagnetism. Best known example is 2+ 3+ Fe3O4 (Fe O,Fe 2O3). 5 μ Moment = (2 x /2 -2) B /formula unit

Magnetism – HT10 - RJ Nicholas 18