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) 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 Energy 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 magnetic dipole moment/unit volume Magnetism – HT10 - RJ Nicholas 2 Langevin Diamagnetism Electrons in an atom 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 = < > = (<x > + <y >) = /3 <r > 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 magnetic energy density Energy due to magnetisation = -m.B χ μ Magnetic moment = V B/ 0 so χ dB μ ∇ 2 χ μ force = -mdB/dx = - V (B /dx)/ 0 = - ½ B V / 0 force = gradient 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 Bohr magneton (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 Saturation behaviour • Curie Law at low field • Saturation at high field Magnetic Resonance 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 spin 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 chemical shift due to magnetic flux from orbital motion 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 dipoles 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 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 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 electron-ion coupling coupling Comes from electrostatic (Coulomb) 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 Metal Ms ( B / atom) gJ 3+ Iron 2.2 Fe 5Tc = 1043 2+ Cobalt 1.7 Co 6Tc = 1388 2+ Nickel 0.6 Ni 6Tc =627 3+ Gd 6.8 Gd 7Tc = 292 3+ Dy 10.2 Dy 10 Tc =88 Rare-earths 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 metals, so we have to remember that just as in Pauli paramagnetism not all of the spins can be aligned.