Magnetic Materials. The inductor Φ B = LI (Q = CV) = L I = N Φ. Power = VI = LI. Energy = Power dt = LIdI = 1 LI 2 = 1 NΦ B capacitor CV 2

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1 Magnetic Materials The inductor Φ B = LI (Q = CV) Φ B 1 B = L I E = (CGS) t t c t EdS = 1 ( BdS )= 1 Φ V EMF = N Φ B = L I t t c t B c t I V Φ B magnetic flux density V = L (recall I = C for the capacitor) t t EdS = E dl (Green's Theorem) I Power = VI = LI V = E dl = 1 Φ B t (explicit Faraday's Law) c t Energy = Power dt = LIdI = 1 LI = 1 NΦ B I 1 capacitor CV 1

2 Insert magnetic material The Inductor 4π 1 E B = J + c c t 4π c 4π c BdS = B dl = J ds = I 4π B = In c N = n length = nl Nφ B N (BA) 4π L = = = n la I I c Magnetic dipoles in material can line-up in magnetic field B = H + 4πχ H = H + 4π M B magnetic induction χ magnetic susceptibility M M = χh = χ μ = 1+ 4πχ H magnetic field strength (applied field) H M magnetization B = 4πM + 1 B = μh

3 H and B H has the possibility of switching directions when leaving the material; B is always continuous H B p M M q B H At p: At q: M=0 B M H 3

4 Maxwell and Magnetic Materials Ampere s law H dl = I = 0 For a permanent magnet, there is no real current flow; if we use B, there is a need for a fictitious current (magnetization current) Magnetic material inserted inside inductor increases inductance Φ B = BA ~ 4πMA = 4πχHA = 4πχ 4π In A c Material Type NΦ B ( ) χ 4π L = = n laχ I c Paramagnetic L increased by ~χ due to magnetic material Diamagnetic Ferromagnetic

5 Microscopic Source of Magnetization No monopoles magnetic dipole comes from moving or spinning electrons Orbital Angular Momentum μ μ is the magnetic dipole moment r r Energy = E = μ H = μ H cosθ A L e- I What is μ? For θ=0, E = μh Φ B I since energy ~ LI and for 1loop L = Φ B I Φ B = H ds ~ HA μh = Φ B I = HAI and μ = IA I = e ω A = πr c π μ = e ωr c 5

6 Microscopic Source of Magnetization Classical mechanics gives orbital angular momentum as: r r r L = p = mr ω e μ L = L QM = eh L Z = μ B L Z Example for l=1: mc mc +μ B H eh μ B = 0 mc E(H=0) L Z = m l = l,...,0,...l -μ B H Spin Moment μ s e Q.M. eh μ s = S = g 0 S z = g 0 μ B S Z mc mc E(H=0) +(1/)μ B g 0 H S Z = m S = ± 1 g 0 = for electron spin -(1/)μ B g 0 H 6

7 Total Energy Change for Bound Electron in Magnetic Field Simple addition of energies if spin-orbital coupling did not exist r r E = μ H = μ B (L Z + g 0 S Z )H = μ T H But spin-orbit coupling changes things such that: μ T μ B (L Z + g 0 S Z ) = μ B J Z QM definitions: μ T = gμ B J Z L = hl Z = hm l 3 1 S(S +1) L(L +1) g = + S = hs Z = hm s J (J +1) r r r r J = L + S E = μ T H = gμ B J H = gμ B J Z H J = hj Z 7

8 Total Energy Change for Bound Electron in Magnetic Field Kinetic energy from Lorentz force has not been included e r r p H = H c Energy change = p = m Lorentz for circular orbit e r r r r (r H ) (r H ) 8mc For the plane perpendicular to H and assuming circular orbit: e e Energy change = r H = (x + y )H 8mc 8mc e ΔE TOT = gμ B HJ Z + H (x + y ) 8mc Numbers: μ B Hfor for H=10-4 Gauss H=10-4 Gauss =10-4 ev =10-9 ev The Lorentz effect is minimal with respect to magnetic moment interaction, if it exists 8

9 Atoms with Filled Shells J=0 (L=0, S=0) e Only Lorentz contribution ΔE = H (x + y ) 8mc Leads to diamagnetism Need to sum over all e- in atom: e ΔE atom = 1 mc H r i 3 i N E χ = - V H (for a spherical shells) 1 E M M = χ = V H H χ = e N e r i = 6 mc V i 6mc V N r Zi e N χ = r Z i ~ mc V 9

10 Atoms with Partially Filled Shells J not zero Need Hund s rules from QM Fill levels with same ms to maximize spin maximize L (first e- goes in largest l) J= L-S for n<=(l+1), J= L+S for n>(l+1) Conventional notation: (S +1) L X J X S P D F G H I J=0 when L and S are not zero is a special case nd order effect--> perturbation theory Partially filled shells give atoms paramagnetic behavior ΔE = gμ B HJ Z ( >10-9 ev for diamagnetic component) 10

11 d-shell (l = ) n l x =, 1, 0, -1, -, S L = Σl x J Symbol f-shell (l = ) 1/ 1 3 3/ 3 5/ 0 3/ / 0 0 3/ 3/ 0 5/ 4 9/ 4 5/ 0 J= L-S J=L+S D 3/ 3 F 4 F 3/ 5 D 4 6 S 5/ 5 D 4 4 F 9/ 3 F 4 D 5/ 1 S 4 Sc Ti V Cr Mn Fe Co Ni Cu Zn n l z = 3,, 1, 0, -1, -, -, S L = Σl z J 1/ 3 5/ F 5/ H 4 3/ 6 9/ J= L-S 4 I 9/ 6 4 5/ 5 5/ 5 I 4 H 5/ F 6 7/ 0 7/ 8 S 7/ / 5 15/ 7 F 6 H 15/ I 8 3/ 6 15/ J=L+S 4 I 15/1 H 6 1/ 3 7/ F 7/ S 0 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Figure by MIT OpenCourseWare. 11

12 Temperature Dependence of Paramagnetism Temperature dependence determined by thermal energy vs. magnetic alignment energy (same derivation as for molecular polarizability in the case of electric dipoles) f = e U pe cosθ U k b T k b T = e k b T for electric dipoles; f = e μ Z = μ fd Z Ω For low H fields and/or low T, fd Ω μ Z = N M = V χ = μ H 3 k T N V b = μ B J H 3 k T b μ B J 3 k T b 1 N χ QM = 3 V μ B J H 3 k T b μ B g J ( J + 1) k T b χ μ H = e k b T 1 T cos θ for magnetic dipoles; Curie s Law 1

13 Effect of De-localized electrons on Magnetic Properties Pauli Paramagnetism dues to the reaction of free e- to magnetic field E(H=0) ΔE = g 0 μ B H μ = g 0 μ B S M = μ B (n + n ) (n + is the density of free electrons parallel to the H field) g + ( E )= 1 g ( E ) g + ( E )= 1 g(e + μ B H ) g( E) For exact solution, need ΔE = g 0 μ B H to expand about E f for n+ 1 = g( E ) E(H=0) g ( E)= 1 g(e μ B H ) and n- Only e- near Fermi surface matter: ΔE = g 0 μ B H B E Note: Pauli paramagnetism has no T dependence, whereas Curie paramagnetism has E F μb H 1/T dependence Δ ( n + n ) = Δn g ( E F ) Δn g( ) M = μ ( ) χ = μ ( ) Hg E F, B g E F 13

14 Effect of De-localized electrons on Magnetic Properties Landau paramagnetism Effect of bands/fermi surface on Pauli paramagnetism F=qvxB for orbits orbit not completed under normal circumstances however, average effect is not zero χ Land = 1 χ Pauli 3 14

15 Ferromagnetism Most important but not common among elements Net magnetization exists without an applied magnetic field To get χ~ as we see in ferromagnetism, most moments in material must be aligned! There must be a missing driving force 1 r r r r 4 μ rˆ)] ~ 10 ev r NOT dipole-dipole interaction: too small E dipole = 3 [μ 1 μ 3(μ 1 rˆ)( Spin Hamiltonian and Exchange H spin = J ij S i S j J ij exchange constant Assuming spin is dominating magnetization, H = 1 S r (R r ) S r r r r (R ')J (R R ') gμ B H S(R r ) r r r R,R' R 15

16 Exchange E~-JS 1 S J negative, E~+S 1 S --> Energy J positive, E~-S 1 S --> Energy if if Fe, Ni, Co ---> J positive! Other elements J is negative Rule of Thumb: r r a interatomic distance > 1.5 (atomic radius) J is a function of distance! 16

17 Ferromagnetism M M ( T ) ( T C T ) β β χ( T ) ( T C T ) γ γ B=H+4πM B r, M s Domain rotation Irreversible boundary displacement normal paramagnet reversible boundary displacement H c H T C T Easy induction, softer Magentic anisotropy hardness of loop dependent on crystal direction comes from spin interacting with bonding 17

18 Domains in Ferromagnetic Materials B N S N S N M S Magnetic energy = 1 B dv 8 N S N S Magnetic domain Domain wall or boundary Flux closure No external field 18

Magnetism in low dimensions from first principles. Atomic magnetism. Gustav Bihlmayer. Gustav Bihlmayer

IFF 10 p. 1 Magnetism in low dimensions from first principles Atomic magnetism Gustav Bihlmayer Institut für Festkörperforschung, Quantum Theory of Materials Gustav Bihlmayer Institut für Festkörperforschung