What Magnetic Measurements tell us What Magnetic Measurements tell - PowerPoint PPT Presentation

What Magnetic Measurements tell us What Magnetic Measurements tell us about magnetism? about magnetism? Viorel Pop Babe ş -Bolyai University, Faculty of Physics, Cluj-Napoca, Romania

Magnetic moment An electrical current, I, is the source of a magnetic field B Idl Current I Magnetic field generated by a single-turn coil far from the origin: r m µ I B = θ 3 0 sin with 2 R r is by definition the magnetic moment of the single-turn coil m I

magnetisation M χ magnetic susceptibility μ magnetic permeability r ∑ r m = M V M χ = H ( ) r r r = µ + B H M B 0 = µ B=µ 0 (H+ χ H)= µ 0 (1+ χ )H= µH H ( ) = + µ µ 1 χ 0 μ 0 = 4 π⋅ 10 -7 H/m

Diamagnetic r M χ (b) m = (a) 0 χ < 0 H χ T C, Cu, Pb, H 2 O, NaCl, SiO 2

Diamagnetic r M χ (b) m = (a) 0 χ < 0 H χ T C, Cu, Pb, H 2 O, NaCl, SiO 2 M 1/ χ T − 1 χ = Paramagnetic C r χ ≠ f(H) m ≠ 0 J ij = 0 χ χ > 0 1/C T Na, Al, CuCl 2 H if χ ( µ B /T ∙ f.u) if χ (emu/mole) 3 k B µ = C µ = µ + eff 1 ⋅ µ µ ( µ = ⋅ µ ( µ = ⋅ J g J ( J ) 8 4 , 466 N ) C ) C 0 eff B eff B eff B

Diamagnetic r M χ (b) m = (a) 0 χ < 0 H χ T C, Cu, Pb, H 2 O, NaCl, SiO 2 M 1/ χ T − 1 χ = Paramagnetic C r χ ≠ f(H) m ≠ 0 J ij = 0 χ χ > 0 1/C T Na, Al, CuCl 2 H if χ ( µ B /T ∙ f.u) if χ (emu/mole) 3 k B µ = C µ = µ + eff 1 ⋅ µ µ ( µ = ⋅ µ ( µ = ⋅ J g J ( J ) 8 4 , 466 N ) C ) C 0 eff B eff B eff B r m ≠ 0 Magnetic ordered J ij ≠ 0 χ >> 0

a) ferromagnetic r r J ij > 0 M s ≠ 0 = − ⋅ 2 S i S H J ij j Fe, Co, Ni, Gd… M T 1 T 2 m = H N M T 1 <T 2 <T c <T 3 M ii Molecular field 1 / χ T 3 approximation θ = T c H C χ = − θ T Curie – Weiss law M s (0) = g J µ B J T c θ T

b) antiferromagnetic Η Η Η=0 Η Η χ J ij < 0 M s =0 χ ⊥ r r = M M A B χ ⎜⎜ MnO, Mn, Cr… T T N 1 / χ θ < 0 + M M C χ = = A B + θ H T θ T T N

c) ferrimagnetism J ij < 0 r r ≠ M M M s ≠ 0 A B Fe 3 O 4 , ferrites, GdCo 5 ,…

T < T c N AA ≈ N BB N AA > N BB N AA < N BB M M M M A M A M A M=M A -M B M 0 M=M A -M B M 0 M 0 M= ⎜ M A -M B ⎜ T c T c 0 0 0 T comp T c T T T M B M B M B M=M A -M B 1/ χ T > T c σ 1 1 T = + − χ χ − θ C T ' 0 θ 0 T T c

M 1 / χ M s (0) C χ = − θ T T c θ T

M T 1 M s (T 1 ) T 2 χ p T 3 M s (T 2 ) = + χ M M H s p M s (T 3 ) T 1 < T 2 < T 3 H

M T 1 M s (T 1 ) T 2 χ p T 3 M s (T 2 ) = + χ M M H s p M s (T 3 ) T 1 < T 2 < T 3 H 6 5 Al 5 Mn 3 Ni 2 Al 5 Mn 3 Ni 2 5 T = 10 K 5.2 4 5 4 M s ( µ B /f.u.) M( µ B /f.u.) 4.8 3 3 M( µ B /f.u.) 4.6 4.4 y = 5.0293 + 0.00043122x R= 0.10622 2 2 4.2 T = 10 K y = 4.7774 + 0.009374x R= 0.96475 T = 100 K 4 T = 200 K y = 4.218 + 0.023918x R= 0.99702 T = 300 K M s = 5.03 µ B /f.u. y = 3.535 + 0.022673x R= 0.99868 1 3.8 1 T c = 401 K 3.6 5 6 7 8 9 10 0 µ 0 H (T) 0 100 200 300 400 500 0 T (K) 0 2 4 6 8 10 µ 0 H (T)

+ + +++ + + + + + + + + + H a H a M s H d M s H d - - - - - - - - - - - - - - r r = = = − H M H N M H N M N ⊥ II d s d s d d

The influence of the demagnetising field on the magnetisation curves r r r r r r = − = = + H M H a = applied field N H H H H d d i a d N dx = N dy = N dz = 1/3. sphere l N d = 0 r r d << l ( ) M d = − − θ ⋅ O 1 H d cos N d = -1 d >> l θ r M ( ) ( ) = χ = χ + = χ − M H H H H M N a d a d χ = M H a + χ 1 N d M M 1/N d H a H

H d ≠ 0 M T 1 M s (T 1 ) T 2 T 3 M s (T 2 ) M s (T 3 ) = + χ M M H s p 5 Al 5 Mn 3 Ni 2 T 1 < T 2 < T 3 T = 10 K 5.2 4 5 M( µ B /f.u.) 4.8 3 M( µ B /f.u.) 4.6 H 4.4 y = 5.0293 + 0.00043122x R= 0.10622 2 4.2 T = 10 K y = 4.7774 + 0.009374x R= 0.96475 T = 100 K 4 T = 200 K y = 4.218 + 0.023918x R= 0.99702 T = 300 K y = 3.535 + 0.022673x R= 0.99868 3.8 1 3.6 5 6 7 8 9 10 µ 0 H (T) 0 0 2 4 6 8 10 µ 0 H (T)

5 Al 5 Mn 3 Ni 2 T = 10 K 5.2 4 5 M( µ B /f.u.) 4.8 3 M( µ B /f.u.) 4.6 4.4 y = 5.0293 + 0.00043122x R= 0.10622 2 4.2 T = 10 K y = 4.7774 + 0.009374x R= 0.96475 T = 100 K 3 4 T = 200 K y = 4.218 + 0.023918x R= 0.99702 T = 300 K y = 3.535 + 0.022673x R= 0.99868 GdCo 4 Si 3.8 1 3.6 5 6 7 8 9 10 T = 4 K µ 0 H (T) 2,5 0 0 2 4 6 8 10 µ 0 H (T) 2 M( µ B /f.u.) = + χ M M H 1,5 s p 1 0,5 0 0 2 4 6 8 10 µ 0 H (T)

⎛ − ⎞ a = + χ ⎜ 1 ⎟ M M H s p ⎝ ⎠ H 3 GdCo 4 Si T = 4 K 2,5 3 2 2,9 M( µ B /f.u.) M s (T) 2,8 1,5 2,7 M( µ B /f.u.) 2,6 2,5 1 2,4 2,3 0,5 2,2 2 3 4 5 6 7 8 9 10 µ 0 *H (T) 0 0 2 4 6 8 10 µ 0 H (T)

Case study: magnetic measurements on plate shape samples NO MAGNETOCRYSTALLINE ANISOTROPY M H H H sat = M s H Magnetic measurements give magnetisation (A/m)

PERPENDICULAR ANISOTROPY M H H sat = H a H Magnetic measurements give magnetocrystalline anisotropy M Magnetic measurements give magnetisation (A/m) H H sat = M s H

M M s (0) 2 2 µ µ + 1 N Ng J ( J ) = ii B o T c 3 k B ? T c T T → 0K M s (0) = g J µ B J 0 For the rare earth (Gd for example): J 0 =J p For 3d transition metals (Fe, Co, Ni…), the orbital moment is blocked by crystalline field: T → 0K M s (0) = g J µ B S 0

Curie temperature evaluation 2 2 ⎛ − ⎞ ⎡ ⎤ + 10 1 T → T c ; T < T c M ( T ) ( J ) T ⎜ ⎟ = ⋅ 1 ⎢ ⎥ ⎜ ⎟ 2 2 0 3 ⎣ ⎦ + + ⎝ ⎠ M ( ) 1 T J ( J ) c 10 ThFe 11 C 1.5 50 8 40 30 6 M 2 (a.u.) M(a.u.) 20 4 10 0 300 350 400 450 500 550 600 650 700 2 T(K) 0 200 400 600 800 1000 1200 O. Isnard, V. Pop, K.H.J. Buschow, T(K) J. Magn. Magn. Mat. 256 (2003) 133

T c (Fe) SmCo 5 +20 wt% Fe_8hMM 10 8 8 7 6 6 5 M (a.u.) M (a.u.) 4 3 4 2 1 0 2 200 400 600 800 1000 1200 T(K) T c = 1119 K 0 200 400 600 800 1000 1200 T(K)

In the low magnetisation region - for example T → T c; T < T c 2 4 M M = + + ⋅ ⋅ ⋅ − µ F m ( M ) a b MH 0 2 4 µ dF m M a 2 0 3 = = − 0 + = µ or M aM bM H 0 dM H b b molecular field approximations: ( ) 2 µ − µ + + 3 2 2 1 N T T ( J J ) T 3 0 0 + = µ ii c M M H 0 ( ) 2 2 + T 10 1 M J C c 0 H m = N ii M ( ) N ii = T c / C T < Tc µ − a < 0 N T T 0 = ii c a T = Tc a = 0 T c a > 0 T > Tc . 2 µ + + 3 2 2 1 J J T ( ) 0 = b ( ) C 2 2 + 10 1 M J 0

T 1 T 1 T 2 T 2 M 2 T 3 M 2 T 3 T 4 T 4 T c T c ⎛ ⎞ ⎛ ⎞ H H H H − + − ⎜ ⎟ ⎜ ⎟ T T 4 3 T 5 T 5 ⎝ ⎠ ⎝ ⎠ M M M M 3 4 = c c T c ⎛ ⎞ H H − ⎜ ⎟ 2 2 ⎝ ⎠ M s M M M 4 3 s T 1 < T 2 < T 3 < T c < T 4 < T 5 . 1/χ 0 H/M 3 H/M c H/M 4 H/M 120 400 K ThFe 11 C 1.5 100 420 K 80 M 2 ( µ B /f.u.) 2 440 K 60 Arrott plot 40 20 0 O. Isnard, V. Pop, K.H.J. Buschow, 0 0.2 0.4 0.6 0.8 1 J. Magn. Magn. Mat. 256 (2003) 133 µ 0 H/M (T*f.u./ µ B )

M 1 / χ M s (0) C χ = − θ T T c θ T T → 0K T > T c µ = g µ + 1 M s (0) = g J µ B J 0 J ( J ) eff B p p For the rare earth (Gd for example): J 0 =J p For 3d transition metals (Fe, Co, Ni…), the orbital moment is blocked by crystalline field: T → 0K T > T c S p = S > 1 µ = g µ + 1 r M s (0) = g J µ B S 0 S ( S ) eff B p p 0

r = 1 local moment limit r → ∞ total delocalisation limit Gd 1 Fe 1 Co 1 ThFe 11 C 1.5 Fe 3 C 3 HoCo 4 Si 4 YCo 3 B 2 2 5 → ∞ r 1.00 1.01 1.32 1.5 1.69 2.03 1 P.R. Rhodes, E.P. Wolfarth, Proc. R. Soc. 273 (1963) 347. 2 O. Isnard, V. Pop, K.H.J. Buschow, J. Magn. Magn. Mat. 256 (2003) 133 3 D. Bonnenberg, K.A. Hempel, H.P.J. Wijn, Landolt-B.orsntein new series, Vol. III, 19a, Springer, Berlin, 1986, p. 142. 4 O. Isnard, N. Coroian, V. Pop (unpublished) 5 R. Ballou, E . Burzo, and V. Pop, J. Magn. Magn. Mat. 140-144 (1995) 945.

M If there are some ferromagnetic impurity Paramagnetic sample = χ + M H cM s M χ = H H

M If there are some ferromagnetic impurity Paramagnetic sample = χ + M H cM s M M M χ = = χ + s c H H H H M H χ 1 H

C χ = T + θ σ 1 1 T = + − χ χ − θ C T ' 1/ χ 0 C χ = T C − θ χ = T θ θ θ T N T c T c T

axial symmetry : 2 E a ≈ θ K 1 sin K1 < 0 K 1 > 0 [100] H H c M 1 [001] H 0,8 H a YCo 5 ∆ M 1 - [110] H m H c [001] [111] 0,6 . A 6 0 1 0,4 , M H a 0,2 [100] [110] 0 H 0 4 8 12 16 6 -1 H, 10 A.m

T < T N , antiferromagnetic materials, χ ⊥ > χ ⎢⎢ Density of energy in magnetic field H, low anisotropy energy E = - χµ 0 H 2 /2 spin – flop transition M H M a M b (z) M a H (z) M b H (a) (z) (b) M a M b 0 H = H H H c a ex E. Du Trémolet de Lacheisserie (editor), Magnetisme, Presses Universitaires de Grenoble, 1999