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Magnetic Materials 1. Three materials are known to have magnetic susceptibilities of 1) 2x10-4 2) -3x10-5 3) 1x105 Indicate which of these materials is paramagnetic, which one is ferromagnetic, and which one is diamagnetic. Explain your answer. 1) Paramagnetic (small positive susceptibility) 2) Diamagnetic (negative susceptibility) 3) Ferromagnetic (very large susceptibility) 2. Calculate the molar susceptibility for a He atom. (Assume an average electron radius of 0.58Å). NZe 2 2 o r Where N is the number of atoms (1 mol of them) Z is the number of 6m electrons per atom, 2 in this case, m the mass of the electron and <r2> the average squared radius of the electrons orbit 2 Henry atoms electrons 4x10 7 6.02 x10 23 2 1.602 x10 19 C 3 2 11 11 m m mol atom 5 . 8 x 10 m 2 . 39 x 10 mol 6 9.1x10 31 Kg 2 2 2 2 (1 Henry=kg m /C =Js /C ) 3. In an H-field of 106A/m, if an atom has a magnetic dipole moment of three Bohr magnetons and the moment rotates from antiparallel to the field to parallel to the field, a. By how much does its energy decreases? U .B o .H When the magnetic moment is antiparallel to the field, the dot produce is negative thus Henry A 3 B 106 m m 2 2 Js Cm C 4x107 2 3 9.2741x10 24 106 1.165 x10 23 7.27 x10 5 eV C m s sm When the moment and the field are parallel, the dot product gives the same numerical value as for the antiparallel case but it is positive, thus U (antiparall el ) o H 4x107 U (antiparall el ) o H 1.165 x10 23 7.27 x10 5 eV The total change in energy is then ΔU=U(parallel)-U(antiparallel)= -1.165x10-23J --1.165x10-23J=-2.33x10-23J=1.454x10-4eV Where the minus means that the energy has decreased b. At T=300K, what fraction of KbT is this energy? U 2.33x1023 J Fraction 5.6 x10 3 0.56% J KbT 1.38 x10 23 300 K K Thus, at this temperature, the orientation is thermally driven 4. Consider a paramagnetic material with a susceptibility of 2x10-4, Calculate the curie constant C at 300K. M C T 300 K 2 x104 mK o o C 47,746.5 B T o 4x107 Henry Henry m 5. Consider a monoatomic H gas and assume that only the 1s level is available to the electrons. A magnetic field of 1T is applied to this gas. Calculate the population of each of the two spin states. B N1 N e B e k BT k BT e B N2 N k BT e B B e k BT The magnetic moment of one electron in an s level (L=0) is μB= 9.2741x10 k BT e 24 B k BT Cm2 s Cm2 kg 1 B B B 9.2741x10 s Cs 2.24 x10 3 J k BT k BT 1.38 x10 23 300 K K Where it was used that 1T=1kg/Cs and that 1J=1kgm2/s2 24 B e k BT 3 e2.24x10 1.0022 e B k BT 3 e 2.24 x10 0.9978 N1 1.0022 N1 0.9978 0.5011 0.4989 N 1.0022 0.9978 N 1.0022 0.9978 This implies that at 300K thermal effect dominates and the material shows no magnetization even at 1T. 6. An H-field of 20 A/m is applied to a ferromagnet with a saturation magnetization of 1.4x106 A/m. If the magnetic field inside the material is 0.5 Testa a. What percentage of the domains are aligned to the field The magnetic field inside the material is equal to the external field plus the field generated inside it due to the magnetization, thus if BT is the total magnetic field, BE= μoH is the external and M is the magnetization B 0.5T A A BT o H M M T H 20 4 x105 Tm o m m 4x10 7 A A M m x100 28.6% % of aligned spins x100 Ms 6 A 1.4 x10 m Where Ms is the saturation magnetization 4 x105 b. What is the magnetic susceptibility A 4 x105 o M M m 2 x104 A B H 20 m c. What H field should be applied to this material to align all the domains to the field For such a field, the magnetization will be equal to the saturation magnetization, thus. A 1.4 x106 M m 70 A or B H 4x10 7 Tm 70 A 8.8 x10 5 T Hs s s o 4 A m 2 x10 m 7. Consider a material for which the anisotropy field is 5x106 A/m with a saturation magnetization of 1.4x106 A/m. a. What is the magnetic anisotropy constant A Kg m A 1.4 x106 4x10 7 5 x106 2K M s o H an m Cs A m H an K M s o 2 2 J m3 b. What is the component of the magnetization parallel to the field for H=2x105 A/m A 2 x105 M || H H m 1.4 x106 A M || Ms A M s H an H an m 5 x106 m A M || 5.6 x104 m c. What for H=5x106 A/m A 5 x106 M || H m 1.4 x106 A 1.4 x106 A M || A M s H an m m 5 x106 m d. What for H=6x106 A/m This field is larger than the anisotropy field, thus the sample has reach saturation and the component of the magnetization in the direction of the field is equal to the A saturation magnetization, thus M || 1.4 x106 m K 4.4 x106 8. For a hypothetical material, the exchange integral is JE=12 meV, the anisotropy constant per unit of volume is K=1.5x104 J/m3 and the wall energy per unit of area is 1x10-3J/m2. What is the wall thickness? S=1/2 N 2JES 2 we have the exchange and anisotropy constant and the spin, but we do not Ka3 have a. However we know that w 2S KJ E and we know the wall energy, thus a can be obtained from this equation a 2 1 2S 2 KJ E a 1x10 3 J w m2 Then 2 0.012 ev1.602 x10 2 N 19 2 J 1.5 x104 J3 0.012ev1.602 x1019 2.846 x1010 m m ev J 1 ev 2 J 1.5 x10 3 2.846 x1010 m 4 3 2 117 Thus the wall thickness is Na=117 x 2.846x10-10m=3.33x10-8m