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)

  1. Calculate the molar susceptibility for a He atom. (Assume an average electron radius of 0.58Å).

WhereN is the number of atoms (1 mol of them) Z is the number of electrons per atom, 2 in this case, m the mass of the electron and <r2> the average squared radius of the electrons orbit

(1 Henry=kg m2/C2=Js2/C2)

  1. 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,
  2. By how much does its energy decreases?

When the magnetic moment is antiparallel to the field, the dot produce is negative thus

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

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

  1. At T=300K, what fraction of KbT is this energy?

Thus, at this temperature, the orientation is thermally driven

  1. Consider a paramagnetic material with a susceptibility of 2x10-4, Calculate the curie constant C at 300K.
  1. 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.

The magnetic moment of one electron in an s level (L=0) is μB=

Where it was used that 1T=1kg/Cs and that 1J=1kgm2/s2

This implies that at 300K thermal effect dominates and the material shows no magnetization even at 1T.

  1. 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
  2. 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

Where Msis the saturation magnetization

  1. What is the magnetic susceptibility
  1. 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.

or

  1. Consider a material for which the anisotropy field is 5x106 A/m with a saturation magnetization of 1.4x106 A/m.
  2. What is the magnetic anisotropy constant
  1. What is the component of the magnetization parallel to the field for H=2x105 A/m
  1. What for H=5x106 A/m
  1. 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 saturation magnetization, thus

  1. For a hypothetical material, the exchange integral is JE=12 meV, the anisotropy constant per unit of volume is K=1.5x104J/m3 and the wall energy per unit of area is 1x10-3J/m2. What is the wall thickness? S=1/2

we have the exchange and anisotropy constant and the spin, but we do not have a. However we know that

and we know the wall energy, thus a can be obtained from this equation

Then

Thus the wall thickness is Na=117 x 2.846x10-10m=3.33x10-8m