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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

4x10 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
 4x107 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  4x107
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
2.33x1023 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 x104
mK
  o  o C 

 47,746.5
B
T
o 4x107 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
4x10 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  4x10 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 4x10 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  2S
KJ E
and we know the wall energy, thus a can be obtained from this equation
a

 2 1
 2S 
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 x1019  2.846 x1010 m
m
ev



J 1
 
ev  2 
J
1.5 x10 3 2.846 x1010
m
4

3
2
 117
Thus the wall thickness is Na=117 x 2.846x10-10m=3.33x10-8m