EMP 6
MAGNETISM – MICROSCOPIC VIEW
ORBITAL MAGNETIC MOMENT
An electron moving in a circular orbit around a nucleus produces an average current I
along its orbit.
I
q e

t T
where T is the orbital period of the electron.
Suppose the electron is moving with a velocity v in a circular orbit with radius R. (This is
a classical picture which is useful but not correct). The orbiting electron possesses an
angular moment L and is related to the period T by
L  mv R
v
2 R
T
T
2 R 2 m
L
As a consequence we can associate a magnetic dipole moment pm with the orbiting
electron.
 Le 
eL
e
pm  I A     R 2   
 R2  


2
2m
T 
 2 R m 
pm  
eL
2m
Magnetic dipole moment and angular momentum point in opposite directions. Since this
magnetic dipole moment is associated with the orbital motion of the electron around the
nucleus it is called the orbital magnetic moment.
The state of an electron in an atom is described by a set of four quantum numbers:
n = 1, 2, 3, ….
principle quantum number (energy – shells)
l = 0, 1, …, (n-1)
angular momentum quantum number (subshells)
ml = 0, 1, …, l
magnetic quantum number (orientation of subshells)
ms = 1/2
spin quantum number
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The angular momentum of the electron is quantized. Assume that there exists an external
magnetic field Bz directed in the +Z direction. Then the components of the angular
momentum in the Z direction can only have values given by
Lz  ml

h
2
and the orbital magnetic moment aligned with the magnetic field can only have values
pm  
eL
e

ml  uB ml
2m
2m
e
 9.27  1024 A.m 2 is called the Bohr magneton. The orbital magnetic
2m
moments of atoms are in the order of a Bohr magneton.
where uB 
When ever a charged particle has angular momentum, the particle will contribute to the
permanent magnetic dipole moment. The total orbital magnetic moment of an atom is
equal to a vector sum of the orbital magnetic moments of each electron. For atoms with
completely filled shells  zero orbital magnetic moment eg
n=2
l = 1 ml = -1, 0, +1
all states occupied  sum of all components of angular momentum cancel.
Transition elements
Incomplete filled inner shells  free atoms do have resultant orbital magnetic
moments.
Solid iron group – magnetic dipoles can not align  no major contribution to
magnetic properties.
Fe Z = 26
1s2 2s22p6 3s23p63d6 4s2
3d subshell incomplete
Orbital angular magnetic moment is not a major contribution to the magnetic properties
of materials.
p163
In a helium atom, one of its electrons is a d-state. The atom is placed in a
strong magnetic field B = 2.00 T. By how much does the magnetic field
change the energy of the electron in its d-state?
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SPIN MAGNETIC MOMENT
Another contribution to the magnetic moment is due to the rotational motion of the
electron. Classically we can regard an electron as a small ball of negative charge spinning
around its axis, hence it possesses spin angular momentum S . Assume that there exists
an external magnetic field Bz directed in the +Z direction. Then the components of the
spin angular momentum in the Z direction can only have values given by
1
2
The two states are often referred to as spin up and spin down  states.
S z  ms
ms  
The spin angular momentum produces a magnetic dipole moment
pm  
eS
m
and the spin magnetic moment aligned with the magnetic field can only have values
e
ml  2uB ml
m
pm   uB
pm  
Fe Z = 26
1s2 2s22p6 3s23p63d6 4s2
3d subshell incomplete
3d subshell (6 electrons)
    
Iron atom spin angular momentum +4
Metallic iron angular momentum +2.2  magnetic properties of iron.
The total magnetic moment of an atom is equal to the vector sum of the orbital magnetic
moments and the spin magnetic moments of all its electrons. The contribution of the
nuclear magnetic moment is small and often can be neglected. Each atom acts like a
magnetic dipole and produces a small, but measurable magnetic field.
Paramagnetism - even though each atom in a material can have a magnetic moment, the
direction of each dipole is randomly oriented and their magnetic fields average to zero. If
the material is immersed in an external magnetic field, the dipoles will tend to align
themselves with the field in order to minimize the torque exerted on them by the external
magnetic field (lowest energy). The atoms in the material will produce an extra magnetic
field in its interior that has the same direction as the external magnetic field. This increase
in strength of the magnetic field can be quantified in terms of the relative permeability of
the material B = r Bfree.
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Ferromagnetism - the alignment of the spins of some of the electrons in a ferromagnetic
material will increase the magnetic field in this material in much the same way as the
alignment of the orbital magnetic dipole moments of atoms increases the field strength in
a paramagnetic material. In a ferromagnetic material the degree of alignment of the
electron spins between neighboring atoms is high as a result of a special force that tends
to lock the spins of these electrons in a parallel direction. This force is so strong that the
spins remain aligned even when the external magnetic field is removed. Materials with
such properties are called permanent magnets. The force that is responsible for the
alignment of the electron spins occurs in only five elements:
Iron
Nickel
Cobalt
Dysprosium
Gadolinium
Although ferromagnetic materials will remain magnetized after the external magnetic
field has been removed, they can also be found in non-magnetized states. On a small
scale (domains with sizes of less than 0.1 - 5 mm) all spins will be perfectly aligned, on a
large scale the domains are oriented randomly, and the net magnetic field is equal to zero.
However, if the material is immersed in an external magnetic field, all dipoles will tend
to align along the external field lines, and the strong spin-spin force will keep the dipoles
aligned even after the external magnetic field has been removed. The increase of the
magnetic field in a ferromagnet can be very large. For iron, the increase in field strength
can be as large as 5000. The degree of alignment of the spins in a ferromagnetic
material after the external magnetic field has been removed depends on the temperature.
An increase in the temperature of the material will increase the chance of random
rearrangement of the magnetic dipoles. Above a certain temperature, called the Curie
temperature TC, the magnetism of the ferromagnet disappears completely and acts like a
paramagnetic material.
permanent magnets
T < Tc
carbon steel
alnico V
platinum-cobalt
high permeability
materials
iron
4% Si-Fe
Mu metal
Supermalloy
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Remanence
magnetism
Br (T)
1
1.25
0.45
Coercivity
HC (A.m-1)
4103
4104
2105
r (max)
Saturation
Bsat (T)
Coercivity
HC (A.m-1)
5103
7103
1105
8105
2.1
2.0
0.65
0.8
80
40
4
0.16
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Ferrimagnetic material – permanent dipoles with anti-parallel orientation of unequal
amounts of spin


Important material because electrical conductivity like a semiconductor or insulator.
DC resistivity many orders of magnitude greater than iron  eddy current problem of
preventing penetration of magnetic flux into material much less severe in ferrites than
iron. Ferrites used in transformer cores upto microwave frequencies.
Structure very important for the magnetic properties of ferrites
X Fe23+ O4where X  Fe2+
Co2+ Mn2+ Zn2+ Cd2+ Mg2+
p357
Under conditions of maximum magnetization, the dipole moment per unit
volume of cobalt is 1.5105 A.m-1. Assuming that this magnetization is due to
completely aligned electrons, how many such electrons are there per unit
volume? How many aligned electrons are there per atom ? The density of
cobalt is 8.9103 kg and the atomic mass is 58.9 g.mol-1.
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SOLUTIONS
s163
d-state l = 2
ml = -2 -1 0 1 2
The angular momentum of the electron is quantized. Assume that the external magnetic
field B = Bz directed in the +Z direction. The orbital magnetic moment of an electron is
given by
eL
e
pm  

ml   uB ml
2m
2m
e
 9.27  1024 A.m 2 is called the Bohr magneton
where uB 
2m
The potential energy of a magnetic dipole in a magnetic field is
U   pm B  ml uB B
Therefore, the energy level is split into 5 separate levels with the state ml = 0
corresponding to the original energy level value.
uBB = (9.2710-24)(2) J = 18.5410-24 J = (18.5410-24/1.60210-19) eV = 1.210-4 eV
The energy levels are
E5 = + 2uBB = + 2.410-4 eV
E4 = + uBB = + 1.210-4 eV
E3 = 0
E2 = - uBB = - 1.210-4 eV
E1 = - 2uBB = - 2.410-4 eV
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s357
magnetization
M = 1.5105 A.m-1
electrons aligned
nm = ? electrons.m-3
number density Co
nCo = ? atoms.m-3
aligned electrons per atom
N = nm / nCo = ?
density of Co
 = 8.9103 kg
molar mass Co
MCo = 58.910-3 kg
Avogadro’s number
NA = 6.021023 mol-1
uB 
e
 9.27  1024 A.m 2
2m
The magnetic dipole moment due to the electron spin
pm = uB = 9.2710-24 A.m2
M  nm pm
nm 
M
1.5  105

electrons.m-3  1.6  1028 electrons.m-3
pm 9.27  1024
nCo  
 6.02  1023 
M Co
 8.9  103  
atoms.m-3  9.1  1028 atoms.m-3
3 
NA
 58.9  10 
Total number of aligned electrons per atom is
N
nm 1.6  1028

 0.18
nCo 9.1  1028
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