Geology Reference
In-Depth Information
z
z
0.4
[Nm x 10
-28
]
H
= 0
f
0.3
θ
U
D
0.2
U
T
H
M
s
M
s
0.1
U
θ
[
°
deg]
0.0
θ
0
0
45
90
135
180
−
0.1
−
0.2
Fig. 6.2
Rotation of the SD grain magnetization after
application of an external field
H
, whose axis forms an
angle
¥
with the easy grain axis. The plot on the right
shows
of
the
rotation
angle ™.
The
curves
have
been
D 45
ı
,
V
D 1.25 10
22
m
3
,
M
S
D 1
traced assuming
¥
A/m,
H
D 0.1
A/m,
and
X
/
Z
D 0.71(
e
D 0.7).
In
this
D 20
ı
the
total
energy
(
dashed
line
)
as a function
example, the total energy minimum is attained for ™
a minimum at an equilibrium angle ™
0
.Forany
rotation angle ™, the total energy is given by
the sum of the potential energy associated with
the external field
H
(Eq.
6.1
) and the internal
demagnetizing energy
U
D
:
For ¥
D
0, Eq. (
6.13
) assumes the form:
H
M
S
.D
x
D
z
/
H
H
c
cos™
0
D
(6.14)
This equation has solutions only for
j
H
j
H
c
.
The quantity
H
c
is called
microscopic coercivity
or simply
micro
-
coercivity
of the SD grain.
For
j
H
j
>
H
c
there is only one minimum for
U
T
, thereby, the grain magnetization undergoes
an irreversible rotation to the unique stable
configuration. Now let us come back to
the Maxwell-Boltzmann distribution of the
magnetization directions for an assemblage of
spheroidal SD grains (Eq.
6.2
). In this instance
¥
D
0, thereby at any time
t
,for
n
grains, we have
that ™
D
0, while
N
-
n
grains have anti-parallel
alignment, so that ™
D
180
ı
. Taking into account
of the demagnetizing energy, the corresponding
energy levels for these states are:
1
2
0
V M
S
H
D
D
D
0
VM
S
H cos.¥
™/
U
T
.™;¥/
D
0
V M
S
H
2
0
VM
S
D
z
C
.D
x
D
z
/ sin
2
(6.11)
1
C
The plot of
U
D
shows that in absence of
external field the demagnetizing field provides an
energy barrier for the complete reversal of the
grain magnetization. To determine the equilib-
rium angle in (
6.11
), we simply set to zero the
first derivative of
U
T
:
LJ
LJ
LJ
LJ
D
0
D
0
VM
S
H sin .¥
™
0
/
@U
T
@™
0
D
1
2
0
VM
S
D
z
U
1
D
0
VM
S
H
C
1
2
0
VM
S
.D
x
D
z
/ sin 2
0
(6.12)
1
2
0
VM
S
D
z
C
U
2
DC
0
VM
S
H
C
(6.15)
Switching from state 1 to state 2 implies a
rotation ™
D
180
ı
. As shown in Fig.
6.3
, this tran-
sition requires passing the energy peak at ™
D
™
0
.
To rotate from state 1 to state 2, the thermal
energy must be in excess of U
12
D
U
max
-
U
1
.
Hence,
1
2
M
S
.D
x
D
z
/ sin 2
0
(6.13)
H sin .¥
0
/
D
