Post on 25-Feb-2016
description
Applied PhysicsDepartment
Fractional Domain Wall Motion
Wesam Mustafa Al-Sharo'a
Dr. Abdalla ObaidatMay, 23, 07
Ferromagnetic materials are divided into a number of small regions called domains. Each domain is spontaneously
magnetized to the saturation value, but the directions of magnetization of the various domains are different .
Applying an external magnetic field convert the specimen from multi-domain state into one state in which it is a single
domain
Domain walls are interfaces between regions in which the spontaneous magnetization has different directions, and then the magnetization must change direction by domain wall motion.
The equation of motion of the wall per unit area is:
HkXdtdXb
dtXdm s22
2
Where X describes the position of the wall
Theory
The fractional calculus is defined by the derivative of order of a function
The fractional equation of motion of one dimensional simple harmonic oscillator is:
Where is a non integer 0
)(tx
t
duuutx
dtdtxD
0
)()1(
1)(
)(2
2
tfkxdtxdb
dtxdm
1
Case one:
By using the Laplace transform relations, we get,
Let with:
)()( ttf
)}({}{}{2}{ 222
2
txdtxd
dtxd
222
2
2)21()(
SSCSXSX
21)( XXsX 2221 2)(
SS
SXSX
222
2
2 221)(
SSCSX
To simplified with the aid of the geometric series as :
Then rewriting and as follows
)2
1)(2()(
22
222
1
SSSS
SXSX
1X
1)(20,
)(21
)2(!!)!()1(
mmn
m
mn
mmnmn
SmnmnXX
Similarly
After that
0,
2)(2
)(22
2 !!)2()!()1()21(
mnmmn
mmmnmn
SmnmnCX
]1)(2
)21(1[]1)(2[!!
)2()!()1()(
0,
)(2)(2
tmmnmmnmn
XtmntX
mn
mmnmmnmmn
If and f (t) =0, the summation will vanish except if m=0
=
The exact solution when is
0
0
22
)12(!)!()1(
)(n
nnn
nnXtn
tX tXtX cos)( 0
1
teX
teteXtXtt
t1
1
)(0
11
)(
1)(
0 sin2
sincos)(
-Case two:
Following the same procedure as in case one, we get
The solution is
tconstf tan)(
)2(2)( 222
022
SSSfCSXSSX
])2)(2)(1)(2(
1)(2)2(1[
]1)(2[!!)()2()!()1(
)(
0
02
0,
)(2
mmnmmnXft
mmnt
mmnmnXtmn
tXmn
mmnmmn
-When , the above equation reduce to :
-And when
-
)cos()cos()( 20
20
0 tff
tXtX
0
1
)sin())((
2))((
sin2
cos)(
121
21
0
221
01
1
)(0
1)(
0
tef
ft
eXteXtX
t
tt
The equation of wall motion was solved by assuming , and then it was plotted for different values of ranging from 0.1 to 1 for and as shown in Figure (1) and Figure (2) respectively
:
Results and Discussions :
)()( ttf
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
Figure (1): shows X as a function t of for and various value of ranging from 0.1 to 1, with, and
1.0 srad /10 mX 1.00 10 f
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
Figure (2): shows X as a function of t for and various value of ranging from 0.1 to 1
5.0
The equation of wall motion was solved
Figure (3) and (4) has been plotted for different values
of ranging from 0.1 to 1 for and
0tan)( ftconstf
1.0 5.0
Figure (3): shows X as a function of t for and various value of Ranging from 0.1 to 1, with
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
1.0 srad /10
mX 1.00
mX 1.00 10 f
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
Figure (4): shows X as a function t of for and various value of ranging from 0.1 to 1
5.0
The following two figures show the variation of the equation when an impulse driving force is applied :
0.2 0.4 0.6 0.8 1
-1
-0.5
0.5
1
1.5
Figure (5) :Shows X(t) as a function of t for and = 0.1, 0.5, 0.7, 2 and 52.0
0.2 0.4 0.6 0.8 1
-0.5
0.5
1
Figure (6) :Shows X(t) as a function of t for and = 0.1, 0.5, 0.7, 2 and 57.0
Conclusion
It is concluded that the series solutions of the equation of motion of the wall is calculated by fractional analysis with a regular oscillatory behavior. The same results, when we affect two kinds of forced oscillator on the system,
and the Figures show the series solution as a function of for =0.1 , =0.5, and various of ranging from 0.1 to 1
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