Read Back Signals in Magnetic Recording - Research Group Fidler

out in in
2 2 1
∂V
Numerical Methods
H − H = H (4.61)
to guarantee (4.50) and (4.53). Again we have an equivalent problem as in Section 4.2, but for
three dimensions. Analogously the field H 2 can be evaluated by
( − ′ )
1 r r n
H () r = ( ′ ) dA
π � ∫ H r
. (4.62)
′
2
4 ∂V
in
1
r−r 3
To save memory we only create an interaction matrix from boundary to boundary, and solve
the Laplace equation (4.60) again to calculate H 2 .
4.4.6 Example: Long Wire
To verify the program and the calculated fields, some relative simple models are simulated,
which can also be calculated analytically. The first example is a long straight Cu-wire with
length l = 200 mm and wire radius r = 2 mm . The conductivity of Cu is
7 -1 -1
σ= 5.961⋅10 Ω m . At least we have to specify the boundary conditions of the wire. At one
side the electric potential was set to V 1 = 0 . The boundary condition of the opposite end was
chosen in a way, that the total inflowing current has a value of I = 1 mA . With these
parameters the total resistance of the Cu-wire can be calculated: R = 267 µ Ω . Using Ohm’s
Law (2.1) we get for the electric potential V 2 = 0.267 µV . This analytically calculated value
matches well with that of the FEM calculation 2
0.278 µV
FEM
FEM
V = . The error ( 2
V is about
4.2% larger than V 2 ) is traced back to the mesh. The mesh has to approximate a cylindrical
surface with triangles. So the cross sectional area of the mesh model is smaller than that of the
realistic model. This leads to a larger resistance and further to a larger voltage.
I
2
The current density of the realistic model is j = = 79.58 A/m . The magnetic field of an
2
r π
infinite long straight wire with constant current can be easily calculated due to the symmetry
in respect of rotations around the wire’s axis. The only non-zero field component is that in
azimuthal direction and can be determined using Ampere’s Law (2.6).
� ∫ H⋅ ds= ∫j×
dA
(4.63)
∂A
A
54