Read Back Signals in Magnetic Recording - Research Group Fidler

Appendix B
Appendix B
To calculate the magnetic field of conductive materials (4.50) has to be solved:
Δ H =−curl
j. (B.1)
In addition boundary conditions are needed to make this problem solvable. The derivation of
the boundary conditions for a surface point r is given here. We have to calculate
∂ ∂
∂n ∂n
out in
( H () r − H () r ) = lim ( H( r+ nε) −H( r−nε) )
ε→0
. (B.2)
n denotes the unit vector normal to the boundary face in point r. We start from the Biot-
Savart Law (2.46)
1 r± nε−r′ Hr ( ± nε ) = ×
4π ∫ j
V r± nε−r′ Now we separate a half sphere
Volume V.
3
dV ′ . (B.3)
1/2
Kη () r with radius η (η � ε ) and midpoint r from the
1 r± nε− r′ 1 r± nε−r′ Hr ( ± nε ) = × dV ′ + × dV ′
4π ∫ j
4π
∫ j
3 3
1/2 1/2
V / Kη ( r) r± nε− r′ Kη
( r)
r± nε−r′ (B.4)
In this half sphere the current density j is assumed to be constant. If j is a continuous function,
1/2
the mean value theorem tells us that there exists a point r ′′ ∈ K for which
± η
1 r± nε− r′ 1 r± nε−r′ Hr ( ± nε ) = × dV ′ + ( ′′ ) ×
dV ′
4π ∫ j jr
4π
∫
. (B.5)
3 ±
3
1/2 1/2
V / Kη ( r) r± nε− r′ Kη
( r)
r± nε−r′ The integrand of the second integral can be expressed as gradient of a scalar term
1 r± nε−r′ 1 1
Hr ( ± nε ) = × dV ′ + ( ′′ ) × ∇′
dV ′
4π ∫ j jr
4π
∫
. (B.6)
r± nε−r′ 3
±
1/2 1/2
V / Kη ( r) r± nε−r′ Kη
( r)
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