Read Back Signals in Magnetic Recording - Research Group Fidler

Appendix B
Moreover the second integral can be transformed into a surface integral by applying the
theorem, proved in Appendix A
1 r± nε−r′ 1 1
Hr ( ± nε ) = × dV ′ + ( ′′ ) ×
d ′
4π ∫ j j r
4π
� ∫
A . (B.7)
r n r r± nε−r′ 3
±
1/2 1/2
V / Kη ( r) ± ε− ′
∂Kη
( r)
Now we are able to calculate the difference in (B.2). Due to the difference and ε→ 0 the first
integral and the rotund surface of the half sphere do not contribute ( η � ε ), which leads to
1 dA′ 1 dA′
Hr ( + nε) −Hr ( −nε ) = jr ( ′′ ) × − ( ′′ ) ×
π ∫ jr
+ ε− ′ π∫ . (B.8)
r n r r−nε−r ′
+ −
4 4
Cη( r) Cη(
r)
Here Cη( r ) denotes a circle with radius η and radius r. Derivation in direction n gives
∂ 1 εdA′
∂n4π C r+ nε−r′ ( Hr ( + nε) −Hr ( −nε ) ) =− ( jr ( ′′ + ) + jr ( ′′ −)
) ×
3
∫
η ( r)
. (B.9)
The substitution ρ= r−r ′ and the simultaneous consideration that n is perpendicular to
( − ′ )
r r lead to
∂ 1 ε
( Hr ( + nε) −Hr ( −nε ) ) =− ( jr ( ′′ + ) + jr ( ′′ −)
) × n
ρdρdϕ. (B.10)
3/2
∂n4π ρ +ε
Calculating the integral yields
2π
η
∫∫ 2 2
0 0(
)
∂ out in jr ( ′′ + ) + jr ( ′′ ) ⎛
− ε ⎞
( H () r − H () r ) = n×
⎜1− ⎟.
(B.11)
∂n 2 ⎜ 2 2
η +ε ⎟
⎝ ⎠
Now the limit can be calculated in (B.2)
∂
∂n
out in j( r′′ + ) + j( r′′
−)
H r − H r = n×
. (B.12)
2
( () () )
And finally, for η→ 0 we get the result
∂
∂n out in
( )
H () r − H () r = n× j() r . (B.13)
In case of an interface between two volumes V 1 and 2
interface j 1 and j 2 respectively, we have a jump for the normal derivation of H
V with different current densities at the
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