Biomedical Engineering Reference
In-Depth Information
the second term on the right-hand side of Eq. (
A.49
) can be rewritten as
r
)
×
r
)
d
3
r
=
r
)
r
)
]
d
3
r
V
j
∇
ʦ
(
G
(
r
,
V
j
∇×[
ʦ
(
G
(
r
,
r
)
r
)
r
)
]
=
S
j
n
(
dS
×[
ʦ
(
G
(
r
,
r
)
r
)
×
r
)
=
S
j
ʦ
(
n
(
G
(
r
,
dS
,
(A.51)
r
)
where
S
j
indicates the surface of
V
j
, and
is the outward unit normal vector of
a surface element on
S
j
. We use the Gauss theorem to derive the right-hand side
of Eq. (
A.51
). Substituting Eq. (
A.51
)into(
A.49
), we can derive the following
Geselowitz formula [5]:
n
(
)
−
μ
0
4
1
(˃
j
−
˃
j
)
r
)
r
)
×
r
)
(
)
=
B
0
(
S
j
ʦ
(
(
(
,
,
B
r
r
n
G
r
dS
(A.52)
ˀ
j
=
˃
j
where
is the conductivity just outside
V
j
. Also, in Eq. (
A.52
),
B
0
(
r
)
is the magnetic
field for the infinite homogeneous conductor in Eq. (
A.40
).
A.2.4
Magnetic Field from a Homogeneous Spherical
Conductor
Here, we assume that
V
j
is spherically symmetric, and we set the coordinate origin
at the center of
V
j
.Wehave
e
r
−
μ
0
4
1
(˃
j
−
˃
j
)
r
)
r
)
×
r
)
·
B
(
r
)
·
e
r
=
B
0
(
r
)
·
S
j
ʦ
(
n
(
G
(
r
,
e
r
dS
,
(A.53)
ˀ
j
=
where
e
r
is the unit vector in the radial direction, which is defined as
e
r
=
r
/
|
r
|
.
Since
e
r
in the case of a spherically symmetric conductor, the second term is
equal to zero, and we have the relationship
n
(
r
)
=
B
(
r
)
·
e
r
=
B
0
(
r
)
·
e
r
.
(A.54)
This equation indicates that the radial component of the magnetic field is not affected
by the volume current and that the radial component is determined solely by the
primary current.
We next derive a closed-form formula for the magnetic field outside a spherically-
symmetric homogeneous conductor. The derivation is according to Sarvas [6]. The
