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an isolated circle inhomogeneity placed into crossed electric
E
and magnetic
B
0
fields. The total dipole moment of the ensemble of such inhomogeneities
is easily derived. Let us suppose that there are
N
1
inhomogeneities of radius
a
and conductivity
σ
1
,
and
N
2
inhomogeneities of radius
b
and conductivity
σ
2
.
Assuming that the whole region is occupied by non-intersecting inhomo-
geneities and denoting
x
1
=
N
1
a
2
,
x
2
=
N
2
b
2
,
we see that
x
1
and
x
2
must
satisfy
x
1
+
x
2
=1
.
The total polarization caused by the inhomogeneities should vanish. After
some algebraic manipulations with expressions in brackets of (10.2), we can
write a system of equations to find
σ
e
P
and
σ
e
H
:
σ
e
P
2
+
σ
e
H
−
σ
Hi
2
σ
Pi
−
x
j
=0
,
(10.35)
σ
e
P
+
σ
Pi
2
+
σ
e
H
−
σ
Hi
2
i
=1
,
2
σ
e
H
σ
Hi
−
x
j
=0
.
(10.36)
σ
e
P
+
σ
Pi
2
+
σ
e
H
−
σ
Hi
2
i
=1
,
2
Here
j
=2
,
1 respectively for
i
=1
,
2.
The 3D case of isotropic media (
σ
H
1
=
σ
H
2
= 0) was studied by Landauer
[20] for spherical inhomogeneities.
Figure 10.3 shows the dependency of the ratio of effective Pedersen con-
ductivity
σ
e
P
on the magnetization
parameter
β
e
.
The conductivities
σ
P
1
(
σ
H
1
)and
σ
P
2
(
σ
H
2
) refer to the local
Pedersen (Hall) conductivities of the inhomogeneities of the 1-st- and 2-nd
kind, respectively. One of the curves with
x
1
=0
.
5 relates to the case when
the areas of the two components are equal. The curve with
x
1
=0
.
1shows
σ
e
P
to average Pedersen conductivity
σ
P
in which 10% of the whole area is occupied by a highly
conductive component with
σ
1
,
whereas the rest of the mixture is
σ
2
=0
.
9
σ
1
.
The effective conductivity is 5 times larger than the average conductivity, even
when the medium is weakly perturbed. The ratio
σ
e
P
(
β
e
)
/
σ
P
(
β
e
)
(
β
e
)
/<σ
P
(
β
e
)
>
can
be broadly estimated from
σ
e
P
(
β
e
)
∼
β
e
ε,
(10.37)
σ
P
(
β
e
)
where
ε
is defined in (10.1).
Using the same approach used to derive (10.3) for an isolated inhomogene-
ity, we can construct 'an effective bounded medium' in a strong magnetic field.
By virtue of the fact that in this case the dependency on magnetic field drops
out, for
we can use (10.35) putting
σ
Hi
=
σ
e
H
σ
eff
=0
.
Equation (10.35)
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