Biomedical Engineering Reference
In-Depth Information
4.2.8
Detailed Theory of Inductive Heating
In the previous section, electric power loss in inductive heating has been
explained on the basis of the ohmic loss produced by eddy currents that arise
in a cylindrical conductor surrounded by a coil. In this section, we shall
describe in detail the method for deriving the electric power loss in the form
of Eqn. (4.11). We consider the case where a sinusoidal alternating magnetic
field is generated by a coil wound around the cylindrical conductor, as in
Figure 4.14
b
, while a high-frequency current
I
0
flows through this coil. In the
present analysis, we consider the ideal case where there is no leakage mag-
netic field, assuming that the coil is closely wound around the conductor. It is
assumed that only the axial
z
component
H
z
of the magnetic field arises in this
case and that
H
z
is distributed circumferentially and symmetrically with
respect to the conductor.
We first solve a magnetic field equation derived from Maxwell's equation
for the magnetic field. Second, the eddy current density of the cylindrical con-
ductor is calculated using the electric field derived from the magnetic field.
Finally, the total electric power loss based on the ohmic loss produced by the
eddy current is obtained. We use cgs units such as
f
(Hz),
A
(cm
2
), and
H
a
(Oe)
because of the small size of the conducting cylinder.
The alternating magnetic field is expressed as
HHe
Z
=
(4.37)
jt
w
0
where w is the angular frequency. This magnetic field must satisfy the follow-
ing equation derived from Maxwell's equations:
—
H
=
j
w psm
4
H
(4.38)
2
expressed in cylindrical coordinates (
r
, j,
z
). When taking the symmetrical dis-
tribution of the EM field in the circumferential direction of the cylindrical con-
ductor into consideration, the following expression is obtained:
∂
2
H
r
1
∂
∂
H
r
0
+
0
-
j
w psm
4
H
=
0
(4.39)
0
∂
r
This equation appears as a Bessel differential equation with respect to zero
order when defining g
2
=-
j
pmws:
dH
dr
2
1
dH
dr
0
+
0
+
g
H
=
0
2
0
r
dH
dr
2
1
dH
dr
0
2
+
()
+=
0
H
0
(4.40)
0
g
r
g
()
g
Therefore, the solution is the linear combination of Bessel functions of the first
and second kinds:
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