Coupling of eddy-current
and circuit problems
Ana Alonso Rodrı́guez*, ALBERTO VALLI*, Rafael Vázquez Hernández**
* Department of Mathematics, University of Trento
** Department of Applied Mathematics, University of Santiago de
Compostela
Coupling of eddy-currentand circuit problems – p.1/28
Time-harmonic eddy-current equations
Starting from Maxwell equations, assuming a sinusoidal
dependence on time and disregarding displacement
currents one obtains the so-called time-harmonic
eddy-current problem
(
curl H − σE = 0
in Ω
(1)
curl E + iωµH = 0 in Ω .
Coupling of eddy-currentand circuit problems – p.2/28
Time-harmonic eddy-current equations
Starting from Maxwell equations, assuming a sinusoidal
dependence on time and disregarding displacement
currents one obtains the so-called time-harmonic
eddy-current problem
(
curl H − σE = 0
in Ω
(1)
curl E + iωµH = 0 in Ω .
Here
H and E are the magnetic and electric fields,
respectively
σ and µ are the electric conductivity and the magnetic
permeability, respectively
ω 6= 0 is the frequency.
Coupling of eddy-currentand circuit problems – p.2/28
Time-harmonic eddy-current equations (cont’d)
[As shown in the previous talk, in an insulator one has
σ = 0, therefore E is not uniquely determined in that region
(E + ∇ψ is still a solution).
Some additional conditions ("gauge" conditions) are thus
necessary: as in the insulator ΩI we have no charges, we
impose
div(ǫE) = 0
in ΩI ,
(2)
where ǫ is the electric permittivity.]
Coupling of eddy-currentand circuit problems – p.3/28
Geometry
The physical domain Ω ⊂ R3 is a “box", and the conductor
ΩC is simply-connected with ∂ΩC ∩ ∂Ω = ΓE ∪ ΓJ , where ΓE
and ΓJ are connected and disjoint surfaces on ∂Ω (“electric
ports"). Notation: Γ = ΩC ∩ ΩI , ∂Ω = ΓE ∪ ΓJ ∪ ΓD ,
∂ΩC = ΓE ∪ ΓJ ∪ Γ, ∂ΩI = ΓD ∪ Γ.
ΓJ
ΓD
Ξ
Γ
ΓE
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Boundary conditions
We will distinguish among three types of boundary
conditions.
Coupling of eddy-currentand circuit problems – p.5/28
Boundary conditions
We will distinguish among three types of boundary
conditions.
Electric (Case A). One imposes E × n = 0 on ∂Ω.
Coupling of eddy-currentand circuit problems – p.5/28
Boundary conditions
We will distinguish among three types of boundary
conditions.
Electric (Case A). One imposes E × n = 0 on ∂Ω.
Magnetic (Case B). One imposes E × n = 0 on ΓE ∪ ΓJ ,
H × n = 0 and ǫE · n = 0 on ΓD .
Coupling of eddy-currentand circuit problems – p.5/28
Boundary conditions
We will distinguish among three types of boundary
conditions.
Electric (Case A). One imposes E × n = 0 on ∂Ω.
Magnetic (Case B). One imposes E × n = 0 on ΓE ∪ ΓJ ,
H × n = 0 and ǫE · n = 0 on ΓD .
No-flux (Case C) [Bossavit, 2000]. One imposes
E × n = 0 on ΓE ∪ ΓJ , µH · n = 0 and ǫE · n = 0 on ΓD .
Coupling of eddy-currentand circuit problems – p.5/28
Voltage and current intensity
When one wants to couple the eddy-current problem with a
circuit problem, one has to consider, as the only external
datum that determines the solution, a voltage V or a current
intensity I0 .
Coupling of eddy-currentand circuit problems – p.6/28
Voltage and current intensity
When one wants to couple the eddy-current problem with a
circuit problem, one has to consider, as the only external
datum that determines the solution, a voltage V or a current
intensity I0 .
Question:
Coupling of eddy-currentand circuit problems – p.6/28
Voltage and current intensity
When one wants to couple the eddy-current problem with a
circuit problem, one has to consider, as the only external
datum that determines the solution, a voltage V or a current
intensity I0 .
Question:
how can we formulate the eddy-current problems when
the excitation is given by a voltage or by a current
intensity?
Coupling of eddy-currentand circuit problems – p.6/28
Voltage and current intensity
When one wants to couple the eddy-current problem with a
circuit problem, one has to consider, as the only external
datum that determines the solution, a voltage V or a current
intensity I0 .
Question:
how can we formulate the eddy-current problems when
the excitation is given by a voltage or by a current
intensity?
This is a delicate point, as eddy-current problems, for the
two cases A and B, have a unique solution already before a
voltage or a current intensity is assigned!
Coupling of eddy-currentand circuit problems – p.6/28
Poynting Theorem (energy balance)
In fact one has:
Uniqueness theorem. In the cases A and B for the solution
of the eddy-current problem (1) the magnetic field H in Ω
and the electric field EC in ΩC are uniquely determined.
[Adding the "gauge" conditions, also the electric field EI in
ΩI is uniquely determined.]
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Poynting Theorem (energy balance)
In fact one has:
Uniqueness theorem. In the cases A and B for the solution
of the eddy-current problem (1) the magnetic field H in Ω
and the electric field EC in ΩC are uniquely determined.
[Adding the "gauge" conditions, also the electric field EI in
ΩI is uniquely determined.]
Proof. Multiply the Faraday equation by H, integrate in Ω
and integrate by parts: it holds
R
R
0 = Ω curl E · H + Ω iωµH · H
R
R
R
= Ω E · curl H + Ω iωµH · H + ∂Ω n × E · H .
Coupling of eddy-currentand circuit problems – p.7/28
Poynting Theorem (energy balance)
In fact one has:
Uniqueness theorem. In the cases A and B for the solution
of the eddy-current problem (1) the magnetic field H in Ω
and the electric field EC in ΩC are uniquely determined.
[Adding the "gauge" conditions, also the electric field EI in
ΩI is uniquely determined.]
Proof. Multiply the Faraday equation by H, integrate in Ω
and integrate by parts: it holds
R
R
0 = Ω curl E · H + Ω iωµH · H
R
R
R
= Ω E · curl H + Ω iωµH · H + ∂Ω n × E · H .
Remembering that curl HI = 0 in ΩI and replacing curl HC
with σEC , one has the Poynting Theorem (energy balance)
Coupling of eddy-currentand circuit problems – p.7/28
Poynting Theorem (energy balance) (cont’d)
R
ΩC
σEC · EC +
R
Ω iωµH · H
=−
R
∂Ω n × E
· H.
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Poynting Theorem (energy balance) (cont’d)
R
ΩC
σEC · EC +
R
Ω iωµH · H
=−
R
∂Ω n × E
· H.
The term on ∂Ω is clearly vanishing in the cases A and B. Coupling of eddy-currentand circuit problems – p.8/28
Poynting Theorem for the case C
In the case C, instead, since divτ (E × n) = −iωµH · n = 0
on ∂Ω, one has
E × n = grad W × n on ∂Ω ,
and therefore
R
R
− ∂Ω n × E · H = − ∂Ω H × n · grad W
R
= ∂Ω div(H × n) W
R
R
= ∂Ω curl H · n W = W|ΓJ ΓJ curl HC · n,
as curl HI = 0 in ΩI , and we have denoted by W|ΓJ the
(constant) value of the potential W on the electric port ΓJ
(whereas W|ΓE = 0).
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Poynting Theorem for the case C (cont’d)
In this case a degree of freedom is indeed still free
(either the voltage W|ΓJ , that will be denoted by V , or
R
else the current intensity ΓJ curl HC · n in ΩC , that will
be denoted by I0 ).
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Poynting Theorem for the case C (cont’d)
In this case a degree of freedom is indeed still free
(either the voltage W|ΓJ , that will be denoted by V , or
R
else the current intensity ΓJ curl HC · n in ΩC , that will
be denoted by I0 ).
We can thus conclude that the only meaningful boundary
value problem is the one with assigned no-flux boundary
conditions: the case C.
Coupling of eddy-currentand circuit problems – p.10/28
The case C: variational formulation
How can we formulate the problem when the voltage or
the current intensity are assigned?
[Alonso Rodríguez, Valli and Vázquez Hernández, 2009]
[Other approaches: Bíró, Preis, Buchgraber and Tičar,
2004; Bermúdez, Rodríguez and Salgado, 2005]
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The case C: variational formulation
How can we formulate the problem when the voltage or
the current intensity are assigned?
[Alonso Rodríguez, Valli and Vázquez Hernández, 2009]
[Other approaches: Bíró, Preis, Buchgraber and Tičar,
2004; Bermúdez, Rodríguez and Salgado, 2005]
This orthogonal decomposition result turns out to be useful:
each vector function vI can be decomposed as
vI = µ−1
I curl qI + grad ψI + αρI ,
where ρI is a harmonic field, namely, it belongs to the space
HµI (ΩI ) := {vI ∈ (L2 (ΩI ))3 | curl vI = 0, div(µI vI ) = 0,
µI vI · n = 0 on ∂ΩI } .
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The case C: variational formulation (cont’d)
The harmonic field ρI is
R known from the data of the
problem, and satisfies ∂ΓJ ρI · dτ = 1; moreover, if the
vector field vI Rsatisfies curl vI = 0, it follows qI = 0 and
therefore α = ∂ΓJ vI · dτ .
Coupling of eddy-currentand circuit problems – p.12/28
The case C: variational formulation (cont’d)
The harmonic field ρI is
R known from the data of the
problem, and satisfies ∂ΓJ ρI · dτ = 1; moreover, if the
vector field vI Rsatisfies curl vI = 0, it follows qI = 0 and
therefore α = ∂ΓJ vI · dτ .
In particular, setting HI = grad ψI + αI ρI , from the Stokes
Theorem one has
Z
Z
Z
HI · dτ = αI ,
HC · dτ =
curl HC · nC =
I0 =
ΓJ
∂ΓJ
∂ΓJ
hence
HI = grad ψI + I0 ρI .
(3)
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The case C: variational formulation (cont’d)
The harmonic field ρI is
R known from the data of the
problem, and satisfies ∂ΓJ ρI · dτ = 1; moreover, if the
vector field vI Rsatisfies curl vI = 0, it follows qI = 0 and
therefore α = ∂ΓJ vI · dτ .
In particular, setting HI = grad ψI + αI ρI , from the Stokes
Theorem one has
Z
Z
Z
HI · dτ = αI ,
HC · dτ =
curl HC · nC =
I0 =
ΓJ
∂ΓJ
∂ΓJ
hence
HI = grad ψI + I0 ρI .
(3)
We want to provide a "coupled" variational formulation, in
terms of EC in ΩC and of HI in ΩI .
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The case C: variational formulation (cont’d)
Inserting the Faraday equation into the Ampère equation in
ΩC we find
R
R
−1
C · curl wC + iω ΩC σEC · wC
ΩC µC curl E
R
(4)
−iω Γ wC × nC · HI = 0 .
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The case C: variational formulation (cont’d)
Inserting the Faraday equation into the Ampère equation in
ΩC we find
R
R
−1
C · curl wC + iω ΩC σEC · wC
ΩC µC curl E
R
(4)
−iω Γ wC × nC · HI = 0 .
Instead, the Faraday equation in ΩI gives
Z
Z
iω
µI HI · grad ϕI + EC × nC · grad ϕI = 0
ΩI
(5)
Γ
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The case C: variational formulation (cont’d)
Inserting the Faraday equation into the Ampère equation in
ΩC we find
R
R
−1
C · curl wC + iω ΩC σEC · wC
ΩC µC curl E
R
(4)
−iω Γ wC × nC · HI = 0 .
Instead, the Faraday equation in ΩI gives
Z
Z
iω
µI HI · grad ϕI + EC × nC · grad ϕI = 0
(5)
Γ
ΩI
and
iω
Z
ΩI
µ I HI · ρ I +
Z
Γ
EC × nC · ρI = V .
(6)
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The case C: variational formulation (cont’d)
Here we have to note that
R
R
× nI · ρI = ΓD grad W × nI · ρI
ΓD EI R
R
= ΓD divτ (ρI × nI )W + V ∂ΓJ ρI · dτ = V .
Coupling of eddy-currentand circuit problems – p.14/28
The case C: variational formulation (cont’d)
Here we have to note that
R
R
× nI · ρI = ΓD grad W × nI · ρI
ΓD EI R
R
= ΓD divτ (ρI × nI )W + V ∂ΓJ ρI · dτ = V .
Using (3) in (4), (5) and (6) one has
R
R
−1
σEC · wC
ΩC µCR curl EC · curl wC + iω ΩC R
(7)
−iω Γ wC × nC · grad ψI − iωI0 Γ wC × nC · ρI = 0
−iω
Z
EC ×nC ·grad ϕI +ω 2
ΩI
Γ
−iωQ
Z
Z
Γ
EC × nC · ρI + ω 2 I0 Q
Z
µI grad ψI ·grad ϕI = 0 (8)
ΩI
µI ρI · ρI = −iωV Q . (9)
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The case C: existence and uniqueness
If V is given, one solves (7), (8), (9) and determines EC ,
ψI and I0 (hence HC and HI ).
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The case C: existence and uniqueness
If V is given, one solves (7), (8), (9) and determines EC ,
ψI and I0 (hence HC and HI ).
If I0 is given, one solves (7), (8) and determines EC and
ψI (hence HC and HI ); then from (9) one can also
compute V .
Coupling of eddy-currentand circuit problems – p.15/28
The case C: existence and uniqueness
If V is given, one solves (7), (8), (9) and determines EC ,
ψI and I0 (hence HC and HI ).
If I0 is given, one solves (7), (8) and determines EC and
ψI (hence HC and HI ); then from (9) one can also
compute V .
Both problems are well-posed, namely, they have a unique
solution, since the associated sesquilinear form is coercive
(thus one can apply the Lax–Milgram Lemma).
Coupling of eddy-currentand circuit problems – p.15/28
The case C: existence and uniqueness
If V is given, one solves (7), (8), (9) and determines EC ,
ψI and I0 (hence HC and HI ).
If I0 is given, one solves (7), (8) and determines EC and
ψI (hence HC and HI ); then from (9) one can also
compute V .
Both problems are well-posed, namely, they have a unique
solution, since the associated sesquilinear form is coercive
(thus one can apply the Lax–Milgram Lemma).
Moreover, it is simple to propose an approximation method
based on finite elements, of "edge" type for EC in ΩC and of
(scalar) nodal type for ψI in ΩI . Convergence is assured by
the Céa Lemma.
Coupling of eddy-currentand circuit problems – p.15/28
The case C: existence and uniqueness
If V is given, one solves (7), (8), (9) and determines EC ,
ψI and I0 (hence HC and HI ).
If I0 is given, one solves (7), (8) and determines EC and
ψI (hence HC and HI ); then from (9) one can also
compute V .
Both problems are well-posed, namely, they have a unique
solution, since the associated sesquilinear form is coercive
(thus one can apply the Lax–Milgram Lemma).
Moreover, it is simple to propose an approximation method
based on finite elements, of "edge" type for EC in ΩC and of
(scalar) nodal type for ψI in ΩI . Convergence is assured by
the Céa Lemma. [However, an efficient implementation
demands to replace the harmonic field ρI with an easily
computable function.]
Coupling of eddy-currentand circuit problems – p.15/28
Physical interpretation
Note: the physical interpretation of equation (9) is that
Z
Z
− EC · dr + iω µI HI · nΞ = V ,
γ
Ξ
where γ = ∂Ξ ∩ Γ is oriented from ΓJ to ΓE , and nΞ is
directed in such a way that γ is clockwise oriented with
respect to it.
In other words, if it is possible to determine the electric field
EI in ΩI satisfying the Faraday equation, it follows that
Z
EI · dr = V ,
γ∗
where γ∗ = ∂Ξ ∩ ΓD is oriented from ΓE to ΓJ : hence (9) is
indeed determining the voltage drop between the electric
ports.
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Physical interpretation (cont’d)
This explains from another point of view why, when the
source is a voltage drop or a current intensity, it is not
possible to assume the electric boundary conditions
E × n = 0 on ∂Ω.
Coupling of eddy-currentand circuit problems – p.17/28
Physical interpretation (cont’d)
This explains from another point of view why, when the
source is a voltage drop or a current intensity, it is not
possible to assume the electric boundary conditions
E × n = 0 on ∂Ω.
In fact, in that case one would have
Z
EI · dr = 0 ,
γ∗
hence from (9)
R
R
R
iω Ξ µI HI · nΞ = V + γ EC · dr = V + γ∪γ∗ E · dr
R
= V + ∂Ξ E · dr ,
with ∂Ξ clockwise oriented with respect nΞ : due to the term
V the Faraday equation would be violated on Ξ!
Coupling of eddy-currentand circuit problems – p.17/28
Numerical results for the Case C
Coming back to the case C and to its variational formulation
(7), (8), (9), we use edge finite elements of the lowest
degree (a + b × x in each element) for approximating EC ,
and scalar piecewise-linear elements for approximating ψI .
Coupling of eddy-currentand circuit problems – p.18/28
Numerical results for the Case C
Coming back to the case C and to its variational formulation
(7), (8), (9), we use edge finite elements of the lowest
degree (a + b × x in each element) for approximating EC ,
and scalar piecewise-linear elements for approximating ψI .
The problem description is the following: the conductor ΩC
and the whole domain Ω are two coaxial cylinders of radius
RC and RD , respectively, and height L. Assuming that σ
and µ are scalar constants, the exact solution for an
assigned current intensity I0 is known (through suitable
Bessel functions), and also the basis function ρI is known,
thus from (9) one easily computes the voltage V , too.
Coupling of eddy-currentand circuit problems – p.18/28
Numerical results for the Case C (cont’d)
We have the following data:
RC
RD
L
σ
µ
ω
=
=
=
=
=
=
0.25 m
0.5 m
0.25 m
151565.8 S/m
4π × 10−7 H/m
2π × 50 rad/s
and
I0 = 104 A
or
V = 0.08979 + 0.14680i
[the voltage corresponds to the current intensity I0 = 104 A].
Coupling of eddy-currentand circuit problems – p.19/28
Numerical results for the Case C (cont’d)
The relative errors (for EC in H(curl; ΩC ) and for HI in
L2 (ΩI )) with respect to the number of degrees of freedom
are given by:
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Numerical results for the Case C (cont’d)
The relative errors (for EC in H(curl; ΩC ) and for HI in
L2 (ΩI )) with respect to the number of degrees of freedom
are given by:
Elements
DoF
eE
eH
eV
2304
1684 0.2341 0.1693 0.0312
18432
11240 0.1132 0.0847 0.0089
62208
35580 0.0750 0.0567 0.0048
147456
81616 0.0561 0.0425 0.0018
Coupling of eddy-currentand circuit problems – p.20/28
Numerical results for the Case C (cont’d)
The relative errors (for EC in H(curl; ΩC ) and for HI in
L2 (ΩI )) with respect to the number of degrees of freedom
are given by:
Elements
DoF
eE
eH
eV
2304
1684 0.2341 0.1693 0.0312
18432
11240 0.1132 0.0847 0.0089
62208
35580 0.0750 0.0567 0.0048
147456
81616 0.0561 0.0425 0.0018
Elements
2304
18432
62208
147456
DoF
1685
11241
35581
81617
eE
0.2336
0.1132
0.0750
0.0561
eH
0.1685
0.0847
0.0566
0.0425
eI 0
0.0274
0.0085
0.0041
0.0024
Coupling of eddy-currentand circuit problems – p.20/28
Numerical results for the Case C (cont’d)
On a graph: for assigned current intensity
0
10
−1
Relative errors
10
−2
10
Rel. error E
−3
C
10
Rel. error H
D
Rel. error V
y = Ch
y=Ch2
−4
10
3
10
4
10
Number of d.o.f.
5
10
Coupling of eddy-currentand circuit problems – p.21/28
Numerical results for the Case C (cont’d)
for assigned voltage
0
10
−1
Relative errors
10
−2
10
Rel. error E
−3
C
10
Rel. error H
D
Rel. error I
y = Ch
y=Ch2
−4
10
3
10
4
10
Number of d.o.f.
5
10
Coupling of eddy-currentand circuit problems – p.22/28
Numerical results for the Case C (cont’d)
A more realistic problem, considered by Bermúdez,
Rodríguez and Salgado, 2005, is that of a cylindrical
electric furnace with three electrodes ELSA [dimensions:
furnace height 2 m; furnace diameter 8.88 m; electrode
height 1.25 m; electrode diameter 1 m; distance of the
center of the electrode from the wall 3 m].
Coupling of eddy-currentand circuit problems – p.23/28
Numerical results for the Case C (cont’d)
A more realistic problem, considered by Bermúdez,
Rodríguez and Salgado, 2005, is that of a cylindrical
electric furnace with three electrodes ELSA [dimensions:
furnace height 2 m; furnace diameter 8.88 m; electrode
height 1.25 m; electrode diameter 1 m; distance of the
center of the electrode from the wall 3 m].
The three electrodes ELSA are constituted by a graphite
core of 0.4 m of diameter, and by an outer part of
Söderberg paste. The electric current enters the electrodes
through horizontal copper bars of rectangular section (0.07
m × 0.25 m), connecting the top of the electrode with the
external boundary.
Coupling of eddy-currentand circuit problems – p.23/28
Numerical results for the Case C (cont’d)
A more realistic problem, considered by Bermúdez,
Rodríguez and Salgado, 2005, is that of a cylindrical
electric furnace with three electrodes ELSA [dimensions:
furnace height 2 m; furnace diameter 8.88 m; electrode
height 1.25 m; electrode diameter 1 m; distance of the
center of the electrode from the wall 3 m].
The three electrodes ELSA are constituted by a graphite
core of 0.4 m of diameter, and by an outer part of
Söderberg paste. The electric current enters the electrodes
through horizontal copper bars of rectangular section (0.07
m × 0.25 m), connecting the top of the electrode with the
external boundary.
Data: σ = 106 S/m for graphite, σ = 104 S/m for Söderberg
paste, σ = 5 × 106 S/m for copper, µ = 4π × 10−7 H/m,
ω = 2π × 50 rad/s, I0 = 7 × 104 A for each electrode.
Coupling of eddy-currentand circuit problems – p.23/28
Numerical results for the Case C (cont’d)
The value of the magnetic "potential" in the insulator: the
magnetic field is the gradient of the represented function
(not taking into account the jump surfaces).
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Numerical results for the Case C (cont’d)
The magnitude of the current density σEC on a horizontal
section of one electrode.
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Numerical results for the Case C (cont’d)
The magnitude of the current density σEC on a vertical
section of one electrode.
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References
A. Alonso Rodríguez, A. Valli and R. Vázquez Hernández:
A formulation of the eddy-current problem in the presence
of electric ports. Numer. Math., 113 (2009), 643–672.
A. Bermúdez, R. Rodríguez and P. Salgado: Numerical
solution of eddy-current problems in bounded domains
using realistic boundary conditions. Comput. Methods Appl.
Mech. Engrg., 194 (2005), 411–426.
O. Bíró, K. Preis, G. Buchgraber and I. Tičar: Voltage-driven
coils in finite-element formulations using a current vector
and a magnetic scalar potential, IEEE Trans. Magn., 40
(2004), 1286–1289.
A. Bossavit: Most general ‘non-local’ boundary conditions
for the Maxwell equations in a bounded region. COMPEL,
19 (2000), 239—245.
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Additional references
A. Alonso Rodríguez and A. Valli: Voltage and current
excitation for time-harmonic eddy-current problem. SIAM J.
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