48 CHAPTER 2 Magnetostatics: "Scalar Potential" Approach
constants ~0 and ~t I such that ~3
(18) ~t 0 < ~t(x) < ~t 1 a.e. in
E 3.
Remark 2.6. The two values in (17) are equal when b = ~th. But notice
we have there two different expressions of the energy, one in terms of h,
the other in terms of b. It's customary to call
energy
of a vector field b
(any vector field b, not necessarily the physical induction) the integral
1 -1 2
• 1 2
~t I bl , and
coenergyof
h the integral -~ ~t I hl . Note that
energy(b) + coenergy(h) > Y b. h, with equality only when b = ~th.
2.4 MODELLING: THE SCALAR POTENTIAL FORMULATION
At last, we found a problem that, first, is relevant to the situation, and
second, makes mathematical sense:
Given ~t and
j,
piecewise smooth, with ~t as in (18)and j with compact
support, find piecewise smooth fields b and h such that
~ b. grad qo = 0 V qo
~ Co~(E3 ),
(19) b = ~t h,
~ h. rot a - ~E3 J" a V a
~ ¢0~(E3 ).
Whether there is a solution and how to get it is another story, but at least
we have, for the first time so far, a
model.
2.4.1 Restriction to a bounded domain
For the moment, let us return to physics, and criticize this model on the
grounds of an element of the situation which has been neglected up to
now: the large value of ~t in the magnetic core M of the apparatus. A
look at Fig. 2.6, where M is the hashed region, shows that flux lines will
arrive almost orthogonally to the "magnetic wall" 3M (the boundary of
M). On the other hand, if ~t is large, h must be small in M, since the
magnetic energy is finite. 14 We are thus entitled to neglect M in the
eventual calculation, and to set
13The abbreviation "a.e." stands for "almost everywhere", meaning "at all points except
those of some negligible set". (The latter notion is discussed in Appendix A, Subsection
A.4.2.) The a.e. clause is a necessary precaution since ~t has no definite value at discontinuity
points.
2.4 MODELLING: THE SCALAR POTENTIAL FORMULATION
49
nx h=0 on 3M
as a boundary condition for a problem that will now be posed in the
complementary domain E 3 - M.
x s"0
• ¢ ,N'q
" '
7D°:
S h
h
S1
FIGURE 2.6. Left: Expected pattern of field lines inside the box, showing the
existence of a horizontal plane on which n . b = 0, an annular part of which,
called S b, will close the box. Right: Persective view of the "computational
domain" D thus delimited, and of its surface. One has S = S h w S b and the
"magnetic wall" S h is in two parts, sh0 and shl.
But one can go farther here, and restrict the domain of interest to the
"central box" of Fig. 2.1, the experimental volume. Fig. 2.6 explains why:
The air region E 3 - M is almost cut in two by the magnetic circuit, and
between the North and South poles of the electromagnet, there is an air
gap in which the flux lines go straight from one magnetic wall to the
opposite one, horizontally. So we can introduce there an artificial boundary
(S b in Fig. 2.6), horizontal, which one can assume is spanned by flux
lines (this is only approximately true, but a legitimate approximation),
and therefore
n.b=0 onS b.
Consequently, let us restrict our computational domain to the part of the
inner box below the plane of S b, and call D this region. Its boundary S
is thus made of S b and of the part of 3M which bounds the inside of the
box, which we shall denote by S h. Hence our boundary conditions,
which, combined with the strong form of the magnetostatics equation,
lead to
14Be wary of this line of reasoning, which is correct in the present case, but can lead to
unexpected trouble in some topologically complex situations [Bo].
50 CHAPTER 2 Magnetostatics: "Scalar Potential" Approach
(20) roth=0 in D, (21) nxh=0 on S h,
(22) b = gh in D,
(23) divb=0 in D, (24) n.b=0 on S b.
We note that S h is in two parts, sh0 and sh~, corresponding to the two
poles of the electromagnet.
This calls for a few remarks. First, let's not forget that materials of
various permeabilities can be put inside D, so we must expect discon-
tinuous fields, and weak formulations are still in order.
The second remark is about the symmetry of the box, and of its
content. In the "Bath cube" experiment, four identical aluminum cubes
(hence the nickname) were put inside the box, symmetrically disposed,
so that it was possible to solve for only a quarter of the region, since the
whole field is then symmetrical with respect to the vertical symmetry
planes, hence n. b = 0 there. The equations are thus the same, provided
D and S b are properly redefined: D as a quarter of the cavity, and S b
as a quarter of the former S b plus the part of the symmetry planes
inside the box. We shall do that in Chapter 6, but we may ignore the
issue for the time being.
Third remark, the above display (20-24) does not say anything about
the source of the field. That was j, the current density in the coil, which
is now out of the picture. This lost information must be reintroduced
into the formulation in some way.
~../; ~ ...7....:~ .(/////////////~
_ _ _ .t.,r~.~.l~,.~, _.
I
~IltllIlIlIIA
FIGURE 2.7. Left: Applying Amp6re's theorem to the path y shows that the
mmf along c is approximately equal to the DC intensity I. Right: Topological
aspects of the situation.
Figure 2.7 suggests how it can be done. Consider a circuit y which,
except for the part c inside D that links opposite poles, is entirely
contained in M. By Amp6re's theorem, the circulation of h along y is
equal to I, the DC intensity in the coil. Is But g being very large in M,
2.4 MODELLING: THE SCALAR POTENTIAL FORMULATION 51
the field h is so small there that the circulation along y is approximately
equal to the circulation along the sub-path c. Since we already assumed
h = 0 in M in this modelling, we consistently set
(25) ~cZ'h=I,
where z is the field of unit tangent vectors along h c (Fig. 2.7).
(Exercise
2.5:
Show that any path c from sh0 to S 1 will give the same
circulation.) Now, common sense says that (20-24) and (25) do uniquely
determine the field, and the mathematical model we are building had
better have this property (which we'll eventually see is the case).
There is another possibility: We could specify the magnetic flux F
through the box instead, like this:
(26)
fS1 h n. b = F.
(Exercise 2.6:
Show that other surfaces
than
Shl can be used in (26) with
the same result. How would you characterize them?) Of course, F is
not known here, but this is not important for a
linear
problem: Just solve
with some value for F, get I, and scale. In fact, since we want to
compute the
reluctance
of the system, which is by definition the ratio R =
I/F, the flux is the objective of the computation if I is known, and the
other way around. We may thus solve (20-24) (25) and then compute F,
using some approximation of formula (26), or solve (20-24) (26), with an
arbitrary nonzero value for F, and then compute I by (25). This alternative
reflects the symmetry between b and h in the problem's formulation.
We shall return to this symmetry (Chapter 6). We now break it by
playing the obvious move in the present situation, which is to introduce
a magnetic potential.
2.4.2 Introduction of a magnetic potential
Indeed, since the field h we want must be curl-free, it is natural to look
for it as the gradient of some function q0. The boundary condition n x h
= 0 on S h is then satisfied by taking q0 equal to a constant there. (This
is general: Magnetic walls are equipotentials for q0 in static contexts.)
Since S h is in two pieces, there are two such constants, one of which can
be 0. The other one must then be equal to I, after (25).
15Notice how the equality of intensities in the energizing coils is necessary in this
reasoning: Otherwise, we could not assume ~t infinite in M without contradiction. This
is a well-known difficulty of the theory of the transformer, which we shall ignore here.
52 CHAPTER 2 Magnetostatics: "Scalar Potential" Approach
All these considerations lead us to the definition of a class of
admissible
potentials: piecewise smooth functions q0, which satisfy all the a priori
requirements we have about q0 (finite energy, being equal to 0 or I on
sH), and we shall select in this class
the
potential which solves the problem.
This is, still grossly sketched,
the functional point of view:
Define a functional
space of eligible candidates, characterize the right one by setting tests it
will have to pass, and hence an
equation,
which one will have to solve,
exactly or approximately.
To define admissible potentials, let's proceed by successive reductions.
First, a broad enough class:
= {all q0's piecewise smooth (over the closure of D)}.
(If D was not, as here, bounded, we should add "such that ~D I grad
q) i 2
is finite", in order to take care of the finite energy requirement. This is
implicit in the present modelling, but should be kept in mind.) Next,
(27) cI)I= {all q0~ ~" q0= 0 on S h and q0= I on shl}
where I is just a real parameter for the moment. In particular, we shall
have cI)°= {q0 ~ cI). q) = 0 on sh}. If q0 is in
(I)I
for
some
value of I, it
means that n x grad q) = 0 on S h, and thus Eqs. (20) and (21) are satisfied
by h = grad q0, if q0 is any of these ~otentials. Last, we select the given
value of I, and now, if (p is in
this
cI), (25) is satisfied.
Eligible potentials thus fulfill conditions (20), (21), and (25). To deal
with the other conditions, we request b (= ~t grad q0) to satisfy (23) by
using the weak solenoidality condition. But since
the set of test functions is
left to our choice,
we may do better and also check (24), all in one stroke:
Proposition 2.3.
If (p ~ ~i is such that
(28)
fD ~t
grad q0. grad q0'= 0 for all test functions q0' in ~0,
then the field
b = ~tgrad q0
verifies
(23)
and
(24).
(Pay attention to the notational shift: Since from now on we shall have
the eligible potentials on the one hand, and the test functions on the
other hand, the latter will be denoted with a prime. This convention will
be used throughout.)
Proof.
Set b = ~t grad q0. This is a piecewise continuous field. Since ~0
contains C0°°(D), we have div b = 0 in the weak sense, as required. But
since there are test functions in cI) ° which do not belong to C0°°(D) (all
those that do not vanish on sD), the implications of (28) may not have
been all derived. Starting from (28), and integrating by parts with formula
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