Chapter 3
Magnetostatics
3.1
Introduction
Magnetostatics is a branch of electromagnetic studies involving magnetic …elds produced by steady
non-time varying currents. Evidently currents are produced by moving charges undergoing translational motion. An e¤ective current (called magnetization current) is also produced if magnetic
dipoles are nonuniformly distributed. To realize steady currents, a large number of charges must
be involved so that collection of charges can be regarded as a continuous ‡uid. In non-time varying
cases (@=@t = 0), the charge conservation principle requires that
@
+ r J = r J =0; in steady state.
@t
This implies that a steady current is transverse. A static magnetic …eld can also be produced by a
nonuniform magnetization (magnetic dipole density in the case of a collection of magnetic dipoles)
M (A/m) which produces a magnetization current
JM = r
M; (A m
2
):
The magnetization current is transverse (divergence-free), since r JM = r (r M) 0:
The magnetic …eld B exerts a force perpendicular to the velocity of charged particles. The force
to act on a charge q is
F = q (E + v B) :
The magnetic …eld does not do any work on charged particles since the rate of work v F is
independent of B;
v F = qv E:
(Note that v (v
B) =0:) In contrast to the electric …eld which is a true (or polar) vector, the
1
magnetic …eld is a pseudo (or axial) vector, or more precisely, constitutes a pseudotensor,
2
0
6
ij
B = 4 Bz
By
Bz
0
Bx
3
By
7
Bx 5 :
0
The magnetic …eld due to a prescribed current distribution can be calculated using the BiotSavart’s law. This law is generic because all other laws in magnetostatics, such as the Ampere’s law
and vanishing divergence of the magnetic …eld r B = 0; follow from it. Calculation of the magnetic
…eld is facilitated by introducing a magnetic vector potential A; which is related to the magnetic
…eld through B = r A: This expression is consistent with r B = 0; since r (r A)
0
identically. It also follows that r A = 0 in static cases.
One important objective of magnetostatics is to derive formulae for the self-inductance and
mutual inductance for given current con…gurations. Use of the vector potential greatly facilitates
calculation of the magnetic ‡ux,
Z
Z
I
= B dS = r A dS = A dl;
and inductance,
L=
3.2
I
:
Biot-Savart’s Law and Vector Potential
Figure 3-1: Geometry to illustrate Biot-Savart’s law.
2
For a given, non-time-varying current density distribution J(r); the magnetic …eld B(r) can be
calculated from the Biot-Savart’s law,
B(r) =
0
4
Z
J(r0 )
jr
(r r0 ) 0
dV ; (T)
r0 j3
(3.1)
where
0
= 4
10
7
(T m A
= 4
10
7
(H m
1
1
)
);
is the vacuum magnetic permeability. As explained in Chapter 1,
introduced to de…ne 1 Ampere current. Noting
r
jr
r0
=
r0 j3
1
r
jr
r0 j
(3.2)
0
is an assigned constant
;
(3.3)
we may rewrite the law in the form
B(r) =
0
4
Z
r
J(r0 )
dV:
jr r0 j
(3.4)
Then, one of the Maxwell’s equations
r B = 0;
(3.5)
immediately follows because of the identity for an arbitrary vector …eld A;
r (r
A)
0:
Physical meaning of r B =0 is that there be no magnetic charges in contrast to the electric …eld
r E = ="0 ; and that the magnetic …eld is transverse without longitudinal component. Eq. (3.4)
implies that the magnetic …eld is a curl of a vector …eld de…ned by
A(r) =
0
4
Z
J(r0 )
dV 0 :
jr r0 j
(3.6)
This vector quantity is called the magnetic vector potential. In terms of the vector potential, the
magnetic …eld is thus given by
B(r) =r A:
(3.7)
The magnetic …eld given in Eq. (3.1) is a pseudo vector because it is invariant against coordinates inversion, r ! r: In contrast, the Coulomb electric …eld,
1
E (r) =
4 "0
Z
(r
3
r0 ) (r0 ) 0
dV ;
jr r0 j3
is a true (or polar) vector, because it changes sign on coordinate inversion.
As will be shown in Chapter 4, for time varying current density, the vector potential is to be
modi…ed by taking into account the retardation due to …nite propagation time of electromagnetic
disturbances,
Z
J(r0 ; t
) 0
A(r;t) = 0
dV ;
4
jr r0 j
where
=
jr
c
r0 j
:
B(r; t) =r A and r B = 0 continue to hold in general.
The curl of the magnetic …eld is
r
B=r
r
A = r (r A)
r2 A:
(3.8)
The divergence of the vector potential vanishes for static …elds, since
Z
1
r
J(r0 )dV 0
4
jr r0 j
Z
1
0
=
r0
J(r0 )dV 0
4
jr r0 j
Z
Z
J(r0 )
1
0
0
0
=
r0
dV
+
r0 J(r0 )dV 0
0
4
jr r j
4
jr r0 j
= 0;
0
r A =
where it is noted that
Z
r0
J(r0 )
jr r0 j
dV 0 =
I
J(r0 )
dS0 = 0;
jr r0 j
and that for steady current ‡ow, r J = 0: Therefore, for static magnetic …elds,
r
B=
r2 A:
r2
Z
4
1
(3.9)
However,
r A =
=
=
Z
0
2
4
0
4
J(r0 )dV 0
jr
r0 j
(r
r0 )J(r0 )dV 0
(3.10)
0 J(r):
This means that the vector potential for a prescribed current distribution J(r)
A(r) =
0
4
Z
4
J(r0 )
dV 0 ;
jr r0 j
is the particular solution for the vector Poisson’s equation
r2 A =
0 J(r);
(3.11)
0 J:
(3.12)
which is identical to the familiar Ampere’s law
r
B=
In summary, the Biot-Savart’s law is generic in the sense other well known laws in magnetostatics
follow from it. In particular, we have derived the following vectorial relationships from the BiotSavart’s law:
r B = 0;
r B = 0 J;
B = r A;
r A = 0;
r2 A =
0J
absence of magnetic monopoles (always)
Ampere’s law for static …elds
always
for static …elds; always in Coulomb gauge (by de…nition)
static Ampere’s law in terms of the vector potential A
The Ampere’s law in Eq. (3.12) is valid only for static …elds. For more general time-varying …elds,
it is generalized as
@E
;
(3.13)
r B = 0 J + "0
@t
where
D = "0
@E
; (A/m2 )
@t
(3.14)
is the displacement current density originally conceived by Maxwell. The divergence of Eq. (3.13)
yields (note that "0 r E = )
r (r
B) = 0 =
0
r J+
@
@t
;
that is, the Maxwell’s equation in Eq. (3.13) is consistent with the charge conservation law,
@
+ r J = 0:
@t
(3.15)
The divergence of the vector potential A can be assigned an arbitrary scalar function without
a¤ecting the electric and magnetic …elds. However, for static …elds, the divergence of the vector
potential identically vanishes. For time varying …elds, the following choice is often made,
r A=
where
1 @
; Lorentz gauge
c2 @t
(3.16)
is the scalar potential. As we will see, this choice for r A allows complete decoupling
5
between
and A in the sense that the potentials satisfy similar inhomogeneous wave equations,
r2
1 @2
c2 @t2
=
1
(r; t);
"0
(3.17)
r2
1 @2
c2 @t2
A=
0 J(r; t):
(3.18)
If the Coulomb gauge, characterized by r A = 0 (absence of longitudinal vector potential), is
used instead, such decoupling cannot be achieved. In Coulomb gauge, the scalar potential C and
transverse vector potential A? satisfy the following,
r2
C
1 @2
c2 @t2
r2
1
(r; t);
"0
=
where J? is the transverse current. Solution of
(r; t) =
0 J? (r; t);
A? =
C
1
4 "0
Z
(3.19)
(3.20)
(r; t) is non-retarded,
(r0 ; t)
dV 0 ;
jr r0 j
(3.21)
and so is the longitudinal electric …eld,
EC (r; t) =
r
C
=
1
4 "0
Z
(r
r0 ) (r0 ; t) 0
dV :
jr r0 j3
(3.22)
As we will see in Chapter 4, this non-retarded electric …eld is cancelled and observable electric …eld
is in fact always retarded. Observable electromagnetic …elds, E and B; should be independent of
the choice of gauge.
In the vector Poisson’s equation for the vector potential,
r2 A =
0 J;
(3.23)
if the current density J is unidirectional, so is the vector potential and A k J holds. For example,
if only Jz exists, the resultant vector potential is also in the z-direction, Az : If the current density
is azimuthal J ; so is the vector potential, and only A exists. It is noted that in noncartesian
coordinates,
r2 A i 6= r2 Ai ;
(3.24)
6
in general. For example, in the spherical coordinates,
r2 A
r(r A)
r (r A)
2
2 @A
2 cot
2 @A
er
=
r 2 Ar
Ar
A
2
2
2
2
r
r @
r
r sin @
1
2 @Ar
2 cos @A
+ r2 A
e
2 A + r2 @
2
r sin
r2 sin2 @
2 cos @A
2 @Ar
1
+
e :
A
+
+ r2 A
2
2
r sin @
r2 sin
r2 sin2 @
(3.25)
If the current is purely azimuthal, J = J e ; the azimuthal component of the vector potential
satis…es
1
r2 A
A =
(3.26)
0J :
r2 sin2
In axially symmetric cases with @=@ = 0; elementary solutions for
1
A = 0;
r2 sin2
r2 A
are
A (r; ) = rl Pl1 (cos ) ;
1
rl+1
Pl1 (cos ) :
(3.27)
(3.28)
Then the lowest order harmonic is l = 1 (dipole) which is consistent with the absence of magnetic
monopoles.
3.3
Boundary Conditions
The vanishing divergence of the magnetic …eld r B = 0 requires that the normal component of
the …eld B be continuous at any boundary,
B1n = B2n :
(3.29)
The static Ampere’s law,
r
B=
(3.30)
0 J;
indicates that the tangential component can be discontinuous in the presence of a surface current,
n
(B1
B2 ) =
0 Js ;
(3.31)
where n is a unit vector normal to the surface and Js (A/m) is the surface current density. The
surface current density Js is the total current density regardless of its origin, including translational
motion of charge (conduction current), collection of magnetic dipoles (magnetization current), etc.
As will be shown later, for practical applications, it is convenient to separate the conduction current
Jcond = v ( (r) the charge density and v = dr=dt the velocity of charge density) and rewrite the
7
Ampere’s law in terms of a vector H,
r
where
H=
H = Jcond ;
B
M; (A m
(3.32)
1
)
(3.33)
0
with M the magnetization vector or the magnetic dipole moment density. The boundary condition
for the tangential component of the vector H is
n
(H1
H2 ) = Jcs ;
where Jcs is the conduction surface current density having dimensions of A m
3.4
(3.34)
1.
Some Examples
In this section, vector potentials will be worked out for some simple current con…gurations.
Example 1 Ring Current
Figure 3-2: Ring current. Because of axial symmetry, …nding the vector potential on the y
plane will su¢ ce.
z
Consider a thin circular ring current I with radius a: The current is unidirectional in the
direction and the current density can be written as
J =I
(r
a)
a
8
(cos ) :
(3.35)
Therefore, the vector potential is also unidirectional and only the azimuthal component A is
nonvanishing. When Ar = A = 0; in Eq. (3.25), the component of the vector Poisson’s equation
r2 A =
(3.36)
0 J(r);
in the spherical coordinates is
1
r2
A =
r2 sin2
where
r2 =
0J
=
0I
@
@2
2 @
1
+
+
@r2 r @r r2 sin @
(r
a)
a
sin
@
@
(cos ) ;
(3.37)
;
(3.38)
is the scalar Laplacian. Note that @=@ = 0 because of axial symmetry of the problem. Since
bounded general solutions to the di¤erential equation
@2
2 @
1
@
+
+
@r2 r @r r2 sin @
@
@
sin
are
1
rl Pl1 (cos );
rl+1
1
r2 sin2
A (r; ) = 0;
Pl1 (cos );
(3.39)
(3.40)
the solution for A (r; ) may be assumed in the form
A (r; ) =
8 X
r
>
>
Al
>
>
a
>
< l
l
Pl1 (cos );
r<a
(3.41)
>
X
>
a
>
>
>
Al
:
r
l+1
Pl1 (cos ); r > a
l
where the continuity of A at r = a is required from the continuity in the radial component of the
magnetic …eld,
1 @
(sin A ):
(3.42)
Br (r; ) =
r sin @
The appearance of Pl1 (cos ); even though the problem is axisymmetric, is due to the nature of the
vector Laplace equation. Eq. (3.39) is formally identical to scalar Laplace equation with m = 1:
The coe¢ cient Al can be determined by multiplying Eq. (3.39) by Pl10 (cos ) sin and integrating
the result over = 0 to ;
X
l
Al
d2
2 d
+
2
dr
r dr
l(l + 1)
r2
gl (r)
Z
1
1
Pl1 ( )Pl10 d =
9
0I
(r
a)
a
Pl10 (0);
(3.43)
where
= cos and
8
>
>
>
<
gl (r) =
Noting
Z
1
1
Pl1 ( )Pl10 ( )d =
Pl1 (0) =
and the singularity at r = a;
>
>
>
:
r
a
l
a
r
l+1
;
r<a
(3.44)
; r>a
2 (l + 1)!
2l + 1 (l 1)!
8
l 1
>
>
>
< ( 1) 2
=
2
l(l + 1)
2l + 1
ll0 ;
l!!
; l odd
(l 1)!!
>
>
>
: 0;
d2
gl (r) =
dr2
ll0
(3.45)
(3.46)
l even
(2l + 1)
(r
a)
a
;
(3.47)
we …nally obtain
Al =
0I
Pl1 (0)
=
2l(l + 1)
0 I(
and
A (r; ) =
0I
1
X
1)
l 1
2
gl (r)
l=1
(l 2)!!
;
2(l + 1)!!
l = 1; 3; 5;
Pl1 (0) 1
P (cos ):
2l(l + 1) l
(3.48)
(3.49)
Note the absence of the monopole potential, l = 0:
Far away from the ring r
a; the vector potential is dominated by the dipole term (l = 1) as
expected,
mz
A (r; ) ' 0 2 sin ;
(3.50)
4 r
or in vector form,
m r
;
4
r3
0
(3.51)
m = a2 I ez ;
(3.52)
A =
where
is the magnetic dipole moment of the ring current,
Z
1
m =
r JdV
2
Z
1
=
( e )rJ r2 sin drd d
2
Z
(r a) (cos ) 3
I
=
(sin ez cos e )
r sin drd d
2
a
=
a2 I ez :
10
The vector potential of the ring current can alternatively be found from the following direct
integration,
Z
J(r0 )
0
A(r) =
dV 0 ;
(3.53)
4
jr r0 j
where for a …lamentary ring current, the volume integral can be replaced with line integral,
J(r0 )dV 0 = I dl = Ia d 0 e 0 :
Because of the axial symmetry, the vector potential can be evaluated at an arbitrary location and
we choose = =2; a point on the y z plane. Then only the x component of Iad 0 e 0 contributes
to the integral,
Z 2
sin 0
p
d 0;
A (r; ) = 0 Ia
2
2
4
r +a
2ar cos
0
where
cos
= cos cos
+ sin sin
2
= sin sin 0 :
By changing the variable from
0
to
cos
2
2
0
through
2 =
0
+
2
;
the integral can be manipulated into the following form,
A (r; ) =
0 Ia
p
r2
+
a2
1
1
(2
+ 2ar sin k 2
where
k2 =
r2
k 2 )K(k 2 )
4ar sin
;
+ a2 + 2ar sin
2E(k 2 ) ;
(3.54)
(3.55)
and K(k 2 ) and E(k 2 ) are the complete elliptic integrals of the …rst and second kind, respectively,
de…ned by
Z =2
Z =2 p
1
2
2
p
K(k ) =
d ; E(k ) =
1 k 2 sin2 d :
(3.56)
0
0
1 k 2 sin2
Using this result, the inductance of a thin conductor ring of loop radius a and wire radius b
with a
b can be estimated. The magnetic ‡ux enclosed by the ring current itself is
Z
I
=
B dS =
A dl:
(3.57)
S
C
11
Figure
H 3-3: A thin conductor ring. The magnetic ‡ux enclosed by the ring can be found from
= A dl where A = A is the vector potential on the surface of the ring.
On the inner surface of the wire, r = a b; and = =2: The argument of the elliptic functions k 2
approaches unity and we may approximate the elliptic functions by
lim K(k 2 ) = ln
p
k2 !1
4
1
= ln
k2
8a
b
lim E(k 2 ) = 1;
;
k2 !1
where
1
k2 = 1
4a(a b)
b)2 + a2 + 2a(a
(a
b)
'
b2
:
4a2
(3.58)
(3.59)
Then, the vector potential on the wire surface is
A '
0I
ln
2
8a
b
2 ;
(3.60)
and the magnetic ‡ux enclosed by the ring is
= 2 (a
b)A
8a
0 aI ln
b
'
2 :
(3.61)
The external self inductance of the ring is therefore given by
Lext =
I
=
0a
ln
8a
b
2 :
(3.62)
External inductance pertains to the magnetic ‡ux outside the conductor. Magnetic …eld also
exists in the conductor. For a straight conducting wire, the internal inductance per unit length is
Li
= 0 ; (H/m).
l
8
12
(3.63)
The internal inductance Li is de…ned by
1
Li I 2 = magnetic energy in the conductor.
2
(3.64)
For a straight wire of radius a carrying a current I uniformly distributed across the cross-section,
the magnetic …eld in the conductor can be found by using Ampere’s law,
2
2
B ( )=
B ( )=
0 2 I;
a
0I
:
(3.65)
):
(3.66)
2 a2
As will be shown, the magnetic energy density is
1
2
B 2 ; (J m
3
0
Then, the magnetic energy stored per unit length of the wire is
Z
0
a
1
2
B 2 ( )2
d =
0
1 0 2
I ; (J m
28
1
),
(3.67)
which is independent of the wire radius a:This de…nes the internal inductance per unit length,
Li
= 0 ; (H m
l
8
1
).
(3.68)
If the current density is not uniform as in strong skin e¤ect, the internal inductance must be
evaluated according to Eq. (3.64).
In the cylindrical coordinates ( ; ; z); the vector potential A ( ; z) satis…es
@2
1 @
@2
+
+
@ 2
@
@z 2
1
2
A ( ; z) =
0I
(
a) (z):
(3.69)
The solution for this equation can readily be found in terms of Fourier transform involving the
modi…ed Bessel functions I1 (k ) and K1 (k ) as
A ( ; z) =
0
8 R1
>
I1 (k )K1 (ka) cos(kz)dk;
Ia < 0
>
: R1
0
I1 (ka)K1 (k ) cos(kz)dk;
< a;
(3.70)
> a:
This expression or its alternative based on Laplace transform,
A ( ; z) =
0 Ia
2
Z
1
J1 (ka)J1 (k )e
kjzj
dk;
0
is useful in calculating the self-inductance of a uniformly wound solenoid.
13
(3.71)
Example 2 Rotating Charged Conducting Sphere
If a conducting sphere of radius a carrying a surface charge density rotates about its axis at
an angular frequency !, the current density becomes singular and is given by
J = !a (r
a) sin :
(3.72)
Solutions for the vector potential satisfying
Figure 3-4: Rotating charged conductor sphere.
2 @
1
@
@2
+
+ 2
2
@r
r @r r sin @
sin
@
@
1
A (r; ) =
r2 sin2
0
!a (r
a) sin ;
(3.73)
can be sought in the form
A (r; ) =
8
P
r
>
>
>
l Al
<
a
>
>
>
: P Al a
l
r
l
Pl1 (cos );
r<a
(3.74)
l+1
Pl1 (cos ); r > a
where Al is a constant. The radial function
gl (r) =
8
>
>
>
<
>
>
>
:
r
a
l
a
r
l+1
;
r>a
(3.75)
14
; r<a
has discontinuity in its derivative at the surface r = a and thus
d2
gl (r) =
dr2
The sin
present,
(2l + 1)
(r
a)
a
:
(3.76)
dependence of the singular current density requires that only the l = 1 harmonic be
P11 (cos ) = sin :
Therefore, the solution for the vector potential is
8 r
>
sin ;
r<a
>
>
< a
2
!a
A (r; ) = 0
>
3
>
a 2
>
:
sin ; r > a
r
The outer …eld is of pure dipole with an e¤ective dipole moment
!a4
!qa2
=
;
3
3
4
mz =
q = 4 a2 :
The exterior magnetic …eld is given by
B = r
A
m
z
0
(2 cos er + sin e ) :
4 r3
=
The interior magnetic …eld is uniform,
B =
=
2
0
!a
3
2
0
!a
3
ez
(cos er
sin e ); r < a:
(3.77)
The polar component of the magnetic …eld B is discontinuous at the surface with a jump
B (r = a + 0)
B (r = a
0) =
0
!a sin =
where
Js = !a sin ;
is the surface current density.
Example 3 Rotating Charged Sphere
15
(A/m)
0 Js ;
(3.78)
If a uniformly charged sphere of radius a and uniform charge density (C m
its axis at an angular frequency ! (rad s 1 ); the current density J (r; ) is
J (r; ) =
The vector potential satis…es
@2
2 @
@
1
+
+ 2
2
@r
r @r r sin @
@
sin
@
3)
rotates about
8
>
< ! r sin ; r < a;
>
:
0;
1
2
r sin2
(3.79)
r > a:
A (r; ) =
8
>
<
>
:
0!
r sin ; r < a;
(3.80)
0;
r > a:
In the region outside the sphere r > a where there is no current, the solution may be assumed to
be
X
a l+1 1
(3.81)
Pl (cos ); r > a:
A (r; ) =
Al
r
l
Interior solution may be sought in the form of a sum of general and particular solutions. The
particular solution can be readily found by assuming
A (r; ) = Br3 sin :
(3.82)
Substituting this into
@2
2 @
1
@
+
+ 2
2
@r
r @r r sin @
sin
@
@
1
A (r; ) =
r2 sin2
0!
r sin ;
(3.83)
we …nd the particular solution,
1
10
AP (r; ) =
r3 sin ;
0!
(3.84)
and the interior solution may be assumed to be
A (r; ) =
X
Bl
l
r
a
l
Pl1 (cos )
1
10
0!
r3 sin ;
r < a;
(3.85)
where the …rst term in the RHS is the general solutions. Continuity of the magnetic …eld at the
sphere surface requires that both A and @A =@r be continuous. Then, only the l = 1 component
is nonvanishing and we easily …nd
A1 =
1
15
0!
a3 ; B1 =
16
1
6
0!
a3 :
(3.86)
As in the preceding example, the …eld outside the sphere is pure dipole. This is due to the particular
angular dependence of the current density J _ sin : Comparing the outer vector potential
A (r; ) =
1
15
a5
0!
sin
;
r2
(3.87)
with the standard dipole …eld, we …nd the dipole moment of the rotating charged sphere
4
! a5 :
15
mz =
(3.88)
The reader should check this result agrees with the dipole moment directly calculated from
Z
1
m=
r J dV:
(3.89)
2 V
Example 4 Long Straight Currents
Let us assume a straight current I directed along the z axis. The cartesian component of the
vector potential Az satis…es scalar Poisson’s equation
r 2 Az =
0 Jz ;
(3.90)
Jz = I (x) (y):
(3.91)
where
The Poisson’s equation is mathematically identical to that for the scalar potential due to a long
line charge,
r2
=
"0
(x) (y);
which has a solution
( )=
where
=
p
2 "0
ln ;
x2 + y 2 is the radial distance in the cylindrical coordinates. By analogy, we …nd
0I
Az ( ) =
2
ln :
(3.92)
A resultant magnetic …eld is the familiar one,
B=r
Az =
17
0I
2
e :
(3.93)
For an in…nite array of identical z directed currents I located in the plane x = 0 at y = 0;
2a;
, the potential is given by
Az (x; y) =
=
=
0I
4
0I
4
0I
4
ln (x2 + y 2 )[x2 + (y
ln
x2
y2
2 x
a
1+
ln cosh
a;
a)2 ][x2 + (y + a)2 ]
x2
x2
1
+
(y a)2
(y + a)2
2 y
cos
+ constant,
a
y 2 (y
1+
a)2 (y + a)2
(3.94)
where the following closed forms of the in…nite products are used,
x2
1+ 2
y
x2
1+
(y a)2
y 2 (y
a)2 (y + a)2
= 1
cos
2 x
a
cosh
x2
1+
(y + a)2
=
1
2 y
a
cos
cos
2 y
a
constant.
2 y
a
;
(3.95)
(3.96)
Calculation of the magnetic …eld B(x; y) is left for exercise.
3.5
Inductance
The magnetic ‡ux enclosed by a closed current circuit is in general proportional to the current
I: The external inductance is de…ned by the ratio,
Le =
I
:
(3.97)
In linear systems without ferromagnetic materials, the inductance is a constant determined solely
by geometrical factors. The magnetic ‡ux can be calculated in terms of either the magnetic …eld
B or the vector potential A since
Z
Z
I
B dS = (r A) dS =
A dl:
(3.98)
=
S
S
C
As an example, consider a parallel wire transmission line with a common wire radius a and
center to center separation distance d: In the absence of skin e¤ect (uniform current density), the
vector potential outside the wires is given by
Az =
0I
2
ln
2
;
1;
2
> a:
(3.99)
1
Then, the magnetic ‡ux enclosed by the unit length of the transmission line is
I
1
=2
Az ( 1 = a; 2 )d ;
l
2
18
(3.100)
Figure 3-5: Parallel wire transmission line. The current in the conductors is assumed to be uniform
(no skin e¤ect).
where
is the azimuthal angle about the center of the current I; and
2
=
The integral needed is
1
2
Z
2
ln
0
"
p
d2 + a2
2
d
a
+1
2ad cos :
#
d
2 cos
a
(3.101)
d = 2 ln
d
a
:
(3.102)
Then the ‡ux per unit length is
l
0I
=
ln
d
a
; (Wb m
d
a
; (H m
1
)
(3.103)
and the external inductance per unit length is
Le
=
l
0
ln
1
).
(3.104)
This holds for an arbitrary ratio of d=a (> 1) as long as the current ‡ows uniformly in both wires
(no skin e¤ect). Each wire has an internal inductance of
Li
= 0 ; (H m
l
8
1
).
(3.105)
Therefore, the total inductance per unit length of the transmission line is
L
=
l
0
ln
d
a
+
1
; (H m
4
19
1
), (uniform current).
(3.106)
If the current is concentrated at the wire surface as in the case of strong skin e¤ect, the internal
inductance vanishes and the external inductance is modi…ed as
!
p
Le
d + d2 a2
0
; (H m 1 ), (strong skin e¤ect),
(3.107)
=
ln
l
2a
which is dual of the capacitance of parallel wire transmission line,
C
=
l
"0
p
d + d2
2a
ln
a2
! ; (F m
1
).
(3.108)
The similarity between the capacitance and inductance is due to the fact that the scalar potential
and the cartesian component of the vector potential Az both satisfy the scalar Laplace equation,
r2
r2 Az = 0;
= 0;
(3.109)
in the case of parallel wire transmission line.
Example 5 Inductance of a Solenoid
The inductance of closely wound solenoid of circular cross-section as shown in Fig.3-6 can be
found using the vector potential of a single current loop of radius a,
A ( ; z) =
0
8 Z
>
>
>
>
Ia <
Z
>
>
>
>
:
1
I1 (k )K1 (ka) cos(kz)dk;
< a;
0
(3.110)
1
I1 (ka)K1 (k ) cos(kz)dk;
> a:
0
Consider a solenoid of radius a and length l with a total number of windings of N: The vector
potential at arbitrary axial location z on the solenoid surface = a is
0 N Ia
A (a; z) =
l
Z
l
dz
0
0
Z
1
I1 (ka)K1 (ka) cos[k(z
z 0 )]dk:
(3.111)
0
A di¤erential segment of length dz has a number of windings N dz=l and the magnetic ‡ux linked
to the segment is
d
= 2 aA
=
0I
N
dz
l
Z l
Z 1
N 2 2
dz 0
I1 (ka)K1 (ka) cos[k(z
2a dz
l
0
0
z 0 )]dk:
(3.112)
z 0 )]dk;
(3.113)
The total ‡ux linked to the solenoid is therefore
=
0I
N
l
2
2a2
Z
0
l
dz
Z
0
l
dz 0
Z
0
20
1
I1 (ka)K1 (ka) cos[k(z
and the inductance is given by
L =
=
(3.114)
I
0
Z
2
N
l
2a2
l
dz
Z
l
dz 0
1
I1 (ka)K1 (ka) cos[k(z
z 0 )]dk; (H).
(3.115)
0
0
0
Z
This is exact as long as windings are dense enough so that only the azimuthal component of the
vector potential is present.
Figure 3-6: Solenoid with uniform winding density.
For a long solenoid l
a; we note
1
lim I1 (x)K1 (x) = ;
x!0
2
and
Z
1
cos[k(z
z 0 )]dk =
(z
(3.116)
z 0 ):
(3.117)
0
Then, the inductance reduces to the familiar form of the inductance of a long solenoid,
L'
0
N2
a2
=
l
0
a2 2
N ; (H) ; (long solenoid l
l
a).
(3.118)
An expression valid to order (a=l)2 can be worked out using the approximation that the magnetic
…eld is axial everywhere inside the solenoid. The magnetic …eld on the axis of a circular loop current
is
I
a2
Bz (z) = 0
:
(3.119)
2 (z 2 + a2 )3=2
21
For a solenoid of winding number density n, length l and radius a; the axial magnetic …eld is
2
0 nIa
Bz (z) =
2
l
dz 0
z)2 + a2 ]3=2
[(z 0
0
0 nI
2
=
Z
l
p
z
z
+p
2
z + a2
z)2 + a2
(l
!
:
(3.120)
For a long solenoid, the magnetic …eld inside the solenoid can be approximated by the axial …eld.
Therefore, the ‡ux linkage is
'
2
0n I
a2
2
=
0n
Z
l
p
0
a I
p
0
N
l
2
a2
0
N
l
2
a2
2 2
l
(l
z
z
+p
2
2
2
z + a2
z) + a
l2 + a2
!
dz
a ;
(3.121)
and the self-inductance is given by
L '
'
p
l2 + a2
l
a+
a ;
1 a2
2 l
l
a;
(3.122)
:
(3.123)
The exact inductance in Eq. (3.114) may be written in the form
L=
0
(aN ) 2
f (a=l);
l
where f (a=l) is the correction factor given by
a
f
l
=
=
Z
Z l
Z 1
2 l
0
dz
dz
I1 (ka)K1 (ka) cos[k(z
l 0
0
0
Z
a
a
8 1 sin2 (x=2)
I1
x K1
x dk:
2
x
l
l
0
z 0 )]dk
This had been tabulated by Nagaoka. The function f (a=l) is shown in Fig.3-7.
3.6
Neumann’s Formula for Mutual Inductance
The mutual inductance among electric circuits can be conveniently calculated using Neumann’s
formula. Consider two closed current loops C1 and C2 : To …nd the mutual inductance of the two
loops, we let the loop 1 carry a current I1 and calculate the magnetic ‡ux linked to the loop 2. The
vector potential created by the current I1 at a point on the loop 2 is
A=
0 I1
4
I
C
22
dl1
;
r12
(3.124)
1
0.8
0.6
0.4
0.2
00
0.5
1
x
1.5
2
Figure 3-7: Nagaoka factor f (a=l) for a solenoid of radius a and length l: In the …gure, x = a=l:
where r12 is the distance between the segment dl1 and the position on loop 2 as illustrated in
Fig.3-8. Then, the magnetic ‡ux enclosed by loop 2 is given by
I
A dl2
12 =
I I
dl1 dl2
0 I1
:
(3.125)
=
4
r12
The mutual inductance between the two loops is de…ned by
Figure 3-8: Geometry for two circuits mutual inductance (Neumann’s formula).
23
12
M12 =
I1
=
0
4
I I
dl1 dl2
:
r12
(3.126)
It is evident that the mutual inductance is reciprocal,
M12 = M21 ;
(3.127)
and depend only on the geometrical shapes and their mutual orientation.
Example 6 Coaxial Parallel Square Loops
Figure 3-9: Coaxial parallel square loops. The mutual inductance can be found from the sum of
contributions from all parallel and anti-parallel pairs.
The mutual inductance between two identical square loops, which are coaxial and whose sides
are parallel, can be calculated as a sum of mutual inductances of parallel wires. If two parallel
wires of length a are a distance d apart, the mutual inductance is
M
=
=
=
0
4
Z
a
dl1
0
0
2
Z
0
2
a
0
a
ln
0
"
Z
a ln
dl2 p
p
1
(l1
l2 )2 + d2
!
x2 + d2
dx
d
!
#
p
p
a + a2 + d2
a2 + d2 + d ; (H).
d
x+
(3.128)
Applying this results to all parallel pairs of the square loops, we …nd
M12 =
2
"
!
p
2 + d2
a
0
a ln
a2 + d2 + d
d
!
#
p
p
p
a + 2a2 + d2
p
a ln
+ 2a2 + d2
a2 + d2 ; (H).
a2 + d2
a+
p
Example 7 Mutual Inductance of Coaxial Circular Loops
24
(3.129)
Figure 3-10: Coaxial circular loops.
If two circular loops of radii a and b are coaxial and their planes are a distance c apart, the
mutual inductance can readily be found from the vector potential at the loop B due to a current I
in the loop A,
1
1
Ia
(2 k 2 )K(k 2 ) 2E(k 2 ) ;
(3.130)
A = 0 p
2
2
2
k
(a + b) + c
where
k2 =
4ab
:
(a + b)2 + c2
(3.131)
The magnetic ‡ux enclosed by the loop B is thus
= 2 bA
1
2 0 abI
(2
= p
(a + b)2 + c2 k 2
k 2 )K(k 2 )
2E(k 2 ) ;
(3.132)
2E(k 2 ) ; (H).
(3.133)
and the mutual inductance is
Mab =
1
2
0
p
k 2 )K(k 2 )
(a + b)2 + c2 (2
An alternative expression for the mutual inductance can be found in terms of spherical harmonic
p
expansion. The vector potential at the loop B is, if a < b2 + c2 ;
A (r =
p
b2 + c2 ;
0)
=
0I
X
l odd
p
where
cos
0
=p
25
a
2
b + c2
b
:
b2 + c2
l+1
Pl1 (0) 1
P (cos
2l(l + 1) l
0 );
(3.134)
(3.135)
If a >
p
b2 + c2 ;
A (r =
p
b2 + c2 ;
0)
=
0I
p
X
l odd
b2 + c2
a
!l
Pl1 (0) 1
P (cos
2l(l + 1) l
0 ):
(3.136)
The ‡ux linked to the loop B is therefore given by
= 2 bA (r =
p
b2 + c2 ;
0 );
(3.137)
and the mutual inductance is
Mab = b
0
X
l odd
and
Mab = b
0
X
p
a
2
b + c2
p
l odd
b2 + c2
a
l+1
!l
Pl1 (0) 1
P (cos
l(l + 1) l
Pl1 (0) 1
P (cos
l(l + 1) l
0 );
if a <
0 ); if a >
p
b2 + c2 ;
p
b2 + c2 :
(3.138)
(3.139)
Figure 3-11: Geometry for calculating the mutual inductance of two circular loops whose axes
intersect at an angle :
If two circular loops are not coaxial but their axes intersect as shown in Fig.3-11, the mutual
inductance can be calculated as follows. First, we note the vector potential due to a circular current
of radius a sin residing on a sphere of radius a is given by (see Eq. (3.49))
A (r; ) =
0I
X a
r
l
l+1
sin
P 1 (cos )Pl1 (cos ):
2l(l + 1) l
26
(3.140)
The radial magnetic …eld is
1 @
(sin A )
r sin @
X a l+1
0 I sin
Pl1 (cos )Pl (cos );
2r
r
Br =
=
(3.141)
l
where use is made of the identity,
1 d
[sin Pl1 (cos )] = l(l + 1)Pl (cos ):
sin d
(3.142)
The magnetic ‡ux linked to the loop B can be found by integrating the radial magnetic …eld over
the incomplete spherical surface at radius b: Denoting the angular location on the sphere by ( 0 ; 0 )
where 0 is measured from the axis of the loop B and 0 is the angle about it, we note that the
argument of the Legendre function Pl (cos ) becomes
cos = cos
0
cos + sin
0
sin cos
0
:
Then the Legendre function can be expanded as,
Pl (cos
0
cos + sin
0
sin cos ) = Pl (cos 0 )Pl (cos ) + 2
1
X
Plm (cos 0 )Plm (cos ) cos m ; (3.143)
m=1
and the magnetic ‡ux through the loop B is
= b
2
Z
d
0
2
d
0
0
=
Z
0 Ib sin sin
0
sin 0 Br (r = b)
X
l
where the integration over
1
a
l(l + 1) b
l+1
Pl1 (cos )Pl1 (cos )Pl (cos );
(3.144)
reduces to
Z
sin 0 Pl (cos 0 )d
0
=
0
sin
P 1 (cos ):
l(l + 1) l
(3.145)
The desired mutual inductance is
Mab =
0 b sin
sin
X
l
3.7
1
a
l(l + 1) b
l+1
Pl1 (cos )Pl1 (cos )Pl (cos ); (H).
(3.146)
Multipole Expansion of Static Vector Potential
The vectorial nature of the vector potential prohibits us from expanding the vector potential directly
in terms of the spherical harmonics in contrast to the case of scalar potential. However, the
vector potential can be indirectly expanded in spherical harmonic functions. In magnetostatics,
27
the Coulomb gauge r A = 0 allows us to write A in terms of another vector function F as
A=r
(3.147)
F;
since r (r F) 0 holds identically. The vector F introduced may be regarded as a generating
function for the vector potential and it may not have any physical meanings. The vector …eld F
can be decomposed into “radial”or “longitudinal”component, and “transverse”component in the
form
F= r +r r ;
(3.148)
where and are scalar functions. If the vector potential satis…es vector Laplace equation r2 A =
0; then and both satisfy scalar Laplace equation,
r2
r2 = 0;
= 0;
(3.149)
and thus can be expanded in spherical harmonics,
;
=
X
Alm rl +
l;m
Blm
rl+1
Ylm ( ; ):
(3.150)
The transverse function does not contribute to the magnetic …eld B: (Why? Note the identity
r (r r ) = 0.) Then, can de discarded and
A=r
r :
(3.151)
The di¤erential operator
r
r=
e
@
1 @
+e
;
sin @
@
(3.152)
is the well known angular momentum operator and operates only on the harmonic function Ylm ( ; ):
Evidently, Eq. (3.151) generates components A and A only, but the remaining radial component
Ar can be calculated from r A = 0: Note that the radial component Ar ; which exists in
r
(r
r );
does not contribute to the magnetic …eld.
Let us revisit the vector potential due to a ring current. Because of the axial symmetry, the
scalar function may be expanded in the form
(r; ) =
8 P
r
>
>
al
>
>
a
< l
>
>
P
a
>
>
:
al
r
l
l
Pl (cos );
r<a
(3.153)
l+1
28
Pl (cos ); r > a
and the vector potential can be calculated as
A = r
=
@
r =e
@
X
e
al gl (r)Pl1 (cos );
(3.154)
l
where
gl (r) =
and use is made of
8
>
>
>
<
r
a
l
a
r
l+1
;
r<a
(3.155)
>
>
>
:
d
Pl (cos ) =
d
; r>a
Pl1 (cos ):
(3.156)
The vector potential is therefore consistent with that worked out in Example 1.
If the current density J(r) is con…ned in a limited spatial region, the vector potential at a
su¢ ciently large distance can be Taylor expanded as
A(r) =
=
where r
0
4
Z
1
4 r
0
Z
J(r0 )
dV 0
jr r0 j
1
J(r )dV +
4 r3
0
0
0
Z
(r r0 )J(r0 )dV 0 +
(3.157)
r0 is assumed. The …rst term in RHS vanishes for static current, since
Z
Z
0
Ji dV =
rri0 JdV 0
Z
Z
0
0
=
r (ri J)dV
ri0 r JdV 0
= 0;
Note that for static currents, r J = 0 holds. The absence of monopole vector potential is consistent
with the absence of magnetic charges (magnetic monopoles), r B = 0: Therefore, the lowest order
far …eld vector potential is the dipole …eld,
Z
1
A(r) ' 0 3 (r r0 )J(r0 )dV 0 :
(3.158)
4 r
Using
r
(r0 J) = r0 (r J)
29
(r r0 )J;
and
Z
ri0 (r
J)dV
0
=
Z
ri0 rj Jj dV 0
= rj
= rj
=
the integral can be reduced to
Z
Z
= rj
Z
ri0 Jj dV 0
ri0 (rrj0 J)dV 0
Z
(ri0 rj0 J)dV 0
r
Z
rj rj0 Ji d0 =
1
2
(r r0 )J(r0 )dV 0 = r
Z
Z
rj
Z
rj0 (rri0 J)dV 0
(r r0 )JdV 0 ;
J(r0 )
r0 dV 0
(3.159)
and the dipole vector potential can be written in the form
A(r) =
0
4
m r
;
r3
(3.160)
where m is the magnetic dipole moment,
1
m=
2
Z
JdV; (A m2 ):
r
(3.161)
V
For a …lamentary loop current, the volume integral reduces to line integral,
JdV = Idl;
and the dipole moment can be found from the contour integral,
I
1
m= I
r dl = IS;
2 C
where
1
S=
2
I
r
dl;
(3.162)
(3.163)
C
is the area enclosed by the loop.
3.8
Magnetic Materials
If magnetic dipoles are continuously distributed, it is convenient to introduce a magnetic dipole
moment density,
M(r) = n(r)m(r); (A m 1 );
(3.164)
where n(r) is the number density of dipoles and m(r) is the dipole moment. Although individual
dipoles do not carry macroscopic current, a collection of many dipoles can e¤ectively produce
30
a macroscopic current. However, such an e¤ective current is divergence-free and thus does not
contribute to charge transfer. The vector potential in the presence of distributed magnetic dipoles
can be approximated by
A(r) =
0
4
Z
Jc (r0 )
dV 0 + 0
0
jr r j
4
Z
M(r0 )
jr
(r r0 ) 0
dV ;
r0 j3
where Jc (r) is the conduction current due to translational motion of charges. The second term in
RHS is the contribution from the distributed magnetic dipoles. Singling out the conduction current
Jc in magnetostatics is similar to separating the electric charges into free and bound charges in
dielectrics,
Z
Z
0
1
P(r0 ) (r r0 ) 0
1
f (r )
0
dV :
dV
+
(r) =
4 "0
jr r0 j
4 "0
jr r0 j3
Noting
r0
1
0
;
3 = r jr
0
r0 j
rj
r
jr
Z
M(r0 )
jr
Z
(r r0 ) 0
dV =
r0 j3
M(r0 )
Z
r0
1
jr
r0 j
dV 0
M(r0 )
r0 M(r0 )
dV 0
jr r0 j
jr r0 j
I
Z
M(r0 )
r0 M(r0 ) 0
=
dS+
dV
r0 j
jr r0 j
S jr
Z
r0 M(r0 ) 0
dV ;
=
jr r0 j
=
r0
(3.165)
we …nd that the curl of the magnetic …eld becomes
r
B=
r2 A =
r2
1
0 Jc
+
0r
M:
(3.166)
where use is made of the identity
jr
=
r0 j
4
(r
r0 ):
Rearranging Eq. (3.166) yields
r
B
M
= Jc :
(3.167)
0
It is customary to introduce a vector H de…ned by
H=
B
0
31
M;
(3.168)
and an alternative form of Ampere’s law is
r
H = Jc :
(3.169)
In vacuum and nonmagnetic media, M = 0; and
B=
holds. The permeability
0 H;
of linear magnetic materials is de…ned by
B = H:
Permanent magnets are highly nonlinear. In permanent magnets, there are no conduction current,
Jc = 0; and thus
r H = 0;
(3.170)
must holds. This means that the vector H can be assigned a scalar potential,
H=
r
m:
(3.171)
In permanent magnets, the two vectors B and H are in general oriented in opposite directions as
shown in the following simple example.
Example 8 Magnetized Sphere
If an iron sphere of radius a is uniformly magnetized with an axial uniform magnetization
M = M0 ez ;the …elds B and H can be found as follows. The scalar potential m satis…es the
Laplace equation,
r2 m = 0:
In the sphere r < a the uniform magnetization produces uniform magnetic …elds B and H;
Bz =
0 (Hz
+ M0 ); r < a;
(3.172)
where Bz and Hz are constants. Then the solution for the scalar potential may be assumed as
m (r;
)=
8
>
>
<
>
>
:
Hz r cos ;
r<a
a3
Hz 2 cos ; r > a
r
(3.173)
where continuity of m ; that is, continuity of the tangential component of H; H ; at the surface
r = a is taken into account. The continuity of the radial component of the magnetic …eld Br yields
0 (M0
+ Hz ) =
32
2
0 Hz ;
(3.174)
or
1
M0 :
3
Hz =
(3.175)
This is the intensity of the vector H inside the sphere. The interior magnetic …eld is
Figure 3-12: Spherical permanent magnet. The B pro…le is shown qualitatively in the left …gure
and H pro…le in the right …gure. Note the discontinuity in the H …eld lines at the surface.
Bz =
2
3
0 M0 :
(3.176)
The …eld outside the sphere is of dipole type,
B=
0H =
In the absence of conduction current, r
3
0 M0 a
3r3
(2 cos er + sin e ) :
H = 0 requires
I
H dl = 0;
(3.177)
(3.178)
along an arbitrary closed path. If the path intersects with the magnetized sphere, the vector H
H
must change its sign at the sphere surface in order to satisfy H dl = 0: The magnetic …eld B
found in this example is mathematically identical to that in Example 2 worked out for a rotating
charged conducting sphere. This is not surprising because an e¤ective current density associated
with a uniform magnetization indeed yields
Jef f
= r
M
= M0 (r
a) sin e ;
which is identical to that due to a rotating surface charge on a conducting sphere.
33
(3.179)
In the absence of conduction current, Jc = 0; the following integral over the entire volume
should vanish,
Z
B HdV = 0:
(3.180)
all space
This is because the LHS can be written in terms of the scalar potential m as
Z
Z
Z
B r m dV =
r (B m )dV +
m r BdV = 0:
(3.181)
The magnetic energy associated with a permanent magnet can thus be calculated from either
Z
1
2
Em =
(3.182)
0 H dV;
2
all space
or
Em =
Z
magnet
1
2
0M
(3.183)
HdV:
In the case of spherical permanent magnet, the total magnetic energy is
Z
1
4 3 1
2
Em =
H
Hr2 + H 2 dV
a
+
2 0 z
3
2 0 r>a
2
4
=
M 2 a3 +
M 2 a3
27 0 0
27 0 0
2
=
M 2 a3 ;
9 0 0
(3.184)
where the …elds in the exterior region r > a
Hr =
2M0 a3
M0 a3
cos
;
H
=
sin ;
3r3
3r3
have been substituted. The energy agrees with that from
Z
1
1
4 3 2
2
a =
0 M HdV =
0 M0
2
6
3
9
magnet r<a
2 3
0 M0 a :
Example 9 Image Currents
If a current I is placed parallel to a ‡at surface of highly permeable bloc, an image current
in the same direction as the current itself appears. This is because of the boundary condition
that the tangential component of the magnetic …eld should vanish at the surface. The image of a
current segment perpendicular to the surface is opposite to the current also because of the boundary
condition.
34
Image currents I 0 in iron (left) and superconductor (right).
If the permeability of the bloc is …nite, the vector potential in air region can be calculated by
summing contributions from the current I itself and an image current in the iron,
0
I0 =
+
I;
0
and the vector potential in the bloc by assuming a current
2
+
I 00 =
I;
0
at the location of the current.
For a superconducting surface, the normal component of the magnetic …eld should vanish.
Therefore, the image of a current parallel to superconducting surface is opposite, while for a vertical
current, the image is in the same direction. Fig. ?? illustrates the two cases.
3.9
Magnetic Force and Stress Tensor
The volume force to act on a current density J is
f =J
Substituting J = r
B=
0;
B; (N m
3
):
(3.185)
we may rewrite f as
f
=
1
(r
B)
B
0
=
1
0
(B r)B
1
rB 2 :
2
(3.186)
However,
r (BB) = (r B)B + (B r)B
= (B r)B:
35
(3.187)
Therefore, the force density can be written in the form
1
f=
0
r
BB
B2
1 ;
2
(3.188)
where 1 is the unit tensor. The magnetic stress tensor is de…ned by
Tm =
1
BB
0
B2
1 :
2
(3.189)
This is reminiscent to the electric stress tensor we have encountered in Chapter 1,
Te = "0 EE
E2
1 :
2
(3.190)
The magnetic energy density is
um =
B2
; (J/m3 = N/m2 )
2 0
(3.191)
and the total magnetic energy is
Um =
=
=
=
Z
B2
dV
2 0
Z
1
B (r A)dV
2 0
Z
Z
1
1
r (A B)dV +
A (r
2 0
2 0
Z
1
(A J)dV; (J).
2
B)dV
(3.192)
For a closed current loop, this may be written as
1
Um = I
2
I
A dl;
C
where the vector potential can be divided into that part due to the current I itself and another
part due to currents in other circuits. Recalling the de…nition of self and mutual inductances, the
magnetic energy associated with a single current loop can be written as
1
1 X
Mi Ii ;
Um1 = LI 2 + I
2
2
(3.193)
i
where Mi is the mutual inductance between the circuit under consideration and other circuits. For
two loops, the total magnet energy stored is given by
1
1
Um = L1 I12 + L2 I22 + M12 I1 I2 ;
2
2
36
(3.194)
because of reciprocity M12 = M21 : The mutual interaction energy M12 I1 I2 can be either negative
or positive depending on the directions of current ‡ows I1 and I2 : However, the total magnetic
energy Um is evidently positive de…nite.
As electric forces act so as to increase self and mutual capacitances, magnetic forces tend to
increase inductances. The force in the direction of geometrical metric factor i contained in the
inductances can be calculated from
Fi=
1
L( i )I 2 ;
2
@
@ i
(3.195)
for the case of self-inductance, and mutual interaction force from
Fi=
@
(M12 ( i )I1 I2 ) :
@ i
(3.196)
For example, the self-inductance of a circular current loop of loop radius a and wire radius
L=
0a
8a
ln
2+
li
;
2
is
(3.197)
where li is the factor of internal inductance due to the magnetic energy stored in the wire. For
uniform current distribution (no skin e¤ect), li = 1=2: If the loop carries a current I; it tends to
expand so as to increase a with a force
1 2 @
I
L(a)
2 @a
1
8a
I 2 ln
2 0
Fa =
=
1+
li
:
2
(3.198)
Example 10 Toroidal Magnet
Consider a toroidal magnet with a major radius R; minor radius a; and number of windings N:
By the Ampere’s law, the magnetic …eld in the magnet can be readily found,
B =
0
NI
;
2
R
a<
< R + a:
The magnetic ‡ux is
=
0N I
Z
R+a
R a
=
0N I R
q
a2
(a
R2
a2 :
p
)2
d
(3.199)
Therefore, the inductance of the magnet is
L=
2
R
0N
37
p
R2
a2 :
(3.200)
The magnet tends to shrink in the major radius direction with a force
FR =
=
1
2
1
2
2
0 (N I)
p
@
R
@R
2
0 (N I)
1
p
R2
R
R2
a2
a2
< 0:
(3.201)
In the minor radius direction, the magnet tends to expand with a force
Fa =
=
1
2
1
2
p
@
R2 a2
R
@a
a
2
> 0:
0 (N I) p 2
R
a2
2
0 (N I)
(3.202)
In large fusion devices, mechanical support structures must be able to handle these magnetic forces
which are enormous.
3.10
Boundary Value Problems
As discussed brie‡y, the Maxwell’s equation for the magnetic …eld B;
r B = 0;
(3.203)
requires that the normal components of B be continuous across a boundary of two magnetic media,
B1n = B2n :
(3.204)
For the …eld H;
r
H = Jc ;
(3.205)
the tangential components may be a provided there exists a surface conduction current Js (A/m)
‡owing on the boundary surface,
n (H1 H2 ) = Js ;
(3.206)
In the absence of conduction surface current, the tangential components of H should be continuous.
On the surface of highly permeable medium such as iron, the magnetic …eld lines falls almost
normal to the surface provided iron body and current circuits are not linked topologically. Linked
and unlinked examples are shown in Fig. 3-13. In unlinked cases, the …eld H in iron should vanish
if iron
0 : Since
B1 cos
1
= B2 cos
2;
or
1 H1 cos 1
=
2 H2 cos 2 ;
(3.207)
and
H1 sin
1
= H2 sin
38
2;
(3.208)
Figure 3-13: Iron rings topologically unlinked (left) and linked (right) to a current loop I: For
unlinked ring, H = 0; while for linked ring, H 6= 0 in iron even if
0 : The magnetic …eld lines
fall normal to iron surface if a current is unlinked.
we …nd
1
tan
=
1
2
tan
:
(3.209)
2
Therefore, if 2
1 ; the angle 1 must approach 0. The …eld H2 in iron should vanish also.
In linked cases, the …eld H remains …nite even in iron and should satisfy
I
H dl = I;
(3.210)
along a path in iron. The magnetic …eld lines do not necessarily fall normal to the iron surface.
For example, if a current is placed at the center of hollow iron cylinder, the magnetic …eld line at
the iron surface is tangential everywhere.
Example 11 Ring Current around a Long Iron Core
We consider a circular ring current I of radius b which is coaxial with a long, highly permeable
iron cylinder of radius a as shown in Fig.3-14. If the iron cylinder has no return legs, it is not
topologically linked to the current ring. Therefore, the boundary condition for the magnetic …eld
at the iron surface is
Bz ( = a; z) = 0:
(3.211)
Because of axial symmetry, the vector potential A ( ; z) can be found from the following equation,
1 @
@2
@2
+
+
@ 2
@
@z 2
1
A ( ; z) =
2
0I
(
b) (z):
(3.212)
Fourier transformation through
A ( ; z) =
1
2
Z
1
1
39
A ( ; k)eikz dk;
(3.213)
Figure 3-14: A ring current coaxial with an iron cylinder.
reduces the problem to one dimensional,
d2
1 d
+
d 2
d
k2
1
A ( ; k) =
2
0I
(
b):
(3.214)
Bounded solutions in each region may be assumed as
A ( ; k) =
8
>
< AI1 (k ) + BK1 (k ); a <
>
:
CK1 (k );
< b;
(3.215)
> b;
where the coe¢ cients A; B; and C may be functions of the Fourier variable k: Continuity of A ( ; k)
at = b yields
AI1 (kb) + BK1 (kb) = CK1 (kb):
(3.216)
The axial magnetic …eld is
Bz =
Since
1 @
( A ):
@
1 d
1 d
[ I1 (k )] = kI0 (k );
[ K1 (k )] =
d
d
(3.217)
kK0 (k );
(3.218)
the boundary condition Bz = 0 at the cylinder surface gives
B=
I0 (ka)
A;
K0 (ka)
40
(3.219)
and
C=
I0 (ka)
I1 (kb)
+
K1 (kb) K0 (ka)
A:
(3.220)
The coe¢ cient A can be found from the discontinuity of the radial derivatives of the vector potential,
d2
A ( ; k)
d 2
I1 (kb) 0
K (kb)
K1 (kb) 1
= Ak
=b
=
A
(
bK1 (kb)
I10 (kb)
(
b)
b);
(3.221)
where use is made of the Wronskian of the modi…ed Bessel functions,
0
Im
(x)Km (x)
0
Im (x)Km
(x) =
1
:
x
Comparing with the RHS of the original equation in Eq. (3.214), we …nd
0 IbK1 (kb);
A=
(3.222)
and the solution of the physical vector potential is
A ( ; z) =
0
8 Z
>
>
>
>
Ib <
Z
>
>
>
>
:
1
K1 (kb) I1 (k ) +
0
I0 (ka)
K1 (k ) cos(kz)dk; a <
K0 (ka)
< b;
(3.223)
1
0
I0 (ka)
K1 (k ) I1 (kb) +
K1 (kb) cos(kz)dk;
K0 (ka)
> b:
The …rst terms indicate the vector potential due to the current ring alone while the second terms are
corrections due to the presence of the iron cylinder. The iron cylinder increases the ring inductance
by an amount
Z1
I0 (ka) 2
L = 2 0 b2
K (kb)dk:
(3.224)
K0 (ka) 1
0
The magnetic …eld can be calculated from
B=r
The radial component is, for a <
B ( ; z) =
=
A :
< b;
@A
@z Z
0 Ib
1
K1 (kb) I1 (k ) +
0
41
I0 (ka)
K1 (k ) k sin(kz)dk;
K0 (ka)
(3.225)
and for
> b;
B ( ; z) =
0 Ib
Z
1
K1 (k ) I1 (kb) +
0
The axial component in the region a <
Bz ( ; z) =
=
(3.226)
< b is
1 @
( A )
@
Z 1
0 Ib
K1 (kb) I0 (k )
0
and for
I0 (ka)
K1 (kb) k sin(kz)dk:
K0 (ka)
I0 (ka)
K0 (k ) k cos(kz)dk;
K0 (ka)
(3.227)
I0 (ka)
K1 (kb) k cos(kz)dk:
K0 (ka)
(3.228)
> b;
Bz ( ; z) =
0 Ib
Z
1
K0 (k ) I1 (kb) +
0
If the permeability of the iron cylinder is not in…nite but …nite , the vector potential in the
region a < < b is modi…ed as
A ( ; z) =
0 Ib
Z1
(K1 (kb)I1 (k ) + f (k)K1 (kb)K1 (k )) cos(kz)dk;
(3.229)
0
where
f (k) =
(
(
0 )kaI0 (ka)I1 (ka)
0 )kaI1 (ka)K0 (ka)
+
:
0
Derivation of this modi…cation is left for an exercise.
Example 12 Magnetic Shielding
Figure 3-15: Cross-section of a cylindrical iron shell placed in an external magnetic …eld. The …eld
inside the shell is greatly reduced.
42
An iron cylinder having inner and outer tic …eld radii a; b and permeability
is placed in
an external magnetic …eld B0 with its axis perpendicular to the …eld. The magnetic …eld inside
the cylinder is greatly reduced if
0 even if the thickness of the cylinder is small. Since no
conduction currents are present, the magnetic …eld H can be generated from a scalar potential m
which satis…es the Laplace equation,
r2 m = 0;
(3.230)
or
@2
1 @
1 @2
+
+
2
@ 2
@
@ 2
m(
; ) = 0:
(3.231)
The external magnetic …eld can be generated from
0
=
H0 cos :
(3.232)
Since the magnetic …eld due to the presence of the iron cylinder should have the same angular
dependence, we may assume the following solutions in each region,
8
>
c1 cos ;
0 < < a;
>
>
>
<
c3
cos ;
a < < b;
c2 +
(3.233)
m( ; ) =
>
>
c
>
4
>
cos
H0 cos ; b < :
:
The boundary conditions are both
m
and @
m =@
be continuous at
c1 a = c2 a +
c2 b +
0 c1
c2
c3
b2
c3
;
a
c3
c4
=
b
b
=
=
(3.234)
H0 b;
c2
0
= a and b: Then,
(3.235)
c3
;
a2
c4
+ H0 :
b2
(3.236)
(3.237)
These simultaneous equations can be readily solved. For the interior magnetic …eld, only c1 is
needed. Its solution is
c1 =
1
4H0 b2
0 2
b
2:
2
a2
1+
0
If
=
0;
we recover c1 =
H0 : If
0;
c1 =
(3.238)
1
0
we …nd
4H0
43
b2
0
b2
a2
;
(3.239)
and thus the interior magnetic …eld
Hi = 4H0
b2
0
b2
a2
H0 :
(3.240)
Example 13 Superconducting Disk in an External Magnetic Field
In this example, we analyze how a superconducting disk disturbs a uniform external magnetic
…eld. The boundary condition for the magnetic …eld at a surface of superconducting body is that
the normal component vanish because magnetic …eld cannot penetrate into superconductors. The
same boundary condition should also apply for ordinary conductors in oscillating magnetic …eld if
the skin depth given by
1
; (m)
(3.241)
=p
f 0
is small compared with conductor dimensions. In both cases, electric currents ‡ow at the conductor
surface in a manner to cancel the external magnetic …eld.
Figure 3-16: Superconducting disk in an external magnetic …eld with its face normal to the …eld.
If the disk is ideally thin, the greatest disturbance to the magnetic …eld occurs when the disk
axis is parallel to the magnetic …eld because in this case, the magnetic dipole moment induced
on the disk surface becomes maximum. The potential of the external uniform magnetic …eld is as
before,
(3.242)
m0 = H0 z = H0 r cos :
It is convenient to use the oblate spheroidal coordinates ( ; ; ): A thin disk is described by = 0:
Because of axial symmetry, @=@ = 0; general solutions to the Laplace’s equation may be assumed
as
X
[Al Pl (i sinh ) + Bl Ql (i sinh )]Pl (cos ):
(3.243)
m( ; ) =
l
Since the external …eld has cos = P1 (cos ) dependence, only the l = 1 harmonic is relevant, and
44
the potential reduces to
m(
; ) = [AP1 (i sinh ) + BQ1 (i sinh )] cos :
(3.244)
Noting z = a sinh cos in the oblate spheroidal coordinates, and P1 (i sinh ) = i sinh ; we …nd
A = iaH0 :
(3.245)
The coe¢ cient B can be found from the boundary condition B = 0 at the disk surface, namely,
@
@
m
= 0 at
= 0:
(3.246)
Since
Q1 (i sinh ) = sinh arccot(sinh )
1;
(3.247)
we readily …nd
B=
2
aH0 ;
where arccot(0) = =2 is noted. The solution for
m(
; ) = aH0
The behavior of the potential at r
sinh +
2
m(
(3.248)
; ) is thus given by
(sinh arccot(sinh )
(3.249)
a can be found using the asymptotic form of Q1 (i sinh );
1
;
3 sinh2
Q1 (i sinh ) !
with result
m(
1) cos :
1; ) '
H0 r cos
1;
(3.250)
2a3 H0
cos :
3 r2
(3.251)
The dipole moment of the disk placed perpendicular to an external magnetic …eld is thus given by
m=
4
2a3
H0 :
3
(3.252)
Let us check if this is consistent with the dipole moment expected from
Z
Z
1
1
m=
r JdV =
r Js dS;
2
2
where
Js = n
=
H
2
H0 p
a2
45
2
e ;
(3.253)
which exists on both sides of the disk. Then,
2
mz =
H0
Z
a
0
8
H0 a3 ;
3
=
2
p
a2
2
2
d
which agrees with that identi…ed from the far …eld potential. Note that
Z
0
1
p
x3
2
dx = :
2
3
1 x
Example 14 Leakage through a Hole in a Superconducting Plate
A uniform magnetic …eld parallel to a superconducting plate exists in one of the regions separated by the plate. The plate has a hole of radius a:We wish to …nd the magnetic …eld in the other
region. This problem is similar to Example 7 in Chapter 2. The leakage …led is expected to be of
dipole nature. Since the unperturbed …eld can be described by the potential
0
=
H0 x =
H0 a cosh sin cos ;
in the oblate spheroidal coordinates ( ; ; ); we assume
m(
; ; )=
(3.254)
m
H0 a cosh sin cos + AQ11 (i sinh ) sin cos ;
where
Q11 (i sinh ) = cosh
sinh
cosh2
arccot(sinh )
:
(3.255)
(3.256)
It is convenient to de…ne the function arccot x without discontinuity in its derivative. The asymptotic forms are:
1
1
arccot x ' +
+ ; x ! 1;
(3.257)
x 3x3
1
1
arccot x '
+
; x ! +1:
(3.258)
x 3x3
? 0 corresponds to z ? 0; respectively. In the region z !
and the coe¢ cient A can thus be readily found,
A=
aH0
1; the uniform …eld should vanish
:
(3.259)
In the upper region, the far …eld potential becomes
m( ; ; ) =
H0 x +
2H0 a3 r x
;
3
r3
46
r
a; z > 0:
(3.260)
while in the lower region
m (r) =
2H0 a3 r x
;
3
r3
r
a; z < 0:
(3.261)
The e¤ective magnetic dipole moments for the region below the hole is
m=
8a3
H0 ; z < 0:
3
(3.262)
In the region z > 0; the e¤ective magnetic dipole moment is
m=
8a3
H0 ; z > 0:
3
(3.263)
The formulae derived here may be conveniently used in analyzing leakage radiation from a small
hole drilled in the wall of waveguides.
Figure 3-17: Leakage of magnetic …eld through a hole in a superconducting plate.
47
Problems
3.1 Find the magnetic …eld at the center of (a) an equilateral triangular loop and (b) square loop
both having side a and carrying current I:
3.2 Find the magnetic …eld at point P:
3.3 A current I ‡ows along a ‡at spiral coil described by
( )=
R
;
2 N
where N is the number of turns and R is the outermost radius of the coil. Show that the
magnetic …eld along the axis is given by
Bz (z) =
0N I
R
3.4 An elliptic loop of axes a and b (a
the center of the loop is given by
"
R+
ln
p
R2 + z 2
z
!
#
R
p
:
R2 + z 2
b) carries a current I: Show that the magnetic …eld at
Bz =
0I
where
l = 4aE 1
Z
0
Discuss the limiting cases of a = b and a
b2
a2
;
ab is its area. E k 2 is the complete elliptic
is the circumference of the loop and S =
integral of the second kind de…ned by
E k2 =
l
;
4S
=2 p
1
k 2 sin2 d :
b:
3.5 Find the mutual interaction force between a circular loop current I1 of radius a and a coplanar
straight long current I2 when the center of the loop is at a distance b from the line current.
48
3.6 Show that an alternative expression for the vector potential due to a circular loop current
(radius a; current I) is
0 Ia
A ( ; z) =
2
Z
1
J1 (ka)J1 (k )e
kjzj
dk:
0
3.7 The loop in the preceding problem is placed with its plane parallel to the surface of a large
magnetic bloc of permeability at a distance d: Using the method of image, show that the
inductance of the loop increases by
L=
0a
0
+
0
2
k
where
k=p
2
E(k 2 ) ;
k
k K(k 2 )
a
:
a2 + d2
3.8 A circular current loop of radius a is concentric with an iron sphere of in…nite permeability
of radius b. Show that the presence of the iron sphere increases the loop inductance by
L=
0b
1
X
l=0
(2l 1)!!
(2l)!!
2
b
a
4l+2
:
3.9 Show that the mutual inductance between two circular loops of radii a and b with parallel
49
axes a distance c apart and loop planes a distance d apart is given by
M=
0 ab
Z
1
J1 (ka)J1 (kb)J0 (kc)e
kd
dk:
0
3.10 Evaluate the Nagaoka factor numerically for a=l = 0:1; 0:5 and 1.
3.11 A current ‡ows along a circular tunnel in a highly permeable body. Do the magnetic …eld lines
fall normal to the tunnel surface? What about the case in which a return current coexists?
Draw qualitatively the magnetic …eld lines for both cases.
3.12 A conductor having a rectangular cross-section a
J0 = J0 ez : Find the magnetic …eld:
50
b carries a uniform axial current density
3.13 A current sheet I with a width 2a is placed on one of iron walls as shown. Show that the
vector potential in the gap is given by
Az =
0I
4 a
Z
a
a
h
ln cosh
d
(x
x0 )
cos
d
y
i
dx0 :
3.14 A current I is placed on the midplane of air gap formed by iron as shown. Find the vector
potential and magnetic …eld in the gap.
3.15 A thin superconducting ring of major radius a and minor (wire) radius b ( a) is placed in
an external magnetic …eld with its plane perpendicular to the …eld. Determine the vector
potential and the magnetic dipole moment induced by the ring.
3.16 If the magnetic …eld and current density are parallel to each other, the Lorentz force J
vanishes and Ampere’s law reduces to
r
B
B = B;
where (m 1 ) is a constant. Such arrangement is pertinent to designing superconducting
wires which are fragile and also to low pressure plasma equilibria such as Reversed Field
Pinch (RFP) and Spheromak. Solve the equation in (a) the cylindrical coordinates assuming
B = 0; @=@z = @=@ = 0; and (b) in spherical coordinates with the boundary condition Br
= 0 at r = a:
51