PIERS ONLINE, VOL. 3, NO. 6, 2007
812
Several Rules about the Magnetic Moment of Rotational Charged
Bodies
Guo-Quan Zhou
Department of Physics, Wuhan University, Wuhan 430072, China
Abstract— A strict and delicate analogy relation between the magnetic moment of a rotational
charged body and the rotation inertia of a rigid body about a fixed axis has been found in this
paper. Based on this analogy relation, many rules and theorems about calculating the rotation
inertia of a rigid body can be transplanted to field of electromagnetism and used to calculate
the magnetic moment of a rotational charged body. Some principles or theorems are extended,
generalized, illustrated, and enumerated. Some related examples are listed in the paper.
DOI: 10.2529/PIERS061120233608
1. PREFACE
Reference [1] gives the analogy relation between the magnetic moment [3–5] of a rotational charged
body and the rotation inertia of a rigid body about a fixed axis, which is very stimulating and
pedagogical. According to this paper, a rotational charged body with an element electric charge of
dq, a constant angular speed of ω (the size of ω
~ ) about a fixed axis, and a rotation radius of r⊥ ,
must have a definite magnetic moment of following magnitude:
¯
¯
¯ ~ ¯
2
¯dPm ¯ = dIπr⊥
2
= (ω/2π) · dq · πr⊥
2
dq
= (ω/2) · r⊥
or
2
dq
d(2Pm /ω) = r⊥
Its direction is parallel or anti-parallel to ω
~ according to the positive or negative sign of dq
respectively. The above expression has a strict and delicate analogy relation to the rotation inertia
2 dm for an element mass of dm about a definite axis:
expression — dJ = r⊥
2Pm /ω ↔ J, dq ↔ dm, q ↔ m
Let us pay more attention to the analogy relation of 2Pm /ω ↔ J, which is more appropriate
than the corresponding one given in the reference [2]. Because only Pm /ω is the physical quantity
that independent of ω and only impressed by the distribution of electric charge and position of
axis. Thus there is a strict analogy relation between Pm /ω and J. Nevertheless, on the same time
we should notice that, charged conductor which is in a state of a electrostatic equilibrium only has
its charge distributed on its surface. So we must treat the rotational charged solid conductor as a
charged shell, analogical to the hollow rigid body of the same shape. Other analogy relationship
between the two circumstances is:
Q(the total charge value) ↔ m(the total mass),
Ce (the center of elecric charge) ↔ Cm (the mass center),
σe (electrical charge areal density) ↔ σm (mass areal density),
ρe (electrical charge body density) ↔ ρm (mass body density),
λe (electrical line density) ↔ λm (mass line density).
It is obvious that many computation rules and property theorem of rotation inertia for a rigid
body can be transplanted to that of the magnetic moment for a rotational charged body. Just as
in reference [6], the rotation inertia for any triangle rigid plane with homogeneous mass distribution about its center-of-mass axis vertical to this rigid plane can be formulated as: JABC(Cm ) =
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813
m(a2 +b2 +c2 )/36. According to the analogy relation given as above, we can immediately deduce the
magnetic moment formula for a rotational charged triangle plate with homogeneous charge distribution about its center-of-charge axis vertical to the plate to be: PmABC(Ce ) = Q(a2 + b2 + c2 )ω/72.
This paper extends and applies the conclusions drawn from references [6–9] to the case for calculation of magnetic moment of a rotational charged body, tabulates and enumerates related examples
of those theorems and makes some further discussion.
2. THE PARALLEL-AXIS THEOREM FOR THE MAGNETIC MOMENT OF A
ROTATIONAL CHARGED BODY
The parallel-axis theorem [6–9] of rotation inertia for a rigid body is universal to all cases of any
shape and any mass distribution. Then there is some reviews are in order: once we know the
rotation inertia of rigid body about its center-of-mass axis, we can calculate its rotation inertia
about any other parallel axis. ZBased on above analogy relation, first of all, we can define the
1
~rdq. And suppose the value of magnetic moment of a rotational
electrical charge center ~rc =
Q
charged body about one of its axis passing through its charge center is Pmc (the corresponding
angular speed is ω), the distance between the two parallel axes is d, the total charge value is Q,
then following conclusion holds:
2
2
Pm =
Pmc + Qd2
(1)
ω
ωc
In need of simplicity and conciseness, we can take ω = ωc (but notice that ω and ωc are
independent of each other), and attain a formula:
Pm = Pmc +
ωQ 2
d
2
(10 )
This is the parallel-axis theorem for the magnetic moment of a rotational charged body with any
shape and any electric charge distribution (a volume, surface, curve or discrete points distribution).
No matter what a kind of charged body it is, the above formula always holds — for a rigid body
or a kind of liquid matter, a conductor or a kind of electrolysis. Its application examples can be
found in Table 1.
There is a very impotent deduction for theorem (10 ): if the total charge value of a rotational
charged body is zero, then Pm = Pmc and it has nothing to do with distance d.
That is to say: for a body with a total zero charge and a nonzero charge density, its magnetic
moment about any axis with any direction is equal to that of a charged body about the center-ofcharge axis.
3. THE ORTHOGONAL-AXES THEOREM FOR THE MAGNETIC MOMENT OF A
ROTATIONAL CHARGED BODY
For a thin slice of rigid plate of zero thickness, its rotation inertia about the x, y, z axes respectively
satisfies the so-called orthogonal-axes theorem [7, 9]. (provided the xoy coordinate plane is located
upon the rigid plate):
Jz = Jx + Jy
Based on the analogy relation, for a thin plate with an arbitrary surface charge distribution,
its magnetic moments about the x, y, z axes with an angular speed of ωx , ωy , ωz respectively will
satisfy the following orthogonal-axes theorem:
Pmx Pmy
Pmz
+
=
ωx
ωy
ωz
(2)
For simplicity and conciseness, without loss of generality, we can also assume ωx = ωy = ωz ,
and attain following equation:
Pmx + Pmy = Pmz
(20 )
Its application examples can be found to be example (3) in Table 1. Now let us pay more
attention to the so-called generalized orthogonal-axes theorem published in one of my papers [7]. For
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814
a column rigid body with an arbitrary shape of transversal section and of uniform mass distribution
along its generatrix, with a height of L, a mass of m, with its Z axis parallel to its generatrix and
the coordinate plane of xoy plumb to its generatrix line, amputating it to be two parts, each with
a length of L1 and L2 respectively (L1 + L2 = L), as Fig. 1 shows. Then the following generalized
orthogonal-axes theorem holds
¢
2 ¡
Jx + Jy = Jz + m L21 − L1 L2 + L22
3
Figure 1: A column with height of L.
According to the analogy relation, for a uniform charged rotational column which has a total
charge value of Q and angular speeds of ωx , ωy , ωz about the x, y, z axes respectively, its corresponding magnetic moments must satisfy the following orthogonal-axes theorem when following
substitution is taken: ω2 Pm ↔ J, Q ↔ m.
1
Pmx /ωx + Pmy /ωy = Pmz /ωz + Q(L21 − L1 L2 + L22 )
3
(3)
For simplicity and conciseness, assuming that ωx = ωy = ωz = ω (notice that ω is not the
resultant angular velocity of ωx , ωy , ωz ), then
Pmx + Pmy = Pmz +
ω
Q(L21 − L1 L2 + L22 )
3
(30 )
When L1 = 0, L2 = L, we have
ω
QL2
3
(The end − face orthogonal − axes theorem)
Pmx + Pmy = Pmz +
When L1 = L2 =
L
2,
(300 )
we have
ω
QL2
12
(The orthogonal − axes theorem on the middle transversal section)
Pmx + Pmy = Pmz +
(3000 )
We can get its application examples such as No. 4, No. 5, No. 6 in Table. 1 by means of analogy
transition from reference [7].
What should be paid more attention to is that when the charged column has a total zero charge
value and a nonzero charge density, the generalized orthogonal-axes theorem (3), (30 ) reduce to the
original orthogonal-axes theorem (2), (20 ). That is to say, its magnetic moment is independent of
the column length. It is very important and useful to calculate the magnetic moment of a rotational
charged body, especially for a charged body with a four-degree symmetry axis (in terms of group
theory). The examples Nos. 5 and 6 in Table 1 are its application.
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Table.1. The table of calculation rules and examples of magnetic moment for rotational charged bodies.
(Provided all matter is evenly charged with total charge value of Q and
with a constant volume (surface, linear) charge density)
Examples
Rules and
theorems
Form of charged
bodies
No. 1.The thin and
straight rod
( linear )
The analogy
relation and
definition
l
No.2.The rectangular
plate
( surface distri. )
l
y
R
a
z
o c
y
x
x
z
The extended
orthogonal
axes theorem
y
R
l
x
No. 6.The thin and
hollow cylinder
(surface distri. and
axial sym.)
The
orthogonalaxes theorem
The extended
orthogonalaxes theorem
b
No. 5.The solid
Cylinder
(volume distri. and
axial sym.)
→
1
Ql 2 ω
24
→
1 2→
Ql ω
6
Pmx →
x
No. 4. The cuboid
( volume distri. )
→
Pm The parallelaxis theorem
z
No. 3. The circular
and thin plate
( surface distri. )
Formula of magnetic moment
→
1
P m QR 2 ωx
8
→
1
→
P mz QR 2 ωz
4
{
→
( Pmy the similar )
→
→
1
Q(a 2 c 2 )[ωx
24
→ →
(Pmy , Pmz the similar)
Pmx →
→
1
1
Pmx ( QR 2 QL2 )ωx
8
24
→
Pmy the similar
z
The extended
orthogonalaxes theorem
y
R
l
→
1
1
Pmx ( QR 2 QL2 ) ωx
4
24
→
z
No. 7.The solid
globe
( volume distri. and
spherical sym.)
y
The originmoment
theorem
→
Pm 1
→
QR 2 ω
5
x
z
No. 8.The thin
Spherical crust
( surface distri. and
spherical sym.)
y
The originmoment
theorem
1
→
Pm QR 2 ω
3
→
x
4. THE ORIGIN-MOMENT THEOREM
For those rigid bodies of arbitrary shape and mass distribution, randomly selected coordinate system
O-XYZ, the rotation inertia Jx , Jy , Jz about the axes X, Y , Z satisfy the following equation
Z
Jx + Jy + Jz = 2 r2 dm(= 2Jo )
Here Jo is defined as the origin moment [8, 9] (also called the center moment). r is the distance
between the mass element and the origin point. The above expression is called origin moment
theorem or center moment theorem, which has been mentioned in many references [8, 9]. And its
latest manifestation can be found in reference [8].
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According to the simple analogy relation given before, we have
Jx →
2
Pmx ,
ωx
Jy →
2
Pmx ,
ωy
Jz →
2
Pmz and dm → dq
ωz
Then we can get the origin-moment theorem related to moments of Pmx , Pmy , Pmz
Z
Pmx Pmy
Pmz
+
+
= r2 dq
ωx
ωy
ωz
For simplicity and conciseness, assuming that ωx = ωy = ωz (= ω).
Then
Z
Pmx + Pmy + Pmz = ω r2 dq
(4)
(40 )
Especially for those rotational charged bodies of spherical symmetry, their volume charge density
and element charge can be expressed as ρe = ρe (r), dq = 4πr2 ρe (r)dr.
Then
Z
Pmx + Pmy + Pmz = 4πω r4 ρe (r)dr
But due to the spherical symmetry Pmx = Pmy = Pmz , we have
Z
4
Pmx = Pmy = Pmz = πω r4 ρe (r)dr
3
(400 )
Its application examples are enumerated in Table 1, or No. 7, No. 8.
5. CONCLUSIONS
Such a perfect and delicate analogy relation between the magnetic moment of a rotational charged
body and the rotation inertia of a rigid body supply us with a powerful tool to calculate precisely
the magnetic moments of rotational charged bodies with all kinds of shape. Formulas such as
(10 )–(40 ) can all be viewed as theorems to deal with all kinds of problem involved in magnetic moment calculation of rotational charged bodies, especially for those rotational bodies of symmetrical
charge distribution. Table 1 gives some concrete illustrations of these formulas. Such calculation is
very necessary and meaningful for research and teaching of the electromagnetic field and even for
investigating of space such as the launching of satellites and spaceships.
ACKNOWLEDGMENT
Supported by National Science Foundation Grant. No. 10375041.
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