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Eur J Phys In11 11990) 9 4 9 8 . Prlnted
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I
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Electrostatic charges in v x B
fields:the Faraday diskandthe
rotatingsphere
Paul Lorrain
Department of Geological Sciences, McGill University, Montreal, Quebec, Canada, H3A 2A7
and Departamento de Fisica, Facultad de Ciencias, Universidad de Murcia, Espinardo, Murcia, Spain
Received 16 June 1989
Abstract. When a conductor moves in a magnetic field,
this vector is not zero, then the divergence of E is not
zero either and the conductor carries electrostatic charges
whose fields are as important as v x B. The Faraday disk
and a conducting sphere rotating in a uniform magnetic
field serve as examples. This little known effect plays a
fundamental role in magnetohydrodynamic phenomena.
RCumB. A l’interieur d’un conducteur en mouvement
dans un champ magnetique, v x B agit comme une source
distribuee. Si la divergence de ce champ n’est pas nulle, la
divergence de E est non-nulle egalement et le conducteur
porte des charges electrostatiques dont le champ est aussi
important que v x B. Le disque de Faraday, et une sphere
conductrice tournant dans un champ magnetique
uniforme, servent a illustrer notre propos. Cet effet peu
connu joue un r d e essentiel dans les phenomenes
magnetohydrodynamiques.
1. Introduction
2. Electrostatic charges in v x 6 fields
Imagine a body that moves ata velocity Y in a region
where there exist an electric field E and a magnetic
field B. Then anelectric charge Q inside the body feels
a force Q(E Y x B). Thus, inside the moving body,
the Y x B field acts like the electric field ofadistributed source.
We are concerned here with conducting media that
move in magnetic fields. We shall see that they carry
electrostatic charges whose
field is just as important as
v x B. Indeed, there are many
cases where the two
fields cancel each other exactly at every point. The
Faraday disk and a conducting sphere rotating in a
magnetic field will serve as examples, but this little
known effect plays a fundamental role in magnetohydrodynamics. Another article will discuss the function ofthese electrostaticcharges in thedynamo
mechanism that generates the Earth’s magneticfield.
We first consider a general velocity v , but most of
our discussion will concerntherotation
of rigid
bodies, either connected toa stationary circuit or not.
We perform our calculations both with respect to a
fixed reference frame and with respect to a rotating
frame.
It is well known that conductors do not support an
electric spacecharge;anyextrachargedeposited
insidemoves outtotheperipheryalmostinstantaneously (Lorrain et al 1988, p 75). However, few
physicists realise that conductors do carry anelectric
space charge when subjected to a v x B field whose
divergence is not equal to zero. If the conductor is
isolated, then it also carries a compensating surface
charge.
v x B acts like a distributed source. If the divergence of
+
2.1. Fixed reference frame
Choose a fixed reference frame S. With respect t o S,
the electric field strength is E and the magnetic flux
density B. If now a point inside a body of conductivity
Q moves a t a velocity Y with respect to S, then the
electric current density at that point, again with respect to S, is
J = u(E + Y x B).
(1)
The body need not be rigid and all the terms can be
both space and time dependent. The Y x B field acts
like the electric field of a distributed source (Lorrain
et a1 1988, p 494).
0143-08071901010094 + 05 $03.50@ 1990 IOP Publlshlng Ltd 8 The European Physlcal Soclety
Electrostatic charges
in
Y
x B fields
95
Nowtakethe
divergenceof both sides. Under
steady-state conditions, the divergenceof J is zero
and, if the medium is homogeneous,
V * E =-V.(U
X
B).
(2)
Thus, if the divergenceof v x B is not zero, there
exists an electrostatic space charge of density
Q = &,%V E = -&,%V
= &,%[v
(V
X
(v
X
B) - B * (V
B)
X V)].
(3)
(4)
Observe that the chargedensity is independent of the
conductivity. The relative permittivity of good conductors is not measurable, butit is presumably of the
order of three, like that of ordinary dielectrics.
We use p for the distance to theaxis of rotation, e
(varrho) for the space charge density, a for the conductivity,and s (varsigma) forthe surface charge
density.
With an idealdielectric, J and a would be both
zero, equation (1) would be meaningless, and these
expressions for e would not apply.
One author argues that, in conducting media such
as plasmas in space, the charge density is in fact zero
because oftheabundance
of chargecarriers.He
concludes that, under steady-state conditions, E = 0.
That is of course a misconception; the above equations
apply in any medium and E # 0 if the divergence of
v x BZO.
We shall see that, if the conductor is isolated, surface charges compensate for the space charge.
If the fields are time dependent, andif the medium
is not uniform, then there still exists an electric space
charge, but there are further terms on the right-hand
side of equation (4).
It follows from equation (1) that
E
= -(v
x B)
+ J/a.
(5)
Thevectors J and B arenotindependent;they
satisfy the Maxwell equation
V
X
B=hJ
(6)
assumingthatthemedium
is non-magnetic,and
under steady-state conditions.
Now consider a rotating body. If a rigid axisymmetric conducting body rotates at an angular
velocity
W about the z axis in any magnetic field, then
v = up$
4
vx
v = 2wi
(7)
where and 2 are the unit vectors pointing, respectively, in the positive 4 and z directions.The
azimuthalcurrentcomesfromtherotation
of the
spacecharge,and
possibly alsofromaninternal
source. If there is no
internally
generated
azimuthal current, then the conductor need not be axisymmetric.
Let us calculate the curl of J for a passive rotating
conductor that is not necessarily axisymmetric. From
equation (l), in a homogeneous medium,
VxJ=a[VxE+Vx(vxB)]
+
= a[-aB/at
v X (W X B)].
Understeady-stateconditionsthetimederivative
vanishes and
V x J = aV x (v x B).
(9)
(10)
(1 1)
Suppose now that our rotating conductor
is not
connected to a stationary circuit through sliding contacts.Supposealsothatthemagnetic
field is both
uniform and parallel to the axis of rotation. Both W
and B are constant. Then
VxJ=aVx(wp$xBi)=O.
(12)
If the magneticfield is not uniform, butaxisymmetric,
then it turns out that the curl is again zero:
V xJ=v(V*B)=O.
(13)
Now, in our isolated rotating conductor, currents
can only flow around closed circuits. Then the line
integral of J around anyclosed circuit is equal tozero
and there are no induced currents:
J=O
E=
-(W
X
B).
(14)
In this instance, the field of the electrostatic charges
cancels the v x B field exactly at every point.
This is not a new result (van Blade1 1984), but it is
important for thefollowing reason. Since there are no
induced currents in the rotating conductor, the magnetic field is not disturbed and the lines of B are not
dragged alongby the moving conductor, as one would
expect from Alfven’s ‘theorem’ (Alfven and Falthammar 1963). Another paper will discuss this ‘theorem’
at some length.
If the applied magnetic field is symmetrical about
an axis that forms an angle with the axis of rotation,
then we can consider it to be composed of an axisymmetric field B,,, plus a transverse field B , f . Then
+
V x J = aV x [v x (B,,, B , i ) ]
= aV x ( w p ~ , $x a)
= aV x
(wpB,cos~i)
(15 )
(16)
(17)
and there are induced currents in the conductor.
2.2. Rotating reference frame
Rotating reference frames are troublesomein relativity because they are non-Euclidian and because clock
rate depends on the distance to the axis of rotation
(Msller 1974). Fortunately, the situationhere is simple
because we need onlycomparethe
valuesof the
various variables at, andin the immediate neighbourhood of, a given point in the two frames.
We use unprimed variables in the inertial reference
frame S, as above. Primed variables at a given point
P’in the rotating reference frame S’ apply in the local
inertial reference frame S” occupied momentarily by
P’ (Msller 1974). Thentherelations
between the
P Lorrain
96
primed and unprimed variables are those of special
relativity (Lorrain et a1 1988, pp 287, 303, 321). We
disregard terms of the order of v2/c2,where c is the
speed of light. Then
E' = E + v
J' = J
X
B =J'/d
a' = a
B' = B.
't
(18)
(19)
Also, in a uniform medium,
v X J' = av X E T= - a a ~ p t t .
(20)
Recalling now that, in an isolated moving conductor,
there areno induced currents if the curl of the current
density is zero, this equation shows that there are
induced currents in an isolated moving conductor
only if the magnetic induction in the reference frame
of the conductor is time dependent. This is in agreement with equations (13) and (14).
The space charge density is the same in the two
reference frames if the azimuthal current results only
from the motion of the space charge.
3. The Faraday disk
Figures 1 and 2 show schematic diagrams of a
homopolar motor and of a homopolar generator, as
viewed from a fixed reference frame (Lorrain et a1
1988, p 399t). Both devices are also called Faraday
disks. We have shown single brushes on the axis and
on the rim, but there are in fact continuous brushes all
around. Also, we have not shown the coils that generate theapplied magnetic fields B. At any point in the
disks, v 2 << c'.
Assume that the axial magnetic field B is that of a
long solenoid, and is thus uniform. This assumptionis
not realistic, but it makes the calculation simple and
instructive. We may set J+ = 0 in equation (8)
because, as we shall see below, the convection current
of the rotating space charge generates an opposing
Flgure 2. Homopolar generator.
axial magnetic field that is negligible. We disregard
the magnetic field of the radial current through the
disks because it is azimuthal, which makes its v x B
equal to zero. Set B, = B in equation (8).
Say B and CO are the same in figures 1 and 2. Then
v x B and Q are the same and, from equation (8), the
space charge density is uniform and negative:
e= ( 2~1E) , Q w B .
This space charge density is independent of the conductivity, except if a were zero, and it is independent
of whether the switch SW in figure 2 is open or closed.
Therotation of this space charge generates an
azimuthal convection current of density ea whose
magnetic field opposes the above B. Since that B is
uniform, it is easy to calculate the magnetic induction
B, of this azimuthal convection current at the centre
of the disk. Setting R and S equal to the outer radius
and to the thickness of the disk, respectively, we find
that, for S << R,
Figure 1. Homopolar motor.
since w2sR is of the order of v 2 , and vz/c2<< 1 by
hypothesis.
Now disregard end effects. This is equivalent to
stretching the disk into a long cylinder. This assumption does not invalidate the above approximation.
Then
Observe that, here,
E, = - ( U X B).
?Note that in this reference B is negative.
The v x B field 'pumps' conduction electrons in the
- j direction, and their space charge cancels the
v x B field exactly at every point in both figures 1 and
2, whether the switch SW of figure 2 is open or closed.
The simplest case is that of the generator, with SW
Electrostatic charges in
Y x
B fields
97
open. Then
J = 0,
E+v
X
B=O
(25)
E = Ep.
A surface charge of density E , E ~ ~ compensates
BR
for
the negative space charge withoutaffecting the electric
field inside the cylinder.
Now close the switch SW in the circuit of the
generator.Conduction electrons escape at the axis
into the external circuit, where the v x B is zero, and
return at the rim. The field J/a in the disk points in the
direction of the rim.
The case of the motor is similar, except that Jpoints
in the - j direction: current flows in the direction of
v x B in the generator, and in the opposite direction
in the motor.
The current in the homoplanar motor depends not
only on the characteristics of the motor, but also on
the source that feeds it and on its mechanical load,
while the current in the generator depends similarly
on the applied torque and on the load resistance.
Let us now calculate the potentials V and A for
the motor. From Lorrain et al (1988, p 349), under
steady-state conditions and for a long solenoid,
aA/at = o
A = (BPI216
(26)
Since J = Z/(2nps), where the current Z is independent
of the radius,
The quantity p. is a constant of integration; we have
arbitrarily set the second term equal to zero at p = po.
In the rotating reference frame, from Lorrain et a1
(1988, pp 287, 303, 321),and from equations (26) and
(2819
Figure 3.
4. Conducting sphere rotating in a uniform
magnetic field
Figure 3 shows a passive conducting sphere of radius
R rotating at anangular velocity w in a uniform
magnetic field B parallel to the axis of rotation. The
sphere is uncharged. As in section 2.1, there are no
induced currents and equations (14) apply. It will be
instructive to discuss this simple case (see also Thomson 1893, Mason and Weaver 1929, van Blade1 1984).
We assume that the sphere is rigid: a further paper
will discuss the effectof differential rotation with
reference to the Earth's core.
We can again disregard theazimuthalcurrent
associated with the rotation of the space charge. Then
equation (21) applies: the space charge is uniform
throughout the sphere, and negative. As we shall see,
positive surface charges maintain the net charge equal
to zero. This negative space charge establishes a radial
electric field
The electrostatic field of the space charge now canfirst
cels the v x B field only to acertainextent,
because the electric field is along - P , while the v x B
field is along fi, and second because the magnitudes of
the two fields are not equal. Equations(14) still apply,
however: the combined electrostatic fields ofthe space
and surface charges cancel the v x B field.
We can find the surface density c of electrostatic
charge as follows. From the Gauss law,
w2p2
=A --A
C2
op2J
+ --In-$.
c2
0
p
Po
-
In this last equation the second term on the right is
proportional to v 2 / c 2and is therefore negligible. The
third term in negligible for the same reason because J
is proportional to the sum of two terms, E v x B,
that are of the same order of magnitude. Thus
A' = A
aA'/at' = o
E' = - V V ' .
(33)
+
c = Er.outside - Er'% Er,ins>de.
(35)
Inside the sphere, in cylindrical coordinates,
E= - ( V X B) = - ~ p B j
(36)
or, in spherical coordinates,
E = - wr
sin
BB(sin
BP
+ cos 06)
(37)
Er,insrde
= - wrB sin2B.
Also, inside the sphere, the electric fieldis radial
P Lorrain
98
and the potential along the axis is uniform. Call that
potential V,. Then, on the surface, from equation
(361,
Vsurface
= V,
+ wBp2/2= V, + wBR2sin28/2.
(39)
Outside the sphere, the space charge density is zero,
Laplace’s equation applies, and we can expand V as a
series of Legendre polynomials (Lorrain et al 1988,
p 225):
show that the net surface charge is equal to minus the
space charge.
The value of the potential inside the sphere follows
from equations (36) and (43):
(47)
The potential is equal to zero on an axial cylindrical
surface whose radius is equal to (2/3)’’2R.
5. Conclusion
The first term, which is the monopole term, implies
the existence of a net charge on the sphere. That
potential is of no interest; it would apply only if the
sphere carried a net charge. The second, dipole, term
implies a charge asymmetry between the upper and
lower halves of the sphere. This term is also zero
because both E and E, are symmetrical about the
equatorial plane. Let us therefore set thethird, or
quadrupole, term of equation (40) equal to the V of
equation (39) at r = R:
V
’
B, 3cos28 - 1
+ +oBR2sin20= R3
2
We have therefore shown that, if aconducting
medium moves in a magnetic field, and if the divergence of the ZJ x B field is not zero, then there exists
an electrostatic space charge and its associated electric
field. If the conductor is isolated, then surface charges
correct for any charge imbalance. If the conductor is
isolated and if the magnetic field isaxisymmetric, then
the fieldof the electrostatic charges cancels ZJ x B
exactly at every point, there are no induced currents,
the rotating conductor does not affect the magnetic
field, and the Alfven ‘theorem’ does not apply.
(41)
Acknowledgment
Upon matching corresponding terms, we find that
B2 = -wBRS/3
V,
=
-wBR2/3.
I thank David Crossley and Oliver Jensen for several
fruitful discussions.
(43)
References
Thus
Finally, fromequations (35), (38), and (45), the
surface charge density is given by
= E,,wBR[(Q
+ E,)sin20 - I].
(46)
The surface charge density is positive in the region of
the equator, and negative near the poles. It is easy to
Alfven H and Falthammar C-G 1963 Cosmical
Electrodynamics (Oxford: Clarendon) p 189
Lorrain P, Corson D R, and Lorrain F 1988
Electromagnetic Fields and Waves 3rd edn (NewYork:
Freeman)Mason M and Weaver W 1929 The Electromagnetic Field
(New York: Dover)
Moller C 1914 The Theory of Relativity 2nd edn (London:
Oxford University Press) ch 8 and 9
Thomson J J 1893 Notes on Recent Researches on
Electricity and Magnetism (Oxford: Clarendon) P 534
van Blade1 J 1984 Relativity and Engineering (Berlin:
Springer) p 287
INSTITUTE OF PHYSICS PUBLISHING
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LETTERS AND COMMENTS
Comment on electrostatic charges in v × B fields
Dragan V Redžić
Faculty of Physics, University of Belgrade, POB 368 11001 Beograd, Yugoslavia
Received 22 May 2000
Abstract
A basic error is rectified in Lorrain’s paper (1990 Eur. J. Phys. 11 94) on
conducting media that move in magnetic fields.
Lorrain (1990) gave an interesting discussion concerning conducting media that move in
magnetic fields, focusing on steady-state conditions and disregarding terms of the order of
v 2 /c2 . It seems, however, that some of his results are either wrong or derived in a wrong way
for the following reason. In his argument the author used the well-known constitutive equation
for a linear isotropic medium at rest, D = ε0 εr E , as if it were valid for the medium in motion
too. That, of course, is not so; in the first-order theory for a moving non-magnetic medium
discussed by Lorrain, the correct constitutive equation is D = ε0 E +ε0 (εr −1)(E + v × B ) (see,
for example, Rosser (1964)). Therefore, one must rectify Lorrain’s results accordingly. For
example, one gets D = ε0 E inside a homogeneous uniformly rotating isolated conductor in
a magnetic field that is symmetrical about the axis of rotation. (Under steady-state conditions
E + v × B = 0 inside the conductor since the conduction current density vanishes, as Lorrain
pointed out.) As one can see, Lorrain’s equations concerning rotating bodies will become
correct after substituting εr = 1 in them and thus the troublesome relative permittivity of good
conductors disappears from final results (compare Jefimenko (1989)). By the way, if Lorrain’s
equation (46) were valid one could, in principle, measure the relative permittivity of good
conductors.
Also, it should be mentioned that Lorrain inadvertently used the same symbol J for
both the conduction and the total current densities. For example, in the Maxwell equation
∇ × B = µ0 J that is valid for a moving homogenous non-magnetic medium under steadystate conditions and when the curl of (E + v × B ) × v is zero (Rosser 1964), J is the total
current density that includes the convection as well as the conduction current densities. The
reader of Lorrain’s paper must therefore be cautious with J .
Especially, it appears that under steady-state conditions the divergence of the conduction
current density in the general case is not zero inside a moving medium. (The divergence of
the total current density is always zero under steady-state conditions.) Consequently, in the
general case the space charge density inside a moving homogeneous not necessarily rigid body
satisfies the equation
= ε0 [v · (∇ × B ) − B · (∇ × v )] − (ε0 εr /σ )∇ · (v )
(1)
at variance with Lorrain’s equation (4). For a rigid axisymmetric conductor that rotates
uniformly in a magnetic field that is symmetrical about the axis of rotation (the z axis), the
divergence of v is zero and the above equation together with ∇ × B = µ0 J imply Lorrain’s
equation (8), without εr , of course. As one can see, this is true for an isolated conductor as
well as for the Faraday disk connected to a stationary circuit through sliding contacts.
The present author believes that the above remarks could add to the pedagogical value of
Lorrain’s fine article.
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Letters and Comments
References
Jefimenko O D 1989 Electricity and Magnetism 2nd edn (Star City: Electret Scientific Company) ch 12
Lorrain P 1990 Eur. J. Phys. 11 94–8
Rosser W G V 1964 An Introduction to the Theory of Relativity (London: Butterworths) p 337
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LETTERS AND COMMENTS
Electrostatic charges in v × B fields
Paul Lorrain
Earth and Planetary Sciences, McGill University, Montréal, Québec, Canada
Received 17 August 2000
Abstract
Professor D V Redžić corrects an error that appeared in my 1990 paper
on electrostatic space charges in v × B fields. These charges are almost
invariably disregarded, despite the fact that they play a crucial role throughout
magnetohydrodynamics.
Many thanks to Professor D V Redžić for his corrections to my 1990 paper. The constitutive
equation for D in a moving medium is indeed the one that he gives, and not simply D = r 0 E ,
as I assumed. The Rosser book that he refers to is a solid reference, except for the fact that it
uses vectors only sparingly. The book by Penfield and Haus (1967) is also a good reference
on moving media, except that it does not state the constitutive equations, and that it makes no
reference to space charges.
Thus, as Professor Redžić correctly points out, if E + (v × B ) = 0, then, surprisingly,
D = 0 E . So the space-charge density inside a conductor that rotates in an axial, steady,
magnetic field is given by
= −0 ∇·(v × B ).
(1)
As to the convection current density v , it is negligible if v 2 c2 , at least in the case of
the Faraday disc as I have shown, and presumably more generally. So my equation (4) is in
general correct, after substituting 1 for r .
Two points are worth noting.
First, these space charges play an essential role whenever conductors move in magnetic
fields. The reason is that
J = σ [E + (v × B )]
(2)
and the E usually comes from . If you disregard the space charge, then you disregard E and
you get into absurd situations.
For example, it has been known since Thomson (1893) that there are no induced currents
in a conducting sphere that rotates in an axial magnetic field. This is extremely mystifying
because J = 0, despite the fact that v × B = 0! Moffatt (1978), who does not know about
the existence of , resorts to the following ‘explanation’: as the sphere rotates, the lines of B ,
inside, are stationary with respect to the sphere, and ‘disconnect and reconnect’ at the surface!
Second, very few authors know about the existence of this space charge and of its
field. Moffatt is just one example. Among the many thousands of papers and books on
the magnetohydrodynamics of the Earth, of the Sun, and of space, no author takes into account
this electric field. Rarely, an author will deduce the expression for (setting r = 1), but then
disregard its electric field.
Parker (1979) finds on page 46 of his book the correct expression for (setting r = 1),
but then goes on to say that, since ‘the universe is saturated with electric charge’, ‘a slight
rearrangement takes care of the induced electric field’. In other words, he is saying that the
Maxwell equation for the divergence of E does not apply! That statement is not too surprising
on his part because, throughout his book, he also disregards the Maxwell equation for the curl
of B .
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Letters and Comments
References
Lorrain P 1990 Eur. J. Phys. 11 94–8
Moffatt H K 1978 Magnetic Field Generation in Electrically Conducting Fluids (Cambridge: Cambridge University
Press)
Parker E N 1979 Comical Magnetic Fields (Oxford: Clarendon)
Penfield P Jr and Haus H 1967 Electrodynamics of Moving Media [Research Monograph No. 40] (Cambridge, MA:
MIT)
Redžić D V 2001 Eur. J. Phys. 22 L1
Rosser W G V 1964 An Introduction to the Theory of Relativity (London: Butterworths) p 337
Thomson J 1893 Notes on Recent Researches on Electricity and Magnetism (Oxford: Clarendon)