Download Current flow patterns in a Faraday disc

INSTITUTE OF PHYSICS PUBLISHING
EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 25 (2004) 171–183
PII: S0143-0807(04)65658-3
Current flow patterns in a Faraday disc
H Montgomery
Department of Physics and Astronomy, University of Edinburgh, Kings Buildings,
Edinburgh EH9 3JZ, UK
Received 7 July 2003
Published 2 December 2003
Online at stacks.iop.org/EJP/25/171 (DOI: 10.1088/0143-0807/25/2/004)
Abstract
A simple model is developed in order to calculate the flow of current in a
Faraday disc when it is delivering power to an external circuit. Several counterintuitive features emerge which are not revealed by a qualitative discussion.
The argument is non-relativistic, and it is conducted in the laboratory frame of
reference.
1. Introduction
In the course of his researches into electromagnetic induction, Faraday constructed a number
of systems similar to that shown in figure 1 [1]. A copper disc was mounted on an axle which
rested on the head of a cylindrical steel magnet, and a galvanometer was connected to the rim
and axle of the disc by brush contacts at points P and A. The disc and magnet could be rotated
independently about their common axis, while the external circuit containing the galvanometer
was held at rest. Faraday’s results can be summarized as follows:
Experiment 1. When the disc was rotated and the magnet was held stationary, a continuous
current was observed in the galvanometer.
Experiment 2. When the disc and magnet were cemented together so that they rotated at the
same speed, the current was again observed in the galvanometer.
Experiment 3. When the disc was held at rest and the magnet was rotated, no current was
observed in the galvanometer.
Hence the production of an induced current requires a relative motion of the disc and the
external circuit, and not as one might expect a relative motion of the disc and the magnet.
These experiments were later confirmed by Weber, who described the phenomenon as unipolar
induction, because he believed mistakenly that only one pole of the magnet was involved in
the process [2].
Faraday was convinced that the current was induced when the material of the disc cut
through lines of magnetic force. In order to explain experiment 2 he was led to suggest that
when the magnet rotated, its lines of force remained stationary [3]. However, other physicists
(including Weber) believed that the lines of force did rotate with the magnet, and that these lines
of force interacted both with the disc and with the external circuit. For example experiment 2
could be explained by saying that the ‘seat of the electromotive force’ had shifted from the
disc to the external circuit. It was impossible to distinguish experimentally between these two
hypotheses, because although it was easy to demonstrate the form of lines of force with the
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171
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H Montgomery
ω
B
B
A
I
Disc
P
N
Magnet
Figure 1. Faraday’s disc.
I
P
v
-eE
Q
-ev×B
B
Re
A
ω
I
Figure 2. Forces on electrons.
help of a sheet of paper and iron filings, there was no way of testing directly whether or not the
lines of force rotated with the magnet. Nowadays most physicists accept that lines of force are
a valuable way of describing a field, but are not real in themselves; a magnetic field is defined
solely by the value of the magnetic vector at each point in space.
Many student texts seem to regard the theory of Faraday’s disc as too elementary to merit
discussion, and they relegate it to a problem at the end of the chapter on electromagnetic
induction [4, 5]. My own experience is that students do not find the topic particularly easy,
and they have a justified scepticism about some of the explanations they are offered. Recently
a detailed account of the subject has appeared in the book by Lorrain et al [6], and a number of
research papers have discussed it from various points of view [7–10]. The present contribution
is a down-to-earth approach, concentrating on what actually happens inside the disc as it rotates.
2. The magnitude of the electromotive force (EMF)
In figure 2 we have a disc whose rim has a radius b, mounted on an axle of radius a, rotating
with angular velocity ω in a magnetic field B. Assume initially that the resistance Re of the
Currents in a Faraday disc
173
external circuit is infinite, so that the system is electrically isolated. Consider the situation
at a general point Q, whose position is described by cylindrical coordinates {r, φ, 0}. The
conduction electrons have taken up the rotational motion, and they have a mean drift velocity
v as shown. The magnetic force −ev × B has a component directed radially inwards, and this
forces conduction electrons to move towards the centre, leaving a positive charge at the rim.
An electrostatic field is set up by these charges, and in the steady state the mean Lorentz force
must be zero:
F = −eE − ev × B = 0.
(1)
(Owing to the low mass of the electrons, they rapidly move to a configuration in which the
total force on each electron averages to zero.)
The radial component of the electric field is given by:
E r = −vφ Bz = −ωBz r.
(2)
Hence the potential V between rim and axle is:
b
b
ω
ω
B
E r dr = +
2πr Bz dr =
V =−
2π a
2π
a
where B is the magnetic flux passing through the disc, not including the flux through the axle
itself.
Now place a resistor across the brush contacts, whose resistance Re greatly exceeds the
resistance of the disc. The effect of Re on the voltage V is negligible, and a current I = V /Re
flows round the circuit. The power dissipated in the system is V I , so that the device has an
EMF which is numerically equal to V :
ω
B .
E=
(3)
2π
Conservation of energy requires that a mechanical torque be applied to the disc in order to
keep it in constant motion.
The above account is correct as far as it goes, but it raises a number of questions. In an
isolated stationary conductor the internal electric field is zero, and any net charge lies on the
surface. However, if an isolated conductor is rotating in a magnetic field the internal Lorentz
force is zero, and the electric field has a non-zero divergence; the charge is distributed partly
as a surface charge and partly as a space charge, spread through the volume of the body. This
point has been emphasized strongly by Lorrain [7], and it is discussed fully by Bringuier [10].
The case of a Faraday disc is even more complicated, as here the Lorentz force is usually not
zero. Fortunately we shall not have to consider these charges explicitly, as our main concern
will be with the electric fields which the charges produce.
However, there is another problem which will concern us directly. When we calculated the
EMF we assumed that the resistance of the disc could be neglected, and this might imply that
the disc resistance is an unimportant detail, whose only effect is to reduce slightly the power
delivered to the external circuit. This is quite untrue: we shall find that the disc resistance
plays a vital role in the transformation of mechanical into electrical energy, and if the disc
were a perfect conductor no current would be induced. A more careful study is needed of the
processes occurring in the disc when it is delivering power to the external circuit.
3. Collision forces in a moving conductor
Unipolar induction is a problem in solid state physics as well as in electromagnetism, and we
need to review some elements of transport theory [11]. Initially we shall adopt a simple model
of a metal, in which the conduction electrons occupy the lower part of an energy band, such
that the energy of each electron varies quadratically with its wavevector k:
h̄ 2 k2
ε(k) =
(4)
2m ∗
where m ∗ is an effective electron mass.
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H Montgomery
ky
E
J
C
kx
Figure 3. Electron gas in field E.
In thermal equilibrium the electrons lie almost entirely inside a spherical Fermi surface,
which is centred on the origin in k space. When a macroscopic electric field E is applied
the conduction electrons become subject to two opposing influences. The electric field tends
to perturb the electron gas away from its state of thermal equilibrium, while collisions with
impurities and thermal vibrations in the crystal tend to restore that state. A dynamic equilibrium
is set up, in which the rates of perturbation and restoration are equal; the Fermi sphere comes
to rest at a displaced position (see figure 3).
Under these conditions an electric current is flowing through the metal, the current density
J being parallel to the applied field E. Each electron state has a group velocity:
h̄k
1 ∂ε(k)
vg (k) =
= ∗
h̄ ∂k
m
and the current density J is given by:
J = −nev
where n is the number of conduction electrons per unit volume, e is the electronic charge and
v is the group velocity averaged over all occupied electron states. (v is the basic definition of
the drift velocity, and it is equal to the group velocity at the point C in figure 3, at the centre of
the displaced Fermi sphere.)
The current density J is proportional to the applied field E:
J = σE
where the conductivity
ne2 τ
.
m∗
τ is the collision relaxation time for electrons at the Fermi surface. (For copper at room
temperature τ ≈ 10−14 s; it is a measure of the time taken to achieve dynamic equilibrium
after the field has been applied.) It is convenient to define a mobility µn :
µn = eτ/m ∗
σ = neµn .
From these equations it follows that:
(5)
v = −µn E.
Hence µn is the drift velocity in unit electric field. For copper at room temperature the mobility
is:
µn = 4.3 × 10−3 m2 V−1 s−1 .
σ =
Currents in a Faraday disc
175
ky
E
J
C′
B
kx
Figure 4. Electron gas in fields E and B.
Thus in an electric field of 1 V m−1 , the drift velocity in copper is 4.3 mm s−1 . This is
much smaller than the mean speed of the conduction electrons, and it is somewhat smaller than
the mechanical velocities found in a typical Faraday disc.
When a current is flowing, each electron experiences a force:
Fe = −eE.
Collisions with the crystal structure tend to counteract this force, and when averaged over the
whole assembly of conduction electrons they give rise to a ‘collision force’ Fc :
Fc = −Fe = +eE.
Using equation (5) we express Fc in the form:
e
Fc = − v.
µn
Now consider the situation in which the conductor is placed in an electric field E with
a magnetic field B at right angles to it, as shown in figure 4. (E and B are both defined in
the laboratory frame, in which the conductor is at rest.) The Fermi sphere is displaced in
both the k x and k y directions; as a result J and E are no longer parallel, and the drift velocity
v corresponds to the group velocity at point C . There is now a third force acting on each
electron; this is the magnetic force Fm , and its mean value is:
Fm = −ev × B.
(6)
So far we have considered the conductor to be stationary, but now imagine that it has
been set in motion with velocity v0 in the laboratory frame. Equation (6) still applies, but v is
defined in the laboratory frame, and so it has been augmented by the velocity v0 . Hence the
motion of the conductor affects the magnitude of Fm . On the other hand the collision force Fc
depends on the value of the drift velocity relative to the velocity of the crystal, and it has to be
written as:
e
Fc = − (v − v0 ).
µn
(Note that Fc depends on the drift velocity of the electrons, relative to the velocity of the crystal.
It does not depend directly on the forces which cause these velocities.)
In equilibrium we have:
Fe + Fm + Fc = 0.
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H Montgomery
I
P
Re
N
S
B
J
ω
I
J
A
ω
Figure 5. Eddy currents in an asymmetric B field.
According to our simple model, the basic field equation for a moving conductor is as
follows:
1
E+v×B+
(v − v0 ) = 0.
(7)
µn
We now have to solve this equation for a rotating disc.
4. Symmetry considerations
In order to simplify the problem we have to make the system as symmetrical as possible. If
the magnetic field were not symmetrical about the axis of rotation, eddy currents would be
generated in the disc as shown in figure 5. A mechanical torque would have to be applied
to the disc even in the absence of an external circuit, and much energy would be dissipated
internally. (This was a serious problem in the 19th century, when engineers were trying to
develop large unipolar generators commercially [2].) To avoid this problem we must ensure
that B has cylindrical symmetry about the axis of rotation, as shown in figure 1. To simplify
things further we will make B uniform over the whole disc.
A further problem arises if the brushes are point contacts as shown in figure 2. Electrons
are injected into the disc at point P, and the current density and electric field inside the disc
do not have cylindrical symmetry. We can avoid this situation by making the external circuit
cylindrically symmetric, making contact with the disc at all points around the rim. (See
figure 17.10b in [6].)
5. Solution of equation (7) for the Faraday disc
We are now able to solve the field equation (7) for a Faraday disc which is connected to an
external circuit. The magnetic field generated by the induced current will be regarded as
negligible compared with the applied field B. The free charge density is time-independent, so
that the electric field E is irrotational and is described by a scalar potential V (r ).
Let the disc have thickness s, inner radius a and outer radius b (see figure 6).
Consider the components of the various vectors at the general point Q, whose position is
r = {r, φ, 0}:
v0 = {0, ωr, 0}
v = {vr , vφ , 0}
B = {0, 0, B}
v × B = {vφ B, −vr B, 0}
Currents in a Faraday disc
177
v0
b
v
r
φ
a
Q
Fm
B
ω
Figure 6. Vectors v, v0 , and Fm at point Q.
curl E = 0
everywhere;
Eφ = 0
E = {E r , 0, 0}.
(Note that all of these components are independent of the azimuthal angle φ.)
We now take the radial and tangential components of equation (7):
1
vr = 0
E r + vφ B +
µn
1
0 − vr B +
(vφ − ωr ) = 0
µn
vφ = ωr + µn Bvr .
The charge densities are time-independent, so div J = −div (nev) = 0. The electron
density n is slightly disturbed by the density of space charge inside the disc, but this is negligible
compared with the total electron density, and n will be taken as constant:
1 ∂
div (v) =
(r vr ) = 0
r ∂r
λ
(8)
vr = −
r
vφ = ωr − µn Bλ/r
(9)
where λ is a positive constant to be determined.
1
vr
E r = −vφ B −
µn
1
= −Bωr + µn B 2 λ/r +
λ/r
µn
λ
= −Bωr +
(1 + (µn B)2 )
µn r
b
E r dr
V = V (b) − V (a) = −
a
λ
Bω(b2 − a 2 )
−
(1 + (µn B)2 ) ln(b/a)
=
2
µn
λ
=E−
(1 + (µn B)2 ) ln(b/a)
from equation (3).
µn
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H Montgomery
ω
Figure 7. Velocity flow line.
To determine λ we have to consider the external circuit. Let I be the total current round
the system:
I = −nevr 2πr s = 2πneλs = 2πσ sλ/µn
µn I
(10)
λ=
2πσ s
V = E − I (1 + (µn B)2 ) ln(b/a)/2πσ s
V = I Re = E − I Ri
where
(1 + (µn B)2 ) ln(b/a)
Ri =
.
(11)
2πσ s
(Ri is the internal resistance of the system.)
E = I (Ri + Re ),
I = E /(Ri + Re ).
Let α = (Ri + Re )/Ri
E
I =
.
(12)
α Ri
Rearranging the variables, we obtain the final equation for E r :
E /r
E r = −Bωr +
.
(13)
α ln(b/a)
Equations (8)–(13) complete the solution of the field equation (7).
We can now calculate flow lines for the charge carriers. Let φv (r ) be the function describing
a particular flow line for the drift velocities, and let φv (b) = 0.
dφv (r )
= vφ /vr = µn B − ωr 2 /λ
r
dr
φv (r ) = µn B ln(r/b) + ω(b2 − r 2 )/2λ.
The general form of this function is shown in figure 7. For the sake of clarity a very large value
for B (100 T) has been chosen, together with α = 2 and µn = 4.3 × 10−3 m2 V−1 s−1 , the
value for copper. At lower fields the arms of the spiral would be much more closely spaced.
Now consider the flow lines for the current density vector J. When a conductor is in
motion there are two contributions to the current—that of the electrons with drift velocity v,
and that of the positive ions of the crystal with velocity v0 . The ions have a charge density +ne
to counterbalance that of the electrons, and the total current density:
J = ne(v0 − v).
(14)
Currents in a Faraday disc
179
ω
Figure 8. Current flow line.
Re = ∞
V(r)
Re = Ri
/2
0
a
rm
r
b
Figure 9. Scalar potential V (r) inside the disc (a = 1 cm; b = 10 cm).
In the case of the rotating disc, J has the following components:
Jr = −nevr = +λne/r
Jφ = +ner ω − nevφ = +λneµn B/r
dφ j (r )
= Jφ /Jr = µn B
r
dr
φ j (r ) = µn B ln(r/a).
(15)
This flow line is shown in figure 8. (In this case we have chosen φ j (a) = 0; the complete
pattern would of course show lines emerging from the axle in all directions.) The magnetic
field used in figure 8 is 100 T, as in figure 7; for smaller and more realistic fields the curvature
of the current flow lines would be much less pronounced, and they would lie nearly parallel to
the radial directions. We shall return to these flow lines later in the discussion.
Finally we should look at the form of the electric field inside the disc. Integrating
equation (13), we obtain the following expression for V (r ):
2
ln(r/a)
(r − a 2 )
V (r ) = E
−
.
(b2 − a 2 ) α ln(b/a)
Figure 9 shows the form of V (r ) for two different values of Re .
When Re is infinite and the system is an open circuit, the force Fe is directed outwards,
and varies linearly with r as indicated in equation (2). However, when Re is progressively
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H Montgomery
reduced so that heavier currents are drawn from the disc, V (r ) develops a minimum at a radius
rm , and inside this radius Fe is directed inwards. When r > rm , Fe is impeding the flow of
electrons and is overcome by the stronger magnetic force Fm ; but when r < rm the electric
and magnetic forces are acting together in the same direction. This implies that the external
circuit disturbs the net charge densities in the disc. From equation (13) we have:
1 ∂
(r E r ) = −2Bω.
r ∂r
The space charge density is determined by the value of div E, and hence it is independent of
Re . Evidently the external circuit disturbs the surface charge on the disc, but not its space
charge [7].
div E =
6. The electromotive force and the internal resistance
The EMF arises from a close interplay between the magnetic and collision forces. When the
disc rotates the collision forces drag the electrons round with it, and as shown in figure 6
the magnetic force directs them towards the axle. (We are excluding the special case where
Re = ∞, and Fm is exactly cancelled by Fe .) This inward motion gives Fm a backward
tangential component which must be balanced by the collision force. As a result v never quite
catches up with v0 , and the tangential component of Fc is not zero.
Now consider the magnitude of this component:
e
Fcφ = − (vφ − ωr ) = +eλB/r.
µn
The total torque on the electrons due to the collisions is equal to:
b
2πσ sλB (b2 − a 2 )
=
ns2πr 2 Fcφ dr =
µn
2
a
(16)
I B
=
.
2π
The angular momentum of the disc is constant, so that a mechanical torque equal to has
to be applied to it. The mechanical power is as follows:
P = ω = I ωB /2π = E I.
This confirms that the EMF E is the amount of mechanical energy converted into electrical
energy per coulomb of charge circulated. (Equation (16) can also be derived by calculating
the torque m exerted by the magnetic forces on the electrons; m depends on the components
{vr }, not {vφ }, and in equilibrium it equals − [6].)
The reason why we used the first procedure was because it gives a vivid picture of
the transformation of energy. Collisions with the moving disc give the electrons additional
momentum in the tangential direction; the magnetic field performs no work on the electrons,
but it diverts their momentum towards the axle so that the current is driven round the circuit.
According to this argument there seems to be no justification for shifting the ‘seat of the EMF’
to the external circuit when one considers Faraday’s second experiment.
Now consider the Ohmic heating in the disc. When a current flows through a conductor
at rest, the rate of heating per unit volume is:
H=
1 2
ne 2
J =
v .
σ
µn
In the case of a conductor moving with velocity v0 , this becomes:
ne
H=
(v − v0 )2 .
µn
Currents in a Faraday disc
181
At the point Q in figure 6 we have:
ne 2
(v + (vφ − ωr )2 )
H=
µn r
ne λ2 λ2 µ2n B 2
+
=
µn r 2
r2
2
σλ
= 2 (1 + (µn B)2 )/r 2 .
µn
The total heat generated in the disc per second is:
b
σ λ2
dr
2
W = 2 (1 + (µn B) )
2πs
µn
r
a
σ I 2 (1 + (µn B)2 )
2πs ln(b/a)
(2πσ s)2
I 2 (1 + (µn B)2 ) ln(b/a)
=
2πσ s
from equation (11).
= I 2 Ri
=
The internal resistance can be written in the form:
Ri =
(1 + (µn B)2 ) ln(b/a)
.
2πneµn s
Usually µn B 1, and the term (µn B)2 is extremely small. If we increase the value of
µn —say by lowering the temperature of the disc—Ri decreases as we would expect. However,
it is possible in principle to have µn B > 1, and this would lead to a paradoxical situation in
which a decrease in the resistivity of the disc material would cause an increase of the internal
resistance of the system. As µn tends to infinity, I tends to zero. In this regime the system
would operate rather like a car with a slipping clutch. As the friction between the clutch plate
and the flywheel weakens, you have to rev the engine harder and harder to maintain a constant
traction on the road. Similarly if there is a progressive increase in the value of the mobility µn ,
you have to rotate the disc faster and faster to maintain a constant current through the circuit,
and this increases the heat generated in the disc. This of course would not apply to the more
usual situation in which µn B < 1.
7. Unipolar induction and the Hall effect
Consider equation (7) for a stationary conductor:
E+v×B+
1
v=0
µn
1
v−v×B
µn
1
1
(17)
E = J + J × B.
σ
ne
The component of E normal to J is the Hall field EH , which is usually written in the form:
E=−
EH = RH B × J
where RH is the Hall coefficient [11]. For the model we have used it follows that:
1
RH = − .
ne
(18)
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H Montgomery
The angle subtended between E and J is the Hall angle θ , where
tan θ = µn B.
(19)
Now return to the case of a conductor moving with velocity v0 .
Let v = v − v0 . J = −nev from equation (14). Equation (7) can be written in the form:
E + v0 × B + v × B + v /µn = 0
1
1
(20)
E + v0 × B = J + J × B.
σ
ne
Equations (17) and (20) are identical apart from the term v0 × B. In (17), J is inclined towards
E at the Hall angle θ : hence in (20) J is inclined towards the vector (E + v0 × B) at the same
Hall angle θ .
This result explains the form of the current flow line in figure 8. At all points inside the
disc, the vector (E + v0 × B) points outwards in the radial direction; J maintains a constant
bearing θ relative to this direction, where θ is given by equation (19). The flow line is the
logarithmic spiral defined by equation (15).
At this point we need to test the validity of our model by comparing equation (18)
with experimental values of RH . For monovalent alkali metals such as Na and K the
agreement is within a few per cent; but for monovalent noble metals such as Cu and Ag
the agreement is less impressive, the experimental value for Cu being about 70% of that given
by equation (18) [4, p 389]. For many polyvalent metals the model fails completely, as the
experimental values of RH are positive.
The trouble, of course, arises from equation (4); in most metals the Fermi surface is far
from being spherical. For example in Cu the Fermi surface has a roughly spherical ‘belly’, but
it touches the hexagonal faces of the Brillouin zone in a series of ‘necks’, and these reduce the
value of RH . This presents a serious problem for our model, as it would be very difficult to
reformulate it so as to deal with a non-spherical Fermi surface. (In this case the Fermi surface
is not merely displaced by the combined E and B fields, it also changes shape.)
However, we can generalize our argument to a considerable extent, by adopting what is
known as the ‘two-band model’ [11]. Here we have two spherical bands, the first which is
nearly empty and has a positive effective mass, and the second which is nearly full and has a
negative effective mass. The charge carriers consist of a density n of negative electrons with
mobility µn , and a density p of positive holes with mobility µp . The conductivity has the form:
σ = neµn + peµp
(21)
and the Hall coefficient:
RH = −
1 (nµ2n − pµ2p )
e (nµn + pµp )2
(22)
while the Hall angle is given by:
tan θ =
(µn B)nµn − (µp B) pµp
.
nµn + pµp
(23)
It is not difficult to apply this model to the Faraday disc. The torque is given correctly by
equation (16), and both electrons and holes gain energy from the collisions. Also the current
flow lines subtend an angle θ to the radial direction which is given in equation (23). If RH
is positive, Jφ is found to be negative. It may seem surprising that a negative value of Jφ
can coexist with a positive value of , but it must be remembered that the collision forces are
determined by the velocities of the carriers, whereas the currents also depend on their charges.
One can generalize this argument still further by using a Galilean transformation [10]. In
an appendix we show that in any Faraday disc the current flow lines are inclined to the radial
direction at the Hall angle, regardless of the electronic structure of the material. This allows
Currents in a Faraday disc
183
us to calculate the current density for the general case. The circulating current I is given by
equation (12). and the components of the current density are as follows:
Jr = I /2πsr ;
Jφ = −σ RH B Jr
where RH is the experimental value of the Hall coefficient. The mechanical torque is always
given by equation (16); it can be derived from Jr , but not in general from Jφ .
8. Conclusions
An attempt has been made to provide an exact solution for the electrodynamics of a Faraday
disc. This attempt is partly successful, although it runs into complications when the material
of the disc has an anomalous Hall coefficient. From an educational point of view, the chief
merit of the present approach is that it gives a clear picture of the physical processes involved.
It is argued that collisions between the electrons and the crystal structure are not merely a
passive process leading to the production of waste heat; they are an active mechanism in the
conversion of mechanical into electrical energy.
Acknowledgment
I would like to thank Mr Stuart Leadstone for his careful reading of the manuscript, and for
several useful suggestions.
Appendix. Galilean transformation for a Faraday disc
Let S be the laboratory frame of reference.
Let S be a frame moving with a uniform velocity v0 relative to S. In general the two
frames are related by a Lorentz transformation, but if v0 c this assumes the limiting form
of a Galilean transformation:
r = r − v0 t;
t = t;
v = v − v0 .
Consider a small element of the disc which is moving with velocity v0 in S. This element
is momentarily at rest in S .
The current density J depends on the mean velocity of the electrons, relative to the velocity
of the crystal; it is invariant under the transformation.
J = J.
The fields transform as follows:
E = E + v0 × B;
B = B.
In S , J is inclined to E at the Hall angle. Hence in S, J is inclined to the vector E + v0 × B
at the Hall angle.
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[10]
[11]
Faraday M 1932 Faraday’s Diary vol 1 (London: Bell & Sons) p 402 para 255
Miller A I 1981 Ann. Sci., NY 38 155–89
Faraday M 1965 Experimental Researches in Electricity vol 1 (New York: Dover) p 64 para 220
Bleaney B I and Bleaney B 1976 Electricity and Magnetism 3rd edn (London: Oxford University Press)
Grant I S and Phillips W R 1990 Electromagnetism 2nd edn (Chichester: Wiley)
Lorrain P, Corson D R and Lorrain F 2000 Fundamentals of Electromagnetic Phenomena (New York: Freeman)
p 305
Lorrain P 1990 Eur. J. Phys. 11 94–8
Montgomery H 1999 Eur. J. Phys. 20 271–80
Guala-Valverde J, Mazzoni P and Achilles R 2002 Am. J. Phys. 70 1052–5
Bringuier E 2003 Eur. J. Phys. 24 21–9
Ziman J M 1964 Principles of the Theory of Solids (Cambridge: Cambridge University Press) (many good
textbooks on solid state physics cover this topic)