1) The experiment involves dropping a magnet through a hollow conductive tube and analyzing the damped fall to study the tube's conductivity.
2) An equivalent circuit model is used to relate the damping coefficient to the tube's geometry, conductivity, and other parameters.
3) By equating the potential energy lost by the falling magnet to the energy dissipated in the tube via Joule heating, an expression is derived relating the induced electromotive force to the magnet's speed and the tube dimensions. This allows calculating the tube's conductivity.
I am William T. I am a Solid Mechanics Assignment Expert at solidworksassignmenthelp.com. I hold a Bachlor's Degree in Engineering from McMaster University, Canada. I have been helping students with their Assignments for the past 10 years. I solve Assignments related to Solid Mechanics.
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I am William T. I am a Solid Mechanics Assignment Expert at solidworksassignmenthelp.com. I hold a Bachlor's Degree in Engineering from McMaster University, Canada. I have been helping students with their Assignments for the past 10 years. I solve Assignments related to Solid Mechanics.
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I am Baddie K. I am a Magnetic Materials Assignment Expert at eduassignmenthelp.com. I hold a Masters's Degree in Electro-Magnetics, from The University of Malaya, Malaysia. I have been helping students with their assignments for the past 12 years. I solve assignments related to Magnetic Materials.
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Bessel light beams satisfy a wave equation but, in addition, they are solutions of Maxwell’s equations. The light power they convey is proportional to the square modulus of the electric field E. In cylindrical coordinates, Bessel light beams depend on the polarization of E and, assuming an axisymmeteric beam, we give the ex-pressions of E when its polarization is, azi-muthal, radial and linear or circular. The energy flux carried by these beams is obtained from the time averaed Poynting vector.
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build-up in the channel. We propose in this frame work an algorithm of simulation based on mathematical expressions obtained previously. We propose a new mobility model describing the electric field-dependent. predictions of the simulator are compared with the experimental data [1] and
have been shown to be good.
Effect of Mobility on (I-V) Characteristics of Gaas MESFET IJECEIAES
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1 ECE 6340 Fall 2013 Homework 8 Assignment.docxjoyjonna282
1
ECE 6340
Fall 2013
Homework 8
Assignment: Please do Probs. 1-9 and 13 from the set below.
1) In dynamics, we have the equation
E j Aω= − −∇Φ .
(a) Show that in statics, the scalar potential function Φ can be interpreted as a voltage
function. That is, show that in statics
( ) ( )
B
AB
A
V E dr A B≡ ⋅ = Φ −Φ∫ .
(b) Next, explain why this equation is not true (in general) in dynamics.
(c) Explain why the voltage drop (defined as the line integral of the electric field, as
defined above) depends on the path from A to B in dynamics, using Faraday’s law.
(d) Does the right-hand side of the above equation (the difference in the potential
function) depend on the path, in dynamics?
Hint: Note that, according to calculus, for any function ψ we have
dr dx dy dz d
x y z
ψ ψ ψ
ψ ψ
∂ ∂ ∂
∇ ⋅ = + + =
∂ ∂ ∂
.
2) Starting with Maxwell’s equations, show that the electric field radiated by an impressed
current density source J i in an infinite homogeneous region satisfies the equation
( )2 2 iE k E E j Jωµ∇ + = ∇ ∇⋅ + .
Then use Ampere’s law (or, if you prefer, the continuity equation and the electric Gauss
law) to show that this equation may be written as
( )2 2 1 i iE k E J j J
j
ωµ
σ ωε
∇ + = − ∇ ∇⋅ +
+
.
2
Note that the total current density is the sum of the impressed current density and the
conduction current density, the latter obeying Ohm’s law (J c = σE).
Explain why this equation for the electric field would be harder to solve than the equation
that was derived in class for the magnetic vector potential.
3) Show that magnetic field radiated by an impressed current density source satisfies the
equation
2 2 iH k H J∇ + = −∇× .
Explain why this equation for the magnetic field would be harder to solve than the
equation that was derived in class for the magnetic vector potential.
4) Show that in a homogenous region of space the scalar electric potential satisfies the
equation
2 2
i
v
c
k
ρ
ε
∇ Φ + Φ = − ,
where ivρ is the impressed (source) charge density, which is the charge density that goes
along with the impressed current density, being related by
i ivJ jωρ∇⋅ = −
Hint: Start with E j Aω= − −∇Φ and take the divergence of both sides. Also, take the
divergence of both sides of Ampere’s law and use the continuity equation for the
impressed current (given above) to show that
1 ii v
c c
E J
j
ρ
ωε ε
∇⋅ = − ∇⋅ = .
Note: It is also true from the electric Gauss law that
vE
ρ
ε
∇⋅ = ,
but we prefer to have only an impressed (source) charge density on the right-hand side of
the equation for the potential Φ. In the time-harmonic steady state, assuming a
homogeneous and isotropic region, it follows that ρv = ρvi. That is, there is no charge
3
density arising from the conduction current. (If there were no impressed current sources,
the total charge density would therefore be ze ...
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Falling magnet
1. INSTITUTE OF PHYSICS PUBLISHING EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 25 (2004) 593–604 PII: S0143-0807(04)76816-6
Study of the conductivity of a metallic
tube by analysing the damped fall of a
magnet
J ´I˜niguez, V Raposo, A Hern´andez-L´opez, A G Flores
and M Zazo
Department of Applied Physics, University of Salamanca, E-37071 Salamanca, Spain
Received 26 February 2004
Published 10 June 2004
Online at stacks.iop.org/EJP/25/593
doi:10.1088/0143-0807/25/5/002
Abstract
The fall of a magnet through a hollow conducting tube is described. Although
this experiment is well known, a detailed treatment by means of a circuit
analysis allows us to relate the conductivity of the tube to the characteristic
parameters of the experiment.
1. Introduction
A very common experiment in undergraduate physics laboratories is the study of the damped
fall of a magnet through a conductive hollow tube [1]. This experiment is usually done as
a qualitative demonstration to show the effects of electromagnetic induction associated with
Faraday’s law, and not much attention is given to a quantitative analysis. Here we discuss a
thorough study of the phenomenon. Although an exact calculation of the currents induced in
the tube, which are responsible for the damping of the falling magnet, is very complicated, a
reasonable approach is to examine the currents in terms of a single generator of electromotive
force (emf) connected to an assembly of resistor circuits that represent the behaviour of the
conducting tube. An analysis of this resistor network in terms of circuit theory allows one to
relate the parameters involved in the problem (the conductivity and the damping coefficient) to
the geometry and electromagnetic characteristics of the apparatus by means of the equivalent
circuit concept. The validity of this approach is analysed for several magnets. Good agreement
between the experimental results and the conductivity reported in the literature is obtained.
This exercise could be proposed as an experimental task for students of an intermediate course
on electromagnetism.
0143-0807/04/050593+12$30.00 c 2004 IOP Publishing Ltd Printed in the UK 593
2. 594 J ´I˜niguez et al
c
b
a
h
zl
p
q N
S
dz
Figure 1. Description and dimensions of the metallic tube. The currents induced above and below
the magnet are shown.
2. The description of the problem
Figure 1 shows a diagram of the experimental system. The induced currents above and below
the falling magnet are shown. Those above attract the magnet whereas those below repel
it, thereby yielding the damping effect that we wish to study. Although a rigorous analysis
is very complicated and would require the use of numerical methods, we shall demonstrate
how the analysis of a simple circuit lets us relate the falling speed and the geometrical and
electromagnetic properties of the system to the conductivity of the tube and the damping
coefficient.
The equation of motion is
M
dv
dt
= Mg − kv, (1)
where k (disregarding the aerodynamic friction due to extremely low speed, as we will see
later) represents the damping coefficient associated with the induced currents and M is the
mass of the magnet. If we integrate equation (1), we obtain
v(t) =
Mg
k
1 − e− k
M t
. (2)
It is observed that the asymptotic speed (Mg/k) is reached almost instantaneously
(τ = M/k 1 s). For this reason, we concluded that, for a tube length l of several
decimetres, the speed can be considered practically uniform. For example, for a copper tube
with dimensions l = 1 m, inner radius b = 6.5 mm, outer radius c = 7.5 mm (see figure 1)
and a SmCo magnet of mass M = 9.3 g, height h = 10 mm and radius a = 6 mm, a falling
time of 12.3 s is obtained. We thus find
v =
l
t
=
Mg
k
≈ 81 mm s−1
, k ≈ 1.12 kg s−1
, τ =
M
k
≈ 8 ms. (3)
3. Study of the conductivity of a metallic tube 595
Φ0
∆t t
t
Φ(t)
(t)ε
Figure 2. The form of the time dependence of the magnetic flux and induced electromotive force.
(See, for example, the third paper quoted in [1].)
It is difficult to relate the damping coefficient k to the geometrical and electromagnetic
parameters of the experiment, but we shall do so approximately by using energy conservation.
It suffices to say that the variation in the potential energy of the magnet is equivalent to the
energy dissipated in the tube through the Joule effect.
When the magnet falls inside the tube, the magnetic flux, (t), linked by ‘the loop’ (a tube
section z in height and resistance R) and the emf, ε(t), induced in it have the form depicted
in figure 2, where 0 is the maximum flux and t is the characteristic time of the process. If
we neglect the self-inductance of ‘the loop’ (an issue that will be addressed later), the average
of the absolute value of the induced emf is related to the mean value of the modulus of the
current as
ε ≈
0
t/2
≈ IR. (4)
It is not easy to evaluate the equivalent resistance, R, that the magnet feels during its fall,
that is, the equivalent circuit. This task is difficult because the emf is induced throughout the
entire tube, with opposite directions above and below the magnet, although its value diminishes
quickly away from the magnet. In fact, the magnetic induction associated with the magnet
decreases at least as 1/r3
, where r is the distance from the magnet [2].
3. The equivalent circuit approximation
The difficulty of the calculation of R and the rapid decrease in the induced emf with the
distance to the magnet suggest that the problem may be simplified by considering a single
falling generator, instead of the magnet, that feeds the equivalent resistance R corresponding
to a length z of the tube. To explain this idea, the equivalent circuit is shown in figure 3.
In this way, we can model ε(z) by a simple resistor network. The series resistance,
α dz, where α is the series resistance per unit length, corresponds to that of a section of
the tube, dz, throughout its length, and serves only to feed the loops that actually form
the tube. The loops correspond to a parallel conductance, β dz, where β is the parallel
conductance per unit length, which describes the same elementary section (dz) throughout its
circumference:
4. 596 J ´I˜niguez et al
z
V(z)
I(z)
I(z+dz)
dR= dz
dG= dz
V(z+dz)
ε
I
q=l-p
p
dI
β
α
Figure 3. Equivalent circuit used in our circuital approximation. The lengths p, q and z are
labelled.
dR = α dz =
1
σπ(c2 − b2)
dz (5a)
dG = β dz =
σ ln(c/b)
2π
dz, (5b)
where σ is the conductivity of the material of the tube.
It is to be remarked that actually the currents above and below the falling magnet flow in
opposite directions around the tube, as is depicted in figure 1. The fact that in our equivalent
circuit the currents circulate in the same direction (see figure 3) is not a problem because the
power balance that we will carry out corresponds to the square of the current (or voltage), all
the directional aspects disappearing in this way.
Using circuit analysis, we evaluate the equivalent resistance, R, which the magnet feels.
For an elementary section of the tube located at a distance z from the magnet, we have
V (z) − V (z + dz) = I(z)α dz (6a)
I(z) − I(z + dz) = −V (z)β dz, (6b)
5. Study of the conductivity of a metallic tube 597
that is, current and voltage in the tube satisfy the relations
dV (z)
dz
= −αI(z) (7a)
dI(z)
dz
= −βV (z). (7b)
If we take the derivative of equations (7) with respect to z, we obtain two second-order
differential equations. After integration we have
V (z) = V+ e−
√
αβz
+ V− e+
√
αβz
(8a)
I(z) = I+ e−
√
αβz
+ I− e+
√
αβz
. (8b)
If we take the derivative again and use equations (7), the result is
V+
I+
= α/β = −
V−
I−
. (9)
Thus, the current and voltage in the tube can be described by
V (z) = V+ e−
√
αβz
+ V− e+
√
αβz
(10a)
I(z) =
1
√
α/β
(V+ e−
√
αβz
− V− e+
√
αβz
), (10b)
where the quantity R0 =
√
α/β plays the role of a characteristic resistance, because for
z = ∞, V− = 0 and V (z)/I(z) =
√
α/β.
In agreement with figure 3, from the magnet (the generator) we recognize two resistors
in parallel, Rp and Rq, associated with the two parts of the tube length l = p + q. Rp and
Rq obviously correspond to tube sections terminating in an open circuit, RL = ∞, which is
why according to the resistance transformation law for transmission lines and distances p and
q (see appendix C), we can write [3, 4]
Rp = R0
RL + R0 tanh(
√
αβp)
R0 + RL tanh(
√
αβp)
=
R0
tanh(
√
αβp)
(11a)
Rq = R0
RL + R0 tanh(
√
αβq)
R0 + RL tanh(
√
αβq)
=
R0
tanh(
√
αβq)
, (11b)
and therefore,
R =
RpRq
Rp + Rq
=
R0
tanh(
√
αβp) + tanh(
√
αβ(l − p))
. (12)
Finally,
R =
√
α/β
tanh(
√
αβl)[1 + tanh(
√
αβp) × tanh(
√
αβ(l − p))]
(13)
is the equivalent resistance with which the conductive tube of length l behaves in terms of the
location, p, of the magnet (see figure 3).
The actual dependence of the equivalent resistance on p is very simple. In fact, for typical
values of α, β, p and l (see the data of our example in section 2 and use equations (5) with
a metallic conductivity value for σ), all the expressions in the argument of the hyperbolic
6. 598 J ´I˜niguez et al
tangent are close to unity, which is why the equivalent circuit almost corresponds to a constant
resistance given by
R ≈
R0
2
=
√
α/β
2
=
1
σ 2(c2 − b2) ln(c/b)
. (14)
However, at the ends of the tube (p = 0 and p = l) the resistance is exactly twice this value.
4. Calculation of the conductivity
According to equation (5b) and using the equivalent resistance defined in equation (14), we
show that the phenomenon occurs roughly as if the induced emf were operating on a single
circular loop, with the current entirely azimuthal, of height z given by
2π
σ ln(c/b)
1
z
=
1
σ 2(c2 − b2) ln(c/b)
. (15)
Thus, we have
z = 2π
2(c2 − b2)
ln(c/b)
. (16)
By using equation (16), and according to equation (4), we can write the induced emf in
terms of the speed, v = z/ t, of the magnet as
ε ≈
0
t/2
=
2 0
z/v
=
0
π 2(c2−b2)
ln(c/b)
v. (17)
We can carry out the energy balance by integrating over the currents in the circular
elements β dz. In fact, in the tube there are only azimuthal induced currents. The series
resistances α dz were introduced only to feed these circular loops, so that the complicated
dependence in ε(z) could be modelled using Ohm’s law applied to our resistor network.
Therefore, power conservation is given by
p
0
V 2
(z)β dz +
q
0
V 2
(z)β dz = Mgv. (18)
In agreement with our analysis and the geometrical and electromagnetic characteristics of the
system, we can consider the copper tube as a kind of infinite transmission line (V− ≈ 0) and,
except when the magnet is close to the ends of the tube, we can write
V (z) ≈ V+ e−
√
αβz
= ε e−
√
αβz
, (19)
for 0 z p and 0 z q.
We substitute equation (19) in equation (18), integrate, and assume (in accordance with
typical values for α, β, p and q) that
e−2
√
αβp
≈ e−2
√
αβq
≈ 0, (20)
and obtain
ε2
√
α/β
= Mgv. (21)
If we substitute ε, α and β (see equations (17) and (14)) in equation (21), we have
2 2
0v2
π2
σ
ln3
(c/b)
2(c2 − b2)
= Mgv, (22)
7. Study of the conductivity of a metallic tube 599
from which follows the expression that relates the conductivity of the tube and the speed of
the falling magnet to the geometrical and electromagnetic characteristics of the experiment:
σ =
√
2
2
π2 Mg
2
0
c2 − b2
ln3
(c/b)
1
v
. (23)
Finally, we use equation (3) to find the expression for the damping coefficient:
k =
Mg
v
=
√
2
π2
2
0σ
ln3
(c/b)
c2 − b2
. (24)
5. Experimental evaluation and discussion
Equations (23) and (24) allow us to measure the conductivity of the tube and the damping
coefficient by analysing the falling magnet. It is necessary to know the tube geometry, the
mass of the magnet, its asymptotic speed and its magnetic flux with great precision.
For the usual dimensions of the magnet (see the data of our example in section 2), an
imprecise knowledge of the demagnetizing field and the difficulties involved in evaluating the
return flux of the magnetic lines of force suggest that we should ignore the manufacturer’s
information or numerical methods to evaluate the magnetic flux 0 and instead measure it
experimentally. By using a simple integrator and a small coil of radius (b + c)/2, we obtained
0 = 92 µWb for the SmCo magnet. (We used a ballistic galvanometer [5] and results
were checked with an integrator based on an operational amplifier.) After carrying out the
calculations for this example, we found z ≈ 88 mm, ε ≈ 170 µV and, using the conductivity
of copper (59.8 × 106 −1
m−1
) [6], we found R ≈ 8.5 µ , the equivalent resistance through
which a current of the order of 20 A would circulate.
The result for z suggests measuring the magnet speed by using two sensing coils located
several centimetres away from the ends of the tube. The conductivity we obtained with our
calculation was 64 × 106 −1
m−1
which differs by about 7% from the published value [6],
an error that can be considered reasonable in the context of our analysis.
We conclude by considering the accuracy of our two approximations: the substitution
of a distributed emf by a single generator and the use of an equivalent circuit to represent
the behaviour of the tube. The discrepancies are essentially associated with these two
approximations. Indeed, instead of calculating the emf, ε(z), created by the magnet during
its fall analytically (bearing in mind that the magnet does not correspond to a point dipole,
see appendix A), we arranged longitudinal resistances, α dz, to feed the circular loops, β dz.
We simply replaced the superposition of the contributions of the form Bnz−n
with n 3
(corresponding to the multipole expansion of the magnet’s induction [2]) by an exponential
dependence of V (z) (recall the power series expansion of the exponential function [7]). This
exponential decay allows us to approximate the induction phenomenon throughout the tube
in a simple and reasonable way, although it introduces a small error in the results for the
conductivity.
On the other hand, no appreciable error is made when the self-inductance of the equivalent
circuit is not considered. The small error introduced by neglecting self-inductance can be
readily evaluated by comparing the inductive and Ohmic contributions. The self-inductance
of a length z of equivalent resistance R is approximately [5]
L =
µ0π(b + c)2
4 z
, (25)
which, although the problem does not correspond to sinusoidal excitation, allows us to evaluate
8. 600 J ´I˜niguez et al
Table 1. Summary of the results for three different magnets.
Magnetic flux Mass Speed Conductivity Relative error
Magnet type (µWb) (g) (mm s−1) (106 −1 m−1) (%)
Sm2Co17 92 9.3 81 64 7.0
NdFeB235 96 8.0 58 70 17.1
Barium ferrite 32 5.3 357 68 13.7
the inductive reactance as the one corresponding to the first harmonic:
ω =
2π
t
=
2π
z
v. (26)
The ratio Lω/R is given by
Lω
R
=
µ0π2
2
b+c
z
2
v
1
σ
√
2(c2−b2) ln(c/b)
=
µ0π2
(b + c)2
16
Mg
2
0
, (27)
which is of the order of 10−3
for our example, fully justifying the approximation.
The error introduced when using the average value of the emf induced during the interval
t/2 also could be taken into account. Because a power balance is considered, it would be
necessary to work with the rms value of the emf, although due to the form of ε(t), it is to be
hoped that the error will be very small.
There are of course other sources of error, such as accurate determination of the geometry
and the measurement of the magnetic flux, 0. In any case, performing the experiment
with several conductive tubes of the same dimensions (copper, aluminium and other non-
magnetic alloys) readily allows one to compare their conductivities. In fact, within the limit
of infinite conductivity (a superconducting loop), one would have an experiment of zero speed
or magnetic levitation.
It is also interesting to do experiments for several magnets and tubes with different wall
thicknesses. The damping of the fall increases dramatically with the tube thickness. We
emphasize that for tubes with very thick walls, the evaluation of the magnetic flux associated
with the return of the induction lines of force demands special attention. We repeated the
experiment with two different magnets with the same geometry. In table 1 we summarize our
results, including those of the example used in our previous calculations.
All these small inaccuracies can be considered by introducing a calibration constant C
into equation (22):
C
2 2
0v2
π2
σ
ln3
(c/b)
2(c2 − b2)
= Mgv, (28)
where C corresponds to an empirical coefficient, σexperimental/σtrue, close to unity (1.07 for our
example, with the SmCo magnet) that, after experimental measurement, will allow us to take
into account such inaccuracies and to obtain a better value for the conductivity.
To obtain satisfactory results, the experiment must be performed using strong magnets.
The relative errors associated with the determination of the magnetic flux and the speed when
operating with the low-field barium ferrite magnet are greater (see table 1). Nickel-coated
magnets (for example, NdFeB magnets) also are inadequate because their exact dimensions
are not well known and their rounded shape at the boundaries is a source of distortion in the
magnetic lines of force. Their true size, which is smaller than the apparent one, increases the
return of magnetic flux, making the precise determination of 0 difficult.
Our expressions can be verified by repeating the experiment with a non-magnetic
additional mass on the magnet. (Note that if changes are made by varying the magnet,
9. Study of the conductivity of a metallic tube 601
0
2
4
6
8
10
12
14
0 0.02 0.04 0.06 0.08 0.1 0.12
1/M (g
-1
)
t(s)
Figure 4. The dependence of the falling time on the mass of the magnet. The fit corresponds to
the experimental data obtained by loading the magnet with different amounts of Pb pellets.
the quantity 0 also will be modified due to the variation in the demagnetizing factor.) An
example of this experiment can be seen in figure 4 for our SmCo magnet. From equation (23),
we can write
t =
√
2
π2g
ln3
(c/b)
c2 − b2
2
0lσ
1
M
, (29)
where t represents the time that the magnet takes to fall through a tube of length l. From
the slope of the time versus inverse mass (see figure 4), the value of the conductivity is
readily obtained. For that purpose, we loaded the magnet with different amounts of Pb pellets
up to 10 g.
Acknowledgment
We wish to thank J Garc´ıa (laboratory technician of the Department of Applied Physics) for
helping us in the manufacture of the experimental equipment developed for this work.
Appendix A. The dipolar contribution
Although it is very complicated to do a rigorous analysis of the distribution of currents induced
in the tube, it is simple to evaluate only the magnetic dipole contribution. In agreement with
the usual dimensions of the experimental device, the magnet cannot be considered as a simple
point dipole, and it is necessary to include higher order multipole contributions.
According to figure 5, the magnitude of the induced azimuthal electric field, E, in a
filamentary loop of radius ρ (b ρ c) at a distance z from the magnet (represented by its
magnetic moment m) in terms of the magnetic vector potential, A, is
E(ρ, z) =
∂A(ρ, z, t)
∂t
= v
∂A
∂z
= v
∂
∂z
µ0
4π
m × r
r3
= v
∂
∂z
µ0mρ
4π(ρ2 + z2)3/2
= v
3µ0m
4π
ρz
(ρ2 + z2)5/2
. (30)
The power dissipated in the tube due to the dipole contribution will be
Pdip =
tube
σE2
dv = σv2 9µ2
0m2
16π2
p
−q
c
b
ρ2
z2
(ρ2 + z2)5
2πρ dρ dz. (31)
10. 602 J ´I˜niguez et al
c
b
a
hm
z
r
ρ
Figure 5. Notation used in the dipolar contribution calculation.
We can do the integral in equation (31) using that b and c are much smaller than p and q, and
obtain
Pdip ≈ σv2 15µ2
0m2
1024
1
b3
−
1
c3
. (32)
Therefore, the balance of energy, Pdip = Mgv, allows us to write
σ =
1024
15
Mg
µ2
0m2
c3
b3
c3 − b3
1
v
, (33)
which is to be compared with equation (23).
To evaluate the accuracy that is obtained by including only the magnetic dipole
contribution, we determined the magnetic moment of the SmCo magnet experimentally. To
do so, we suspended it from a very thin cotton thread and measured its oscillating period
in the presence of the Earth’s magnetic field. The value of its horizontal component in our
laboratory, measured with a tangent compass, is Bh = 25.2 µT, yielding a period T = 0.447 s.
From the expression [8]
T = 2π
J
mBh
, (34)
where J represents the inertial moment of the magnet (J = Ma2
/4 + Mh2
/12), the magnetic
moment was found to be 1.26 A m2
. If we substitute this result in equation (33), the conductivity
is found to be 24 × 106 −1
m−1
, which corresponds to approximately 40% of the published
value [6], thus illustrating the need to include higher order multipole contributions, or to use
the method presented in section 3.
Appendix B. Other approximations
Instead of proceeding to carry out a standard multipole expansion, we would prefer to first try
some other options in order to reach better approximation.
11. Study of the conductivity of a metallic tube 603
The most convenient option corresponds to considering a collection of n equal point
dipoles, of magnetic moment m/n, uniformly spaced at distances h/n in the symmetry axis of
the tube representing the falling magnet. In this way, the calculation of the induced azimuthal
electric field, E, corresponds to a simple superposition of n contributions as in equation (30).
But according to equation (31), the dissipated power would be a complicated polynomial
expression with (n2
+ n)/2 contributions. Except for small values of n it constitutes a tedious
problem without major interest.
Another possibility would be the substitution of the magnet by two opposite surface
densities of magnetic pole strength at a distance h apart. The calculation is very difficult
because this model involves sources out of the symmetry axis, and it does not contribute
anything to the better understanding of the problem.
Appendix C. Derivation of expressions (5), (11) and (25)
To obtain the differential longitudinal resistance, dR in equation (5a), we will consider a
uniform current distribution in a tube section of length dz and cross-sectional area π(c2
− b2
).
If σ represents its conductivity, we have
dR =
1
σπ(c2 − b2)
dz = α dz, (35)
where α corresponds to the series resistance per unit length of the tube.
The expression for the azimuthal differential conductance, dG in equation (5b), is derived
taking into account the circular currents in the loop:
dG =
dI
V
, (36)
where dI represents the azimuthal current around a section of height dz and V corresponds to
the potential difference throughout the loop. According to the symmetry, we can write
dI =
c
b
J · dS = σ
c
b
E · dS =
σV
2π
dz
c
b
dr
r
=
σV
2π
ln
c
b
dz. (37)
And therefore
dG =
σ ln(c/b)
2π
dz = β dz, (38)
β being the parallel conductance per unit length.
Equations (11) are obtained from expressions (10) according to circuit theory of
transmission lines, considered as a distributed parameter network [3–5]. For that purpose, we
start writing equations (10) in terms of the distance p − z or q + z from the ends (the load RL)
of the tube (see figure 1). For the bottom end, at a distance p from the magnet, we have
V (p) = V+ e−
√
αβp
+ V− e+
√
αβp
= RLI(p)
V+ e−
√
αβp
− V− e+
√
αβp
= R0I(p),
(39)
and in a similar way we can find the expressions for the other end of the tube at a distance q
from the magnet.
Combining both expressions, one has
V+ = 1
2
(RL + R0) e+
√
αβp
I(p)
V− = 1
2
(RL − R0) e−
√
αβp
I(p),
(40)
12. 604 J ´I˜niguez et al
and equations (10) can be rewritten as
V (z) =
I(p)
2
[(RL + R0) e+
√
αβ(p−z)
+ (RL − R0) e−
√
αβ(p−z)
]
I(z) =
I(p)
2R0
[(RL + R0) e+
√
αβ(p−z)
− (RL − R0) e−
√
αβ(p−z)
].
(41)
It helps us to define a new quantity ζ = p − z which allows us to find the resistance along
the tube (looking towards the terminating impedance RL):
R(ζ) =
V (ζ)
I(ζ)
= R0
(RL + R0) e+
√
αβζ
+ (RL − R0) e−
√
αβζ
(RL + R0) e+
√
αβζ − (RL − R0) e−
√
αβζ
. (42)
Finally, using the hyperbolic functions, the resistance along the tube is
R(ζ) = R0
RL + R0 tanh(
√
αβζ)
R0 + RL tanh(
√
αβζ)
. (43)
For distances ζ = p or ζ = q, and taking into account that RL = ∞, we find the desired
equations:
Rp =
R0
tanh(
√
αβp)
Rq =
R0
tanh(
√
αβq)
.
(44)
Expression (25) can be easily justified using the self-inductance per unit length of an
infinite coil of average radius (b + c)/2. Assuming a uniform current distribution, J, and
neglecting the edge effects, the magnetic induction inside the tube is
B ≈ µ0
I
z
= µ0J(c − b), (45)
where I/ z represents the azimuthal current density per unit length in the conductive tube.
For a height z, the self-induction will be
L =
I
=
B · S
I
≈
µ0
z
π
b + c
2
2
. (46)
References
[1] See, for example, Hahn K D, Johnson E M, Brokken A and Baldwin S 1998 Eddy current damping of a magnet
moving through a pipe Am. J. Phys. 66 1066–76
Kingman R, Rowland S C and Popescu S 2002 An experimental observation of Faraday’s law of induction Am.
J. Phys. 70 595–8
Singh A, Mohapatra Y N and Kumar S 2002 Electromagnetic induction and damping: quantitative experiments
using a PC interface Am. J. Phys. 70 424–7
Sawicki C A 2000 A Lenz’s law experiment revisited Phys. Teach. 38 439–41
Seealso
http://store.pasco.com/pascostore/showdetl.cfm?&DID=9&Product II=51814&Detail=1
http://web.mit.edu/jbelcher/www/java/falling/falling.html
http://www.du.edu/∼jcalvert/phys/eddy.htm
http://128.148.60.98/physics/demopages/Demo/em/demo/5k2025.htm
http://www.tshankha.com/eddy currents.htm
[2] Jackson J D 1999 Classical Electrodynamics (New York: Wiley)
[3] Ramo S, Whinnery J R and Van Duzer T 1984 Fields and Waves in Communications Electronics (New York:
Wiley)
[4] Collin R E 1966 Foundations of Microwave Engineering (New York: McGraw-Hill)
[5] Popovi´c B D 1971 Introductory Engineering Electromagnetics (Reading, MA: Addison-Wesley)
[6] See http://www.efunda.com/materials/elements/element info.cfm?Element ID=Cu
[7] Morse P M and Feshbach H 1953 Methods of Theoretical Physics (New York: McGraw-Hill)
[8] Tipler P A 1998 Physics for Scientists and Engineers (New York: Freeman)