\problem1*
C vc2 + 1\over 2 Li2.
Argue that w
is the energy stored in the inductor and capacitor, while i2 R is
the power dissipated in the resistor.
and located between x = x1 and x = x2, (1) becomes
| (a) | Show that for a volume having area A in any y - z plane |
| (b) | Take the limit where x1 - x2 =  x  0 and
show that the one-dimensional form of (3) results.
|
| (c) | Based on (a), argue that Sx is the power flux density in the x direction. |
y/a2)Vd dVd/dt.
| (a) | Show that the power flux density is S = iy (- o
|
| (b) | Using S, show that the power input is d(
CVd2)/dt, where C = o b w/a.
|
| (c) | Evaluate the right-hand side of (11.1.1) to show that if the magnetic energy storage is neglected, the same result is obtained. |
| (d) | Show that the magnetic energy storage is
indeed negligible if b/c is much shorter than times of interest.
|
| (a) | Determine S. |
| (b) | From S, find the input power. |
| (c) | Evaluate the right-hand side of (11.1.1) for a volume enclosing the region between the electrodes, and show that if the electric energy storage is neglected, it is indeed equal to the left-hand side. |
| (d) | Under what conditions is the electric energy storage negligible? |
| (a) | Show that |
| (b) | Show that
 |
| (c) | Using these results, show that (11.1.3) is indeed
satisfied.
|
| (d) | Now, using the alternative EQS power theorem, evaluate S as given by (23) and again show that (11.1.3) is satisfied. |
| (e) | Observe that the latter evaluation is much simpler to carry out and that the latter power flux density is easier to picture. |
in the annulus b < r <
a, as shown in Fig. P11.3.2. The length l is large compared to
a. A voltage source v drives the system at the left, while the
electrodes are ``open'' at the right. Assume that v(t) is so slowly
varying that the voltage can be regarded as independent of z.
| (a) | Determine E,  , and H
|
| (b) | Evaluate the Poynting power flux density S [as given by (3)] in the annulus. |
| (c) | Use S to evaluate the total power dissipation by integration over the surface enclosing the annulus. |
| (d) | Show that the same result is obtained by integrating Pd over the volume. |
| (e) | Evaluate S as given by (23), and use that distribution of the power flux density to determine the total power dissipation. |
| (f) | Make sketches of the alternative distributions of S. |
| (g) | Show that the input power
is vi, where i is the total current from the voltage source.
|
and uniform permittivity
, while the surrounding region, where b < r < a, is free space. A
distributed voltage source v(t) constrains the potential difference
between the outer edges of the electrodes. Assume that the system is EQS.
| (a) | Show that the Poynting power flux density is |
| (b) | Integrate this flux density over a surface enclosing the region between the plates, and show that it is equal to the sum of the rate of change of electric energy storage and the power dissipation. |
| (c) | Now show that the alternative power flux density given by
(23) is
 |
| (d) | Carry out part (b) using this distribution of S, and show that the result is the same. |
| (e) | Show that the power input is equal to vi,
where i is the total current from the voltage source.
|
| (a) | Determine the distribution of Poynting power |
| (b) | Determine the alternative S given by (23). |
| (c) | Find the power dissipation density Pd in and around the rod. |
| (d) | Show that the differential
energy conservation law [(11.1.3) with  W/ t = 0]
is satisfied at each point in and around the rod using either of
these distributions of S.
|
Li2 to show that L is as given by
(8.5.20).
L i2 to find L.
the total coenergy wm'.
| (a) | Determine the magnetic coenergy density Wm', and hence |
| (b) | By writing wm' in the form of (11.4.24), determine L11, L12, and L22. |
a = o and
the region where B = b H now
filled with a material having the constitutive law
| (a) | Determine B and H in each region. |
| (b) | Find the coenergy
density in each region and hence the total coenergy wm' as a
function of the driving current i.
|
t)]. The sinusoidal steady state has been
established. Show that the time average power dissipation in the
lossy dielectrics is
| (a) | Find the time average power dissipation density in each |
| (b) | What is the total time average power dissipated in the sphere? |
and conductivity
is in the plane x = -b and makes contact with the perfectly
conducting plates above and below. At their left edges, in the plane
x = -(a + b), a source of surface current density, K(t), is
connected to the plates. The regions to left and right of the
resistive sheet are free space, and w is large compared to a, b, and
d.
| (a) | Show that the total power dissipation and magnetic energy |
| (b) | Show that the integral on the left in (11.1.1) over the
surface
indicated by the dashed line in the figure gives the same result as
found in part (a).
|
t) and sinsuoidal steady state conditions prevail. Determine the
time average power dissipation in the conducting sheet.
and conductivity , and is filled by a material
having permeability . Both the shell and the solenoid have a
length d perpendicular to the paper that is large compared to a.
current i1 = io cos t and the sinusoidal steady state has been
established, integrate the time average power dissipation density over
the volume of the shell to show that the total time-average power
dissipation is
| (a) | Given that the terminals of the solenoid are driven by the |
| (b) | In the sinusoidal steady state, the time average Poynting
flux through a surface enclosing the shell goes into the time average
dissipation. Use this fact to obtain (a).
|
| (a) | Show that in a region where there is no macroscopic current |
| (c) | Given that the spherical shell of Prob. 10.4.3 comprises each
element in the cubic array of Fig. P11.5.6, each sphere with spacing
s such that s \gg R, what is the complex permeability
defined such that \hat B = \hat \hat H?
|
| (d) | A macroscopic material composed of this array of spheres is placed in the one-turn solenoid of rectangular cross-section shown in Fig. P11.5.6. This configuration is long enough in the z direction so that fringing fields can be ignored. At their left edges, the perfectly conducting plates composing the top and bottom of the solenoid are driven by a distributed current source, K(t). With the fringing fields in the neighborhood of the left end ignored, the resulting fields take the form H = Hz (x, t) iz and E = Ey (x, t)iy. Use an evaluation of the Poynting flux to determine the total time average power dissipated in the length l, width d, and height a of the material. |
is small compared to the
length b, the magnetic field distribution in the conductor of Fig.
10.7.2 is given by (10.7.15). Show that (per unit y - z area) the
time average power dissipation associated with the current flowing in
the ``skin'' region is |Ks|2/2 watts/m2.
| (a) | Determine the total time average power dissipation. |
| (b) | Show that in the case \ll b this expression
reduces to that obtained in Prob. 11.5.7, while in the limit
\gg b, the result is i2R where R is the dc resistance of the
slab and i is the total current.
|
o N12 w2 /8R in
series with a resistance Rm = o 
N12 w2/8R.
R/N1) cos ( t).
| (a) | Draw a dimensioned plot of B(t). |
| (b) | Find the terminal voltage v(t) and also make a dimensioned plot. |
| (c) | Compute the time average power input, defined as
 /.
|
| (d) | Show that the result of part (c) can also be
found by recognizing that, during one cycle, there is an
energy/unit volume dissipated which is equal to the area enclosed by
the B-H characteristic.
|
, have a
fixed spacing a as shown in Fig. P11.6.1. Show that the force of
electrical origin acting on the lower electrode in the direction is
f = - o v2 d/2a.
is free to slide in and out of the
annular region between electrodes.
( - o
)/ln (a/b).
| (a) | Show that the force of electric origin acting on the |
| (b) |
Show that if the electrical terminals are constrained by the circuit
shown, R is very small and the plunger suffers the displacement
(t) the output voltage is vo = -2 RV( - o )(d /dt)/ln
(a/b).
|
.
| (a) | Ignore the fringing field and determine the |
| (b) | For
the energy conversion cycle of Demonstration 11.6.1, but for this
transducer, make dimensioned plots of the cycle in the (q, v) and
(f, ) planes (analogous to those of Fig. 11.6.5).
|
| (c) | By calculating both, show that the electrical energy input in one cycle is equal to the work done on the external mechanical system. |
) acting in the x direction on the
plunger of the magnetic circuit shown in Fig. P9.7.6.
having outer
and inner radii a and b can suffer a displacement into the
annular gap of a magnetic circuit otherwise made of infinitely
permeable material. The coil has N turns. Assume that the left end
of the plunger is well within the magnetic circuit, so that fringing
fields can be ignored, and determine the force f(i, ) acting to
displace the plunger in the direction.
\ll R. The system has depth d \gg \Delta into
the paper. Assume that 0 < 
< 
, as shown, and show that
the torque caused by passing a current i through the two N-turn
coils is 
= - o Rd N2 i2/.
| (a) | Determine the coenergy wm' (ia, ib, ).
|
| (b) | Find the torque on the rotor, (ia, ib, ).
|
| (c) | With ia = I cos ( t) and ib = I sin (
t), where I and are given constants, argue that the
magnetic axis produced by the stator rotates with the angular
velocity .
|
| (d) | Using these current constraints together with ir = Ir
and = \Omega t - , where Ir, and \Omega
are constants, show that under synchronous conditions (where =
\Omega), the torque is = MI Ir sin ( ).
|
o R3
E. Provided that R is short compared to
distances over which the field varies, this gives a good approximation
to p, even where the field is not uniform. Such a particle is
shown at the location x = X, y = Y in Fig. P11.8.1, where it is
subject to the field produced by a periodic potential = Vo cos
(
x) imposed in the plane y = 0.
Vo cos ( x) exp (- y).
| (a) | Show that the potential imposed in the region 0 < y is |
| (b) | Show that, provided that the particle has no net charge,
the force on the particle is
 |
x)iy, where Mo and are given positive constants.
the upper half-space is
| (a) | Show that the resulting magnetic potential in |
| (b) | A small infinitely permeable particle having the radius
R is located at x = X, y = Y. Show that the magnetization
force on the particle is as given by (a) of Prob. 11.8.1, with Vo
(Mo/2 ) and o \rightarrow o.
|
acting on the shell Tr = o K(Ho + Hi)/2. (Note that the
thin-shell model implies that H varies in an essentially linear
fashion with R inside the shell.)
| (a) | Show that there is a radial magnetic force per unit area |
| (b) | Specifically, show that
 |
shell is Tr = o K(H
o + Hi)/2. (Note that
according to the thin-shell model, H has an essentially linear
dependence on r within the shell.)
| (a) | Show that the radial force per unit area acting on the |
| (b) | Determine Tr ( ,t) and
relate the result to Demonstration 10.4.1.
|
The third objective of this chapter has been to see the quasistatic approximations from the perspective of electrodynamics. This began in Sec. 12.2 with the consideration of the turn-on transient of an electric dipole. It continued in Sec. 12.5 with a study of the sinusoidal steady state standing waves on the parallel plate transmission line. In the low frequency limit
@eq[s=0.2,n=15] an "open-circuit" termination resulted in a capacitor while with the "short-circuit" termination the system took on the characteristics of an inductor. These limiting cases were found in Sec. 12.6 by respectively making the EQS and MQS approximations at the outset. That section concluded with a formal procedure for making the quasistatic approximations in systems composed of perfect conductors and perfect insulators. The EQS and MQS approximations were identified as the lowest order fields in a time-rate-parameter expansion.
In materials of finite conductivity, additional processes that
depend on the time-rate-of-change contribute to the distribution of
the fields. Because fields are not only influenced by material
properties, length scales and time scales but by the topology as well,
there is no simple procedure for identifying quasistatic systems or
subsystems. However, for systems having all relevant dimensions of
the same order, typically of length @s[L], it was shown in Sec. 12.8
that a rough idea of the relevant physical processes that could occur
on a time scale @g[t] could be obtained by considering the subsystems
position in the @s[L]-@g[t] plane.
A summary of regimes for electromagnetic subsystems having one
characteristic length @s[L] and composed of linear materials is shown
in Fig. 12.9.1. The length scale has been normalized to the matching
length @s[L]@+[@m[8]] defined by Eq. 13 while the characteristic time
@g[t] has been normalized to the charge relaxation time /@g[s].
Thus, the lines demarking the regimes are
@eq[s=0.6,n=15] For a simple system having one characteristic length to be quasistatic we must have @g[t]@-[em]@s[ @s[L] is large compared to @s[L]@+[@m[8]] and EQS if @s[L] is small compared to @s[L]@+[@m[8]].
-here But surely the difference between the magnetic force density of Lorentz and Kelvin
and H 
H respectively playing the roles of dL/d and i2,
the magnetization term in (14) takes a form found for the
force on a magnetizable material in Sec. 11.7.)
In Example 11.9.2 (where J = 0), we found the force density of
(13) to be confined to the fringing field. By contrast, (14)
gives no force density in the fringing region (where is uniform)
but rather puts it all at the interface. According to this latter
equation, through the agent of a surface force density (a force
density that is a spatial impulse at the interface) the field pulls
upward on the interface.
Even though the force densities of (13) and (14) have very
different distributions, they predict the same height of rise of the
liquid! This is because the liquid deformations being considered are
essentially incompressible, in the sense that, (with v denoting
the liquid velocity) v = 0
\footnote*Solids can
also deform in
an essentially incompressible fashion. An example is the low
frequency motion of jello or muscle.. In the force equation
representing an incompressible material, there will always be another
force density taking the form p, where the pressure p assumes
whatever distribution it must to insure that the deformations are
incompressible. As a result, contributions to the force density that
take the form will have no effect on the incompressible
deformations. The contribution of is simply balanced by that
due to p.
For a magnetically linear material (o M = ( = \mo )H)
the force densities of
(13) and (14 do indeed only differ by a term taking the form
. To see this, use a vector identity
A = (\nabla x A) x A + 
A
A to write (13) as
x H in this expression, which then becomes
x (\Psi A) = \Psi \nabla
A + A \cdot
where
= o on S and is zero everywhere on S. We
conclude that integration of either (13) or (14) over the volume
will give the same total force. In summary, if an object is
surrounded by free space, integration over its volume of two force
densities that differ by the gradient of a scalar that is zero in free
space will result in the same net force.
A block of conducting material having permeabilityand is returned to the source in the -x direction at the left edge of the upper electrode. The thickness a of the block is small compared to its other two dimensions, so the magnetic field between the electrodes is z directed and dependent only on x. From Ampère's law it follows that in the conducting block. The alternative force densities, (13) and (14), have very different distributions in the block. Yet, we must find that the net force on the block found by integrating each over its volume is the same. To see that this is so consider first the sum of the Lorentz and Kelvin force densities, (13).
There is no x component of the magnetic field intensity, so for this particular configuration, the magnetization term makes no contribution to (13). Evaluation of the first term using (22) and (23) then gives
Integration of this force density over the volume amounts to a multiplication by the cross-sectional area ab and integration on x.
Now, the force density given by (14) is evaluated. The permeability Note that Hz is K at x = -b and is zero at x = 0. Integration of (26) over the volume of the block therefore gives
Note that Hz is constant through the interface at x = -b. So, the integration of the last term can be carried out. Simplification of this expression gives the same total force as found before, (25).
The distributions of the force densities given by (13) and (14)
are generally different, even very different. It is natural to
therefore ask which of the two is the ``right'' one. In general, until
the ``other'' force densities acting on the medium in question are
specified, this question cannot be answered. Here, where a discussion
of continuum mechanics is beyond our purvue, we have identified a
class of mechanical continua (namely incompressible materials), where
these force densities are equally valid. In fact so would any other
force density differing from these by a term having the form
. The combined Lorentz and Kelvin force densities have the
advantage of a satisfying physical interpretation. However, the
derivation has the weakness of making an ``ad hoc'' use of the
macroscopic fields. Force densities resulting from an energy argument
have the advantage of dealing rigorously with the macroscopic fields.
The form illustrated by (14) also has the advantage of concentrating
the magnetization force density at interfaces. For example, it is
then clear that the height to which the liquid rises in the experiment
of Fig. 11.9.4 depends on the field intensity at the interface and not
on the details of the fringing field!
-here
A macroscopic force density f(r ) is the force per unit volume acting on a medium in the neighborhood of r. Fundamentally, the electromagnetic force density is the result of forces acting on those microscopic particles embedded in the material that are charged, or that have electric or magnetic dipole moments. The understanding is that the forces acting on these individual particles is passed along through interparticle forces to the macroscopic material as a whole. In the limit where that volume becomes small, the force density can then be regarded as the sum of the microscopic forces over a volume element
Of course, the linear dimensions of Strictly, the forces in this sum should be evaluated using the microscopic fields. However, we can gain insight concerning the form taken by the force density by using the macroscopic fields in this evaluation. This is the basis for the following discussions of the force densities associated with unpaired charges and with conduction currents (the Lorentz force density) and with the polarization and magnetization of media (the Kelvin Force Density). To be certain that the usage of macroscopic fields in describing the force densities is consistent with that implicit in the constitutive laws already introduced to describe conduction, polarization and magnetization, the electromagnetic force densities should be derived using energy arguments. These derivations are extensions of those of Secs. 11.6 and 11.7 for forces. We end this section with a discussion of the results of such derivations and of circumstances under which they will predict the same total forces or even material deformations as those derived here.
The Lorentz Force Density
Without restricting the generality of the resulting force density, suppose that the electrical force on a material is due to two species of charged particles. One has N+ particles per unit volume, each with a charge q+, while the other has density N- and a charge equal to -q-. With v denoting the velocity of the macroscopic material and vrepresenting the respective velocities of the carriers relative to that material, the Lorentz force law gives the force on the individual particles.
Note that q- is a positive number. In typical solids and fluids, the charged particles are either bonded to the material or migrate relative to the material, suffering many collisions with the neutral material during times of interest. In either case, the inertia of the particles is inconsequential so that on the average the forces on the individual particles is passed along to the macroscopic material. In either case, the force density on the material is the sum of (2) and (3) respectively multiplied by the charged particle densities.
Substitution of (2) and (3) into this expression gives the Lorentz force density.
where u is the unpaired charge density (7.1.6) and J is the current density.
Because the material is in motion, with velocity v, the current density J has not only the contribution familiar from Sec. 7.1 (7.1.4) due to the migration of the carriers relative to the material, but one due to the net charge carried by the moving material as well. In EQS systems, the first term in (5) usually outweighs the second, while in MQS systems (where the unpaired charge density is negligible) the second term dominates.
The derivation and Fig. 11.9.1 suggest why the electric term is proportional to the net charge density. In a given region, the force density resulting from the positively charged particles tends to be canceled by that due to the negatively charged particles and the net force density is therefore proportional to the difference in absolute magnitudes of the charge densities. We exploited this fact in Chap. 7 to let electrically induced material motions evidence the distribution of the unpaired charge density. For example, in Demonstration 7.5.1, the unpaired charge density was restricted to an interface and as a result the motion of the fluid was suppressed by constraining the interface. A more recent example is the force on the upper electrode in the capacitor transducer of Example 11.6.1. Here again, the force density is confined to a thin region on the surface of the conducting electrode.
The magnetic term in (6), pictured in Fig. 11.9.2 as acting on a current carrying wire, is also familiar. It was this force density that was responsible for throwing the metal disk into the air in the experiment described in Sec. 10.2. The force responsible for the levitation of the pan-cake coil in Demonstration 11.7.1 was also the net effect of the Lorentz force density, either acting over the volume of the coil conductors, or over that of the conducting sheet below. In MQS systems, where the contribution of the ``convection'' current \footnote*Indeed, it is fortunate that the carriers do not have the same velocity as the material, for if they did it would not be possible to use the magnetic Lorentz force density for electromechanical energy conversion. If we recognize that the rate at which a force f does work on a particle that moves at the velocity v is v
The Kelvin Polarization Force Density
If microscopic particles carrying a net charge were the only contributors to a macroscopic force density, it would not be possible to explain the forces on polarized materials that are free of unpaired charge. Example 11.6.2 and Demonstration 11.6.2 highlighted the polarization force. The experiment was carried out in such a way that the dielectric material did not support unpaired charge, so the force is not explained by the Lorentz force density.In these EQS cases where
This is often called the Kelvin polarization force density.
Example 11.11.1. Force on a Dielectric Material
In Fig. 11.9.4, the cross-section of a pair of electrodes that are dipped into a liquid dielectric is shown. The picture might be of a cross-section from the experiment of Demonstration 11.6.2. With the application of a potential difference to the electrodes, the dielectric rises between the electrodes. According to (7), what is the distribution of force density causing this rise?For the liquid dielectric, the polarization constitutive law is taken as linear (6.4.2) so that with the understanding that By using a vector identity \footnote* A second vector identity \footnote** converts this expression into one that will now prove useful in picturing the distribution of force density.
Provided that the interface is well removed from the fringing fields at the top and bottom edges of the electrodes, the electric field is uniform not only in the dielectric and gas above and below the interface between the electrodes, but through the interface as well. Thus, throughout the region between the electrodes there is no gradient of E, and hence, according to (7), no Kelvin force density. The Kelvin force density is therefore confined to the fringing field region where the fluid surrounds the lower edges of the electrodes. In this region, Although there is no physical basis for doing so, it might seem reasonable to take the force density caused by polarization as being
The Kelvin Magnetization Force Density
Probably forces caused by magnetization are the most commonly experienced of electromagnetic forces. They account for the attraction between a magnet and a piece of iron. In Example 11.7.1, it is this force density that acts on the disk of magnetizable material.Given that the magnetizable material is made up of microscopic dipoles, each experiencing a force of the nature of (11.8.6), and that the magnetization density M is the number of these per unit volume multiplied by m, it follows that the force density due to magnetization is
This is sometimes called the Kelvin magnetization force density.
Example 11.11.2. Force Density in a Magnetized Fluid
With the dielectric liquid replaced by a ferrofluid having a uniform permeability\footnote*Rosensweig, R.E., Magnetic Fluids, Scientific American, Oct. 1982, pp136-145., the magnetization results from an essentially permanent suspension of magnetized particles. Each particle comprises a magnetic dipole and passes its force on to the liquid medium in which it is suspended. Provided that the magnetization obeys a linear law, the discussion of the distribution of force density given in Example 11.9.1 applies equally well here.
Alternative Force Densities
We now return to comments used to introduce this section. The fields used to express the Lorentz and Kelvin force densities are macroscopic. To assure consistency between the averages implied by these force densities and those already inherent to the constitutive laws, an energy principle can be used. The approach is a continuum version of that exemplified for lumped-parameter systems in Secs. 11.7 and 11.8. In the lumped parameter systems, electrical terminal relations were used to determine a total energy and energy conservation was used to determine the force in turn. In the continuum system,\footnote*Melcher, J.R., Continuum Electromechanics, M.I.T. Press, 1981, Chap. 3. the electrical constitutive law is used to find an energy density and energy conservation used in turn to find a force density. This energy method, like the one exemplified in Secs. 11.7 and 11.8 for lumped parameter systems, describes systems that are loss-free. In making practical use of the result it is assumed that it will be applicable even if there are losses. A more general method, which invokes a principle of virtual power, \footnote**Penfield, P. and Haus, H.A., Electrodynamics of Moving Media, M.I.T. Press, Cambridge, Mass., 1967 allows for dissipation but requires more empirical information than the polarization or magnetization constitutive law as a starting point.
Force densities derived from more rigorous arguments then given here can have very different distributions from the superposition of the Lorentz and Kelvin force densities. We expect that the arguments used here prove inadequate when the microscopic particles become so densely packed that the field experienced by one is significantly altered by its nearest neighbor. But surely the difference between the magnetic force density of Lorentz and Kelvin
we have derived here and the Korteweg-Helmholtz force density for incompressible media is not due to interactions between microscopic particles. This latter force density is often obtained for an incompressible material energy arguments. (Note that, with - Even though the force densities of (13) and (14) have very different distributions, they predict the same height of rise of the liquid! This is because the liquid deformations being considered are essentially incompressible, in the sense that, (with v denoting the liquid velocity)
For a magnetically linear material (
AA
The MQS form of Ampère's law makes it possible to substitute J for The second term in this expression is then expanded using a second vector identity \footnote* This expression differs from (14) by the last term, which indeed takes the form Thus, although they have very different distributions, the force densities of (13) and (14) will predict the same incompressible deformations of a material. Will the integration of these two force densities over the volume of an object result in two different total forces? Provided that the object is surrounded by media having the properties of free space, the answer is no. This is because the integral over the volume of the force density by which the two differ is zero. To see this, consider the x component (say) of that integral.
With the objective of converting this volume integral to one over the enclosing surface, we write this expression in the equivalent form It follows from Gauss' Theorem that the volume integral on V is equivalent to one over the enclosing surface S.
With the surface S taken as enclosing an object surrounded by free space, Example 11.11.3. Magnetic Force on a Magnetizable Current Carrying Material
A block of conducting material having permeabilityand is returned to the source in the -x direction at the left edge of the upper electrode. The thickness a of the block is small compared to its other two dimensions, so the magnetic field between the electrodes is z directed and dependent only on x. From Ampère's law it follows that in the conducting block. The alternative force densities, (13) and (14), have very different distributions in the block. Yet, we must find that the net force on the block found by integrating each over its volume is the same. To see that this is so consider first the sum of the Lorentz and Kelvin force densities, (13).
There is no x component of the magnetic field intensity, so for this particular configuration, the magnetization term makes no contribution to (13). Evaluation of the first term using (22) and (23) then gives
Integration of this force density over the volume amounts to a multiplication by the cross-sectional area ab and integration on x.
Now, the force density given by (14) is evaluated. The permeability Note that Hz is K at x = -b and is zero at x = 0. Integration of (26) over the volume of the block therefore gives
Note that Hz is constant through the interface at x = -b. So, the integration of the last term can be carried out. Simplification of this expression gives the same total force as found before, (25). The distributions of the force densities given by (13) and (14) are generally different, even very different. It is natural to therefore ask which of the two is the ``right'' one. In general, until the ``other'' force densities acting on the medium in question are specified, this question cannot be answered. Here, where a discussion of continuum mechanics is beyond our purvue, we have identified a class of mechanical continua (namely incompressible materials), where these force densities are equally valid. In fact so would any other force density differing from these by a term having the form
A macroscopic force density F(r ) is the force per unit volume acting on a medium in the neighborhood of r. Fundamentally, the electromagnetic force density is the result of forces acting on those microscopic particles embedded in the material that are charged, or that have electric or magnetic dipole moments. The forces acting on these individual particles are passed along through interparticle forces to the macroscopic material as a whole. In the limit where that volume becomes small, the force density can then be regarded as the sum of the microscopic forces over a volume element
Of course, the linear dimensions of Strictly, the forces in this sum should be evaluated using the microscopic fields. However, we can gain insight concerning the form taken by the force density by using the macroscopic fields in this evaluation. This is the basis for the following discussions of the force densities associated with unpaired charges and with conduction currents (the Lorentz force density) and with the polarization and magnetization of media (the Kelvin Force Density). To be certain that the usage of macroscopic fields in describing the force densities is consistent with that implicit in the constitutive laws already introduced to describe conduction, polarization and magnetization, the electromagnetic force densities should be derived using energy arguments. These derivations are extensions of those of Secs. 11.6 and 11.7 for forces. We end this section with a discussion of the results of such derivations and of circumstances under which they will predict the same total forces or even material deformations as those derived here.
The Lorentz Force Density
Without restricting the generality of the resulting force density, suppose that the electrical force on a material is due to two species of charged particles. One has N+ particles per unit volume, each with a charge q+, while the other has density N- and a charge equal to -q-. With v denoting the velocity of the macroscopic material and v
Note that q- is a positive number. In typical solids and fluids, the charged particles are either bonded to the material or migrate relative to the material, suffering many collisions with the neutral material during times of interest. In either case, the inertia of the particles is inconsequential so that on the average the forces on the individual particles is passed along to the macroscopic material. In either case, the force density on the material is the sum of (2) and (3) respectively multiplied by the charged particle densities.
Substitution of (2) and (3) into this expression gives the Lorentz force density.
where
Because the material is in motion, with velocity v, the current density J has not only the contribution familiar from Sec. 7.1 (7.1.4) due to the migration of the carriers relative to the material, but one due to the net charge carried by the moving material as well. In EQS systems, the first term in (5) usually outweighs the second, while in MQS systems (where the unpaired charge density is negligible) the second term tends to dominate.
The derivation and Fig. 11.9.1 suggest why the electric term is proportional to the net charge density. In a given region, the force density resulting from the positively charged particles tends to be canceled by that due to the negatively charged particles and the net force density is therefore proportional to the difference in absolute magnitudes of the charge densities. We exploited this fact in Chap. 7 to let electrically induced material motions evidence the distribution of the unpaired charge density. For example, in Demonstration 7.5.1, the unpaired charge density was restricted to an interface and as a result the motion of the fluid was suppressed by constraining the interface. A more recent example is the force on the upper electrode in the capacitor transducer of Example 11.6.1. Here again, the force density is confined to a thin region on the surface of the conducting electrode.
The magnetic term in (6), pictured in Fig. 11.9.2 as acting on a current carrying wire, is also familiar. It was this force density that was responsible for throwing the metal disk into the air in the experiment described in Sec. 10.2. The force responsible for the levitation of the pan-cake coil in Demonstration 11.7.1 was also the net effect of the Lorentz force density, either acting over the volume of the coil conductors, or over that of the conducting sheet below. In MQS systems, where the contribution of the ``convection'' current
f then it follows from the Lorentz force,
(1.1.1), that the rate of doing work on individual particles through the
agent of the magnetic field is v (v x o H).
The cross-product is perpendicular to v, so this rate of doing
work must be zero.The Kelvin Polarization Force Density
If microscopic particles carrying a net charge were the only contributors to a macroscopic force density, it would not be possible to explain the forces on polarized materials that are free of unpaired charge. Example 11.6.2 and Demonstration 11.6.2 highlighted the polarization force. The experiment was carried out in such a way that the dielectric material did not support unpaired charge, so the force is not explained by the Lorentz force density.
In these EQS cases where
This is often called the Kelvin polarization force density.
Example 11.12.1. Force on a Dielectric Material
In Fig. 11.9.4, the cross-section of a pair of electrodes that are dipped into a liquid dielectric is shown. The picture might be of a cross-section from the experiment of Demonstration 11.6.2. With the application of a potential difference to the electrodes, the dielectric rises between the electrodes. According to (7), what is the distribution of force density causing this rise?For the liquid dielectric, the polarization constitutive law is taken as linear (6.4.2) and (6.4.4) so that with the understanding that By using a vector identity
A= (\nabla (A
and invoking the EQS approximation whereA second vector identity
A ) = A
Provided that the interface is well removed from the fringing fields at the top and bottom edges of the electrodes, the electric field is uniform not only in the dielectric and gas above and below the interface between the electrodes, but through the interface as well. Thus, throughout the region between the electrodes there is no gradient of E, and hence, according to (7), no Kelvin force density. The Kelvin force density is therefore confined to the fringing field region where the fluid surrounds the lower edges of the electrodes. In this region, Although there is no physical basis for doing so, it might seem reasonable to take the force density caused by polarization as being
The Kelvin Magnetization Force Density
Probably forces caused by magnetization are the most commonly experienced of electromagnetic forces. They account for the attraction between a magnet and a piece of iron. In Example 11.7.1, it is this force density that acts on the disk of magnetizable material.Given that the magnetizable material is made up of microscopic dipoles, each experiencing a force of the nature of (11.8.28), and that the magnetization density M is the number of these per unit volume multiplied by m, it follows that the force density due to magnetization is
This is sometimes called the Kelvin magnetization force density.
Example 11.12.2. Force Density in a Magnetized Fluid
With the dielectric liquid replaced by a ferrofluid having a uniform permeability
Alternative Force Densities
We now return to comments used to introduce this section. The fields used to express the Lorentz and Kelvin force densities are macroscopic. To assure consistency between the averages implied by these force densities and those already inherent to the constitutive laws, an energy principle can be used. The approach is a continuum version of that exemplified for lumped-parameter systems in Secs. 11.7 and 11.8. In the lumped parameter systems, electrical terminal relations were used to determine a total energy and energy conservation was used to determine the force in turn. In the continuum system[2], the constitutive laws are used to find an energy density and energy conservation used in turn to find a force density. This energy method, like the one exemplified in Secs. 11.7 and 11.8 for lumped parameter systems, describes systems that are loss-free. In making practical use of the result it is assumed that it will be applicable even if there are losses. A more general method, which invokes a principle of virtual power[3], allows for dissipation but requires more empirical information than the polarization or magnetization constitutive law as a starting point.Force densities derived from more rigorous arguments then given here can have very different distributions from the superposition of the Lorentz and Kelvin force densities. We expect that the arguments used here prove inadequate when the microscopic particles become so densely packed that the field experienced by one is significantly altered by its nearest neighbor. But surely the difference between the magnetic force density of Lorentz and Kelvin
we have derived here and the Korteweg-Helmholtz force density for incompressible media is not due to interactions between microscopic particles. This latter force density is often obtained for an incompressible material by energy arguments. (Note that, with - Even though the force densities of (13) and (14) have very different distributions, they predict the same height of rise of the liquid! This is because the liquid deformations being considered are essentially volume conserving (incompressible), in the sense that, (with
p, where the pressure p assumes
whatever distribution it must to insure that the deformations are
incompressible. As a result, contributions to the force density that
take the form will have no effect on the incompressible
deformations. The contribution of is simply balanced by that
due to p.
For a magnetically linear material (o M = ( - \mo )H)
the force densities of
(13) and (14 do indeed only differ by a term taking the form
. To see this, use a vector identity
A
The MQS form of Ampère's law makes it possible to substitute J for The second term in this expression is then expanded using a second vector identity
x (\Psi A ) = \Psi \nablaThis expression differs from (14) by the last term, which indeed takes the form Thus, although they have very different distributions, the force densities of (13) and (14) will predict the same incompressible deformations of a material. Will the integration of these two force densities over the volume of an object result in the same total force? Provided that the object is surrounded by media having the properties of free space, the answer is yes. This is because the integral over the volume of the force density by which the two differ is zero. To see this, consider the x component (say) of that integral.
With the objective of converting this volume integral to one over the enclosing surface, we write this expression in the equivalent form It follows from Gauss' Theorem that the volume integral on V is equivalent to one over the enclosing surface S.
With the surface S taken as enclosing an object surrounded by free space, Example 11.12.3. Magnetic Force on a Magnetizable Current Carrying Material
A block of conducting material having permeabilityand is returned to the source in the -x direction at the left edge of the upper electrode. The thickness a of the block is small compared to its other two dimensions, so the magnetic field between the electrodes is z directed and dependent only on x. From Ampère's law it follows that in the conducting block. The alternative force densities, (13) and (14), have very different distributions in the block. Yet, we must find that the net force on the block found by integrating each over its volume is the same. To see that this is so consider first the sum of the Lorentz and Kelvin force densities, (13).
There is no x component of the magnetic field intensity, so for this particular configuration, the magnetization term makes no contribution to (13). Evaluation of the first term using (22) and (23) then gives
Integration of this force density over the volume amounts to a multiplication by the cross-sectional area ab and integration on x.
Now, the force density given by (14) is evaluated. The permeability Note that Hz is K at x = -b and is zero at x = 0. Integration of (26) over the volume of the block therefore gives
Note that Hz is constant through the interface at x = -b. So, the integration of the last term can be carried out. Simplification of this expression gives the same total force as found before, (25).
The distributions of the force densities given by (13) and (14)
are generally different, even very different. It is natural to
therefore ask which of the two is the ``right'' one. In general, until
the ``other'' force densities acting on the medium in question are
specified, this question cannot be answered. Here, where a discussion
of continuum mechanics is beyond our purvue, we have identified a
class of mechanical deformations (namely those that are volume
conserving or ``incompressible''), where
these force densities are equally valid. In fact so would any other
force density differing from these by a term having the form
. The combined Lorentz and Kelvin force densities have the
advantage of a satisfying physical interpretation. However, the
derivation has the weakness of making an ``ad hoc'' use of the
macroscopic fields. Force densities resulting from an energy argument
have the advantage of dealing rigorously with the macroscopic fields.
The form illustrated by (14) also has the advantage of concentrating
the magnetization force density at interfaces. For example, it is
then clear that the height to which the liquid rises in the experiment
of Fig. 11.9.4 depends on the field intensity at the interface and not
on the details of the fringing field!
The basis for a self-consistent macroscopic description of any continuum subsystem is a power-flow statement having the forms identified in Sec. 11.1. Describing a volume V enclosed by a surface S, the integral conservation of energy statements takes the form (11.1.1)Implied by this is the differential form of the conservation of energy statement. Poynting's Theorem, the subject of Sec. 11.2, is obtained starting from the laws of Faraday and Ampère to obtain an expression of the form of (2). For materials that are Ohmic (J = Of course, taking the free space limit where In Sec. 11.3, we found that in EQS systems an alternative to Poynting's vector is (11.3.24)
This expression is of practical importance because it can be evaluated without determining H, which is generally not of interest in EQS systems. Perhaps a more important reason for introducing the alternative EQS forms of S is that it emphasizes that S is not unique. In Sec. 11.3 we saw that Poynting's flux density and (6) assumed entirely different spatial distributions. Yet, evaluation of the integral of S An important application of the integral form of the energy conservation statement is to lumped parameter systems. In these cases, the surface S of (1) encloses a system that is connected to the outside world through terminals. It is then convenient to describe the power flow in terms of the terminal variables. It was shown in Sec. 11.3 (11.3.29) that the net power into the system represented by the left hand side of (1) becomes
provided that the magnetic induction is negligible on the surface S, and the electric displacement current is parallel to the surface S. This set the stage for the application of the integral form of the energy conservation theorem to lumped parameter systems. In Sec. 11.4, attention focused on the energy storage term, the first terms on the right in (1) and (2). The energy density concept was broadened to include materials having constitutive laws relating the flux densities to the field intensities that were single valued and collinear. With E, D, H and B representing the field magnitudes, the energy density was found to be the sum of electric and magnetic energy densities.
Integrated over the volume V of a system, this function leads to the total energy w. For quasistatic lumped parameter systems, the total electric or magnetic energy is often conveniently found following a different route. First, the terminal relations are determined and then the total energy is found by adding up the increments of energy put into the system as it is energized. In the case of an n terminal pair EQS system, where the relation between terminal voltage vi and associated charge qi is vi (q1, q2, \ldots qn ), the increment of energy is vi dqi and the total electric energy is (11.4.9) The line integration in an n-dimensional space representing the n independent qi's was illustrated by Example 11.4.2. Similarly, for an n terminal pair MQS system where the current ii is related to the flux linkage i by ii = ii (
Note the analogy between these expressions for the total energy of EQS and MQS lumped parameter systems and the respective electric and magnetic energy densities of (8). The transition from the field picture afforded by the energy densities to the lumped parameter characterization is made by E Especially in using the energy to evaluate forces of electrical origin, we found it convenient to define coenergy density functions.
It followed that these functions were natural when it was desirable to use E and H as the independent variables rather than D and B.
The total coenergy functions for lumped parameter EQS and MQS systems could be found either by integrating these densities over the volume or by again viewing the system in terms of its terminal variables. With the total coenergy functions defined by it followed that the coenergy functions could be determined from the terminal relations by again carrying out line integrations, but this time with the voltages and currents as the independent variables. For EQS systems,
while for MQS systems Again, note the analogy to the respective terms in (12). The remaining sections of the chapter developed some of the possible implications of the ``dissipation'' term in the energy conservation statement, the last terms in (1) and (2). In Sec. 11.5, coupling to a thermal subsystem was the point. In this section, the disparity between the power input and the rate of increase of the energy stored was accounted for by heating. In addition to Ohmic heating, caused by collisions between the migrating carriers and the neutral media, we considered losses associated with the dynamic polarization and magnetization of materials.
In Secs. 11.6-11.9 we considered coupling to a mechanical subsystem as a second mechanism by which energy gain or energy loss from the electromagnetic system could occur. With the displacement of an object denoted by
or from the coenergy functions ((11.7.7) and the analogous expression for electric systems).
In Sec. 11.8, where the Lorentz force on a particle was generalized to account for electric and magnetic dipole moments, one objective was a microscopic picture that would lend physical insight to the forces on polarized and magnetized materials. The Lorentz force was generalized to indicate the force on stationary electric and magnetic dipoles, respectively The total macroscopic forces resulting from microscopic forces had already been encountered in the previous two sections. The force density describes the interaction between a volume element of the electromagnetic subsystem and a mechanical continuum. A rigorous approach to finding the force density is based on a generalization of the energy method introduced in Secs. 11.6 and 11.7. Such an approach has to include information on the effect of thermodynamic variables (such as density and entropy) on the constitutive laws[1,2].
