A Dash of Maxwell’s
Copyright 2000, 2005 Ampyx LLC
In Chapter Two, I introduced Maxwell’s Equations in their “integral form.” Simple in concept, the integral form can be devilishly difficult to work with. To overcome that, scientists and engineers have evolved a number of different ways to look at the problem, including this, the “differential form of the Equations.” The differential form makes use of vector operations.
A physical phenomena that has the both the attributes of magnitude and direction may be described by a vector. Velocity can be drawn in vector form; it has the attributes of both direction and speed. Vectors can be illustrated graphically as shown in Figure 1(a) – the length of the vector represents its magnitude and its angle from the x axis defines its direction. However, we will be using Cartesian coordinates. In this system, vectors are described as a sum of “unit” denominated vectors. A unit vector along the x axis is simply a vector aligned with the x axis that is one unit long (one meter long in the MKS system). We’ll denote a unit vector in the x direction as i. Similarly, unit vectors in the y and z directions will be denoted j and k respectively.
The vector shown in Figure 1 has a magnitude of 5 and is angled away from the x axis by 30 degrees. It can also be described as the sum of two unit denominated vectors one 3j units long and the other 4i units long.

Figure 1: A vector can be described in terms of its length and angle, or it can be described in terms of unit vectors.
Vectors are manipulated through the use of “vector operations.” We have already seen one of these, the dot product which illustrated in Figure 2.

Figure 2:
The dot product illustrated.
Use of the dot product allows
computation of the component of Vector A which is aligned with Vector B. It is expressed as:
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However, the dot product can
also be expressed in terms of unit vectors.
I’ll skip the proof and just give you the formula:
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Where:
A =
Ax i + Ay j + Az k
B =
Bx i + By j + Bz k
i,
j, k = Unit vectors in the x, y and z
directions respectively.
The dot product does not have
any direction associated with it. It’s
a scalar, not a vector.

Figure 3: An electric charge moving through a magnetic field will “feel” a force perpendicular to the plane formed by the direction of the field and the direction of travel. This effect can be used to deflect an electron beam. (Note that the charge shown here is negative.) The force can be calculated using the cross product.
The second vector operation we’ll need to use is known as the “cross product.” The cross product is best illustrated by a real world example, the deflection of an electron beam within a Cathode Ray Tube.
Figure 3 shows a beam of
electrons moving through a vacuum exposed to a magnetic field. The field here points out of the page. (By convention, a magnetic field that points
out of the page is designated by a dot in a circle, and one that point into the
page by an x within a circle.) As the
charge moves through the field it is acted on by a force known as the Lorentz
Force. The Lorentz Force operates
through the “right hand rule,” the charge “feels” a force perpendicular to the
plane formed by the field and the direction of charge movement. The Lorentz Force is equal to:
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Where:
F =
Lorentz Force in Newtons.
e =
Positive electric charge in Coulombs.
v =
Velocity of the charge in m/s.
B =
Magnetic flux density in Teslas
The ´ in this equation is not simply a multiplier. Both v and B are vectors, and in this context ´ denotes a vector operation known as the cross product. The cross product (v ´ B) is equal to:
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The
symbol ^denotes the direction of
the force, perpendicular to the plane formed by the vectors v and
B.
In
Cartesian coordinates the cross product is equal to:
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Unlike the dot product, the
cross product is a vector itself.
Cross products can be solved for easily by using a "determinant." Take two vectors, A and B, and set them up in a matrix form as follows:

The determinant is used (expanded) as shown in Figure 4.
In what is known as “determinant form,” the cross product of two vectors A and B can be expressed simply as:


Figure 4:
Calculation of a cross product is made easy through the use of a
determinant. To “expand” the
determinant, positive components are computed moving from the upper left to
lower right. Then negative components
are calculated by moving from the lower left to the upper right.
In our Chapter II, we
introduced Maxwell’s Equations in their integral form:

Where:
D=
Electric flux density = e0E
E=
Electric field in Volts/meter
B=
Magnetic flux density = m0H
H=
Magnetic field in Amps/meter
e0= Free
space permittivity= 8.85 x 10-12
m 0= Free space permeability= 4p x 10-7
We also adapted Maxwell’s
Equations for the conditions of “free space.” Operating in free space leaves us
only with us only two equations to consider:

We did find one combination
of electric and magnetic fields that satisfies these two equations. The combination consists of two sinusoidal
fields set perpendicular to each other, forming what is known as a “plane
wave.”

Figure 5: The plane wave.
This plane wave consists of an electric field component in the form of:
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and a magnetic field
component in the form of:
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The plane wave is easy to visualize, but it is hardly a general solution for Maxwell’s Equations.
We’ll make the solution more
general by considering a composite
magnetic field that itself is composed of two magnetic field components. These components are:
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The first component, Hy=H0cos(wt-kz) is a wave traveling in the z direction
consisting of magnetic field vectors oriented in the y direction. The second component, Hz=H0cos(wt-ky) is a wave consisting of field vectors oriented
in the z direction and traveling in the y direction.

Figure 6: Two magnetic field waves are shown, one
consisting of magnetic field vectors oriented in the z direction and traveling
in the y direction, and the other consisting of vectors oriented in the y
direction and traveling in the z direction.
A small rectangular movement in the y-z plane can be used to compute a
loop integral of the magnetic field.
This must be equal to the current, plus the change in the electric flux
density, passing through and normal to the loop.
The equation we’ll need to
satisfy is:
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We’ll assume that the
derivative of the electric field with respect to time is uniform over the small
rectangular surface shown in Figure 6.
In that case, the right side of this equation can be simplified as
follows:
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We’ll use the same technique
we used before to solve for the line integral of the magnetic field. We’ll solve for it pictorially by
multiplying the length traveled around the perimeter of the small rectangle in
Figure 6 by the field in the direction of travel.
H0 is the magnetic
field along Segments 1 and 4 of our perimeter.
As for the magnetic fields along the other two segments, they are:

Remembering that we have to
move counter clockwise to preserve the right hand rule, the line integral is:

Therefore the fourth of
Maxwell’s Equations for the fields illustrated in Figure 6 can be expressed as:

We can also define the “current density” J as being equal to I /DyDz, so:
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We could extend this analysis
to situations involving magnetic fields, electric fields and currents in three
dimensions, but the analysis will become ever more complex. Fortunately, there is a short cut. Things are made vastly easier because the
line integral is, in fact, a cross product.
To see that, we’ll start where we left off with vectors,
with the determinant form of the cross product of two vectors A and B.

.
Now for a bit of a
trick. We substitute for Vector A in
the determinant:
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and for Vector B:
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Our determinant becomes:

Then we note that there is no
current and no change in the electric field in the y or z directions. We also note that there is no magnetic field
component in the x direction. Replacing
these elements in our determinant with zeros yields the following:

Expanding the determinant, we
find:

More generally, if we have
components of magnetic fields, electric fields and currents in the three
dimensions, Maxwell’s fourth equation becomes:

Determinants may make a
solution easier to obtain, but the solution obtained is still unwieldy. Here, mathematicians have come to the
rescue. The have defined a vector operator known as the “del” which is equal to
(¶/¶x)i + (¶/¶y)j + (¶/¶z)k and is drawn
as an upside down triangle Ń. Using it, we can state an
equivalent to the fourth of Maxwell’s Equations by writing simply:
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Similarly, it can be shown
that:
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The technical name for Ń´E is the “curl” of the electric field. It has very real physical properties. An
electric field circulating in a loop is linked to the changing magnetic flux
through it. A gyroscope serves as a
useful mechanical analog. The spinning
ring of the gyroscope creates an angular momentum, which, through the
conservation of angular momentum, keeps the ring spinning in the same plane. Angular momentum is describe by a vector
perpendicular to the plane of the spinning ring. The two are linked, the spinning ring creates an angular momentum
which, in turn, keeps the ring spinning in the same plane. Similarly, a circulating electric field
creates a vector perpendicular to the plane of circulation equal to the change
in electric flux through and normal to it.
The two are inexorably
linked. The same is true of the
circulating magnetic field and its normal vector, the conduction and
displacement currents.
Another mechanical analog is
also useful. Figure 7 shows two air
streams. The air stream at the left has
a velocity uniform across its breadth.
The one at the right has a velocity which peaks at the sides of the
stream. We can use a paddlewheel as a
tool to measure the curl. The air
stream at the left exhibits no curl.
Insert the paddlewheel into that air stream and it doesn’t turn. Insert the paddlewheel into the air stream
at the right and it will turn, except in the dead center. To either the left or right of dead center,
the stream exhibits curl which causes a change in the angular momentum of the
paddlewheel.

Figure 7: Curl is illustrated. A paddlewheel can be used to measure the curl of an air
stream. The air stream at the left
exhibits no curl, the paddlewheel will not turn when inserted into the
stream. The stream at the right will
cause the paddlewheel to turn, except when it is place in the dead center of
the stream. (After Ref. 1.)

Figure 8: By measuring the net flux out of a small cube, “divergence” can be calculated.
The first two of Maxwell’s
equations can also be expressed in their “differential” form. Again, we’ll describe what is meant by these
equations pictorially in order to then derive their differential expression. We’ll begin with the first of Maxwell’s equations:
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In Figure 9, we calculate the
net flux into a small cube. We’ll adapt
the convention that flux into the cube is negative and out of it,
positive. The flux into the cube from the
left (positive x direction) is:
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Likewise, the flux out of the
cube in the x direction is:
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That makes the net flux in
the x direction equal to:
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Likewise, the flux
contributions from the electric flux density in the y and z directions are:

That makes the total flux,
the sum of the net flux along all three axes, equal to:
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Since DxDyDz = Dvolume, we can define charge per unit volume as the
charge density, r. So,
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Once again, we can use the
del operator to convert the left side of this equation into a vector
expression, in this case one utilizing the dot product:

Similarly, it can be shown
that:
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The latter equation is known as the “divergence” of the magnetic flux density (B). It is the measure of the flux out of a small volume of space. In the case of electric fields, the divergence of the electric flux density is equal to the charge density at a given point in space. Divergence is a scalar value, not a vector. In the case of magnetic fields, the divergence of B is always zero.
We’re now in a position to
state Maxwell’s equations in their differential form. Here they are:
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References
1. J.D. Kraus, Electromagnetics, 4th Edition,
McGraw-Hill Inc., 1991
2. R. Olenick, T. Apostol, and D. Goodstein, Beyond the
Mechanical Universe: From Electricity
to Modern Physics, Cambridge University Press, 1986