Lesson 8 – Ampère’s Law and Differential Operators
© Lawrence B. Rees 2007. You may make a single copy of this document for personal use without written permission.
8.0 Introduction
There are significant differences between the electric and magnetic field lines we have
studied so far. Electric field lines spread outward or converge inward. Magnetic field lines
always form closed loops. Electric field lines naturally represent the q/ r2 dependence of the
electric field of a point charge. Magnetic field contours naturally represent the i / r dependence
of the magnetic field of a long wire. We found that electric field lines could be mathematically
represented by flux and Gauss’s law of electricity followed as a simple consequence. It would
stand to reason that we would be able to make similar sorts of arguments about the magnetic
field contours that would lead to another useful mathematical relationship. And this is exactly
what we are going to do in this chapter.
8.1 Ampère's Law
As you recall, a field contour is a set of surfaces which are everywhere normal to the field
lines. Magnetic field contours originate on current-carrying wires. The number of contours is
chosen to be proportional to the current flowing in the wire. The contours have a direction which
is taken to be the direction of the magnetic field at any point on them. When there are multiple
wires, the field near each wire, the near field, is the same as it would be for a single wire.
Think About It
A current-carrying wire alone in space has perpendicular surfaces coming uniformly
outwardly from it. As a second, parallel wire with current flowing in the same direction is
brought nearer to it, what happens to the field contours? Does it make any difference what the
magnitude of the second current is? What happens when the second wire has current flowing in
the opposite direction to the first wire?
Let’s imagine some perpendicular surfaces near a wire. Now imagine a closed loop in
space near the wire. The loop can be of any shape or size; the only requirement is that it be
closed. A rubber band is a good example of a closed loop that can vary in shape and size. Such a
loop is called an “Amperian loop,” as shown in Fig. 8.1. Note that the
1
Figure 8.1 An Amperian loop and the field contour of a
wire carrying current out of the screen.
Amperian loop has a direction which we have indicated with an arrow. We will take the direction
of the Amperian loop to be counterclockwise in our drawings. Now we want to define the
concept of “net number of surfaces” crossed by the loop. To do this, we go around the loop in the
direction of the arrow. As the loop passes through a perpendicular surface, we either add +1 or
add −1 to the net number of surfaces. We add +1 if the loop is generally in the same direction as
the surface, and we add −1 if the loop is in the opposite direction. By “generally same,” we mean
more precisely that there is an acute angle between the vector direction of the surface and the
vector direction of the loop as it passes through the surface. In Fig. 8.1, the loop crosses a surface
in the +1 sense seventeen times and it passes through a surface in the –1 sense once. This gives
us a net number of +16 contours crossed by the Amperian loop. Note that this concept is
analogous to the "net number of field lines" passing through a Gaussian surface.
Figure 8.2 Amperian loops with current passing through them.
2
If we take a number of different Amperian loops around the same wire, as shown in Fig. 8.2, the
net number of surfaces crossed by each loop is always +16.
Think About It
How would these results differ if the current in the wire were going into the screen?
If the current were going into the screen, the arrow on each perpendicular surface
would be reversed. This would in turn change every +1 into a −1 and every -1 into a +1, so the
net number of surfaces pierced by the loop would be −16. As long as we are careful to go around
our Amperian loop in a general counterclockwise sense, we can establish a convenient sign
convention:
Sign Convention
If the number of surfaces pierced by an Amperian loop is positive, the current comes out
of the screen. If it is negative, current goes into the screen.
Figure 8.3 An Amperian loop with no current passing through it.
If, however, we choose our Amperian loop in such a way that no current passes through
it, as in Fig. 8.3, the net number of perpendicular surfaces pierced by the loop must be zero. If
the current in the wires doubles, the number of perpendicular surfaces also doubles. It should be
evident that we can generalize these results to give us a law for magnetic field contours:
3
Ampère’s Law
Let a set of perpendicular surfaces form a field contour. The net number of surfaces
pierced by an Amperian loop is proportional to the current passing through the loop.
Ampère's law is the third of Maxwell's equations. As Gauss's law is a geometrical way of
stating Coulomb's law, Ampère's law is a geometrical way of describing the magnetic field of
current-carrying wires. Later we will find mathematical ways of expressing this concept and we
will use it quantitatively to determine the magnetic field of current-carrying objects with high
degrees of symmetry. First, however, we will apply Ampère’s conceptually to cylindrical wires.
Things to remember
• Amperian loops are closed loops.
• Ampère’s law: The net number of surfaces pierced by an Amperian loop is proportional to the
current passing through the loop.
• We always go around Amperian loops in a counterclockwise direction. We also define positive
current to be “out” and negative current to be “in.”
•Ampère’s law is equivalent to the relation B = µ 0 i /(2πr ) .
8.2 Current Density
Even as it was necessary to define charge density in the electrical case, it is useful to
define current density in conjunction with magnetic fields. To define current density, let us think
of cutting a wire perpendicularly to its length and placing a “gate” in the wire. The gate isn’t a
hole in the wire or a physical object; it’s just a loop through which electrons pass. We then just
count the number of electrons passing through this gate. The current passing through the wire is
equal to:
Current = (charge per electron)
× (number of electrons passing through the gate in one second).
Now let us make the gate smaller than the cross section of the wire, as in Fig. 8.4. The current
density is then:
Current Density= (charge per electron) ×
× (number of electrons passing through the gate in one second)
÷ (cross sectional area of the gate)
4
small gate
Figure 8.4 A wire with a small gate of area dA.
We use the letter j to denote current density and measure current density in units of amperes /
square meter ( A / m 2 ).
Think About It
Describe in words the following current density:
3r sin θ , r ≤ 1
j (r ,θ ) = 
 0, otherwise
To find the total current passing through a wire, we can integrate over the cross-section of
the wire in much the same way that we integrated over a surface charge density to find the total
charge on a circular disk. That is:
I = ∫ j dA
Things to remember:
• Current density j is the current per unit area passing through a wire. Current density may vary
from region to region in a wire; however, in typical wires, current density is quite uniform.
• We can integrate current density to get total current. I = ∫ j dA
8.3 The Fields of Current Distributions with Radial Cylindrical Symmetry
Consider a wire with a current density having radial cylindrical symmetry. Recall that by
our definition of radial symmetry, the current density may vary with distance from the axis of the
wire, r, but it may not vary with angle. Because of this symmetry, the surfaces of the field
contour outside the wire must come radially outward from the axis and be uniformly spaced.
Therefore, the magnetic field in this region must be the same as the field of a thin wire having
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the same total current flowing through it. Thus the magnetic field outside a wire does not depend
upon the wire's radius.
But how do we find the magnetic field within the wire having a radially symmetric
current density? We can be guided by the method we used to find the electric field inside a
spherical charge distribution. There, we divided the spherical charge into two regions: a hollow
sphere and a core. We used symmetry along with Gauss’s law to prove that the electric field
inside the hollow sphere is zero. Finally, we concluded that the electric field was just the electric
field of the core – which in turn was the electric field of a point charge having the same charge as
the core alone. Now let’s apply the same logic to current in a wire.
First, we construct an Amperian loop at a radius r within the wire. We then divide the
wire into two pieces, a hollow cylinder, and a cylindrical core, as shown in Fig. 8.5. First, let’s
consider the hollow core. Since no current is passing through the core, Ampère’s law tells us that
the net number of perpendicular surfaces pierced by the Amperian loop must be zero. Because of
the radial symmetry of the hollow wire, any surfaces within the hollow part of the wire must be
radial and they all must point in the same direction (as the surfaces in Fig. 8.1 do). However, this
contradicts the conclusion that the Amperian loop pierces zero net surfaces – unless there are no
surfaces at all in the wire. We therefore conclude that there can be no magnetic field inside a
hollow wire that has a cylindrical current density. The entire magnetic field at radius r must then
be the magnetic field produced by the current in the core, the current that passes through the
Amperian loop of radius r.
r
r
Figure 8.5 Dividing a wire into a hollow conductor and a cylindrical core.
This leads us to the conclusion that
B(r ) =
µ 0 ienc
.
2π r
where:
B(r) is the magnetic field at r of a wire with a charge density that is radially symmetric.
r is the radius of an Amperian loop.
r or R
ienc is the total current passing through the Amperian loop. ienc =
∫ j (r ) 2π r dr .
0
6
Things to remember:
• The magnetic field is zero inside a hollow wire with radially symmetric current density.
• The magnetic field B(r) inside a solid wire with radially symmetric current density is the
magnetic field of the core (the part of the wire with radius < r) alone.
8.4 The Magnetic Line Integral
In the same way that we used flux to create an integral form for Gauss's law, we now
want to do something similar for the magnetic field to develop an integral equation for Ampère's
law. We previously expressed Ampère's law in the following manner: "The net number of
perpendicular surfaces pierced by an Amperian loop is proportional to the current passing
through the loop." We built the geometry of the magnetic field into the field contours in such a
way that the number of perpendicular surfaces per unit length is proportional to the magnetic
field strength. We must now put this in mathematical form.
Let us begin with the magnetic field of an infinitely-long, current-carrying wire. The
magnetic field lines pass around the wire in concentric circles and the field contour is composed
of a set of half-planes originating on the wire. Let’s use a magnetic field line at radius r as our
Amperian loop. We then know that
B∝
number of surfaces crossed
l
where l = 2 r is the length of the loop. We may solve this for the number of perpendicular
surfaces crossed by the loop:
π
number of perpendicular surfaces crossed ∝ B l.
Let us express this as an equality:
number of perpendicular surfaces crossed = kBl
where k is a constant that depends on how many surfaces we associate with each ampere of
current. We call the quantity Bl the "magnetic line integral" and denote it with an upper case
lambda, . We know then that
Λ
number of perpendicular surfaces crossed = k .
Λ
Λ
Since = Bl, the line integral is a measure of how much field “lies along” the Amperian loop.
The units of the line integral are Tm (tesla meters).
The Magnetic Line Integral for Segment of a Field Line
(8.1)
Λ = Bl
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In a more general case, the Amperian loop need not lie along a magnetic field line. To
calculate how much of the field tends to lie along the loop, we need to take the dot product of the
magnetic field with the direction of the loop:
r r
B⋅l
r
r
The problem we now have is that the relative orientation of B and l usually changes as we go
around the loop. We need then to look at the contribution to the total line integral from one small
segment of the wire:
r r
dΛ = B ⋅ d l .
The total line integral around the loop is then just the sum over all such contributions.
The Magnetic Line Integral
r r
Λ = ∫ B ⋅ dl
(8.2)
The symbol
∫
denotes an integration around the entire Amperian loop.
This type of integral is called a "line integral" or "path integral.” We have previously
used line integrals when we have calculated the work done by a force. The magnetic line integral
differs from other one-dimensional integrals in that it is a sum of the integrand along a path
which is always closed and often irregularly shaped. That is, it usually cannot reduce to a simple
integral over dx, for example. In general we have to slice the path into small sections, calculate
r
B for each section, calculate the length of the path segment and the direction of the path to get
r
d l , take the dot product of these two vectors, and then continue this process for each segment
around the entire path. As you may be hoping, we will always evaluate the line integral in cases
where it reduces to a simple form; however, this prescription for doing a line integral can be
applied to numerical calculations for integrals that are not easily done by hand.
The key to understanding the behavior of the line integral is to think of the dot product
r
between the field and each segment d l of the loop. If the path is parallel to the loop dΛ = + B dl ,
if the path is opposite the loop dΛ = − B dl , and if the path is perpendicular to the loop dΛ = 0 .
Things to remember:
• The line integral is a quantity that is proportional to the number of surfaces (belonging to a
field contour) pierced by a line segment.
• When we choose a magnetic field line for the line segment and the magnetic field is constant
on the field line, the line integral reduces to Λ = Bl .
r r
• The general form for the line integral is Λ = ∫ B ⋅ d l .
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8.5 The Integral Form of Ampère's Law
Now that we have a mathematical expression for the line integral, we may easily
formulate Ampère's law in terms of it. Since the line integral is proportional to the net number of
perpendicular surfaces crossed by the Amperian loop, we know it is also proportional to the
current flowing through the loop. Thus:
Λ = cienc
where c is a constant, and i enc is the total current passing through the loop.
To evaluate the constant, let us apply Ampère's law to the case of a simple currentcarrying wire. Since we know the magnetic field of such a wire, we can directly evaluate the line
integral. We again use a magnetic field line at radius r as the Amperian loop. The line integral is
r r
easy to evaluate because the magnetic field is parallel to the path everywhere. Therefore B ⋅ d l
is just B dl. We then can put the expression for the magnetic field of a long wire into the
integral:
µi
Λ = ∫ Bdl = Bl = 0 2πr = µ 0 i .
2πr
But, by Ampère's law = c i enc. Comparing the two results, we see that the constant c =
Knowing this constant, we can write Ampère's law in its integral form:
Λ
0
.
Ampère’s Law
(8.3)
r r
Λ B = ∫ B ⋅ d l = µ 0 ienc
where:
Λ B is the magnetic line integral around an arbitrary Amperian loop. It has units of tesla
meters (Tm).
r
d l is the (vector) length of a small segment of the Amperian loop in units of m.
r
r
B is the magnetic field on d l in units of tesla (T).
µ 0 is the permeability of free space, = 4π × 10 −7 Tm / A .
ienc is the current passing through the Amperian loop. It has units of amperes (A).
As with Gauss's law, Ampère's law is always true, even if we do not know how to
evaluate the integral. If a current i0 passes through a loop of any shape or size, the line integral is
always 0 i0. If you are asked to find a line integral around a loop, you need not evaluate a complicated integral as long as you know the total current passing through the loop.
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Think About It
When we discussed Ampère's law earlier, we defined the net number of perpendicular
surfaces intersected as the "number of perpendicular surfaces pierced in the same general
direction as the loop minus the number crossed in the opposite direction." How does our
definition of the line integral incorporate this concept?
In working with line integrals, we apply the same conventions we did earlier:
• Go around the Amperian loop in a counterclockwise direction.
• Currents coming out of the page have positive line integrals; those going into the page have
negative line integrals.
To use Ampère's law to find magnetic fields there are two conditions that must be met:
r
• We must choose the Amperian loop such that the magnetic field is parallel to the loop ( B is
r
parallel to d l ). This allows us to simplify the integrand:
r r
B ⋅ d l = Bl
Note that this implies that the Amperian loop is chosen to be a magnetic field line.
• The problem must have symmetry such that the magnitude of the magnetic field is constant
over the entire loop (or the entire part of the loop where the field is non-zero). Then the integral
becomes
r r
B
∫ ⋅ d l = B ∫ d l = Bl .
Ampère's law then simplifies to
Ampère’s Law – Practical Form
(8.4)
B=
µ 0 ienc
l
where:
B is the magnetic field on an Amperian loop (B) must be a constant.
l is the length of the entire Amperian loop.
µ 0 is the permeability of free space, = 4π × 10 −7 Tm / A .
r or R
ienc is the total current passing through the Amperian loop. ienc =
∫ j (r ) dr .
0
10
Things to remember:
r r
• Ampère’s law in integral form is Λ B = ∫ B ⋅ d l = µ 0 ∫ jdA =µ 0 ienc .
• In practical applications this reduces to Bl = µ 0 ∫ jdA
8.6 Applying Ampère's Law
Now we wish to apply Ampère's law to find magnetic fields for a few special cases.
A. Current Distributions with Radial Cylindrical Symmetry
Let us now find the magnetic field of an infinitely long wire with a radially symmetric
distribution of current. The magnetic field lines are clearly circles centered on the axis of the
wire, so we will use a circular Amperian loop of radius r. Symmetry also requires the magnitude
of the magnetic field to be constant on the loop, so we may directly apply Eq. (8.4):
(8.5. B of a cylindrical wire)
B(r ) =
µ 0 ienc
µ 0 r or a
=
∫ j (r )2π r dr
2π r
2π r 0
where
r is the radius of the Amperian loop (the radius at which we wish to find the field)
a is the radius of the wire.
The integral is taken over the area bounded by the Amperian loop. Thus, the upper limit in this
integral is dependent on whether we wish to find the magnetic field inside the wire or outside the
wire.
Example 8.1 Magnetic field inside a cylindrical wire.
A cylindrical wire of radius R has a current I passing through it. Find the magnetic field at radius
r inside the wire.
In a standard wire, the charge density is essentially uniform, so we can easily solve for it:
j=
I
Atotal
=
I
.
πa 2
From here, we can find the enclosed current in one of two different ways:
I.
Geometrically, we know that
ienc = jAenc = jπ r 2 = I
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π r2
r2
=
I
π a2
a2
II.
We can solve for the enclosed current by explicitly integrating over the current density:
r
ienc = ∫ jdA = ∫ j 2πr dr =
0
r
I
I
r2
r2
=
=
r
dr
I
2
2
π
π
.
∫
2
a2
πa 2 0
πa 2
Putting this in Ampère’s law, we have:
B(r ) =
µ 0 ienc
µ
µ Ir
r2
= 0 I 2 = 0 2.
2π r
2π r a
2π a
Example 8.2. Magnetic field outside a proton beam.
A proton beam of radius a has a current density of j = αr 2 where α is a constant. Find the
magnetic outside the beam.
We proceed in essentially the same way as in Example 8.1, except for two things: 1) We are
forced to integrate to obtain the enclosed current, and 2) the integral over current density must
have a as its upper limit since the there are no protons at r>a.
a
r
πα a 4
ienc = ∫ jdA = ∫ j 2πr dr = 2πα ∫ r 3 dr =
.
2
0
0
µi
µ πα a 4 µ 0α a 4
B(r ) = 0 enc = 0
=
2π r
2π r 2
4r
B. Infinite Plane of Wires
Let us now take an array of wires stacked one on top of another and each carrying current
I out of the screen as shown in the Fig. 8.6. If the plane of wires continues infinitely, we may
use Ampere's law to find the magnetic field.
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B3 B2 B1
d
1
P
2
3
Figure 8.6 The magnetic field of a planar array of wires.
The current is out of the screen.
First we need to qualitatively understand the field. At point P in Fig. 8.6, the total magnetic field is the sum of the magnetic fields from each wire. The field from wire 2 is directed upward. The field from wire 1 has components both upward and to the right. The field from wire 3
has components both upward and to the left. When we add the three vectors together, the net
result is upward. On the left-hand side of the array of wires the net field must be downward. Let
us then draw an Amperian loop as shown in the figure. The magnetic field will vary somewhat
along the left and right sides of the loop; however, if the wires are sufficiently closely spaced,
this variation is small. We will take the field to be a constant along these segments. The fields on
the top and bottom segments of the loop may not zero; however, we do know that the field on the
top must be the same as the field on the bottom, and since we are traversing the loop in opposite
directions on top and bottom, the net contribution to the line integral from the top plus the
bottom must be zero. (In any case, we can let the length of the top and bottom be small enough
that Bl on these segments is small.)
The total path integral is then = 2Bd. The enclosed current is the current in each wire,
i, times the number of wires N. Ampère's law then gives:
Λ
Λ = 2 Bd = µ 0 Ni
B=
µ0 N
2d
=
13
1
µ 0 ni
2
Think About It
If two planes of wires are parallel to each other, what is the magnetic field in each region of
space? How does the direction of the current in each plane affect your result?
If there are two planes of charge with current going in opposite directions, the magnetic
field has much in common with the electric field of a capacitor. On the outsides of the planes, the
field is zero. between the planes the fields add to give B = µ 0 ni .
Based on the magnetic fields of wires and planes of wires, we may draw some
conclusions as we did with the electric field. (The dependence on path length for the plane
includes only the parts of the path for which the integral is non-zero.)
Source of
Field
Dimensionality
of Source
Dependence of
Path Length
Dependence
of Field
Line
1
r1
r -1
Plane
2
r0
r0
C. Solenoid
A solenoid is a coil of wire wrapped around a cylindrical core. We will assume that the
solenoid has a length much larger than its radius, so that effects caused by the non-uniformity of
the field near the ends may be ignored. In some ways we can think of a solenoid as similar to two
planes of wires with current going in opposite directions. Ignoring end effects, the field outside
the solenoid is zero. Thus, a solenoid concentrates magnetic field lines inside the coil much as
parallel-plate capacitors concentrate electric field lines between the plates. Solenoids are used in
many practical applications where large magnetic fields are needed. Often a permanent magnet is
placed inside the solenoid coil so that when current is turned on in the solenoid, the permanent
magnet is pulled into the solenoid or pushed out of it. Such a solenoid can do work, such as start
a car engine turning.
We may treat solenoids much as the plane of current in the above example. We draw an
Amperian loop of length d with one side placed inside the solenoid and one outside. On the
outside, there is no field, so the contribution to the line integral is zero. On the top and bottom,
the magnetic field inside the solenoid is perpendicular to the loop, so the contribution is again
zero. Hence, the line integral is just the contribution from the left side of the loop: Λ = Bd .
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d
r
B=0
1
r
B
2
3
Figure 8.7 A solenoid
Note that, as with the infinite plane of wires, the field strength does not depend on position; we
get exactly the same result no matter where we place the left side of the Amperian loop inside the
solenoid. Thus the field is not only concentrated inside solenoids, but it is uniform within.
Applying Ampère’s law, we have:
Bd = µ 0 Ni
B = µ0
N
i
d
If we call the number of turns per unit length in the solenoid n, this becomes:
Magnetic Field in a Solenoid
(8.6)
B = µ 0 ni
where:
B is the magnetic field anywhere within the solenoid. The units are tesla (T).
µ 0 is the permeability of free space, = 4π × 10 −7 Tm / A .
n is the number of coils per meter in the solenoid.
I is the current passing through the solenoid.
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D. Torus
A torus is a solenoid which is formed in a doughnut shape so that the ends are joined. It
has similar characteristics to a solenoid, but is somewhat different because the wires are farther
apart from each other on the outside the torus than on the inside. By symmetry, we know the
magnetic field lines will be circular loops within the torus. We use one field line of radius r for
an Amperian loop as shown in Fig. 8.7. By symmetry we know the field must be constant on this
loop. The path integral is then just = B 2 r. If N is the total number of turns in the torus,
Ampere's law gives:
Λ
π
(8.7 B of a torus)
Λ = B 2 π r = µ0 N i
µ Ni
.
B= 0
2π r
r
r
B
Figure 8.8 A torus with an Amperian loop (dotted).
Things to remember:
• Be able to use Ampère’s law to find the magnetic field inside and outside a wire with radially
symmetric current density.
• Be able to use Ampère’s law to find the magnetic field in a solenoid and in a torus.
• You do not need to memorize the formulas for these magnetic fields.
8.7 Finding Fields with Direct Integration
I’m attaching two sections at the end of this chapter as a brief introduction into other
applications of electromagnetic theory. This section deals with direct integration over charge and
current distributions to find electric and magnetic fields. The basic idea here is that, if we know
the location and velocity of all the charges in a region, we can simply add up all the contributions
to the field from these charge to obtain the total electric and magnetic fields. We already applied
this sort of method to find the electric and magnetic fields of a current-carrying wire in Lesson 2.
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In this section we will do essentially the same thing, but in terms of electric and magnetic fields
rather than in terms of threads and stubs and their forces.
First we’ll begin with electric potential. We know that the electric potential of a point
charge is
V (r ) =
1
q
.
4πε 0 R
This equation assumes that the source charge is located at the origin of a coordinate
system. With an extended source, we need to make an adjustment in our notation. The following
diagram explains the symbols we will use.
P
r
R
r
r
r
r′
Figure 8.9 Integrating over a charge distribution.
We wish to find the electric potential V(r) at a point P which is located at coordinates (x, y, z). We
slice the charge distribution into a lot of small regions, such as the cube shown in Fig. 8.9. The
coordinates of the cube are ( x ′, y ′, z ′ ). We then define the vectors from the origin to the field point P
and to the cube to be :
r
r = xxˆ + yyˆ + zzˆ
r
r ′ = x ′xˆ + y ′yˆ + z ′zˆ
Now, in our equation for the electric potential, the r that appears in the denominator is the vector from the
charge to the field point. This is:
r r r
R = r − r′
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In terms of this vector, the contribution to the total electric potential from the small volume dv,
the cube, is
1 dq
1 ρ dv
1 ρ dx ′dy ′dz ′
dV =
.
=
=
4πε 0 R 4πε 0 R
4πε 0
R
The total electric potential is then
(8.8)
where
V = ∫ dV =
1
4πε 0
∫∫∫
ρ (r ′) dx ′dy ′dz ′
R
r r r
R = r − r ′ = ( x − x ′)xˆ + ( y − y ′) yˆ + ( z − z ′) zˆ
R=
(x − x′)2 + ( y − y ′) 2 + ( z − z ′) 2
and the integral is over the entire charge distribution.
A similar equation holds for the electric field of the charge distribution, based on
Coulomb’s law:
r
dE =
(8.9)
r
r
E (r ) = ∫ dE =
r
R dq
4πε 0 R 3
1
r
1
4πε 0
∫∫∫
ρ (r ′) R dx ′dy ′dz ′
R3
.
Example 8.3 The charged rod.
Find the electric field at a distance r from an infinitely-long, thin rod of uniform charge
density .
y
r
r
r
R
r
r′
z
Geometry for Example 8.3
18
x
dx ′
We basically have to use Eq. (8.9) as a recipe, but let’s carefully write down each step to be sure
everything is correct.
1. Choose a field point.
We want the field at a point located a distance r from the rod. We can put our axes any
r
place we want, so let’s put the point of interest on the y axis. Then r = yyˆ .
2. Slice the distribution.
Rods are easy to slice – we just take slices along the rod, in the x direction.
3. Find the coordinates of the slice.
We need to remember to put primes on all the coordinates associated with the source so
r
they don’t become confused with the coordinates of the field point. r ′ = x ′xˆ .
4. Find the vector from the slice to the field point.
r r r
R = r − r ′ = yyˆ − x ′xˆ .
5. Find the magnitude of this vector.
R = (y 2 + x ′ 2 )
1/ 2
6. Find the charge on the slice.
The length of the slice is dx ′ (We always prime the variables that appear in the slices of
the source distribution, too.) so that dq = λ dx ′ .
7. Now plug everything in to the equations, using care to get the limits of integration correct.
I’ve done the integration itself using Maple.
r
E=
λ
4πε 0
( yyˆ − x′xˆ ) dx′
∞
∫ (z
−∞
λy
= yˆ
4πε 0
2
+ x′ 2
∞
∫ (y
−∞
)
3/ 2
dx ′
2
+ x′
)
2 3/ 2
λ
− xˆ
4πε 0
λy 2
λ
0
− xˆ
2
4πε 0 y
4πε 0
λ
E (r ) =
2πε 0 r
∞
∫ (y
−∞
x ′ dx ′
2
+ x′2
)
3/ 2
= yˆ
Note that we can also solve this problem using Gauss’s law:
EA =
qenc
ε0
, E=
λL
λ
, E=
2π rLε 0
2πε 0 r
19
Example 8.4 The short charged rod.
Find the electric field and the electric potential at a radial distance r from the end of a thin
rod of uniform charge density ad length L.
r
r
y
r
R
x
r
r′
z
dx ′
L
Geometry for Example 8.4
Again, we use Eqs. (8.8) and (8.9). Steps 1 through 6 are identical to Example 8.3.
7. Now plug everything in to the equations, using care to get the limits of integration correct.
I’ve done the integration itself using Maple.
dx ′
L
λ L
λ
arcsinh
=
∫ 2
1
/
2
4πε 0 0 ( y + x ′ 2 )
4πε 0
y
r
λ L ( yyˆ − x ′xˆ ) dx ′
E=
∫
4πε 0 0 ( y 2 + x ′ 2 )3 / 2
V=
= yˆ
x ′ dx ′
λy L
dx ′
λ L
ˆ
x
−
∫ 2
∫ 2
3/ 2
2
4πε 0 0 ( y + x ′ )
4πε 0 0 ( y + x ′ 2 )3 / 2
y 2 + L2 − y
λy
L
λ
− xˆ
4πε 0 y 2 y 2 + L2
4πε 0 y y 2 + L2
1
λ
=
L yˆ − y 2 + L2 − y xˆ
4πε 0 y y 2 + L2
= yˆ
[ (
)]
The second type of direct integration problem we want to consider is that of finding the
magnetic field of a segment of thin wire carrying a current i. To do this, we start with Eq. (2.17)
for a slowly moving particle to obtain the magnetic field of a moving point charge.
r 1 r r 1 r r
B = β s × E = 2 vs × E .
c
c
20
If we let the charge move slowly, the electric field is essentially just the Coulomb field. We also
use the expression c 2 = 1 / µ 0 ε 0 to simplify the relationship:
r
r
r  1 qs R 
B = µ 0ε 0 vs × 
3 
 4πε 0 R 
r r
µ 0 qs vs × R
.
=
4π
R3
What we really want now, however, is an expression that involves current rather than the
velocity of a single charge. Using a little sleight of hand, we can transform this equation as:
r
dq r r
d
l r
r
r r
r µ 0 dqvrs × R µ 0 dq dt × R µ 0 dt d l × R µ 0 i d l × R
=
=
=
dB =
4π
4π
4π
4π R 3
R3
R3
R3
Integrating over the wire segment, we get a result suggested by Biot and Savart in the 19th
century:
r r
r
r µ 0i d l × R
B = ∫ dB =
4π ∫ R 3
(8.10. The Biot-Savart Law)
where
r
B is the magnetic field of a segment of current carrying wire, measured in tesla (T).
µ 0 is the permeability of free space, = 4π × 10 −7 Tm / A .
i is the current flowing through the wire.
r
d l is a slice of the wire of length dl and pointing in the direction of the current.
r
R is the vector distance from the slice of the wire to the field point.
Example 8.5 the magnetic field of a wire segment
A length of wire extends from 0 to L along the x axis. it carries current i flowing in the −x
direction. Find the magnetic field at a point along the z axis.
r
r
z
y
r
R
r
r′
x
dx ′
i
L
Geometry for Example 8.5
21
Let’s again keep track of the solution step by step.
1. Choose a field point.
r
We’re given r = yyˆ .
2. Slice the distribution.
Slice the wire along the x axis.
3. Find the coordinates of the slice.
r
Again, we put primes on all the coordinates associated with the source: r ′ = x ′xˆ .
4. Find the vector from the slice to the field point.
r r r
R = r − r ′ = yyˆ − x ′xˆ .
5. Find the magnitude of this vector.
(
R = y 2 + x′2
)
1/ 2
r
6. Find d l .
The length of the slice is dx ′ and the direction of the current is in the −x direction.
r
Therefore d l = −dx ′ xˆ .
7. Take the cross product.
r
d l × R = (− dx ′xˆ ) × ( yyˆ − x ′xˆ ) = − ydx ′zˆ as xˆ × xˆ = 0, xˆ × yˆ = zˆ.
8. Now plug everything in to the equations, using care to get the limits of integration correct.
I’ve done the integration itself using Maple.
r
r µ 0 i L − ydx ′zˆ
B = ∫ dB =
4π ∫0 y 2 + x ′ 2 3 / 2
µi
L
= − yzˆ 0
4π y 2 y 2 + L2
(
B(r ) =
)
µ 0i
L
4π r r 2 + L2
Example 8.6. A current loop.
A current loop of radius a is placed in the x-y plane with its center at the origin of a coordinate
system. Find the magnetic field on the z axis. The current flows counterclockwise when viewed
from the +z axis.
22
z
P
i
θ′
a
y
dθ ′
y
x
x
Geometry for Example 8.4.
Again, we go step by step.
1. Choose a field point.
r
We’re given r = zzˆ .
2. Slice the distribution.
Slice the wire in sections along its length – as shown in the figure..
3. Find the coordinates of the slice.
r
it’s most convenient to work in polar coordinates here: r ′ = a cos θ ′ xˆ + a sin θ ′ yˆ .
4. Find the vector from the slice to the field point.
r r r
R = r − r ′ = zzˆ − a cos θ ′ xˆ − a sin θ ′ yˆ .
5. Find the magnitude of this vector.
(
R = a2 + z2
)
1/ 2
r
6. Find d l .
The length of the slice is a dθ ′ (using the formula for the length of an arc). The direction
is just a bit tricky. We know the direction is perpendicular to the r ′ vector, so its slope is the
negative reciprocal of the slope of r ′ . We also note that at the segment shown, the x component
r
of d l is negative while the y component is positive. Thus:
r
d l = − ad θ ′ sin θ ′ xˆ + ad θ ′ cos θ ′ yˆ .
7. Take the cross product.
23
r
d l × R = (− adθ ′ sin θ ′ xˆ + adθ ′ cos θ ′ yˆ ) × ( zzˆ − a cos θ ′ xˆ − a sin θ ′ yˆ )
(
)
= adθ ′ az sin θ ′ yˆ + a sin 2 θ ′ zˆ + az cos θ ′ xˆ + a cos 2 θ ′ zˆ = a 2 dθ ′( z sin θ ′ yˆ + z cos θ ′ xˆ + zˆ )
8. Now plug everything in to the equation, using care to get the limits of integration correct.
The integrations themselves are easy:
r
r µ 0 ia 2
B = ∫ dB =
4π
2π
∫
z sin θ ′ yˆ + z cos θ ′ xˆ + zˆ
(a
0

2π zˆ
0 + 0 + 2
a + z2

µ 0 ia 2
zˆ
B(r ) =
3/ 2
2 a2 + z2
=
µ 0 ia 2
4π
(
(
2
)
+ z2
3/ 2
)
3/ 2
dθ ′



)
Things to remember:
• Be able to find integral expressions for the electric field and the electric potential of straight
rods and charged, circular loops on the axis.
• Be able to find integral expressions for the magnetic fields of straight wire segments and
circular wire loops on the axis.
• You do not need to know how to evaluate the integrals that arise.
8.8 The Differential Form of Gauss’s Law and Ampère’s Law
There is one very fundamental difference between the electric fields of static charges and
the magnetic fields of currents:
Electric field lines always start or end on electric charges, or else go off to infinity.
Magnetic field lines always form closed loops.
This difference is evident in Gauss's law and Ampère's law. Flux is a measurement of the
net number of field lines passing through a surface. If a Gaussian surface contains net charge,
there must be net flux through the surface. The magnetic flux through a Gaussian surface must
be zero because magnetic field lines form loops, and every field line which passes into a
Gaussian surface must pass back out again. The line integral is a measure of how much a field
line tends to form loops. Magnetic fields around wires have line integrals proportional to the
current. The line integrals of electric fields from static charges are always zero.
We can say that these electric field spread and these magnetic fields loop. The difference
between spreading and looping fields can be described by two mathematical concepts:
divergence and curl.
24
A. Divergence
When light rays from the sun pass through a focusing lens, they form a small image of
the sun. We say the light rays "converge." If light from the sun passes through a defocusing lens,
the light rays spread out and we say the rays "diverge." Whenever electric field lines are
produced by finite objects (so we can ignore special cases such as infinite planes), the electric
field lines diverge if the object has positive charge and converge if it has negative charge. This,
however, is not what we mean mathematically when we use the term divergence.
Before we define divergence, let us review a few concepts about fields. The electric field
is a vector field. That is, at every point in space we can define an electric field vector. The
electric potential is a scalar field. At every point in space, the electric potential, a scalar quantity,
is defined. (The two are closely related; the electric field gives the magnitude and direction of the
change in the electric potential.) Divergence is similar to electric potential in that it is a scalar
field. A scalar quantity, the divergence, is defined at every point in space.
The divergence is a measurement of how much a field diverges from a given point in
space. That is, it is a measurement of how many field lines begin or end near that point. We can
tell if a field has divergence at a point because the flux through a small surface around the point
will not be zero. Thus, to measure divergence, we surround a point with a Gaussian surface and
measure the flux through it. But to define the divergence at a point, we must let the Gaussian
surface get small. We then have by Gauss's law:
ΦE =
qenc
ε0
=
1
ε0
r
r
∫ ρ (r )dv →
ρ (r )
∆v as ∆v → 0
ε0
r
We use this flux then to define the divergence. Letting " div E " represent the divergence of the
electric field:
The Definition of Divergence
(8.11)
r
r
Φ E ρ (r )
=
div E = lim
∆v →0 ∆v
ε0
This may seem just a bit complicated; however, the essential idea is that we define divergence at
a point to be the flux per unit volume through a small Gaussian surface located around that point.
Think About It
A single point charge is in empty space. Where is the divergence of the electric field
zero? Where is it non-zero? What if we replace the point charge with a uniformly charged
sphere?
25
Thus, the divergence of an electric field at point in space is proportional to the charge
density at that point. The greater the charge density, the more field lines begin or end near that
point. The source of any electric field with divergence is electric charge.
This equation is, in the end, just a different way of writing Gauss's law. We call it the
"differential form" of Gauss's law as the divergence can be expressed in terms of derivatives.
Thus, this is really a differential equation for the electric field. By using the methods of partial
differential equations, we can solve for the electric field at every point in space as long as we
know the charge density at every point in space. (Unfortunately, it still is difficult!)
A similar equation can be obtained for magnetic field; however, we know that no
magnetic charge exists so the right-hand side of the equation is zero. In summary:
Gauss’s Law of Electricity – Differential Form
(8.12)
where
r
r r
ρ (r )
div E (r ) =
ε0
r r
r
div E (r ) is the divergence of the electric field at a point r .
r
r
ρ (r ) is the charge density at a point r in units of C / m 3 .
ε 0 is the permittivity of free space. It equals 8.85 × 10 −12 C 2 / Nm 2 .
Gauss’s Law of Magnetism – Differential Form
(8.13)
where
r r
div B(r ) = 0
r r
r
div B (r ) is the divergence of the electric field at a point r .
B. Curl
Divergence is a measurement of how much field spreads away from (or in toward) a
point source. Divergence is a scalar field. Because the field lines of a point particle tend to
emanate uniformly in all directions, there is no particular direction associated with the
divergence.
Curl is a measurement of how much a field loops around a line. Curl is a vector field
because at any point we can measure how much the field loops around lines pointing in the x, y,
or z directions. Since the curl is a vector, we can either express curl at a point in space in terms of
three components or in terms of a magnitude and direction. The direction of the curl of the
magnetic field is the direction current flows at that point. For now, we will assume that the
current flows in the x direction. To measure curl, we clearly must rely on the line integral. To
calculate the curl, we 1) choose a point in space, 2) take the line which passes through the point
26
in the direction of the current, 3) construct an Amperian loop around the line, and 4) calculate the
line integral around the Amperian loop. To get a meaningful quantity for the curl; however, we
must let the Amperian loop get small. If we let ∆a be the area of the loop, we have by Ampere's
law:
r
r
Λ = µ 0 ienc = µ 0 ∫ j x (r ) da → µ 0 j x (r )∆a as ∆a → 0
Here, jx is the current density with the x subscript simply emphasizing the fact that the current is
flowing in the x direction. We then use this relationship to define the x component of the curl of
r
B:
(8.14)
Definition of Curl
r r
Λ
r
curl B(r ) x = lim B = µ 0 j x (r )
∆a →0 ∆a
[
]
This then is the differential form of Ampère's law. It simply states that electrical currents are a
source of magnetic fields with curl. Notice that it is only in the region where there is current that
r
the curl of the B field is non-zero. However, this differential equation can be used to solve for
magnetic fields throughout all space. We may generalize this to currents which flow in arbitrary
directions by defining a vector current density which points in the direction of the current at a
given point in space:
Ampère’s Law for Currents
(8.15)
where:
r r
r r
curl B(r ) = µ 0 j (r )
r r
r
curl B(r ) is the curl of the magnetic field at r .
µ 0 is the permeability of free space, = 4π × 10 −7 Tm / A .
r r
r
j (r ) is the current density at r in units of A / m 2 . It points in the direction of the current.
C. The Gradient Operator
Mathematically speaking, an operator is something which does something to something
else. With that vague of a definition, just about anything could be considered an operator. And
that is true. Operators include multiplication, square roots, derivatives, etc. Curl and divergence
are operators. However, before we discuss the curl and divergence operators, let us start from a
more fundamental operator, the gradient operator. The gradient is written as ∇ and often
pronounced "del." As with other operators, we need to know what the gradient can act on and
what is produced after it has acted on something. The gradient operator acts on a scalar field and
produces a vector field. The gradient tells how much the scalar field changes in each of the three
directions, x, y, and z. Therefore, it is much like a derivative in three dimensions. The gradient
operator in Cartesian coordinates can be written as:
27
∇ = xˆ
∂
∂
∂
+ yˆ
+ zˆ
∂z
∂x
∂y
Even if you are not used to the term "gradient," you have used gradient operators already. We
have seen such an expression when we discussed the relationship between electric field and
electric potential. In shorthand, we can write:
Electric Field is the Gradient of Electric Potential
r
E ( x, y, z ) = −∇V .
(8.16)
In other words:
r
∂V
∂V
∂V
+ yˆ
+ zˆ
E ( x, y, z ) = −∇V = − xˆ
∂x
∂y
∂z
What this means, then, is that the electric field is a vector that tells how rapidly and what
direction the electric potential decreases.
D. Divergence and Curl as Differential Operators
Divergence and curl can be expressed in terms of the gradient operator. Many times
students think of the definition of divergence and curl as these differential operators; however, it
is best to remember that they are defined in terms of the flux and the line integral.
The differential form of these operators is typified by the following expressions:
r
r ∂ Ex ∂ E y ∂ Ez
+
div E = ∇ ⋅ E =
+
∂x
∂y
∂z
r
r
∂ By 
∂ B
curl B = ∇ × B = xˆ  z −
+
∂z 
 ∂y
xˆ
∂
=
∂x
Bx
yˆ
∂
∂y
By
 ∂ B y ∂ Bx 
∂ B
∂ Bz 
yˆ  x −
−

 + zˆ 
∂y 
∂x 
 ∂z
 ∂x
zˆ
∂
∂z
Bz
E. Examples
1. A sphere has uniform charge density ρ . By Gauss’s law, we can find the electric
field inside the sphere:
28
EA =
q enc
ε0
4π r 2 E =
⇒E=
4π r 3 ρ
3ε 0
ρr
3ε 0
Since the field is radial, we can write the electric field in vector form as
r ρ rr
ρ
E=
=
(xxˆ + yyˆ + zzˆ )
3ε 0 3ε 0
Then we can take the gradient of the electric field:
r
ρ
ρ
∇⋅E =
(1 + 1 + 1) =
3ε 0
ε0
which is just the differential form of Gauss’s law.
2. A wire has uniform current density j. By Ampère’s law, we can find the
magnetic field inside the wire:
Bl = µ 0 ienc
2π rB = µ 0π r 2 j
⇒B=
µ0 j r
2
Since the field lines circle the wire, it takes a bit of work to find the field direction.
We see that the magnetic field id µ 0 j / 2 times a vector of length r pointing in the
tangent direction. In Fig. 8.10, we draw this vector, assuming current goes in the z
direction. We can then write the magnetic field as:
r µ j
µ j
B = 0 (− r sin θ xˆ + r cos θ yˆ ) = 0 (− yxˆ + xyˆ )
2
2
29
− r sin θ xˆ
r cos θ yˆ
r
r
θ
Figure 8.10. Finding a vector of length r in the tangential direction.
Finally, we can take the curl of the magnetic field:
r µ j
r
∇ × B = 0 ( zˆ + zˆ ) = µ 0 jzˆ = µ 0 j
2
This is just the differential form of Ampère’s law.
Things to remember:
r
Φ
• The definition of divergence: div E = lim E
∆v →0 ∆v
r r
Λ
• The definition of curl: curl B(r ) x = lim B , x
∆a →0 ∆a
r ρ
• Gauss’s law of Electricity: ∇ ⋅ E =
[
]
ε0
r
• Gauss’s law of Magnetism: ∇ ⋅ B = 0
r
r
• Ampère’s law: ∇ × B = µ 0 j
∂
∂
∂
+ yˆ
+ zˆ
• Gradient operator: ∇ = xˆ
∂x
∂y
∂z
r
r ∂ Ex ∂ E y ∂ Ez
• Divergence operator: div E = ∇ ⋅ E =
+
+
∂x
∂y
∂z
r
r
• Curl operator: curl B = ∇ × B
30