lesson 8 – ampère’s law and differential operators · ... what happens to the ... • current...
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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
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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.
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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:
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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 )2/(0 riB πµ= .
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)
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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:
≤
=otherwise
rrrj
,0
1,sin3),(
θθ
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:
∫= dAjI
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. ∫= dAjI
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
small gate
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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.
Figure 8.5 Dividing a wire into a hollow conductor and a cylindrical core.
This leads us to the conclusion that
r
irB enc
πµ2
)( 0= .
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.
enci is the total current passing through the Amperian loop. ∫=Rorr
enc drrrji0
2)( π .
r r
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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
l
crossedsurfacesofnumberB ∝
where l = 2π r is the length of the loop. We may solve this for the number of perpendicular surfaces crossed by the loop: 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,
Λ. We know then that
number of perpendicular surfaces crossed = k
Λ .
Since
Λ = 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) Bl=Λ
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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:
lrr
⋅B
The problem we now have is that the relative orientation of Br
and lr
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:
lrr
dBd ⋅=Λ . The total line integral around the loop is then just the sum over all such contributions.
The Magnetic Line Integral
(8.2) ∫ ⋅=Λ lrr
dB
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 Br
for each section, calculate the length of the path segment and the direction of the path to get
lr
d , 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
between the field and each segment lr
d of the loop. If the path is parallel to the loop ldBd +=Λ , if the path is opposite the loop ldBd −=Λ , and if the path is perpendicular to the loop 0=Λd .
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 lB=Λ .
• The general form for the line integral is ∫ ⋅=Λ lrr
dB .
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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:
encci=Λ
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 current-carrying 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
easy to evaluate because the magnetic field is parallel to the path everywhere. Therefore lrr
dB ⋅ is just B dl. We then can put the expression for the magnetic field of a long wire into the integral:
irr
iBBd 0
0 22
µππ
µ=∫ ===Λ ll .
But, by Ampère's law
Λ = c i enc. Comparing the two results, we see that the constant c = � 0 .
Knowing this constant, we can write Ampère's law in its integral form:
Ampère’s Law
(8.3) encB idB 0µ=⋅=Λ ∫ lrr
where: BΛ is the magnetic line integral around an arbitrary Amperian loop. It has units of tesla meters (Tm).
lr
d is the (vector) length of a small segment of the Amperian loop in units of m.
Br
is the magnetic field on lr
d in units of tesla (T). 0µ is the permeability of free space, = ATm /104 7−×π .
enci 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 com-plicated 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: • We must choose the Amperian loop such that the magnetic field is parallel to the loop (B
r is
parallel to lr
d ). This allows us to simplify the integrand:
llrr
BdB =⋅ 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
lllrr
BdBdB ==⋅∫ ∫ .
Ampère's law then simplifies to
Ampère’s Law – Practical Form
(8.4) l
enciB 0µ
=
where: B is the magnetic field on an Amperian loop (B) must be a constant. l is the length of the entire Amperian loop.
0µ is the permeability of free space, = ATm /104 7−×π .
enci is the total current passing through the Amperian loop. ∫=Rorr
enc drrji0
)( .
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Things to remember:
• Ampère’s law in integral form is encB ijdAdB 00 µµ∫ ∫ ==⋅=Λ lrr
.
• In practical applications this reduces to ∫= jdAB 0µl
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) ∫=ar
00 drrrj
i
=rBor
0
enc 2)(r 2r 2
)( ππ
µπ
µ
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:
2a
I
A
Ij
total π== .
From here, we can find the enclosed current in one of two different ways: I. Geometrically, we know that
2
2
2
22
a
rI
a
rIrjjAi encenc ====
πππ
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II. We can solve for the enclosed current by explicitly integrating over the current density:
2
22
20
20 2
222a
rI
r
a
Idrr
a
IdrrjjdAi
rr
enc ==∫=∫=∫= ππ
ππ
π .
Putting this in Ampère’s law, we have:
20
2
200
222)(
a
Ir
a
rI
rr
irB enc
πµ
πµ
πµ
=== .
Example 8.2. Magnetic field outside a proton beam. A proton beam of radius a has a current density of 2rj α= 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.
222
4
0
3
0
adrrdrrjjdAi
ra
enc
παπαπ =∫=∫=∫= .
r
aa
rr
irB enc
4222)(
40
400 αµπα
πµ
πµ
===
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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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 mag-
netic field is the sum of the magnetic fields from each wire. The field from wire 2 is directed up-ward. 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
Λ = 2Bd. The enclosed current is the current in each wire,
i, times the number of wires N. Ampère's law then gives:
nid
NB
NiBd
00
0
2
1
2
2
µµµ
==
==Λ
B2
1 2 3
d B1 B3
P
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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 niB 0µ= .
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
r 1
r -1
Plane
2
r 0
r 0
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: Bd=Λ .
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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:
id
NB
NiBd
0
0
µ
µ
=
=
If we call the number of turns per unit length in the solenoid n, this becomes:
Magnetic Field in a Solenoid
(8.6) niB 0µ=
where: B is the magnetic field anywhere within the solenoid. The units are tesla (T).
0µ is the permeability of free space, = ATm /104 7−×π .
n is the number of coils per meter in the solenoid. I is the current passing through the solenoid.
1 2 3
d
0=Br
Br
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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
Λ = B 2π r. If N is the total number of turns in the torus,
Ampere's law gives:
i N = r 2 B = 0µπΛ
(8.7 B of a torus) .r 2
i N = B 0
πµ
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.
r
Br
17
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
R
qrV
04
1)(
πε= .
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.
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 ( zyx ′′′ ,, ). We then define the vectors from the origin to the field point P and to the cube to be :
zzyyxxr
zzyyxxr
ˆˆˆ
ˆˆˆ
′+′+′=′++=
r
r
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:
rrR ′−= rrr
P
rr
r ′r
Rr
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In terms of this vector, the contribution to the total electric potential from the small volume dv, the cube, is
R
zdydxd
R
dv
R
dqdV
′′′=== ρ
περ
πεπε 000 4
1
4
1
4
1.
The total electric potential is then
(8.8) ∫∫∫∫′′′′
==R
zdydxdrdVV
)(
4
1
0
ρπε
where ( )
( ) 222 )()(
ˆ)(ˆ)(ˆ
zzyyxxR
zzzyyyxxxrrR
′−+′−+′−=
′−+′−+′−=′−= rrr
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:
304
1
R
dqREd
rr
πε=
(8.9) ∫∫∫∫′′′′
==3
0
)(
4
1)(
R
zdydxdRrEdrE
rrr ρ
πε.
Example 8.3 The charged rod.
Find the electric field at a distance r from an infinitely-long, thin rod of uniform charge density .
Geometry for Example 8.3
y
x
z r ′rrr R
r
xd ′
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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 place we want, so let’s put the point of interest on the y axis. Then yyr ˆ=r
. 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 they don’t become confused with the coordinates of the field point. xxr ˆ′=′r . 4. Find the vector from the slice to the field point.
xxyyrrR ˆˆ ′−=′−= rrr.
5. Find the magnitude of this vector.
( ) 2/122 xyR ′+= 6. Find the charge on the slice. The length of the slice is xd ′ (We always prime the variables that appear in the slices of the source distribution, too.) so that xddq ′= λ . 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.
( )( )
( ) ( )
rrE
xy
yy
xy
xdxx
xy
xdyy
xz
xdxxyyE
0
02
0
2/3220
2/3220
2/3220
2)(
04
ˆ2
4ˆ
4ˆ
4ˆ
ˆˆ
4
πελ
πελ
πελ
πελ
πελ
πελ
=
−=
′+
′′−
′+
′=
′+
′′−=
∫∫
∫∞
∞−
∞
∞−
∞
∞−
r
Note that we can also solve this problem using Gauss’s law:
rE
rL
LE
qEA enc
000 2,
2,
πελ
επλ
ε===
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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.
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.
( )( )( )
( ) ( )
( )[ ]xyLyyLLyy
Lyy
yLyx
Lyy
Lyy
xy
xdxx
xy
xdyy
xy
xdxxyyE
y
L
xy
xdV
LL
L
L
ˆˆ1
4
4ˆ
4ˆ
4ˆ
4ˆ
ˆˆ
4
sinharc44
22
220
22
22
0222
0
02/322
002/322
0
02/322
0
002/122
0
−+−+
=
+
−+−
+=
∫′+
′′−∫
′+
′=
∫′+
′′−=
=∫′+
′=
πελ
πελ
πελ
πελ
πελ
πελ
πελ
πελ
r
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.
Evc
Ec
B ss
rrrrr×=×=
2
11 β .
y
x
z r ′rrr R
r
xd ′
L
21
If we let the charge move slowly, the electric field is essentially just the Coulomb field. We also use the expression 00
2 /1 εµ=c to simplify the relationship:
.4
4
1
30
30
00
R
Rvq
R
RqvB
ss
ss
rr
rrr
×=
×=
πµ
πεεµ
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:
30
30
30
30
4444 R
Rdi
R
Rddt
dq
R
Rdt
ddq
R
RvdqBd s
rlr
rlrrl
r
rrr ×=
×=
×=
×=
πµ
πµ
πµ
πµ
Integrating over the wire segment, we get a result suggested by Biot and Savart in the 19th century:
(8.10. The Biot-Savart Law) ∫∫×==3
0
4 R
RdiBdB
rlr
rr
πµ
where B
r is the magnetic field of a segment of current carrying wire, measured in tesla (T).
0µ is the permeability of free space, = ATm /104 7−×π .
i is the current flowing through the wire.
lr
d is a slice of the wire of length ld 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.
Geometry for Example 8.5
y
x z r ′r
rr R
r
xd ′
L
i
22
Let’s again keep track of the solution step by step. 1. Choose a field point. We’re given yyr ˆ=r
. 2. Slice the distribution. Slice the wire along the x axis. 3. Find the coordinates of the slice. Again, we put primes on all the coordinates associated with the source: xxr ˆ′=′r . 4. Find the vector from the slice to the field point.
xxyyrrR ˆˆ ′−=′−= rrr.
5. Find the magnitude of this vector.
( ) 2/122 xyR ′+=
6. Find lr
d . The length of the slice is xd ′ and the direction of the current is in the −x direction.
Therefore xxdd ˆ′−=lr
. 7. Take the cross product.
( ) ( ) zxydxxyyxxdRd ˆˆˆˆ ′−=′−×′−=×lr
as .ˆˆˆ,0ˆˆ zyxxx =×=× 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.
( )
22
0
222
0
02/322
0
4)(
4ˆ
ˆ
4
Lrr
LirB
Lyy
Lizy
xy
zxydiBdB
L
+=
+−=
′+
′−== ∫∫
πµ
πµ
πµrr
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.
23
Geometry for Example 8.4.
Again, we go step by step. 1. Choose a field point. We’re given zzr ˆ=r
. 2. Slice the distribution. Slice the wire in sections along its length – as shown in the figure.. 3. Find the coordinates of the slice. it’s most convenient to work in polar coordinates here: yaxar ˆsinˆcos θθ ′+′=′r . 4. Find the vector from the slice to the field point.
yaxazzrrR ˆsinˆcosˆ θθ ′−′−=′−= rrr.
5. Find the magnitude of this vector.
( ) 2/122 zaR +=
6. Find lr
d . The length of the slice is θ ′da (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
of lr
d is negative while the y component is positive. Thus:
.ˆcosˆsin yadxadd θθθθ ′′+′′−=lr
7. Take the cross product.
P
z
yx
x
θ ′ay
θ ′d
i
24
( ) ( )
( ) ( )zxzyzdazaxazzayazad
yaxazzyadxadRd
ˆˆcosˆsinˆcosˆcosˆsinˆsin
ˆsinˆcosˆˆcosˆsin222 +′+′′=′+′+′+′′=
′−′−×′′+′′−=×
θθθθθθθθθθθθθθl
r
8. Now plug everything in to the equation, using care to get the limits of integration correct. The integrations themselves are easy:
( )
( )
( ) zza
iarB
za
zia
dza
zxzyziaBdB
ˆ2
)(
ˆ200
4
ˆˆcosˆsin
4
2/322
20
2/322
20
2
02/322
20
+=
+++=
′+
+′+′== ∫∫
µ
ππ
µ
θθθπ
µ πrr
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.
25
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:
0)(
)(1
000
→∆∆→==Φ ∫ vasvr
dvrqenc
E ερρ
εε
rr
We use this flux then to define the divergence. Letting " Edivr
" represent the divergence of the electric field:
The Definition of Divergence
(8.11) 0
0
)(lim
ερ r
vEdiv E
v
rr
=∆Φ
=→∆
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?
26
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) 0
)()(
ερ r
rEdivr
rr=
where
)(rEdivrr
is the divergence of the electric field at a point rr
.
)(rrρ is the charge density at a point r
r in units of 3/ mC .
0ε is the permittivity of free space. It equals 2212 /1085.8 NmC−× .
Gauss’s Law of Magnetism – Differential Form
(8.13) 0)( =rBdivrr
where
)(rBdivrr
is the divergence of the electric field at a point rr
. 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
27
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:
0)()( 000 →∆∆→==Λ ∫ aasarjdarji xxenc
rr µµµ
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 Br
:
Definition of Curl
(8.14) [ ] )(lim)( 00
rja
rBcurl xB
ax
rrrµ=
∆Λ=
→∆
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 the curl of theB
r 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) )()( 0 rjrBcurlrrrr
µ=
where:
)(rBcurlrr
is the curl of the magnetic field at rr
.
0µ is the permeability of free space, = ATm /104 7−×π .
)(rjrr
is the current density at rr
in units of 2/ mA . 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:
28
zz
yy
xx
∂∂+
∂∂+
∂∂=∇ ˆˆˆ
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
(8.16) VzyxE −∇=),,(r
.
In other words:
z
Vz
y
Vy
x
VxVzyxE
∂∂+
∂∂+
∂∂−=−∇= ˆˆˆ),,(
r
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:
zyx
xyzxyz
zyx
BBBzyx
zyx
y
B
x
Bz
x
B
z
By
z
B
y
BxBBcurl
z
E
y
E
x
EEEdiv
∂∂
∂∂
∂∂=
∂∂
−∂
∂+
∂∂
−∂
∂+
∂∂
−∂
∂=×∇=
∂∂
+∂
∂+
∂∂
=⋅∇=
ˆˆˆ
ˆˆˆrr
rr
E. Examples
1. A sphere has uniform charge density ρ . By Gauss’s law, we can find the electric field inside the sphere:
29
0
0
32
0
3
3
44
ερ
ερππ
ε
rE
rEr
qEA enc
=⇒
=
=
Since the field is radial, we can write the electric field in vector form as
( )zzyyxxr
E ˆˆˆ33 00
++==ερ
ερ r
r
Then we can take the gradient of the electric field:
00
)111(3 ε
ρερ =++=⋅∇ E
r
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:
2
2
0
20
0
rjB
jrrB
iB enc
µπµπ
µ
=⇒
=
=l
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 2/0 jµ 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:
)ˆˆ(2
)ˆcosˆsin(2
00 yxxyj
yrxrj
B +−=+−=µθθµr
30
Figure 8.10. Finding a vector of length r in the tangential direction.
Finally, we can take the curl of the magnetic field:
jzjzzj
Brr
000 ˆ)ˆˆ(2
µµµ==+=×∇
This is just the differential form of Ampère’s law.
Things to remember:
• The definition of divergence: v
Ediv E
v ∆Φ=
→∆ 0lim
r
• The definition of curl: [ ]a
rBcurl xB
ax ∆
Λ=
→∆
,
0lim)(
rr
• Gauss’s law of Electricity: 0ε
ρ=⋅∇ Er
• Gauss’s law of Magnetism: 0=⋅∇ Br
• Ampère’s law: jBrr
0µ=×∇
• Gradient operator: z
zy
yx
x∂∂+
∂∂+
∂∂=∇ ˆˆˆ
• Divergence operator: z
E
y
E
x
EEEdiv zyx
∂∂
+∂
∂+
∂∂
=⋅∇=rr
• Curl operator: BBcurlrr
×∇=
θr
xr ˆsinθ−yr ˆcosθ
r