8.2 Sources of Magnetic Fields Continued-
Definition of Ampere's Law
Andre Marie Ampere was not in agreement with Biot and Savart
as to their experimental determination of the B field.
First of all, isolated "current elements" such as appear in
the law are difficult to justify "physically". Currents are
always part of a complete circuit and hence identifying them
as individual, isolated "elements" denies their true nature.
In his own experimental and theoretical studies, Ampere
determined an alternative law that can be shown to be in
agreement with Biot and Savart's Law for the case of
an infinitely long, infinitesimally thin wire.
Ampere's Law
is stated as:
| (8.2.0.16) |
where Ienc is the current penetrating the area
enclosed by the path over which we integrate.
The obvious example to apply this to is the long, thin wire.
Here, the wire is carrying current into the page and we wish to
find the magnetic field at a point P a distance r away. As in
all examples of using Ampere's Law, we wish to find a closed
path, i.e. one that bounds an area, that allows us to separate
B from the integral. To do this, we
need a path in which B is constant in
magnitude and parallel, antiparallel, or perpendicular to
dl for every part of the path. The reasons
for this are:
- In order to be mathematically sensible, the path must
define an area through which the current passes, i.e.
a non-closed path does not define a definitive area
through which we can unambiguously state what current
is contributing to the B field
along the path. Therefore, the path must be closed.
- Parts of the path must be parallel (or anti-parallel)
to the B field so that we have
something to evaluate for
B·dl.
If, for part of the path, we have
B perpendicular to
dl, then, for those parts,
B·dl = 0.
In the end, we wish to know the magnitude of
B. Therefore, we must have
B·dl
be known over the whole path and known in such a way
that all of B is known, not
just a component of it.
- Clearly, for this to work, you can only use Ampere's
Law for cases in which you already know a lot about
the behavior of B. You need to know
its direction and something about how its magnitude
changes over the possible paths you might choose.
In addition, the B field must have
enough symmetry to make it possible for you to
complete steps 1 and 2 above.
8.2.1 Applying Ampere's Law to a Long, Thin Wire
The obvious path for the long, thin wire is shown below as
the dashed line labeled dl.
We use this circular path of radius r since we know, from
the Biot-Savart Law, that the magnetic field around the
wire is tangent to the circle in the clockwise direction
and that it's magnitude depends only on r, not on position
parallel to the wire. Thus, we use Ampere's Law to derive
To be sure that Ampere's Law is consistent with the
Biot-Savart Law, let's see if it works for some path
other than the circular one we chose. For example,
look at the path below which has two circular arc
sections centered on the wire carrying current into
the page. One arc has radius a and the other radius
b. They are connected by straight sections.
Figure 8.8: Ampere's Law applied to a straight current into the
page and an Amperian path which is not circular.
Note that the path is closed and that we do know the
characteristics of the B field. The
circled numbers indicate that we have identified 4 pieces of
the path for the application of Ampere's Law. These four
pieces give
|
|
|
| |
| ó
(ç)
õ
|
1
|
|
®
B
|
1
|
·d |
®
l
|
+ | ó
(ç)
õ
|
2
|
|
®
B
|
2
|
·d |
®
l
|
+ | ó
(ç)
õ
|
3
|
|
®
B
|
3
|
·d |
®
l
|
+ | ó
(ç)
õ
|
4
|
|
®
B
|
4
|
·d |
®
l
|
|
|
|
| |
B1 |
æ
ç
è
|
3
4
|
ö
÷
ø
|
(2pa) + 0 + B3 |
æ
ç
è
|
1
4
|
ö
÷
ø
|
(2pb) + 0 |
|
|
| (8.2.1.18) |
|
where we need to know what B1 and B3 are
and we have used the fact that, for the straight paths 2 and 4,
the B field is perpendicular to the path
dl at every point along the path. The
Biot-Savart Law also tells us that
If you plug these into our previous result, you find that
Ampere's Law is verified. In practical terms, this means that
Ampere's Law is always valid. It may not always be useful
for finding the B field. In this respect,
it is exactly like Gauss's Law in that you can only apply it
to learn about the B field if you happen to
recognize a symmetry of the field that allows you to bring
B outside the integral of B·dl.
8.2.2 Ampere's Law Applied to a Non-Uniform Current Distribution
Let's now try a test of Ampere's Law on something which would
be very difficult to do with the Biot-Savart Law.
- Problem 1:
- An infinitely long wire with a radius
a = 1 cm has a current non-uniformly distributed
over its cross section. The density of current,
j, varies with radius according to:
where j0 and a = radius of wire are considered
as known quantities.
- Derive expressions for the magnetic field B as a function
of radius inside (r < a) and outside (r > a) the wire.
- What is the force per unit length on a long thin parallel
wire carrying a current I (in the same direction) at
a distance d > a from the center of the conductor of
radius a in part (a)? Is the force attractive or
repulsive?
Figure 8.9: A wire carrying a non-uniform current out of
the page. The current is non-uniform in its
density across the wire.
- Solution:
-
- To get the magnetic field, we use Ampere's Law. Choose
a circular Amperian loop with radius r where r can be
either greater than or less than a as shown in
the figure below.
Figure 8.10: A wire carries a current out of the page. We define
Amperian paths for use of Ampere's Law.
To find the current enclosed by each loop, we need
to integrate over a ring of infinitesimal thickness,
dr¢, and radius r¢ where we let
r¢ go from 0 to r in the case where
r < a and from 0 to a in the case
where r > a as in
figure 8.11.
Figure 8.11: We break the current distribution into cylindrical
shells of radius r¢ and infinitesimal
thickness dr¢ in order to apply Ampere's Law.
The integration can now proceed.
|
|
|
| |
|
|
m0 |
ó
õ
|
r
0
|
j0 |
r¢2
a2
|
2pr¢ dr¢ |
| |
|
| |
|
| (8.2.2.20) |
|
Now, we can plug in our two cases. If r < a, then
If r > a, then our solution becomes
- No matter where the new wire is with respect to the
original one, if the current through it is in the same
direction as through the original one, then the force
is attractive, i.e. toward the original wire. The
force per unit length is
|
Fnet/L = IBr > a = |
m0j0Ia2
4r
|
|
| (8.2.2.23) |
8.2.3 Applying Ampere's Law to a Uniform Current Distribution
- Problem 2:
Do the above problem in the case where the
current density is uniform.
- Solution:
In this case, nothing changes about our
analysis except that j = j0 so
the total current in the wire is
In other words, the integration of j over
the cross-sectional area is trivial since
j is constant. It comes outside the integral
and we integrate over the area to get pa2
where a is the radius of the wire. So, outside
the wire the field is
|
Br > a(2pr) = m0j0pa2 ÞBr > a = |
m0j0pa2
2pr
|
= |
m0I
2pr
|
|
| (8.2.3.25) |
Our analysis of the Ampere integral
is also the same for r < a in that we get
B (2pr). Therefore, Ampere's Law gives us
where I is j0pa2 as we just showed.
Send comments to larryg@upenn5.hep.upenn.edu.
This page was last modified on 03/04/2003 at 18:15:12 (EST).
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