# MIT 8.02 Capacitance

```MIT: 8.02 Electricity and Magnetism
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Course  PSet 7 (Due Apr 4)  Problems 
Ampere's Law in Action
A long coaxial cable consists of two concentric conductors. The inner conductor is a
, and it carries a current uniformly distributed over its cross
section. The outer conductor is a cylindrical shell with inner radius
. It carries a current that is also uniformly distributed over its cross section, and that is
opposite in direction to the current of the inner conductor. Calculate the magnetic eld
as a function of the distance from the axis. Use the coordinate system described in the
gure below.
Symbolic Check
For the symbolic check, write your answer using some or all of the following: , , R_1 for
, R_2 for
, R_3 for
and mu_0 for
.
For
:
(mu_0*I)/(2*pi*(R_1)^2)*r
Direction: φ^

For
:
(mu_0*I)/(2*pi*r)
Direction: φ^
For
:
mu_0*I*(R_3^2-r^2)/((2*pi*r)*(R_3^2-R_2^2))
Direction: φ^
For
:
0

Direction: None, B is zero
 Answer: None, B is zero
Solution:
Consider the diagram below, showing a cross-section through the cable. Assume the
current
ows “out of the page" in the core and “into the page" in the shell. The cylindrical
symmetry of the system tells us that the magnetic eld lines will be circles centered on the
cable axis, and the magnetic eld strength will depend only on distance from the axis. So
we may construct an Amperian loop of radius centered on the cable axis as shown,
knowing that
will always be tangent to the loop (in the
constant magnitude around the loop.
direction) and will have a
Applying Ampere's law to our loop with counterclockwise circulation, we nd
where
is the component of
zero component of
in the counterclockwise azimuthal direction (the only non-
). Our task now is to nd
for the various regions of interest.
(a) The current in the core is distributed uniformly throughout its cross-sectional area of
. An Amperian loop of radius
encloses an area of
, and will therefore
enclose a fraction of this total current given by the ratio of these two areas:
. Plugging this into our expression for , we have
(Note that we have regarded the enclosed current as positive, since it ows in the same
direction as the normal to the surface bounded by our Amperian loop, given our choice of
counterclockwise circulation.)
(b) An Amperian loop of radius in the region
will enclose all of the current
in the core and none of the current in the shell, so we simply have
(c) The outer shell carries a current
(negative because it is directed opposite to the
normal to the surface bounded by our counterclockwise Amperian loop) which is
distributed uniformly throughout its cross-sectional area of
. A loop in the
region
will enclose
of shell cross-section, and thus will enclose
a fraction of the shell current given by
the full
to get
. The loop also encloses
of the core, so in this region
. Plugging in
:
(d) Finally, in the region
, our loop encloses the full
the shell. The enclosed current is thus
, giving
of the core and the full
of
(Note: when we say “loop
encloses current ", strictly speaking we mean “current
passes through an open surface bounded by loop ".)
Just to get a better feel for the dependence of on , we can take
,
and plot
, as shown below. is plotted in units of
and in units of
.
Note that
is continuous everywhere. Discontinuities in only come from sheets of
current, just as discontinuities in only come from sheets of charge.
 Answers are displayed within the problem