Example 19-5 Magnetic Field Due to a Coaxial Cable
A coaxial cable consists of a solid conductor of radius R1 surrounded by
insulation, which in turn is surrounded by a thin conducting shell of radius
R2 made of either fine wire mesh or a thin metallic foil (Figure 19-23). The
combination is enclosed in an outer layer of insulation. The inner conductor
carries current in one direction, and the outer conductor carries it back in the
opposite direction. For a coaxial cable that carries a constant current i, find
expressions for the magnetic field (a) inside the inner conductor at a distance
r 6 R1 from its central axis, (b) in the space between the two conductors, and
(c) outside the coaxial cable. Assume that the moving charge in the inner
conductor is distributed uniformly over the volume of the conductor.
Figure 19-23 ​A coaxial cable Equal amounts of current flow in opposite directions in
Outer conductor, radius R2
Inner conductor, radius R1
Insulation
the inner and outer conductors of this cable.
Set Up
Both the inner conductor separately and the
­coaxial cable as a whole have the same
cylindrical symmetry as a long, straight wire
(Figure 19-17). So we expect that the field
lines are ­circles concentric with the axis of the
cable. Just as for the long, straight wire, this
means that it’s natural to choose circular paths
concentric with the cable axis as the Amperian
loops. To find the field in the three regions,
we’ll choose the radius r of the Amperian loop
to be less than R1 in part (a), between R1 and
R2 in part (b), and greater than R2 in part (c).
Ampère’s law:
a B} / = m0 i through (19-13)
III
i
II
i
I
R1
R2
Dashed circles labeled I, II, and III: Amperian loops
for parts (a), (b), and (c) respectively
Solve
(a) Find the field inside the inner conductor by
using an Amperian loop of radius r 6 R1.
Inside the inner conductor, the magnetic field has magnitude Binner
and points tangent to the Amperian loop, so B|| = Binner. The left-hand
side of the Ampère’s law equation is
a B} / = Binner a / = Binner 12pr2
The Amperian loop encloses area pr2, which is less than the crosssectional area pR 21 of the inner conductor. The current through the
loop is therefore a fraction 1pr 2 2 > 1pR 21 2 of the total current i in the
inner conductor:
i through = ia
pr 2
r2
b
=
i
pR 21
R 21
Insert these into Equation 19-13 and solve for Binner:
Binner 12pr2 = m0i
Binner = m0i
(b) Find the field between the conductors by
using an Amperian loop of radius r, where
R1 6 r 6 R2.
r2
R 21
m0ir
r2
=
2prR 21
2pR 21
Between the two conductors the magnetic field of magnitude
Bbetween also points tangent to the Amperian loop, so as in part (a)
the left-hand side of Equation 19-13 is
a B} / = Bbetween 12pr2
The Amperian loop encloses the entire inner conductor, so ithrough = i.
Insert these into Equation 19-13 and solve for Bbetween:
Bbetween 12pr2 = m0i
m 0i
Bbetween =
2pr
(c) Find the field outside the cable by using
an Amperian loop of radius r 7 R2.
Just as in parts (a) and (b), outside the outer conductor the magnetic
field of magnitude Bouter points tangent to the Amperian loop, so
a B} / = Bouter 12pr2
The Amperian loop encloses both conductors, each of which carries
current i. Since the currents flow in opposite directions, the net current through the loop is ithrough = 0. So Equation 19-13 tells us
Bouter(2pr) = m0(0)
Bouter = 0
Reflect
Our result from (a) says that the magnetic field is zero at the center
of the ­inner conductor (r = 0), then increases in direct proportion to r
with increasing distance from the center. The field reaches its maximum
value at the outer ­surface of the inner conductor (r = R1). Between the
conductors the field is inversely proportional to r, so the magnitude
decreases with increasing distance from the center of the cable. Outside
the outer conductor there is zero magnetic field.
Coaxial cables are often referred to as “shielded” cables. The arrangement of the two conductors eliminates the presence of stray magnetic fields
outside the cable. The shielding also serves to isolate the inner conductor
from external electromagnetic signals. You’ll find a coaxial cable connected to the back of most television sets (it’s the “cable” in the term “cable
TV”); the signal carried by this cable involves an alternating current and
hence a varying magnetic field rather than a steady one, but the shielding
principle is the same.
B
B=
m 0ir
B=
2pR21
m 0i
2pr
B=0
R1
R2
r