S E C T I O N 3 2 . 4 • Mutual Inductance
1013
32.4 Mutual Inductance
Very often, the magnetic flux through the area enclosed by a circuit varies with time
because of time-varying currents in nearby circuits. This condition induces an emf
through a process known as mutual induction, so called because it depends on the inter-
action of two circuits.
Consider the two closely wound coils of wire shown in cross-sectional view in Figure
32.14. The current I
1
in coil 1, which has N
1
turns, creates a magnetic field. Some of
the magnetic field lines pass through coil 2, which has N
2
turns. The magnetic flux
caused by the current in coil 1 and passing through coil 2 is represented by $
12
.
In analogy to Equation 32.2, we define the
mutual inductance M
12
of coil 2 with
imagine a thin radial slice of the coaxial cable, such as the
light blue rectangle in Figure 32.13. If we consider the inner
and outer conductors to be connected at the ends of the
cable (above and below the figure), this slice represents one
large conducting loop. The current in the loop sets up a
magnetic field between the inner and outer conductors that
passes through this loop. If the current changes, the magnetic
field changes and the induced emf opposes the original
change in the current in the conductors. We categorize this
situation as one in which we can calculate an inductance, but
we must return to the fundamental definition of inductance,
Equation 32.2. To analyze the problem and obtain L, we must
find the magnetic flux through the light blue rectangle in
Figure 32.13. Ampère’s law (see Section 30.3) tells us that the
magnetic field in the region between the shells is due to the
inner conductor and its magnitude is B " '
0
I/2)r, where r is
measured from the common center of the shells. The
magnetic field is zero outside the outer shell (r - b) because
the net current passing through the area enclosed by a circu-
lar path surrounding the cable is zero, and hence from
Ampère’s law,
$ B&ds " 0. The magnetic field is zero inside
the inner shell because the shell is hollow and no current is
present within a radius r * a.
The magnetic field is perpendicular to the light blue
rectangle of length ! and width b # a, the cross section of
interest. Because the magnetic field varies with radial
position across this rectangle, we must use calculus to
find the total magnetic flux. Dividing this rectangle into
strips of width dr, such as the dark blue strip in Figure 32.13,
we see that the area of each strip is ! dr and that the flux
through each strip is B dA " B! dr. Hence, we find the total
flux through the entire cross section by integrating:
Using this result, we find that the self-inductance of the
cable is
(B)
Calculate the total energy stored in the magnetic field
of the cable.
Solution Using Equation 32.12 and the results to part (A)
gives
To finalize the problem, note that the inductance increases
if ! increases, if b increases, or if a decreases. This is consis-
tent with our conceptualization—any of these changes
increases the size of the loop represented by our radial slice
and through which the magnetic field passes; this increases
the inductance.
'
0
!
I
2
4)
ln
"
b
a
#
U "
1
2
L
I
2
"
'
0
!
2)
ln
"
b
a
#
L "
$
B
I
"
$
B
"
!
B dA "
!
b
a
'
0
I
2)r
!
dr "
'
0
I !
2)
!
b
a
dr
r
"
'
0
I!
2)
ln
"
b
a
#
I
!
b
dr
B
r
I
a
Figure 32.13 (Example 32.5) Section of a long coaxial cable.
The inner and outer conductors carry equal currents in oppo-
site directions.
Coil 1
Coil 2
N
1
I
1
N
2
I
2
Figure 32.14 A cross-sectional
view of two adjacent coils. A
current in coil 1 sets up a magnetic
field and some of the magnetic
field lines pass through coil 2.