Example 22-3 A Magnetic Field Due to a Changing Electric Field
A parallel-plate capacitor like that shown in Figure 22-8b has circular plates 5.00 cm in diameter. The electric field between
the plates increases by 8.00 * 105 V>m in 1.00 s, inducing a magnetic field. What is the magnitude of that magnetic field
around a loop 3.50 cm in diameter in the space between the capacitor plates?
Set Up
No charge actually moves through the loop
in question, so ithrough = 0 in Equation 22-8.
However, the electric flux through the loop
changes as the electric field changes, so the
term E > t in Equation 22-8 is not zero
and a circulating magnetic field will result.
This example is very similar to Example 22-2,
except that now we need to calculate the
change in electric flux to determine the
magnitude of the circulating magnetic field.
Equation 16-6 tells us the electric flux
through the 3.50-cm diameter loop; the
s is perpendicular to the plane
electric field E
of the loop, so u = 0. The magnetic field has
the same magnitude B all the way around the
loop and is tangent to the loop, so B } = B in
Equation 22-8.
Electric flux:
E = AE# = AE cos u
i
(16-6)
Maxwell-Ampère law:
E
a B / = m0 ai through + e0 t b
(22-8)
E
Permittivity of free space:
e0 = 8.85 * 10
-12
A#s
V#m
E
B
B
E
increasing
electric
field
B
B
loop
(22-9)
i
Permeability of free space:
m0 = 4p * 10-7 T # m>A
Area of a circle of radius r:
A = pr2
Circumference of a circle of radius r:
C = 2pr
Solve
Find the displacement current e0 1E > t2
associated with the change in electric flux
through the loop.
The change in electric flux in a time t = 1.00 s is equal to the area
of the loop multiplied by the change in electric field in that time (recall
that u = 0). The loop has radius r = 11>22 * 13.50 cm2 = 1.75 cm =
1.75 * 10-2 m, so
E = A1E2cos 0 = A1E2 112
= 1p2 11.75 * 10-2 m2 2 18.00 * 105 V>m2 112
= 7.70 * 102 V # m
The displacement current through the loop is
e0
E
7.70 * 102 V # m
A#s
= a8.85 * 10-12 # b a
b
t
V m
1.00 s
= 6.81 * 10-9 A
Use Equation 22-8 to determine the magnitude of the circulating magnetic field.
Since ithrough = 0, the circulation of the magnetic field is equal to
m0 times the displacement current:
E
Circulation of the magnetic field = a B / = m0 ae0
b
t
= a4p * 10-7
T#m
b 16.81 * 10-9 A2 = 8.56 * 10-15 T # m
A
Since B } = B and B has the same value at all points around the circle,
we can write the circulation as
a B / = a B / = B a /
The sum a / is the total distance around the loop, equal to the loop
circumference 2pr. So
B12pr2 = 8.56 * 10-15 T # m
B =
Reflect
8.56 * 10-15 T # m
= 7.78 * 10-14 T
2p 11.75 * 10-2 m2
The induced magnetic field is very weak (7.78 * 10214 T) because the displacement
current that produces it is very small (6.81 * 1029 A). If the electric field between
the plates were to change more rapidly, the displacement current would be greater
and the induced magnetic field stronger.
The displacement current is in the same direction as the current that brings positive charge to the lower plate of the capacitor. The right-hand rule for using Ampère’s
law (Section 19-7) tells us that the induced magnetic field is in the direction shown.
E increasing
B
B
B
B
The magnetic field due to an increasing
E circulates in the same direction as the
electric field due to a decreasing B
(see Example 22-2).