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MAXWELL’S EQUATIONS AND
ELECTROMAGNETIC WAVES
29
EXERCISES
Section 29.2 Ambiguity in Ampère’s Law
13.
INTERPRET
In this problem we are asked to find the displacement current through a surface.
DEVELOP
As shown in Equation 29.1, Maxwell’s displacement current is
00
0
()
E
d
d
dEA
dE
IA
dt
dt
dt
εε
ε
Φ
==
=
EVALUATE
The above equation gives
2
12
2
2
0
(8.85 10
C /N m )(1 cm )(1.5 V/m
s)
1.33 nA
d
dE
dt
εμ
−
×
⋅
⋅
=
ASSESS
Displacement current arises from changing electric flux and has units of amperes (A), just like ordinary
current.
14.
The electric field is approximately uniform in the capacitor, so
0
( / ) , and
/
ED
EA
V d A
I
t
E
φ
εφ
=
∂
∂
=
12
2
0
(
/ )
/
(8.85 10
F/m)(10 cm) (220 V/ms)/(0.5 cm)
3.89
A.
Ad dVdt
μ
−
=×
=
Section 29.4 Electromagnetic Waves
15.
INTERPRET
We are given the electric and magnetic fields of an electromagnetic wave and asked to find the
direction of propagation.
DEVELOP
The direction of propagation of the electromagnetic wave is the same as the direction of the cross
product
GG
.
EB
×
EVALUATE
When
E
G
is parallel to
and
ˆ
j
B
G
is parallel to
the direction of propagation is parallel to
ˆ
,
i
,
×
or
ˆ
늿
ji
k
×
.
=−
ASSESS
For electromagnetic waves in vacuum, the directions of the electric and magnetic fields, and of wave
propagation, form a righthanded coordinate system.
16.
(a)
The peak amplitude is the magnitude of
which is
늿
(
Ei j
+
)
,
2.
E
Note that
늿
ˆ
2,
ij
n
+=
where
is a unit vector
45
°
between the positive
x
and
y
axes.
(b)
When
ˆ
n
E
G
is parallel to
(for
ˆ
n
sin(
)
kz
t
ω
−
positive)
points 45
°
into the
second quadrant (so that
Thus,
B
G
, and
is in the
direction).
z
⊥×
+
B
G
is parallel to the unit
vector
늿
/
2
.
−+
0
Section 29.5 Properties of Electromagnetic Waves
17.
INTERPRET
This problem is about measuring the distance between the Sun and the Earth using lightminutes.
DEVELOP
A lightminute (abbreviated as cmin) is approximately equal to
81
1 cmin
(3 10 m/s)(60 s)
1.8 10 m
= ×
On the other hand, the mean distance of the Earth from the Sun (an Astronomical Unit) is about
11
1.5 10 m.
SE
R
EVALUATE
In units of cmin,
can be rewritten as
SE
R
11
10
1cmin
(1.5 10 m)
8.33 cmin
SE
R
=
×
29.1
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Chapter 29
ASSESS
The result implies that it takes about 8.33 minutes for the sunlight to reach the Earth.
18.
Assuming the satellite is approximately overhead, we can estimate the roundtrip travel time by
/
tr
c
Δ=Δ =
5
(2 36,000 km)/(3 10 km/s)
0.24 s.
××
=
19.
INTERPRET
In this problem we want to deduce the airplane’s altitude by measuring the travel time of a radio
wave signal it sends out.
DEVELOP
The speed of light is
and the total distance traveled is
8
31
0 m
/
s
c
=×
2.
rh
Δ
=
EVALUATE
Since
(for waves traveling with speed
c
), the altitude is
2
c
Δ= =Δ
t
8
(3 10 m/s)(50
s)
7.5 km
22
ct
h
μ
Δ×
==
=
ASSESS
The airplane is flying lower than the typical cruising altitude of 12,000 m (35,000 ft) for commercial
jet airplanes.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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