ch30 - respostas do livro de eletromagnetismo
TRANSCRIPT
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Chapter 302830
(b) The electromagnetic wave equation can be derived from Maxwells
equations.(c) Electromagnetic waves are transverse waves.
(d) The electric and magnetic fields of an electromagnetic wave in free space
are in phase.
(a) False. Maxwells equations apply to both time-independent and time-
dependent fields.
(b) True. One can use Faradays law and the modified version of Amperes law to
derive the wave equation.
(c) True. Both the electric and magnetic fields of an electromagnetic wave
oscillate at right angles to the direction of propagation of the wave.
(d) True.
4 Theorists have speculated about the existence ofmagnetic monopoles,
and several experimental searches for such monopoles have occurred. Supposemagnetic monopoles were found and that the magnetic field at a distance rfrom a
monopole of strength qmis given byB= (0/4)qm/r2. Modify the Gausss law for
magnetism equation to be consistent with such a discovery.
Determine the Concept Gausss law for magnetism would become
insidem,0S n
qdAB =
where qm, insideis the total magnetic charge inside the
Gaussian surface. Note that Gausss law for electricity follows from the existence
of electric monopoles (charges), and the electric field due to a point charge
follows from the inverse-square nature of Coulombs law.
5 (a) For each of the following pairs of electromagnetic waves, which
has the higher frequency: (1) visible light or X rays, (2) green light or red light,(3) infrared waves or red light. (b) For each of the following pairs of
electromagnetic waves, which has the longer wavelength: (1) visible light or
microwaves, (2) green light or ultraviolet light, (3) gamma rays or ultraviolet
light.
Determine the Concept Refer to Table 30-1 to rank order the frequencies and
wavelengths of the given electromagnetic radiation.
(a) (1) X rays (2) green light (3) red light
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Maxwells Equations and Electromagnetic Waves 2831
(b) (1) microwaves (2) green light (3) ultraviolet light
6 The detection of radio waves can be accomplished with either an
electric dipole antenna or a loop antenna. True or false:
(a) The electric dipole antenna works according to Faradays law.
(b) If a linearly polarized radio wave is approaching you head on such that itselectric field oscillates vertically, to best detect this wave the normal to a loopantennas plane should be oriented so that it points either right or left.
(c) If a linearly polarized radio wave is approaching you such that its electric field
oscillates in a horizontal plane, to best detect this wave using an dipoleantenna the antenna should be oriented vertically.
(a) False. A dipole antenna is oriented parallel to the electric field of an incoming
wave so that the wave can induce an alternating current in the antenna.
(b) True. A loop antenna is oriented perpendicular to the magnetic field of anincoming wave so that the changing magnetic flux through the loop can induce a
current in the loop. Orienting the loop antennas plane so that it points either right
or left satisfies this condition.
(c) False. The dipole antenna needs to be oriented parallel to the electric field of
an incoming wave so that the wave can induce an alternating current in the
antenna.
7 A transmitter emits electromagnetic waves using an electric dipole
antenna oriented vertically. (a) A receiver to detect these waves also uses anelectric dipole antenna that is one mile from the transmitting antenna and at the
same altitude. How should the receivers electric dipole antenna be oriented for
optimum signal reception? (b) A receiver to detect these waves uses a loopantenna that is one mile from the transmitting antenna and at the same altitude.
How should the loop antenna be oriented for optimum signal reception?
Determine the Concept
(a) The electric dipole antenna should be oriented vertically.
(b) The loop antenna and the electric dipole transmitting antenna should be in the
same vertical plane.
8 Show that the expression 0BErr
for the Poynting vector(Equation 30-21) has units of watts per square meter (the SI units for
electromagnetic wave intensity).
Sr
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Chapter 302832
Determine the ConceptWe can that 0BErr
has units of W/m2by substituting
the SI units of Er
, andBr
0 and simplifying the resulting expression.
2222 m
W
m
s
J
m
s
mN
m
s
C
C
mN
m
A
C
N
Am
C
N
AmT
TC
N
==
=
===
9 [SSM] If a red light beam, a green light beam, and a violet light
beam, all traveling in empty space, have the same intensity, which light beam
carries more momentum? (a) the red light beam, (b) the green light beam, (c) theviolet light beam, (d) They all have the same momentum. (e) You cannot
determine which beam carries the most momentum from the data given.
Determine the ConceptThe momentum of an electromagnetic wave is directly
proportional to its energy ( cUp= ). Because the intensity of a wave is its energy
per unit area and per unit time (the average value of its Poynting vector), waves
with equal intensity have equal energy and equal momentum. ( )d is correct.
10 If a red light plane wave, a green light plane wave, and a violet light
plane wave, all traveling in empty space, have the same intensity, which wave hasthe largest peak electric field? (a) the red light wave, (b) the green light wave, (c)
the violet light wave, (d) They all have the same peak electric field. (e) You
cannot determine the largest peak electric field from the data given.
Determine the ConceptThe intensity of an electromagnetic wave is given by
0
00
av 2
BEI == S
r.
The intensity of an electromagnetic
wave is given by:0
00
av 2
BEI == S
r
BecauseE0= cB0:
0
2
0
av 2 c
E=S
r
This result tells us that 20av
ESr
independently of the wavelength of the
electromagnetic radiation. Thus ( )d is correct.
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Maxwells Equations and Electromagnetic Waves 2833
11 Two sinusoidal plane electromagnetic waves are identical except that
wave A has a peak electric field that is three times the peak electric field of wave
B. How do their intensities compare? (a) IA
= 13I
B(b) I
A= 1
9I
B(c) I
A= 3I
B
(d) (e) You cannot determine how their intensities compare from the data
given.
IA
= 9IB
Determine the ConceptThe intensity of an electromagnetic wave is given by
0
00
av 2
BEI == S
r.
Express the intensities of the twowaves:
0
A,0A,0
A2
BEI = and
0
B,0B,0
B2
BEI =
Dividing the first of these equations
by the second and simplifying yields:
B,0B,0
A,0A,0
0
B,0B,0
0
A,0A,0
B
A
2
2
BE
BE
BE
BE
I
I
==
( )( )9
33
B,0B,0
B,0B,0
B
A ==BE
BE
I
I BA 9II =
( )d is correct.
Because wave A has a peak electric
field that is three times that of wave
B, the peak magnetic field of A is
also three times that of wave B.Hence:
Estimation and Approximation
12 In laser cooling and trapping, the forces associated with radiationpressure are used to slow down atoms from thermal speeds of hundreds of meters
per second at room temperature to speeds of a few meters per second or slower.
An isolated atom will absorb only radiation of specific frequencies. If the
frequency of the laser-beam radiation is tuned so that the target atoms will absorbthe radiation, then the radiation is absorbed during a process called resonant
absorption. The cross-sectional area of the atom for resonant absorption is
approximately equal to 2, where is the wavelength of the laser light.
(a) Estimate the acceleration of a rubidium atom (molar mass 85 g/mol) in a laser
beam whose wavelength is 780 nm and intensity is 10 W/m2. (b) About how long
would it take such a light beam to slow a rubidium atom in a gas at roomtemperature (300 K) to near-zero speed?
Picture the ProblemWe can use Newtons 2ndlaw to express the acceleration of
an atom in terms of the net force acting on the atom and the relationship between
radiation pressure and the intensity of the beam to find the net force. Once weknow the acceleration of an atom, we can use the definition of acceleration to find
the stopping time for a rubidium atom at room temperature.
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Chapter 302834
(a) Apply Newtons 2nd
law to the
atom to obtain:maF =r (1)
where Fris the radiation force exerted
by the laser beam.
The radiation pressure Prand
intensity of the beamIare related
according to:c
I
A
FP == rr
Solve for Frto obtain:
c
I
c
IAF
2
r
==
Substitute for Frin equation (1) toobtain: ma
c
I=
2
mc
Ia
2=
Substitute numerical values and evaluate a:
( )( )
( )
25
25
8
23
22
m/s104.1
m/s1044.1
m/s10998.2particles106.022
mol1
mol
g85
nm780W/m10
=
=
=a
(b) Using the definition of
acceleration, express the stopping
time tof the atom:a
vvt initialfinal
=
Because 0:finalv
a
vt initial
Using the rms speed as the initial
speed of an atom, relate to the
temperature of the gas:
initialv
m
kTvv
3rmsinitial ==
Substitute in the expression for the
stopping time to obtain: m
kT
at
31=
Substitute numerical values and evaluate t:
( )( )ms1.2
particles106.022
mol1
mol
g85
K300J/K1038.13
m/s1044.1
1
23
23
25 =
=
t
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Maxwells Equations and Electromagnetic Waves 2835
13 [SSM] One of the first successful satellites launched by the United
States in the 1950s was essentially a large spherical (aluminized) Mylar balloonfrom which radio signals were reflected. After several orbits around Earth,
scientists noticed that the orbit itself was changing with time. They eventually
determined that radiation pressure from the sunlight was causing the orbit of this
object to changea phenomenon not taken into account in planning the mission.
Estimate the ratio of the radiation-pressure force by the sunlight on the satellite tothe gravitational force by Earths gravity on the satellite.
Picture the Problem We can use the definition of pressure to express the
radiation force on the balloon. Well assume that the gravitational force on the
balloon is approximately its weight at the surface of the Earth, that the density ofMylar is approximately that of water and that the area receiving the radiation from
the sunlight is the cross-sectional area of the balloon.
The radiation force acting on the
balloon is given by:APF rr=
whereAis the cross-sectional area of
the balloon.
Because the radiation from the Sun isreflected, the radiation pressure is
twice what it would be if it were
absorbed:
c
IP
2r=
Substituting for PrandAyields: ( )c
Id
c
dIF
2
2 2241
r
==
The gravitational force acting on theballoon when it is in a near-Earth
orbit is approximately its weight at
the surface of Earth:
gtAgVgmwF
ballonsurface,Mylar
MylarMylarballoonballoong
====
where tis the thickness of the Mylarskin of the balloon.
Because the surface area of the
balloon is :224 dr =
gtdF 2Mylarg =
Express the ratio of the radiation-
pressure force to the gravitational
force and simplify to obtain: gct
I
gtd
c
Id
F
F
Mylar2Mylar
2
g
r
2
2
==
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Chapter 302836
Assuming the thickness of the Mylar skin of the balloon to be 1 mm, substitute
numerical values and evaluate Fr/Fg:
( )
7
8
23
3
2
g
r
102
s
m10998.2mm1
s
m81.9
m
kg1000.12
m
kW35.1
=F
F
14 Some science fiction writers have described solar sails that could
propel interstellar spaceships. Imagine a giant sail on a spacecraft subjected to
radiation pressure from our Sun. (a) Explain why this arrangement works better if
the sail is highly reflective rather than highly absorptive. (b) If the sail is assumedhighly reflective, show that the force exerted by the sunlight on the spacecraft is
given by ( )crAP 2S 2 where PSis the power output of the Sun (3.8 1026W),Ais the surface area of the sail, mis the total mass of the spacecraft, ris the distance
from the Sun, and cis the speed of light. (Assume the area of the sail is muchlarger than the area of the spacecraft so that all the force is due to radiation
pressure on the sail only.) (c) Using a reasonable value forA, compare the force
on the spacecraft due to the radiation pressure and the force on the spacecraft dueto the gravitational pull of the Sun. Does the result imply that such a system will
work? Explain your answer.
Picture the Problem(b) We can use the definition of radiation pressure to show
that the force exerted by the sunlight on the spacecraft is given by ( )crAP 2S 2 where PSis the power output of the Sun (3.8 10
26W), Ais the surface area of
the sail, mis the total mass of the spacecraft, ris the distance from the Sun, and cis the speed of light.
(a) If the sail is highly reflective rather than highly absorptive, the radiation force
is doubled.
(b) Because the sail is highly
reflective: c
IAAPF
2rr ==
whereAis the area of the sail.
The intensity of the solar radiation on
the sail is given by2
s
4 r
PI
= .
Substituting forI yields:
cr
AP
cr
APF
2
s
2
sr
24
2
==
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Chapter 302838
Differentiate this expression with
respect to time to obtain an
expression for the rate of change of
the electric field strength:
A
I
dt
dQ
AA
Q
dt
d
dt
dE
000
1
==
=
Substitute numerical values and evaluate dE/dt:
( ) ( )
sV/m104.3
sV/m1040.3m023.0mN/C10854.8
A0.5
14
14
22212
=
=
= dt
dE
(b) Express the displacement current
Id: dt
dI e0d
=
Substitute for the electric flux toobtain:
[ ]dtdEAEA
dtdI 00d ==
Substitute numerical values and evaluateId:
( ) ( )( ) A0.5sV/m1040.3m023.0mN/C10854.8 1422212d == I
16 In a region of space, the electric field varies with time as
E= (0.050 N/C) sin (t), where = 2000 rad/s. Find the peak displacementcurrent through a surface that is perpendicular to the electric field and has an area
equal to 1.00 m2.
Picture the Problem We can express the displacement current in terms of the
electric flux and differentiate the resulting expression to obtainIdin terms of
dE/dt.
The displacement currentIdis
given by: dt
dI e0d
=
Substitute for the electric flux to
obtain: [ ] dt
dEAEA
dt
dI
00d
==
Because :( ) tE 2000sinN/C050.0=
( )[ ]
( ) ( ) tA
tdt
dAI
2000cosN/C050.0s2000
2000sinN/C050.0
0
1-
0d
=
=
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Maxwells Equations and Electromagnetic Waves 2839
Idwill have its maximum value
when cos 2000t= 1. Hence:
( ) ( )N/C050.0s2000 0-1maxd, AI =
Substitute numerical values and evaluateId,max:
( ) ( ) nA89.0C
N050.0m00.1mN
C10854.8s2000 22
2121
maxd, =
= I
17 For Problem 15, show that the magnetic field strength between the
plates a distance rfrom the axis through the centers of both plates is given by
B= (1.9 103T/m)r.
Picture the Problem We can use Amperes law to a circular path of radius r
between the plates and parallel to their surfaces to obtain an expression relatingB
to the current enclosed by the amperian loop. Assuming that the displacement
current is uniformly distributed between the plates, we can relate the displacementcurrent enclosed by the circular loop to the conduction currentI.
Apply Amperes law to a circular
path of radius rbetween the plates
and parallel to their surfaces to
obtain:
IIrBd 0enclosed0C
2 === lrr
B
Assuming that the displacement
current is uniformly distributed: 2d
2 R
I
r
I
= d2
2
IR
rI=
whereRis the radius of the circular
plates.
Substituting forI yields:d2
2
02 IR
rrB
= d2
0
2I
R
rB
=
Substitute numerical values and
evaluateB:( )
( )( )( )
r
rrB
=
=
m
T109.1
m023.02
A0.5A/N104
3
2
27
18 The capacitors referred to in this problem have only empty space
between the plates. (a) Show that a parallel-plate capacitor has a displacement
current in the region between its plates that is given by Id= CdV/dt, where Cis
the capacitance and Vis the potential difference between the plates. (b) A 5.00-nF
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Chapter 302840
parallel-plate capacitor is connected to an ideal ac generator so the potential
difference between the plates is given by V= V0cos t, where V0= 3.00 V and
= 500rad/s. Find the displacement current in the region between the plates asa function of time.
Picture the ProblemWe can use the definitions of the displacement current and
electric flux, together with the expression for the capacitance of an air-core-
parallel-plate capacitor to show thatId= C dV/dt.
(a) Use its definition to express the
displacement currentId: dt
dI e0d
=
Substitute for the electric flux to
obtain:[ ]
dt
dEAEA
dt
dI 00d ==
BecauseE= V/d:dtdV
d
A
dV
dtdAI 00d ==
The capacitance of an air-core-
parallel-plate capacitor whose plates
have areaAand that are separated by
a distance dis given by:
d
AC 0
=
Substituting yields:
dt
dVCI =d
(b) Substitute in the expression derived in (a) to obtain:
( ) ( )[ ] ( )( )( )
( ) t
ttdt
dI
500sinA6.23
500sins500V00.3nF00.5500cosV00.3nF00.5 1d
=
==
19 [SSM] There is a current of 10 A in a resistor that is connected in
series with a parallel plate capacitor. The plates of the capacitor have an area of
0.50 m2, and no dielectric exists between the plates. (a) What is the displacement
current between the plates? (b) What is the rate of change of the electric field
strength between the plates? (c) Find the value of the line integral C dB lrr
, where
the integration path Cis a 10-cm-radius circle that lies in a plane that is parallel
with the plates and is completely within the region between them.
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Maxwells Equations and Electromagnetic Waves 2841
Picture the ProblemWe can use the conservation of charge to findId, the
definitions of the displacement current and electric flux to find dE/dt, and
Amperes law to evaluate lrr
dB around the given path.
(a) From conservation of charge we
know that:
A10d ==II
(b) Express the displacement current
Id:[ ]
dt
dEAEA
dt
d
dt
dI 00
e0d
===
Substituting for dE/dtyields:
A
I
dt
dE
0
d
=
Substitute numerical values and
evaluate dE/dt:
( )
sm
V103.2
m50.0mN
C1085.8
A10
12
2
2
212
=
=dt
dE
(c) Apply Amperes law to a circular
path of radius rbetween the plates
and parallel to their surfaces to
obtain:
enclosed0C
Id = lrr
B
Assuming that the displacement
current is uniformly distributed and
lettingArepresent the area of the
circular plates yields:
A
I
r
I d2
enclosed =
d2
enclosed IA
rI
=
Substitute for to obtain:enclosedId
2
0
CI
A
rd
= l
rrB
Substitute numerical values and evaluate C lrr
dB :
( ) ( ) ( )mT79.0
m50.0
A10m10.0A/N1042
227
C=
=
lrr
dB
20 Demonstrate the validity of the generalized form of Ampres law
(Equation 30-4) by showing that it gives the same result as the BiotSavart law
(Equation 27-3) in a specified situation. Figure 30-13 shows two momentarily
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Chapter 302844
Substitute forI+Idfrom (d) to
obtain:
22
0
22
0
1
2
2
aRR
Ia
aR
aI
RB
+=
+=
Maxwells Equations and the Electromagnetic Spectrum
21 The color of the dominant light from the Sun is in the yellow-green
region of the visible spectrum. Estimate the wavelength and frequency of thedominant light emitted by our Sun.HINT: See Table 30-1.
Picture the Problem We can find both the wavelength and frequency of thedominant light emitted by our Sun in Table 30-1.
Because the radiation from the Sun is
yellow-green dominant, the dominant
wavelength is approximately:
nm580green-yellow =
The corresponding frequency is:
Hz1017.5
nm580
m/s10998.2
14
8
green-yellow
green-yellow
=
=
=
cf
22 (a) What is the frequency of microwave radiation that has a 3.00-cm-long wavelength? (b) Using Table 30-1, estimate the ratio of the shortestwavelength of green light to the shortest wavelength of red light.
Picture the ProblemWe can use c=fto find the frequency corresponding to
the given wavelength.
(a) The frequency of an
electromagnetic wave is the ratio of
the speed of light in a vacuum to the
wavelength of the wave:
cf =
Substitute numerical values and
evaluatef:
GHz10.0
Hz1000.1m1000.3
m/s10998.2 102
8
=
=
=
f
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Maxwells Equations and Electromagnetic Waves 2845
(b) The ratio of the shortest
wavelength green light to the shortest
wavelength red light is:
84.0nm620
nm520
redshortest
greenshortest =
23 (a) What is the frequency of an X ray that has a 0.100-nm-long
wavelength? (b) The human eye is sensitive to light that has a wavelength equal to550 nm. What is the color and frequency of this light? Comment on how thisanswer compares to your answer for Problem 21.
Picture the ProblemWe can use c=fto find the frequency corresponding to
the given wavelengths and consult Table 30-1 to determine the color of light with
a wavelength of 550 nm.
(a) The frequency of an X ray with a
wavelength of 0.100 nm is:
Hz1000.3
m10100.0
m/s10998.2
18
9
8
=
==
cf
(b) The frequency of light with a
wavelength of 550 nm is:Hz1045.5
nm550
m/s10998.2 148
=
=f
Consulting Table 30-1, we see that the color of light that has a wavelength of
550 nm is yellow-green. This result is consistent with those of Problem 21 and is
close to the wavelength of the peak output of the Sun. Because we see naturally
by reflected sunlight, this result is not surprising.
Electric Dipole Radiation
24 Suppose a radiating electric dipole lies along thezaxis. LetI1be the
intensity of the radiation at a distance of 10 m and at angle of 90. Find theintensity (in terms ofI1) at (a) a distance of 30 m and an angle of 90, (b) a
distance of 10 m and an angle of 45, and (c) a distance of 20 m and an angle of
30.
Picture the ProblemWe can use the intensityI1at a distance r= 10 m and at an
angle = 90to find the constant in the expression for the intensity of radiationfrom an electric dipole and then use the resulting equation to find the intensity at
the given distances and angles.
Express the intensity of radiation as a
function of rand :( ) 2
2sin,
r
CrI = (1)
where Cis a constant.
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Chapter 302848
Use the given data to obtain:
( )
( )2
2
2
212
km00.4
90sinkm00.4
W/m104
C
C
=
=
Solving for Cyields: ( )( )W1040.6
W/m1000.4km00.4
5
2122
=
=C
Substitute in equation (1) to obtain:( ) 2
2
5
sinW1040.6
,r
rI
= (2)
For a point at sea level and 1.50 km
from the transmitter:=
= 1.53
km1.50
km00.2tan 1
EvaluateI(53.1,1.50 km):
( )( )
22
2
5
pW/m2.181.53sinkm50.1
W1040.6km5.1,1.53 =
=
I
27 [SSM] A radio station that uses a vertical electric dipole antennabroadcasts at a frequency of 1.20 MHz and has a total power output of 500 kW.
Calculate the intensity of the signal at a horizontal distance of 120 km from the
station.
Picture the ProblemThe intensity of radiation from an electric dipole is given byC(sin
2)/r
2, where Cis a constant whose units are those of power, r is the distance
from the dipole to the point of interest, and is the angle between the antenna
and the position vector .rr
We can integrate the intensity to express the total power
radiated by the antenna and use this result to evaluate C. Knowing C we can find
the intensity at a horizontal distance of 120 km.
Express the intensity of the signal as
a function of rand :( )
2
2sin,
rCrI
=
At a horizontal distance of 120 km
from the station:( )
( )
( )2
2
2
km120
km120
90sin90,km120
C
CI
=
=
(1)
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Maxwells Equations and Electromagnetic Waves 2849
IdAdP= From the definition of intensity wehave: and
( )dArIP = ,tot where, in polar coordinates,
ddrdA sin2=
Substitute for dA to obtain:( )
ddrrIP sin, 22
0 0
tot =
Substitute forI(r,):
ddCP =2
0 0
3
tot sin
From integral tables we find that:
( )] 34
2sincossin 02
31
0
3
=+=
d
Substitute and integrate with respect
to to obtain:[ ] CCdCP
3
8
3
4
3
4 20
2
0
tot
===
Solving for Cyields:tot
8
3PC
=
Substitute for Ptotand evaluate Cto
obtain: ( ) kW68.59kW50083
== C
Substituting for Cin equation (1)
and evaluatingI(120 km, 90):( )
( )2
2
W/m14.4
km120
kW68.5990,km120
=
=I
28 Regulations require that licensed radio stations have limits on their
broadcast power so as to avoid interference with signals from distant stations.
You are in charge of checking compliance with the law. At a distance of 30.0 km
from a radio station that broadcasts from a single vertical electric dipole antennaat a frequency of 800 kHz, the intensity of the electromagnetic wave is
2.00 1013W/m2. What is the total power radiated by the station?
Picture the ProblemThe intensity of radiation from an electric dipole is given by
C(sin2)/r2, where Cis a constant whose units are those of power, r is the distance
from the dipole to the point of interest, and is the angle between the electric
dipole moment and the position vector .rr
We can integrate the intensity to express
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Chapter 302850
the total power radiated by the antenna and use this result to evaluate C. Knowing
Cwe can find the total power radiated by the station.
IdAdP= From the definition of intensity wehave: and
( )dArIP = ,tot
where, in polar coordinates,
ddrdA sin2=
Substitute for dA to obtain:( )
ddrrIP sin, 22
0 0
tot = (1)
Express the intensity of the signal as
a function of rand :( )
2
2sin,
rCrI
= (2)
Substitute forI(r,) in equation (1) to
obtain:
ddCP =2
0 0
3
tot sin
From integral tables we find that:( )]
3
42sincossin
0
2
31
0
3 =+=
d
Substitute and integrate with respect
to to obtain:[ ] CCdCP
3
8
3
4
3
4 20
2
0
tot
===
From equation (2) we have: ( )
2
2
sin
, rrIC=
Substitute for C in the expression for
Ptotto obtain:
( )
2
2
totsin
,
3
8 rrIP =
or, because = 90,
( ) 2tot3
8rrIP
=
Substitute numerical values and
evaluate Ptot:( )( )
mW51.1
km0.30W/m1000.23
8 2213tot
=
=
P
29 A small private plane approaching an airport is flying at an altitude of2.50 km above sea level. As a flight controller at the airport, you know your
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Maxwells Equations and Electromagnetic Waves 2851
system uses a vertical electric dipole antenna to transmit 100 W at 24.0 MHz.
What is the intensity of the signal at the planes receiving antenna when the planeis 4.00 km from the airport? Assume the airport is at sea level.
Picture the ProblemThe intensity of radiation from the airports vertical dipole
antenna is given by C(sin2)/r
2, where C is a constant whose units are those of
power, r is the distance from the dipole to the point of interest, and is the angle
between the electric dipole moment and the position vector .rr
We can integrate the
intensity to express the total power radiated by the antenna and use this result to
evaluate C. Knowing C we can find the intensity of the signal at the planes
elevation and distance from the airport.
Express the intensity of the signal as
a function of rand :( )
2
2sin,
rCrI
= (1)
IdAdP= From the definition of intensity wehave: and( )dArIP = ,tot
where, in polar coordinates,
ddrdA sin2=
Substitute for dA to obtain:( )
ddrrIP sin, 22
0 0
tot =
Substituting forI(r,) yields:
ddCP =2
0 0
3
tot sin
From integral tables we find that:( )]
3
42sincossin
0
2
31
0
3 =+=
d
Substitute and integrate with respect
to to obtain:[ ] CCdCP
3
8
3
4
3
4 20
2
0
tot
===
Solving for C yields:tot
8
3PC
=
Substitute for C in equation (1) to
obtain:( )
2
2
tot sin
8
3,
r
PrI
=
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Chapter 302852
At the elevation of the plane:=
= 0.58
m2500
m4000tan 1
and
( ) ( ) m4717m4000m2500 22 =+=r
Substitute numerical values and
evaluateI(4717 m, 58):( )
( )( )
2
2
2
nW/m386
m4717
0.58sin
8
W10030.58,m4717
=
=
I
Energy and Momentum in an Electromagnetic Wave
30 An electromagnetic wave has an intensity of 100 W/m2. Find its
(a) rms electric field strength, and (b) rms magnetic field strength.
Picture the ProblemWe can use Pr=I/cto find the radiation pressure. The
intensity of the electromagnetic wave is related to the rms values of its electric
and magnetic field strengths according toI=ErmsBBrms/0, where BrmsB =Erms/c.
0
rmsrms
BEI=
or, becauseB
(a) Relate the intensity of the
electromagnetic wave toErmsand
BBrms:Brms=Erms/c,
c
EcEEI
0
2
rms
0
rmsrms
==
Solving forErmsyields: cIE 0rms =
Substitute numerical values and evaluateErms:
( )( )( ) V/m194W/m100m/s10998.2N/A104 2827rms == E
(b) ExpressBBrmsin terms ofErms:
c
EB rmsrms=
Substitute numerical values and
evaluateBBrms:nT647
m/s10998.2
V/m1948rms
=
=B
31 [SSM] The amplitude of an electromagnetic waves electric field is
400 V/m. Find the waves (a) rms electric field strength, (b) rms magnetic fieldstrength, (c) intensity and (d) radiation pressure (Pr).
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Chapter 302854
Substitute numerical values and
evaluateBBrms:
T33.1
T334.1m/s10998.2
V/m4008rms
=
=
=B
(b) The average energy density uav
is
given by: c
BEu
0
rmsrmsav=
Substitute numerical values and
evaluate uav:
( )( )( )( )
33
827av
J/m42.1J/m417.1
m/s10998.2N/A104
T334.1V/m400
==
= u
(c) Express the intensity as the
product of the average energy
density and the speed of light in a
vacuum:
cuI av=
Substitute numerical values and
evaluateI:
( )( )2
83
W/m425
m/s10998.2J/m417.1
=
= I
33 (a) An electromagnetic wave that has an intensity equal to 200 W/m2
is normal to a black 20 cm by 30 cm rectangular card absorbs 100 percent of the
wave. Find the force exerted on the card by the radiation. (b) Find the force
exerted by the same wave if the card reflects 100 percent of the wave.
Picture the Problem We can find the force exerted on the card using the
definition of pressure and the relationship between radiation pressure and the
intensity of the electromagnetic wave. Note that, when the card reflects all the
radiation incident on it, conservation of momentum requires that the force is
doubled.
(a) Using the definition of pressure,
express the force exerted on the card
by the radiation:
APF rr=
Relate the radiation pressure to the
intensity of the wave: c
IP =r
Substitute for Prto obtain:
c
IAF =r
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Maxwells Equations and Electromagnetic Waves 2855
Substitute numerical values and
evaluate Fr:
( )( )( )
nN40
m/s10998.2
m30.0m20.0W/m2008
2
r
=
=F
(b) If the card reflects all of theradiation incident on it, the force
exerted on the card is doubled:
nN80r=F
34 Find the force exerted by the electromagnetic wave on the card in Part
(b) of Problem 33 if both the incident and reflected rays are at angles of 30 to the
normal.
Picture the Problem Only the normal component of the radiation pressure exerts
a force on the card.
Using the definition of pressure,
express the force exerted on the card
by the radiation:
cos2 rr APF =
where the factor of 2 is a consequence
of the fact that the card reflects the
radiation incident on it.
Relate the radiation pressure to the
intensity of the wave: c
IP =r
Substitute for Prto obtain:
c
IAF
cos2r=
Substitute numerical values and evaluate Fr:
( )( )( )nN69
m/s10998.2
30cosm30.0m20.0W/m20028
2
r =
=F
35 [SSM] (a) For a given distance from a radiating electric dipole, at
what angle (expressed as and measured from the dipole axis) is the intensity
equal to 50 percent of the maximum intensity? (b) At what angle is the intensity
equal to 1 percent of the maximum intensity?
Picture the Problem At a fixed distance from the electric dipole, the intensity of
radiation is a function alone.
(a) The intensity of the radiation
from the dipole is proportional to
sin2:
( ) 20 sinII = (1)
whereI0is the maximum intensity.
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Maxwells Equations and Electromagnetic Waves 2857
Substitute numerical values and
evaluateErms:
MV/m245MV/m9.244
mN/C10854.8
kJ/m9.5302212
3
rms
==
=
E
UseBB
rms=Erms/cto findBrmsB
: T817.0m/s10998.2
MV/m9.2448rms =
=B
37 [SSM] An electromagnetic plane wave has an electric field that is
parallel to they axis, and has a Poynting vector that is given by
( ) ( ) [ iS cosW/m100, 22 tkxtx = ]r
, wherex is in meters, k = 10.0 rad/m,
= 3.00 109rad/s, and t is in seconds. (a) What is the direction of propagationof the wave? (b) Find the wavelength and frequency of the wave. (c) Find the
electric and magnetic fields of the wave as functions of xand t.
Picture the ProblemWe can determine the direction of propagation of the wave,its wavelength, and its frequency by examining the argument of the cosine
function. We can find Efrom cE 02 =S
rand Bfrom B= E/c. Finally, we can
use the definition of the Poynting vector and the given expression for to findSr
Er
and .Br
(a) Because the argument of the cosine function is of the form tkx , the wavepropagates in the +xdirection.
(b) Examining the argument of thecosine function, we note that the
wave number kof the wave is:
1m0.102 ==k m628.0=
Examining the argument of the
cosine function, we note that the
angular frequency of the wave
is:
19 s1000.32 == f
Solving forf yields:
MHz4772
s1000.3 19
=
=
f
(c) Express the magnitude of Sr
in
terms ofE: c
E
0
2
=S
r S
rcE 0=
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Chapter 302858
Substitute numerical values and evaluateE:
( )( )( ) V/m1.194W/m100m/s10998.2N/A104 2827 == E
( ) ( ) [ ] jE cosV/m194, tkxtx =r
Because
and
( ) ( ) [ ] iS cosW/m100, 22 tkxtx =r
BESrrr
=0
1
:
where k = 10.0 rad/m and
= 3.00 109rad/s.
UseB=E/cto evaluateB:nT4.647
m/s10998.2
V/m1.1948
=
=B
Because BESrrr
=0
1
, the direction
of must be such that the cross
product of
Br
Er
with is in the
positivexdirection:
Br
( ) ( ) [ ]kB cosnT647, tkxtx =r
where k = 10.0 rad/m and
= 3.00 109rad/s.
38 A parallel-plate capacitor is being charged. The capacitor consists of apair of identical circular parallel plates that have radius band a separation
distance d. (a) Show that the displacement current in the capacitor gap has the
same value as the conduction current in the capacitor leads. (b) What is the
direction of the Poynting vector in the region between the capacitor plates?(c) Find an expression for the Poynting vector in this region and show that its flux
into the region between the plates is equal to the rate of change of the energy
stored in the capacitor.
Picture the Problem We can use the expression for the electric field strength
between the plates of the parallel-plate capacitor and the definition of thedisplacement current to show that the displacement current in the capacitor is
equal to the conduction current in the capacitor leads. In (b) we can use the
definition of the Poynting vector and the directions of the electric and magnetic
fields to determine the direction of the Poynting vector between the capacitor
plates. In (c), well demonstrate that the flux of Sr
into the region between the
plates is equal to the rate of change of the energy stored in the capacitor byevaluating these quantities separately and showing that they are equal.
(a) The displacement current is
proportional to the rate at which the
flux is changing between the plates:
( )dt
dEAAE
dt
d
dt
dI 00
e0d
===
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Chapter 302860
Substitute for andBr
Er
in equation
(1) and simplify to obtain:
( )
R
ji
jiS
2
2
2
1
0
0
00
0
Rdt
dEE
Rdt
dEE
dt
dERE
=
=
=
r
x
y
B
E
S
R
R
whereRb,Eis the electric field
strength between the plates,Ris the
radial distance from the line joining the
centers of the plates, R is a unit vectorpointing radially outward from the line
joining the centers of the plates, and b
is the radius of the plates.
The rate at which energy is stored
in the capacitor is: ( ) ( )
dt
dEEAd
Edt
d
VuVdt
d
dt
dU
0
2
0
=
==
BecauseA
QE
0= :
IA
Qd
dt
dQ
A
Qd
A
Q
dt
d
A
QAd
dt
dU
00
00
0
==
=
Consider a cylindrical surface of
length dand radius b. Because
points inward, the energy flowing
into the solenoid per unit time is:
Sr
( )
( )
dt
dEdEb
bddt
dEEb
bdSdAS
2
0
02
1
n
2
2
=
=
=
Substituting forEand simplifying
yields:
IA
QddtdQ
Adb
bQ
A
Q
dt
ddb
A
QdAS
00
2
2
0
2
0
0n
=
=
=
Because dtdUdAS = n , weve proved that the flux of Sr
into the region
between the capacitor is equal to the rate of change of the energy stored in the
capacitor.
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Maxwells Equations and Electromagnetic Waves 2861
39 [SSM] A pulsed laser fires a 1000-MW pulse that has a 200-ns
duration at a small object that has a mass equal to 10.0 mg and is suspended by afine fiber that is 4.00 cm long. If the radiation is completely absorbed by the
object, what is the maximum angle of deflection of this pendulum? (Think of the
system as a ballistic pendulum and assume the small object was hanging vertically
before the radiation hit it.)
Picture the Problem The diagram
shows the displacement of the
pendulum bob, through an angle , as a
consequence of the complete absorption
of the radiation incident on it. We can
use conservation of energy (mechanical
energy is conserved afterthe collision)
to relate the maximum angle of
deflection of the pendulum to the initial
momentum of the pendulum bob.
Because the displacement of the bob
during the absorption of the pulse is
negligible, we can use conservation of
momentum (conserved during the
collision) to equate the momentum of
the electromagnetic pulse to the initial
momentum of the bob.
h
m
LL cos
0g =U
Apply conservation of energy to
obtain:
0ifif
=+ UUKK
or, because Ui= Kf= 0 andm
pK
2
2
ii = ,
02
f
2
i =+ Um
p
Ufis given by: ( )cos1f == mgLmghU
Substitute for Uf: ( ) 0cos12
2
i =+ mgLm
p
Solve for to obtain:
=
gLm
p2
2
i1
21cos
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Chapter 302862
Use conservation of momentum to
relate the momentum of the
electromagnetic pulse to the initial
momentumpiof the pendulum bob:
iwaveem pc
tP
c
Up =
==
where tis the duration of the pulse.
Substitute forpi: ( )
=
gLcm
tP22
221
21cos
Substitute numerical values and evaluate :
( ) ( )
( )( ) ( )( )
degrees1010.6
m0400.0m/s81.9m/s10998.2mg0.102
ns200MW10001cos
3
2282
22
1
=
=
Remarks: The solution presented here is valid only if the displacement of thebob during the absorption of the pulse is negligible. (Otherwise, the
horizontal component of the momentum of the pulse-bob system is not
conserved during the collision.) We can show that the displacement during
the pulse-bob collision is small by solving for the speed of the bob after
absorbing the pulse. Applying conservation of momentum (mv = P(t)/c) and
solving for vgives v= 6.67 107
m/s. This speed is so slow compared toc,we can conclude that the duration of the collision is extremely close to 200 ns
(the time for the pulse to travel its own length). Traveling at 6.67 107
m/s
for 200 ns, the bob would travel 1.33 1013
ma distance 1000 times
smaller that the diameter of a hydrogen atom. (Because 6.67107
m/s is the
maximum speed of the bob during the collision, the bob would actually travel
less than 1.33 1013
m during the collision.)
40 The mirrors used in a particular type of laser are 99.99% reflecting.(a) If the laser has an average output power of 15 W, what is the average power of
the radiation incident on one of the mirrors? (b) What is the force due to radiation
pressure on one of the mirrors?
Picture the ProblemWe can use the definitions of pressure and the relationship
between radiation pressure and the intensity of the radiation to find the force dueto radiation pressure on one of the mirrors.
(a) Because only about 0.01 percent
of the energy inside the laser "leaks
out", the average power of the
radiation incident on one of the
mirrors is:
W105.1101.0
W15 54
=
=
P
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Maxwells Equations and Electromagnetic Waves 2863
(b) Use the definition of radiation
pressure to obtain: A
FP rr=
where Fris the force due to radiation
pressure andAis the area of the mirror
on which the radiation is incident.
The radiation pressure is also related
to the intensity of the radiation: Ac
P
c
IP
22r ==
where Pis the power of the laser and
the factor of 2 is due to the fact that the
mirror is essentially totally reflecting.
Equate the two expression for the
radiation pressure and solve for Fr: Ac
P
A
F 2r = c
PF
2r=
Substitute numerical values andevaluate Fr:
( ) mN0.1m/s10998.2W105.12
8
5
r ==F
41 [SSM] (a) Estimate the force on Earth due to the pressure of the
radiation on Earth by the Sun, and compare this force to the gravitational force ofthe Sun on Earth. (At Earths orbit, the intensity of sunlight is 1.37 kW/m
2.)
(b).Repeat Part (a) for Mars which is at an average distance of 2.28 10
8km
from the Sun and has a radius of 3.40 103km. (c) Which planet has the larger
ratio of radiation pressure to gravitational attraction.
Picture the ProblemWe can find the radiation pressure force from the definition
of pressure and the relationship between the radiation pressure and the intensity ofthe radiation from the Sun. We can use Newtons law of gravitation to find the
gravitational force the Sun exerts on Earth and Mars.
(a) The radiation pressure exerted on
Earth is given by:A
FP
Earthr,
Earthr, = APF Earthr,Earthr, =
whereAis the cross-sectional area of
Earth.
Express the radiation pressure in
terms of the intensity of theradiationI from the Sun:
c
IP =Earthr,
Substituting for Pr, EarthandAyields:
c
RIF
2
Earthr,
=
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Chapter 302864
Substitute numerical values and
evaluate Fr:( )( )
N1083.5
N10825.5
m/s10998.2
m1037.6kW/m37.1
8
8
8
262
Earthr,
=
=
=
F
The gravitational force exerted
on Earth by the Sun is given by: 2earthsun
Earthg,r
mGmF =
where ris the radius of Earths orbit.
Substitute numerical values and evaluate Fg, Earth:
( )( )( )( )
N10529.3m1050.1
kg1098.5kg1099.1kg/mN10673.6 22211
24302211
Earthg, =
=
F
Express the ratio of the force due toradiation pressure Fr, Earth to thegravitational force Fg, Earth:
14
22
8
Earthg,
Earthr,1065.1
N10529.3
N10825.5 ==
F
F
or
( ) Earthg,14Earthr, 1065.1 FF =
(b) The radiation pressure exertedon Mars is given by:
A
FP
Marsr,
Marsr, = APF Marsr,Marsr, =
whereAis the cross-sectional area of
Mars.
Express the radiation pressure
on Mars in terms of the intensityof the radiationIMarsfrom the
sun:
c
IP MarsMarsr, =
Substituting for Pr, MarsandAyields:
c
RIF
2
MarsMarsMarsr,
=
Express the ratio of the solar
constant at Earth to the solar
constant at Mars:
2
Mars
earth
earth
Mars
=
r
r
I
I
2
Mars
earth
earthMars
=
r
rII
Substitute for to obtain:MarsI
2
Mars
earth
2
MarsearthMarsr,
=
r
r
c
RIF
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Maxwells Equations and Electromagnetic Waves 2865
Substitute numerical values and evaluate Fr, Mars:
( )( )N1018.7
m1028.2
m1050.1
m/s10998.2
km1040.3kW/m37.1 72
11
11
8
232
Marsr, =
=
F
The gravitational force exerted onMars by the Sun is given by:
( )2
Earthsun
2
MarssunMarsg,
11.0
r
mGm
r
mGmF ==
where ris the radius of Mars orbit.
Substitute numerical values and evaluate Fg
( )( )( )( )( )
N1068.1
m1028.2
kg1098.511.0kg1099.1kg/mN10673.6
21
211
24302211
Marsg,
=
=
F
Express the ratio of the force due to
radiation pressure Fr, Mars to the
gravitational force Fg, Mars:
14
21
7
Marsg,
Marsr,1027.4
N1068.1
N1018.7 =
=F
F
or
( ) Marsg,14Marsr, 1027.4 FF =
(c) Because the ratio of the radiation pressure force to the gravitational force is
1.65 1014for Earth and 4.27 1014 for Mars, Mars has the larger ratio. Thereason that the ratio is higher for Mars is that the dependence of the radiation
pressure on the distance from the Sun is the same for both forces (r
2
), whereasthe dependence on the radii of the planets is different. Radiation pressure varies asR
2, whereas the gravitational force varies asR
3(assuming that the two planets
have the same density, an assumption that is nearly true). Consequently, the ratio
of the forces goes as .Because Mars is smaller than Earth, the ratio
is larger.
132 / =RRR
The Wave Equation for Electromagnetic Waves
42 Show by direct substitution that Equation 30-8ais satisfied by the
wave function Ey
=E0sin kxt( )=E0 sink x ct( )where c= /k.
Picture the ProblemWe can show that Equation 30-8ais satisfied by the wave
functionEyby showing that the ratio of 2Ey/x
2to 2Ey/t
2is 1/c
2where c= /k.
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Chapter 302868
Use the second result obtained in (a)
to obtain: 2
2
00002
2
t
E
t
E
tx
E zzz
=
=
or, because 00= 1/c2,
2
2
22
2 1
t
E
cx
E zz
=
.
=
t
E
xx
B
x
zy
00 Using the second result obtained in
(a), find the second partial derivative
ofBBywith respect tox: or
=
x
E
tx
Bzy
002
2
2
2
00002
2
t
B
t
B
tx
Byyy
=
=
or, because 00= 1/c2,
2
2
22
21
t
B
cx
Byy
=
.
Use the first result obtained in (a) to
obtain:
45 [SSM] Show that any function of the formy(x, t) =f(x vt) or
y(x, t) = g(x+ vt) satisfies the wave Equation 30-7
Picture the Problem We can show that these functions satisfy the wave
equations by differentiating them twice (using the chain rule) with respect to xand tand equating the expressions for the second partial of fwith respect to u.
u
f
u
f
x
u
x
f
=
=
Let u = x vt. Then:
and
u
fv
u
f
t
u
t
f
=
=
Express the second derivatives of
fwith respect toxand tto obtain: 2
2
2
2
u
f
x
f
=
and 2
22
2
2
u
f
vt
f
=
Divide the first of these equations by
the second to obtain:
2
2
2
2
2
1
v
t
f
x
f
=
2
2
22
2 1
t
f
vx
f
=
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Maxwells Equations and Electromagnetic Waves 2869
Let u = x + vt. Then:
u
f
u
f
x
u
x
f
=
=
and
u
fv
u
f
t
u
t
f
=
=
Express the second derivatives of
fwith respect toxand tto obtain: 2
2
2
2
u
f
x
f
=
and2
22
2
2
u
fv
t
f
=
Divide the first of these equations by
the second to obtain:
2
2
2
2
2
1
v
t
f
x
f
=
2
2
22
2 1
t
f
vx
f
=
General Problems
46 An electromagnetic wave has a frequency of 100 MHz and is traveling
in a vacuum. The magnetic field is given by ( ) ( ) ( )itkztzB cosT1000.1, 8 = r
.
(a) Find the wavelength and the direction of propagation of this wave. (b) Find
the electric field vector ( tzE , )r
. (c) Determine the Poynting vector, and use it to
find the intensity of this wave.
Picture the ProblemWe can use c= fto find the wavelength. Examination of
the argument of the cosine function will reveal the direction of propagation of the
wave. We can find the magnitude, wave number, and angular frequency of theelectric vector from the given information and the result of (a) and use these
results to obtain Er
(z, t). Finally, we can use its definition to find the Poynting
vector.
(a) Relate the wavelength of the
wave to its frequency and the speed
of light:
f
c=
Substitute numerical values and
evaluate : m00.3MHz100
m/s10998.2 8
=
=
From the sign of the argument of the cosine function and the spatial dependence
onz,we can conclude that the wave propagates in the +zdirection.
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Chapter 302870
(b) Express the amplitude of Er
: ( )( )V/m00.3
T10m/s10998.288
=
== cBE
Find the angular frequency and
wave number of the wave:
( ) 18 s1028.6MHz10022 === fand
1m09.2m00.3
22 ===
k
Because is in the positivezdirection,Sr
Er
must be in the negativeydirection in
order to satisfy the Poynting vector expression:
( ) ( ) ( ) ( )[ ]jtztzE s1028.6m09.2cosV/m00.3, 181 =r
(c) Use its definition to express and evaluate the Poynting vector:
( ) ( )( ) ( ) ( )[ ]( )ijtzBEtzS s1028.6m09.2cos
N/A104
T10V/m00.31,
1812
27
8
0
==
rrr
or
( ) ( ) ( ) ( )[ ]ktztzS s1028.6m09.2cosmW/m9.23, 18122 =r
The intensity of the wave is the
average magnitude of the Poynting
vector. The average value of the
square of the cosine function is 1/2:
( )2
2
21
mW/m9.11
mW/m9.23
=
== Sr
I
47 [SSM] A circular loop of wire can be used to detect electromagneticwaves. Suppose the signal strength from a 100-MHz FM radio station 100 km
distant is 4.0 W/m2
, and suppose the signal is vertically polarized. What is themaximum rms voltage induced in your antenna, assuming your antenna is a
10.0-cm-radius loop?
Picture the ProblemWe can use Faradays law to show that the maximum rms
voltage induced in the loop is given by ,20rms BA = whereAis the area of
the loop,BB0is the amplitude of the magnetic field, and is the angular frequency
of the wave. Relating the intensity of the radiation toB0will allow us to express
rms as a function of the intensity.
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Maxwells Equations and Electromagnetic Waves 2871
The emf induced in the antenna
is given by Faradays law:( ) ( )
( )
ttBR
tBdt
dR
dt
dBA
BAdt
dA
dt
d
dt
d
coscos
sin
peak0
2
0
2
m
==
==
=== nBr
where 02
peak BR = andRis the
radius of the loop antenna..
rms equals peak divided by the
square root of 2: 22
0
2peak
rms
BR == (1)
0
00
2
BEI=
The intensity of the signal is given
by:
or, because 00 cBE = ,
0
2
0
0
00
22
cBBcBI ==
Solving forBB0yields:
c
IB 00
2=
Substituting forBB0and in
equation (1) and simplifying yields:( )
c
IfR
c
IfR
022
02
rms
2
2
22
=
=
Substitute numerical values and evaluate rms:
( ) ( ) ( )( )
mV.62m/s10998.2
W/m0.4N/A104MHz100m100.02
8
22722
rms =
=
48 The electric field strength from a radio station some distance from theelectric dipole transmitting antenna is given by
( ) ( )trad/s1000.1cosN/C1000.1 64 , where tis in seconds. (a) What peakvoltage is picked up on a 50.0-cm long wire oriented parallel with the electric
field direction? (b) What is the maximum voltage that can be induced by thiselectromagnetic wave in a conducting loop of radius 20.0 cm? What orientation of
the loop does this require?
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Chapter 302872
Picture the ProblemThe voltage induced in the piece of wire is the product of
the electric field and the length of the wire. The maximum rms voltage induced in
the loop is given by ,0BA= where A is the area of the loop, BB0 is the
amplitude of the magnetic field, and is the angular frequency of the wave.
(a) BecauseE is independent ofx: lEV= where lis the length of the wire.
( ) ( )( ) t
tV
6
64
10cosV0.50
m500.010cosN/C1000.1
=
=
and V0.50peak =V
Substitute numerical values and
evaluate V:
(b) The maximum voltage induced in
a loop is given by:
AB0=
whereAis the area of the loop andBB0 is
the amplitude of the magnetic field.
EliminateB0in favor of E0and
substitute forA to obtain: c
RE2
0=
Substitute numerical values and evaluate :
( )( ) ( )nV9.41
m/s10998.2
m200.0N/C1000.1s1000.18
2416
=
=
The loop antenna should be oriented so the transmitting antenna lies in the plane
of the loop.
49 A parallel-plate capacitor has circular plates of radius athat areseparated by a distance d. In the gap between the two plates is a thin straight wire
of resistanceRthat connects the centers of the two plates. A time-varying voltage
given by V0sin tis applied across the plates. (a) What is the current drawn bythis capacitor? (b) What is the magnetic field as a function of the radial distance r
from the centerline within the capacitor plates? (c) What is the phase anglebetween the current drawn by the capacitor and the applied voltage?
Picture the Problem Some of the charge entering the capacitor passes through
the resistive wire while the rest of it accumulates on the upper plate. The totalcurrent is the rate at which the charge passes through the resistive wire plus the
rate at which it accumulates on the upper plate. The magnetic field between the
capacitor plates is due to both the current in the resistive wire and the
displacement current though a surface bounded by a circle a distance rfrom the
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Maxwells Equations and Electromagnetic Waves 2873
resistive wire. The phase difference between the current drawn by the capacitor
and the applied voltage may be calculated using a phasor diagram.
(a) The current drawn by the
capacitor is the sum of the
conduction current through the
resistance wire and dQ/dt, where Q
is the charge on the upper plate ofthe capacitor:
dt
dQII += c (1)
Express the conduction currentIcin
terms of the potential difference
between the plates and the resistance
of the wire:
tR
V
R
VI sin0c ==
Because :CVQ=
tCV
dt
dVC
dt
dQ cos0==
Substitute in equation (1):tCVt
R
VI cossin 0
0 += (2)
The capacitance of a parallel-plate
capacitor with plate areaAand plate
separation dis given by:d
a
d
AC
2
00 ==
Substituting for Cequation (2) gives:
+= t
d
at
R
VI
cossin1 20
0
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Chapter 302874
(b) Apply the generalized form of
Amperes law to a circular path of
radius rcentered within the plates of
the capacitor, where is the
displacement current through the flatsurface Sbounded by the path andI
dI'
c
is the conduction current through the
same surface:
( )dc0C
I'Id += lrr
B
By symmetry the line integral isB
times the circumference of the circle
of radius r:
( ) ( )dc02 I'IrB += (3)
In the region between the capacitor
plates there is a uniform electric fielddue to the surface charges +Qand
Q. The associated displacement
current through Sis:
( )
dt
dEr
dt
dEA'
A'E
dt
d
dt
dI'
2
00
0e
0d
==
==
provided ( )ar
0=E , where ( )2aQAQ == so
2
0 a
QE
=
To evaluate the displacement current
we first must evaluateEeverywhere
on S. Near the surface of a
conductor, where is the surface
charge density:
Substituting forEin the equation
for gives:dI'
( )
tVd
r
tVdt
d
d
r
dt
dV
d
r
d
V
dt
dr
dt
dErI'
cos
sin
0
2
0
0
2
0
2
0
2
0
2
0d
=
==
==
Solving forBin equation (3) and substituting forIcand yields:dI'
( ) ( )
+=
+=
+=
td
rt
Rr
V
tVd
rt
R
V
rr
I'IrB
cossin1
2
cossin22
2
000
0
2
000dc0
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Maxwells Equations and Electromagnetic Waves 2875
dc III +=
or
( )
tI
tItI
cos
sinsin
maxd,
maxc,max
+
=+
(c) Both the charge Qand the
conduction currentIcare in phase
with V. However, dQ/dt, which is
equal to the displacement currentId
through Sfor r a, lags Vby 90.(Mathematically, cos t lags behind
sin tby 90.) The voltage Vleads
the currentI=Ic+Idby phase angle
. The current relation is expressed
in terms of the current amplitudes:
The values of the conduction and
displacement current amplitudes
are obtained by comparison with
the answer to Part (a):
R
VI 0maxc, =
and
dVaI 0
2
0maxd, =
A phasor diagram for adding the
currentsIcandIdis shown to the
right. The conduction currentIcis in
phase with the voltage Vacross the
resistor andIdlags behind it by 90:
maxd,I
maxc,I
maxI
dI
cI
I
V
d
aR
RV
d
aV
I
I
2
0
0
2
00
maxc,
maxd,tan
=
==
From the phasor diagram we have:
so
=
d
aR 201tan
Remarks: The capacitor and the resistive wire are connected in parallel.
The potential difference across each of them is the applied voltage V0sin t.
50 A 20-kW beam of electromagnetic radiation is normal to a surface that
reflects 50 percent of the radiation. What is the force exerted by the radiation on
this surface?
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Chapter 302876
Picture the ProblemThe total force on the surface is the sum of the force due to
the reflected radiation and the force due to the absorbed radiation. From the
conservation of momentum, the force due to the 10 kW that are reflected is twice
the force due to the 10 kW that are absorbed.
Express the total force on thesurface:
aFFF += rtot
Substitute for Frand Fa to obtain: ( )c
P
c
P
c
PF
2
32 21
21
tot =+=
Substitute numerical values and
evaluate Ftot:
( )( )
mN10.0m/s10998.22
kW2038tot
=
=F
51 [SSM] The electric fields of two harmonic electromagnetic waves of
angular frequency 1and 2are given byr
E1=E1,0 cos k1x1t( )j andby
r. For the resultant of these two waves, find (a) the
instantaneous Poynting vector and (b) the time-averaged Poynting vector.
E2=E2,0 cos k2x2t+ ( j)
(c) Repeat Parts (a) and (b) if the direction of propagation of the second wave is
reversed so thatr
E2=E2,0 cos k2x+2t+ ( )j
Picture the ProblemWe can use the definition of the Poynting vector and the
relationship between andBr
Er
to find the instantaneous Poynting vectors for each
of the resultant wave motions and the fact that the time average of the cross
product term is zero for 12, and for the square of cosine function to findthe time-averaged Poynting vectors.
(a) Because both waves propagate in
thexdirection:
iBE 0S=rr
kB B=r
ExpressBin terms ofE1andE2: ( )211
EEc
B +=
Substitute forE1andE2 to obtain:
( ) ( ) ([ ]kB coscos1, 220,2110,1 ++= txkEtxkEc
txr
)
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Maxwells Equations and Electromagnetic Waves 2877
The instantaneous Poynting vector for the resultant wave motion is given by:
( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )( )
( )[ ( )
( ) ( ]) i
kj
k
jS
coscos
cos2cos1
coscos1
coscos1
coscos1
,
22
22
0,222
110,20,111
22
0,1
0
2
220,2110,1
0
220,2110,1
220,2110,1
0
+++
+=
++=
++
++=
txkEtxk
txkEEtxkEc
txkEtxkEc
txkEtxkEc
txkEtxkEtxr
(b) The time average of the cross
product term is zero for 12, andthe time average of the square of thecosine terms is :
[ ] iS 2
1 20,2
2
0,1
0
av EE
c
+=
r
(c) In this case kB 2 B=r
because the wave with k= k2propagates in the
direction. The magnetic field is then:i
( ) ( ) ([ ]kB coscos1, 220,2110,1 ++= txkEtxkEc
txr
)
The instantaneous Poynting vector for the resultant wave motion is given by:
( ) ( ) ( )( )
( ) ( )( )
( ) ( )[ ] i
k
jS
coscos1
coscos1
coscos1
,
22
22
0,211
22
0,1
0
220,2110,1
220,2110,1
0
++=
++
++=
txkEtxkEc
txkEtxkEc
txkEtxkEtxr
The time average of the square of the
cosine terms is : [ ] iS
2
1 20,2
2
0,10
av EEc =
r
52 Show thatBzx
= o0Ent
(Equation 30-10) follows from
=S
y
CdA
t
EdB 00lrr
(Equation 30-6dwithI= 0) by integrating along a
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Chapter 302878
suitable curve Cand over a suitable surface Sin a manner that parallels the
derivation of Equation 30-9.
Picture the ProblemWell choose the
curve with sides xand zin thexyplane shown in the diagram and apply
Equation 30-6dto show that
t
E
x
B yz
=
00 .
Because xis very small, we canapproximate the difference inBBz
at the pointsx1andx2by:
( ) ( ) xx
BBxBxB zzz
= 12
Then: zxt
Ed yC
00lrr
B
The flux of the electric field through
this curve is approximately:
yxEdAE yS
= n
zxt
Ezx
x
B yz
=
00
or
t
E
x
B yz
=
00
Apply Faradays law to obtain:
53 For your backpacking excursions, you have purchased a radio capable
of detecting a signal as weak as 1.00 1014
W/m2. This radio has a 2000-turn
coil antenna that has a radius of 1.00 cm wound on an iron core that increases themagnetic field by a factor of 200. The broadcast frequency of the radio station is
1400 kHz. (a) What is the peak magnetic field strength of an electromagnetic
wave of this minimum intensity? (b) What is the peak emf that it is capable of
inducing in the antenna? (c) What would be the peak emf induced in a straight2.00-m long metal wire oriented parallel to the direction of the electric field?
Picture the ProblemWe can use the relationship between the average value of
the Poynting vector (the intensity), E0, and BB0 to find B0B . The application of
Faradays law will allow us to find the emf induced in the antenna. The emf
induced in a 2.00-m wire oriented in the direction of the electric field can be
found using lE= and the relationship betweenEandB.
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Maxwells Equations and Electromagnetic Waves 2879
(a) The intensity of the signal is
related the amplitude of the magnetic
field in the wave:0
2
0
0
00av
22
cBBEIS ===
c
IB 00
2=
Substitute numerical values and evaluateBB0:
( )( )T1016.9
m/s10998.2
W/m1000.1N/A1042 158
21427
0
=
=
B
(b) Apply Faradays law to the
antenna coil to obtain:( ) ( ) ( )
t
tABNK
tBNKdt
dABA
dt
dt
cos
cos
sin
peak
0m
0m
=
=
==
where 0mpeak ABNK=
Substitute numerical values and evaluate peak :
( ) ( )( ) ( )[ ] mV101kHz14002T1016.9m0100.02002000 152peak ==
(c) The voltage induced in the wire is
the product of its length land the
amplitude of electric fieldE0:
lE=
RelateEtoB: tcBcBE sin0==
ttBc sinsin peak0 == l
where 0peak Bcl= Substitute forEto obtain:
Substitute numerical values and evaluate peak :
( )( )( ) V49.5T1016.9m00.2m/s10998.2 158peak ==
54 The intensity of the sunlight striking Earths upper atmosphere is1.37 kW/m
2. (a) Find the rms values of the magnetic and electric fields of this
light. (b) Find the average power output of the Sun. (c) Find the intensity and theradiation pressure at the surface of the Sun.
Picture the ProblemWe can useI=ErmsBBrms/0andBrmsB =Erms/cto expressErms
in terms ofI. We can then useBBrms=Erms/cto findBrmsB . The average power output
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Chapter 302880
of the Sun is given by whereRis the Earth-Sun distance. The
intensity and the radiation pressure at the surface of the sun can be found from the
definitions of these physical quantities.
IRP 2av 4=
(a) The intensity of the radiation
is given by: 0
2
rms
0
rmsrms
c
EBE
I == IcE 0rms =
Substitute numerical values and evaluate :rmsE
( )( )( )
V/m718
V/m4.718kW/m37.1N/A104m/s10998.2 2278rms
=
== E
Use cEB rmsrms= to evaluate :rmsB
T40.2m/s10998.2
V/m4.7188rms =
=B
(b) Express the average power
output of the Sun in terms of the
solar constant:
IRP 2av 4= whereRis the earth-sun distance.
Substitute numerical values and
evaluate Pav:( ) ( )
W1087.3
W10874.3
kW/m37.1m1050.14
26
26
2211
av
=
=
= P
(c) Express the intensity at thesurface of the Sun in terms of the
suns average power output andradius r:
2
av
4 rPI
=
Substitute numerical values (seeAppendix B for the radius of the
Sun) and evaluateIat the surface
of the Sun:
( )
27
27
28
26
W/m1036.6
W/m10363.6
m1096.64
W10874.3
=
=
=
I
Express the radiation pressure in
terms of the intensity: c
IP =r
Substitute numerical values andevaluate Pr: Pa212.0m/s10998.2
W/m10363.68
27
r =
=P
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Maxwells Equations and Electromagnetic Waves 2881
55 [SSM] A conductor in the shape of a long solid cylinder that has a
lengthL, a radius a, and a resistivity carries a steady currentIthat is uniformlydistributed over its cross-section. (a) Use Ohms law to relate the electric field
Er
in the conductor toI, , and a. (b) Find the magnetic field Brjust outside the
conductor. (c) Use the results from Part (a) and Part (b) to compute the Poynting
vector
( )0
BESrrr
= at r= a(the edge of the conductor). In what direction isrS?
(d) Find the flux dASn through the surface of the cylinder, and use this flux toshow that the rate of energy flow into the conductor equalsI
2R, whereRis the
resistance of the cylinder.
Picture the ProblemA side view of the cylindrical conductor is shown in the
diagram. Let the current be to the right (in the +x direction) and choose a
coordinate system in which the +y direction is radially outward from the axis of
the conductor. Then the +z direction is tangent to cylindrical surfaces that are
concentric with the axis of the conductor (out of the plane of the diagram at the
location indicated in the diagram). We can use Ohms law to relate the electricfield strength E in the conductor to I, , and a and Amperes law to find the
magnetic field strength B just outside the conductor. Knowing Er
and we can
find and, using its normal component, show that the rate of energy flow into the
conductor equalsI
Br
Sr
2R, whereR is the resistance.
E
I
a
x
y
Axis of the conductor
B
ELLa
I
A
LIIRV ====
2
where2a
IE
= .
(a) Apply Ohms law to the
cylindrical conductor to obtain:
Because Er
is in the same direction
asI:iE
2a
I
=
rwhere is a unit vector
in the direction of the current.
i
(b) Applying Amperes law to a
circular path of radius aat the
surface of the cylindrical conductor
yields:
( ) IIaBdC
0enclosed02 === lrr
B
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Chapter 302882
Solve for the magnetic field strength
Bto obtain: a
IB
2
0=
Apply a right-hand rule to determine
the direction of at the point of
interest shown in the diagram:
Br
2
0
a
I=B
rwhere is a unit vector
perpendicular to and tangent to the
surface of the conducting cylinder.
i
(c) The Poynting vector is given
by:BESrrr
=0
1
Substitute for Er
and and simplify
to obtain:
Br
j
kiS
2
2
1
32
2
0
2
0
a
I
a
I
a
I
=
=
r
Letting be a unit vector directed
radially outward from the axis of the
cylindrical conductor yields.
r
rS 2 32
2
a
I
=
rwhere is a unit
vector directed radially outward away
from the axis of the conducting
cylinder.
r
(d) The flux through the surface of
the conductor into the conductor is:
( )aLSdAS 2n =
Substitute for Sn, the inward
component ofr
, and simplify to
obtain:
S
( )2
2
32
2
n 22 a
LIaL
a
IdAS
==
Because2
a
L
A
LR
== : RIdAS
2
n =
Remarks: The equality of the two flow rates is a statement of the
conservation of energy.
56 A long solenoid that has nturns per unit length carries a current that
increases linearly with time. The solenoid has radiusR, lengthL, and the currentI
in the windings is given byI= at. (a) Find the induced electric field at a distance r
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Maxwells Equations and Electromagnetic Waves 2883
the flux dASn into the region inside the solenoid, and show that this flux equalsthe rate of increase of the magnetic energy inside the solenoid.
Picture the ProblemAn end view of the solenoid is shown in the diagram. Let
the current be clockwise and choose a coordinate system in which the +x direction
is tangent to cylindrical surfaces concentric with the axis of the solenoid. Notethat the +z direction is out of the plane of the diagram. We can use Faradays law
to express the induced electric field at a distance r
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Maxwells Equations and Electromagnetic Waves 2885
Express the magnetic energy in the
solenoid:( )
( ) ( )
2
2
2
222
0
2
2
0
2
0
2
0
2
m
tLaRn
LRnat
LRB
VuUB
=
=
==
Evaluate dUB/dt:B
==
=
dAStLaRn
tLaRn
dt
d
dt
dUB
n
22
0
2
222
0
2
2
Remarks: The equality of the two flow rates is a statement of the
conservation of energy.
57 Small particles are be blown out of the solar system by the radiationpressure of sunlight. Assume that each particle is spherical, has a radius r,has a
density of 1.00 g/cm3, and absorbs all the radiation in a cross-sectional area of
r2. Assume the particles are located at some distance dfrom the Sun, which has
a total power output of 3.83 1026W. (a) What is the critical value for the radiusrof the particle for which the radiation force of repulsion just balances thegravitational force of attraction to the Sun? (b) Do particles that have radii larger
than the critical value get ejected from the solar system, or is it only particles that
have radii smaller than the critical value that get ejected? Explain your answer.
Picture the Problem We can use a condition for translational equilibrium to
obtain an expression relating the forces due to gravity and radiation pressure thatact on the particles. We can express the force due to radiation pressure in terms of
the radiation pressure and the effective cross sectional area of the particles and the
radiation pressure in terms of the intensity of the solar radiation. We can solve the
resulting equation for r.
(a) Apply the condition for
translational equilibrium to the
particle:
0gr =FF
or, since Fr= PrAand Fg= mg,
02
sr =
R
mGMAP (1)
The radiation pressure Prdepends on
the intensity of the radiationI: c
IP =r
The intensity of the solar radiation at
a distanceRis: 24 R
PI
=
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8/10/2019 Ch30 - Respostas do livro de eletromagnetismo
58/60
Chapter 302886
Substitute forI to obtain:
cR
PP
2r 4=
Substitute for Pr,A, and min
equation (1):( ) 0
4 2s
3
34
2
2 =
R
GMrr
cR
P
Solve for rto obtain:
s16
3
GMc
Pr
=
Substitute numerical values and evaluate r:
( )( )( )( )( )
nm574
kg1099.1kg/mN10673.6m/s10998.2g/cm00.116
W1083.3330221183
26
=
=
r
(b) Because both the gravitational and radiation pressure forces decrease as the
square of the distance from the Sun, it is then a comparison of grain mass to grain
area. Since mass is proportional to volume and thus varies with the cube of the
radius, the larger grains have more mass and thus experience a stronger
gravitational than radiation-pressure force. The critical radius is an upper limit
and so particles smaller than that radius will be blown out.
58 When an electromagnetic wave at normal incidence on a perfectly
conducting surface is reflected, the electric field of the reflected wave at thereflecting surface is equal and opposite to the electric field of the incident wave at
the reflecting surface. (a) Explain why this assertion is valid. (b) Show that the
superposition of incident and reflected waves results in a standing wave. (c) Arethe magnetic fields of the incident waves and reflected waves at the reflecting
surface equal and opposite as well? Explain your answer.
Picture the Problem
(a) At a perfectly conducting surface 0=Er
. Therefore, the sum of the electric
fields of the incident and reflected wave must add to zero, and so ri EErr
= .
( )kxtEE y = cos0i
and
( )kxtEE y += cos0r
(b) Let the incident and reflected
waves be described by:
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8/10/2019 Ch30 - Respostas do livro de eletromagnetismo
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Maxwells Equations and Electromagnetic Waves 2887
Use the trigonometric identity cos(+ ) = coscossinsinto obtain:
( ) ( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
[ ]wave.standingaofequationthe,sinsin2
sinsincoscossinsincoscos
sinsincoscossinsincoscos
coscoscoscos
0
0
0
000ri
kxtE
kxtkxtkxtkxtE
kxtkxtkxtkxtE
kxtkxtEkxtEkxtEEE
y
y
y
yyy
=++=
+=
+=+=+