ch30 - respostas do livro de eletromagnetismo

8/10/2019 Ch30 - Respostas do livro de eletromagnetismo

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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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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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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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10/60

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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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.


13/60


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

=


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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15/60


16/60

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 =


evaluatef:

GHz10.0

Hz1000.1m1000.3

m/s10998.2 102

8

=

=

=

f


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(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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20/60

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)


21/60


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


22/60

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=


ddrrIP sin, 22

0 0

tot = (1)



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



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

=


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


23/60


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.



2

2sin,

rCrI

= (1)

IdAdP= From the definition of intensity wehave: and( )dArIP = ,tot

where, in polar coordinates,

ddrdA sin2=


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



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

=


24/60

Chapter 302852

At the elevation of the plane:=

= 0.58

m2500

m4000tan 1

and

( ) ( ) m4717m4000m2500 22 =+=r


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=


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).


25/60


26/60

Chapter 302854


evaluateBBrms:

T33.1

T334.1m/s10998.2

V/m4008rms

=

=

=B

(b) The average energy density uav

is

given by: c

BEu

0

rmsrmsav=


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=


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


27/60



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.


28/60


29/60



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=


30/60

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

===


31/60


32/60

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.


33/60


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


34/60

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


35/60


(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, =


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,

=


36/60

Chapter 302864


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, =


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


37/60


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.


38/60


39/60


40/60

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

=


41/60


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=


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.


42/60

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.


43/60


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?


44/60

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


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


45/60


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


46/60

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


47/60


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?


48/60

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 =+=


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

)


49/60


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


50/60

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.


51/60


(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


52/60

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.


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


53/60


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


54/60

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


55/60


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


56/60


57/60


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

=


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:


59/60


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

=++=

+=

+=+=+

ch30 - respostas do livro de eletromagnetismo

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