chapter 34 electromagnetic waves

Chapter 34

Electromagnetic Waves

CHAPTE R OUTL I N E

34.1 Maxwell’s Equations andHertz’s Discoveries

34.2 Plane Electromagnetic Waves

34.3 Energy Carried byElectromagnetic Waves

34.4 Momentum and RadiationPressure

34.5 Production of ElectromagneticWaves by an Antenna

34.6 The Spectrum ofElectromagnetic Waves

1066

! Electromagnetic waves cover a broad spectrum of wavelengths, with waves in variouswavelength ranges having distinct properties. These images of the Crab Nebula showdifferent structure for observations made with waves of various wavelengths. The images(clockwise starting from the upper left) were taken with x-rays, visible light, radio waves, andinfrared waves. (upper left—NASA/CXC/SAO; upper right—Palomar Observatory; lowerright—VLA/NRAO; lower left—WM Keck Observatory)

1067

The waves described in Chapters 16, 17, and 18 are mechanical waves. By definition,the propagation of mechanical disturbances—such as sound waves, water waves, andwaves on a string—requires the presence of a medium. This chapter is concerned withthe properties of electromagnetic waves, which (unlike mechanical waves) can propa-gate through empty space.

In Section 31.7 we gave a brief description of Maxwell’s equations, which form thetheoretical basis of all electromagnetic phenomena. The consequences of Maxwell’sequations are far-reaching and dramatic. The Ampère–Maxwell law predicts that atime-varying electric field produces a magnetic field, just as Faraday’s law tells us that atime-varying magnetic field produces an electric field.

Astonishingly, Maxwell’s equations also predict the existence of electromagneticwaves that propagate through space at the speed of light c. This chapter begins with adiscussion of how Heinrich Hertz confirmed Maxwell’s prediction when he generatedand detected electromagnetic waves in 1887. That discovery has led to many practicalcommunication systems, including radio, television, radar, and opto-electronics. On aconceptual level, Maxwell unified the subjects of light and electromagnetism by devel-oping the idea that light is a form of electromagnetic radiation.

Next, we learn how electromagnetic waves are generated by oscillating electriccharges. The waves consist of oscillating electric and magnetic fields at right angles toeach other and to the direction of wave propagation. Thus, electromagnetic waves aretransverse waves. The waves radiated from the oscillating charges can be detected atgreat distances. Furthermore, electromagnetic waves carry energy and momentum andhence can exert pressure on a surface.

The chapter concludes with a look at the wide range of frequencies covered byelectromagnetic waves. For example, radio waves (frequencies of about 107 Hz) areelectromagnetic waves produced by oscillating currents in a radio tower’s transmittingantenna. Light waves are a high-frequency form of electromagnetic radiation (about1014 Hz) produced by oscillating electrons in atoms.

34.1 Maxwell’s Equations and Hertz’s Discoveries

In his unified theory of electromagnetism, Maxwell showed that electromagnetic wavesare a natural consequence of the fundamental laws expressed in the following fourequations (see Section 31.7):

(34.1)

(34.2)! B !d A " 0

! E!d A "q#0

James ClerkMaxwellScottish Theoretical Physicist(1831–1879)

Maxwell developed theelectromagnetic theory of lightand the kinetic theory of gases,and explained the nature ofSaturn’s rings and color vision.Maxwell’s successfulinterpretation of theelectromagnetic field resulted inthe field equations that bear hisname. Formidable mathematicalability combined with greatinsight enabled him to lead theway in the study ofelectromagnetism and kinetictheory. He died of cancer beforehe was 50. (North Wind PictureArchives)

Maxwell’s equations

(34.3)

(34.4)

In the next section we show that Equations 34.3 and 34.4 can be combined to obtain awave equation for both the electric field and the magnetic field. In empty space, whereq " 0 and I " 0, the solution to these two equations shows that the speed at whichelectromagnetic waves travel equals the measured speed of light. This result ledMaxwell to predict that light waves are a form of electromagnetic radiation.

The experimental apparatus that Hertz used to generate and detect electromag-netic waves is shown schematically in Figure 34.1. An induction coil is connected to atransmitter made up of two spherical electrodes separated by a narrow gap. The coilprovides short voltage surges to the electrodes, making one positive and the other neg-ative. A spark is generated between the spheres when the electric field near eitherelectrode surpasses the dielectric strength for air (3 $ 106 V/m; see Table 26.1). In astrong electric field, the acceleration of free electrons provides them with enoughenergy to ionize any molecules they strike. This ionization provides more electrons,which can accelerate and cause further ionizations. As the air in the gap is ionized, itbecomes a much better conductor, and the discharge between the electrodes exhibitsan oscillatory behavior at a very high frequency. From an electric-circuit viewpoint, thisis equivalent to an LC circuit in which the inductance is that of the coil and the capaci-tance is due to the spherical electrodes.

Because L and C are small in Hertz’s apparatus, the frequency of oscillation is high,on the order of 100 MHz. (Recall from Eq. 32.22 that for an LC circuit.)Electromagnetic waves are radiated at this frequency as a result of the oscillation (andhence acceleration) of free charges in the transmitter circuit. Hertz was able to detectthese waves using a single loop of wire with its own spark gap (the receiver). Such areceiver loop, placed several meters from the transmitter, has its own effective induc-tance, capacitance, and natural frequency of oscillation. In Hertz’s experiment, sparkswere induced across the gap of the receiving electrodes when the frequency of thereceiver was adjusted to match that of the transmitter. Thus, Hertz demonstrated thatthe oscillating current induced in the receiver was produced by electromagnetic wavesradiated by the transmitter. His experiment is analogous to the mechanical phenome-non in which a tuning fork responds to acoustic vibrations from an identical tuningfork that is oscillating.

% " 1/√LC

! B !d s " &0I ' &0#0 d (E

dt

! E!d s " )d (B

dt

1068 C H A P T E R 3 4 • Electromagnetic Waves

Input

Transmitter

Receiver

Inductioncoil

q –q

+ –

Figure 34.1 Schematic diagram of Hertz’s apparatus forgenerating and detecting electromagnetic waves. The trans-mitter consists of two spherical electrodes connected to aninduction coil, which provides short voltage surges to thespheres, setting up oscillations in the discharge between theelectrodes. The receiver is a nearby loop of wire containinga second spark gap.

Heinrich RudolfHertzGerman Physicist (1857–1894)

Hertz made his most importantdiscovery of electromagneticwaves in 1887. After finding thatthe speed of an electromagneticwave was the same as that oflight, Hertz showed thatelectromagnetic waves, like lightwaves, could be reflected,refracted, and diffracted. Hertzdied of blood poisoning at theage of 36. During his short life,he made many contributions toscience. The hertz, equal to onecomplete vibration or cycle persecond, is named after him.(Hulton-DeutschCollection/CORBIS )

Additionally, Hertz showed in a series of experiments that the radiation generatedby his spark-gap device exhibited the wave properties of interference, diffraction,reflection, refraction, and polarization, all of which are properties exhibited by light,as we shall see in Part 5 of the text. Thus, it became evident that the radio-frequencywaves Hertz was generating had properties similar to those of light waves and differedonly in frequency and wavelength. Perhaps his most convincing experiment was themeasurement of the speed of this radiation. Waves of known frequency were reflectedfrom a metal sheet and created a standing-wave interference pattern whose nodalpoints could be detected. The measured distance between the nodal points enableddetermination of the wavelength *. Using the relationship v " *f (Eq. 16.12), Hertzfound that v was close to 3 $ 108 m/s, the known speed c of visible light.

34.2 Plane Electromagnetic Waves

The properties of electromagnetic waves can be deduced from Maxwell’s equations. Oneapproach to deriving these properties is to solve the second-order differential equationobtained from Maxwell’s third and fourth equations. A rigorous mathematical treatmentof that sort is beyond the scope of this text. To circumvent this problem, we assume thatthe vectors for the electric field and magnetic field in an electromagnetic wave have aspecific space–time behavior that is simple but consistent with Maxwell’s equations.

To understand the prediction of electromagnetic waves more fully, let us focus ourattention on an electromagnetic wave that travels in the x direction (the direction ofpropagation). In this wave, the electric field E is in the y direction, and the magneticfield B is in the z direction, as shown in Figure 34.2. Waves such as this one, in whichthe electric and magnetic fields are restricted to being parallel to a pair of perpendicu-lar axes, are said to be linearly polarized waves. Furthermore, we assume that at anypoint in space, the magnitudes E and B of the fields depend upon x and t only, and notupon the y or z coordinate.

Let us also imagine that the source of the electromagnetic waves is such that a waveradiated from any position in the yz plane (not just from the origin as might be sug-gested by Figure 34.2) propagates in the x direction, and all such waves are emitted inphase. If we define a ray as the line along which the wave travels, then all rays for thesewaves are parallel. This entire collection of waves is often called a plane wave. Asurface connecting points of equal phase on all waves, which we call a wave front, asintroduced in Chapter 17, is a geometric plane. In comparison, a point source of radia-tion sends waves out radially in all directions. A surface connecting points of equalphase for this situation is a sphere, so this is called a spherical wave.

We can relate E and B to each other with Equations 34.3 and 34.4. In empty space,where q " 0 and I " 0, Equation 34.3 remains unchanged and Equation 34.4 becomes

(34.5)

Using Equations 34.3 and 34.5 and the plane-wave assumption, we obtain the followingdifferential equations relating E and B. (We shall derive these equations formally laterin this section.)

(34.6)

(34.7)

Note that the derivatives here are partial derivatives. For example, when we evaluate+E/+x, we assume that t is constant. Likewise, when we evaluate +B/+t, x is heldconstant. Taking the derivative of Equation 34.6 with respect to x and combining the

+B+x

" )&0 #0 +E+t

+E+x

" )+B+t

! B !d s " #0 &0 d (E

dt

S E C T I O N 3 4 . 2 • Plane Electromagnetic Waves 1069

y

EE

c

Bzc

xB

Figure 34.2 An electromagneticwave traveling at velocity c in thepositive x direction. The electricfield is along the y direction, andthe magnetic field is along the z di-rection. These fields depend onlyon x and t.

! PITFALL PREVENTION 34.1 What Is “a” Wave?A sticky point in these types ofdiscussions is what we mean by asingle wave. We could define onewave as that which is emitted by asingle charged particle. Inpractice, however, the word waverepresents both the emissionfrom a single point (“wave radiatedfrom any position in the yzplane”) and the collection ofwaves from all points on thesource (“plane wave”). Youshould be able to use this term inboth ways and to understand itsmeaning from the context.

result with Equation 34.7, we obtain

(34.8)

In the same manner, taking the derivative of Equation 34.7 with respect to x and com-bining it with Equation 34.6, we obtain

(34.9)

Equations 34.8 and 34.9 both have the form of the general wave equation1 with thewave speed v replaced by c, where

(34.10)

Taking &0 " 4, $ 10)7 T ! m/A and #0 " 8.854 19 $ 10)12 C2/N ! m2 in Equation34.10, we find that c " 2.997 92 $ 108 m/s. Because this speed is precisely the same asthe speed of light in empty space, we are led to believe (correctly) that light is an elec-tromagnetic wave.

The simplest solution to Equations 34.8 and 34.9 is a sinusoidal wave, for which thefield magnitudes E and B vary with x and t according to the expressions

(34.11)

(34.12)

where Emax and Bmax are the maximum values of the fields. The angular wave numberis k " 2,/*, where * is the wavelength. The angular frequency is % " 2,f, where f isthe wave frequency. The ratio %/k equals the speed of an electromagnetic wave, c :

where we have used Equation 16.12, v " c " *f, which relates the speed, frequency,and wavelength of any continuous wave. Thus, for electromagnetic waves, thewavelength and frequency of these waves are related by

(34.13)

Figure 34.3a is a pictorial representation, at one instant, of a sinusoidal, linearlypolarized plane wave moving in the positive x direction. Figure 34.3b shows how theelectric and magnetic field vectors at a fixed location vary with time.

Taking partial derivatives of Equations 34.11 (with respect to x) and 34.12 (withrespect to t), we find that

+B+t

" %B max sin(kx ) %t)

+E+x

" )k E max sin(kx ) %t)

* "cf

"3.00 $ 108 m/s

f

%

k"

2,f2,/*

" *f " c

B " B max cos(kx ) %t)

E " E max cos(kx ) %t)

c "1

√&0#0

+2B+x

2 " &0#0 +2B+t2

+2E+x

2 " &0#0 +2E+t2

+2E+x

2 " )+

+x " +B

+t # " )+

+t " +B

+x # " )+

+t ")&0#0

+E+t #


Speed of electromagneticwaves

1 The general wave equation is of the form (+2y/+x2) " (1/v2)(+2y/+t2) where v is the speed of thewave and y is the wave function. The general wave equation was introduced as Equation 16.27, and itwould be useful for you to review Section 16.6.

Sinusoidal electric andmagnetic fields

Substituting these results into Equation 34.6, we find that at any instant

Using these results together with Equations 34.11 and 34.12, we see that

(34.14)

That is, at every instant the ratio of the magnitude of the electric field to themagnitude of the magnetic field in an electromagnetic wave equals the speed oflight.

Finally, note that electromagnetic waves obey the superposition principle (whichwe discussed in Section 18.1 with respect to mechanical waves) because the differentialequations involving E and B are linear equations. For example, we can add two waveswith the same frequency and polarization simply by adding the magnitudes of the twoelectric fields algebraically.

Let us summarize the properties of electromagnetic waves as we have described them:

• The solutions of Maxwell’s third and fourth equations are wave-like, with both Eand B satisfying a wave equation.

• Electromagnetic waves travel through empty space at the speed of light

• The components of the electric and magnetic fields of plane electromagnetic wavesare perpendicular to each other and perpendicular to the direction of wave propa-gation. We can summarize the latter property by saying that electromagnetic wavesare transverse waves.

c " 1/√#0 &0.

E max

B max"

EB

" c

E max

B max"

%

k" c

k E max " %B max


! PITFALL PREVENTION 34.2 E Stronger Than B?Because the value of c is so large,some students incorrectly inter-pret Equation 34.14 as meaningthat the electric field is muchstronger than the magnetic field.Electric and magnetic fields aremeasured in different units, how-ever, so they cannot be directlycompared. In Section 34.3, wefind that the electric and mag-netic fields contribute equally tothe energy of the wave.

(b)

y

z

y

z

y

zB

E

y

z

y

z

y

z

y

z

y

z

y

z

y

z

y

z

y

z

E

B

c

y

x

z

(a)

Active Figure 34.3 Representation of a sinusoidal, linearly polarized planeelectromagnetic wave moving in the positive x direction with velocity c. (a) The wave atsome instant. Note the sinusoidal variations of E and B with x. (b) A time sequence(starting at the upper left) illustrating the electric and magnetic field vectors at a fixedpoint in the yz plane, as seen by an observer looking in the negative x direction. Thevariations of E and B with t are sinusoidal.

At the Active Figures linkat http://www.pse6.com, youcan observe the wave in part(a) and the variation of thefields in part (b). In addition,you can take a “snapshot” ofthe wave at an instant of timeand investigate the electric andmagnetic fields at that instant.

Properties of electromagneticwaves

• The magnitudes of E and B in empty space are related by the expressionE/B " c.

• Electromagnetic waves obey the principle of superposition.


Quick Quiz 34.1 What is the phase difference between the sinusoidaloscillations of the electric and magnetic fields in Figure 34.3? (a) 180° (b) 90° (c) 0(d) impossible to determine.

Example 34.1 An Electromagnetic Wave

(B) At some point and at some instant, the electric field hasits maximum value of 750 N/C and is along the y axis. Cal-culate the magnitude and direction of the magnetic field atthis position and time.

Solution From Equation 34.14 we see that

Because E and B must be perpendicular to eachother and perpendicular to the direction of wave propa-gation (x in this case), we conclude that B is in the zdirection.

(C) Write expressions for the space–time variation of thecomponents of the electric and magnetic fields for thiswave.

Solution We can apply Equations 34.11 and 34.12 directly:

where

k "2,

*"

2,

7.50 m" 0.838 rad/m

% " 2,f " 2,(4.00 $ 107 s)1) " 2.51 $ 108 rad/s

B " B max cos(kx ) %t) " (2.50 $ 10)6 T) cos(kx ) %t)

E " E max cos(kx ) %t) " (750 N/C) cos(kx ) %t)

2.50 $ 10)6 TB max "E max

c"

750 N/C3.00 $ 108 m/s

"

A sinusoidal electromagnetic wave of frequency 40.0 MHztravels in free space in the x direction, as in Figure 34.4.

(A) Determine the wavelength and period of the wave.

Solution Using Equation 34.13 for light waves and giventhat f " 40.0 MHz " 4.00 $ 107 s)1, we have

The period T of the wave is the inverse of the frequency:

2.50 $ 10)8 s- "1f

"1

4.00 $ 107 s)1 "

7.50 m* "cf

"3.00 $ 108 m/s4.00 $ 107 s)1 "

B c x

y

E = 750j N/C

z

ˆ

Figure 34.4 (Example 34.1) At some instant, a plane electro-magnetic wave moving in the x direction has a maximumelectric field of 750 N/C in the positive y direction. Thecorresponding magnetic field at that point has a magnitude E/cand is in the z direction.

Explore electromagnetic waves of different frequencies at the Interactive Worked Example link at http://www.pse6.com.

Interactive

Derivation of Equations 34.6 and 34.7

To derive Equation 34.6, we start with Faraday’s law, Equation 34.3:

Let us again assume that the electromagnetic wave is traveling in the x direction, withthe electric field E in the positive y direction and the magnetic field B in the positive zdirection.

! E!d s " )d (B

dt

Consider a rectangle of width dx and height ! lying in the xy plane, as shown inFigure 34.5. To apply Equation 34.3, we must first evaluate the line integral of E ! dsaround this rectangle. The contributions from the top and bottom of the rectangle arezero because E is perpendicular to ds for these paths. We can express the electric fieldon the right side of the rectangle as

while the field on the left side2 is simply E(x, t). Therefore, the line integral over thisrectangle is approximately

(34.15)

Because the magnetic field is in the z direction, the magnetic flux through the rectangleof area ! dx is approximately (B " B ! dx. (This assumes that dx is very small comparedwith the wavelength of the wave.) Taking the time derivative of the magnetic flux gives

(34.16)

Substituting Equations 34.15 and 34.16 into Equation 34.3 gives

This expression is Equation 34.6.In a similar manner, we can verify Equation 34.7 by starting with Maxwell’s fourth

equation in empty space (Eq. 34.5). In this case, the line integral of B !ds is evaluatedaround a rectangle lying in the xz plane and having width dx and length !, as in Figure34.6. Noting that the magnitude of the magnetic field changes from B(x, t) to B(x ' dx, t)over the width dx and that the direction in which we take the line integral is as shown inFigure 34.6, the line integral over this rectangle is found to be approximately

(34.17)

The electric flux through the rectangle is (E " E! dx, which, when differentiated withrespect to time, gives

(34.18)

Substituting Equations 34.17 and 34.18 into Equation 34.5 gives

which is Equation 34.7.

+B+x

" )&0#0 +E+t

)! " +B+x # dx " &0#0 ! dx " +E

+t #

+ (E

+t" ! dx

+E+t

! B !d s " [B(x, t)]! ) [B(x ' dx, t)]! $ )! " +B+x # dx

+E+x

" )+B+t

! " +E+x # dx " )! dx

+B+t

d (B

dt" ! dx

dBdt %x constant

" ! dx +B+t

! E!d s " [E(x ' dx, t)]! ) [E(x, t)]! $ ! " +E+x # dx

E(x ' dx, t) $ E(x, t) 'd Edx %

t constantdx " E(x, t) '

+E+x

dx


E + dEE

dx

!

y

xzB

Figure 34.5 At an instant when aplane wave moving in the ' xdirection passes through arectangular path of width dx lying inthe xy plane, the electric field in they direction varies from E to E ' d E.This spatial variation in E gives riseto a time-varying magnetic fieldalong the z direction, according toEquation 34.6.

2 Because dE/dx in this equation is expressed as the change in E with x at a given instant t, dE/dx isequivalent to the partial derivative +E/+x. Likewise, dB/dt means the change in B with time at a particu-lar position x, so in Equation 34.16 we can replace dB/dt with +B/+t.

B

E

B + dB

dx

z

y

x

!

Figure 34.6 At an instant when aplane wave passes through arectangular path of width dx lyingin the xz plane, the magnetic fieldin the z direction varies from B to B ' d B. This spatial variation in Bgives rise to a time-varying electricfield along the y direction,according to Equation 34.7.

34.3 Energy Carried by Electromagnetic Waves

Electromagnetic waves carry energy, and as they propagate through space they cantransfer energy to objects placed in their path. The rate of flow of energy in an electro-magnetic wave is described by a vector S, called the Poynting vector, which is definedby the expression

(34.19)

The magnitude of the Poynting vector represents the rate at which energy flowsthrough a unit surface area perpendicular to the direction of wave propagation. Thus,the magnitude of the Poynting vector represents power per unit area. The direction ofthe vector is along the direction of wave propagation (Fig. 34.7). The SI units of thePoynting vector are J/s ! m2 " W/m2.

As an example, let us evaluate the magnitude of S for a plane electromagnetic wavewhere & E ! B & " EB. In this case,

(34.20)

Because B " E/c, we can also express this as

These equations for S apply at any instant of time and represent the instantaneous rateat which energy is passing through a unit area.

What is of greater interest for a sinusoidal plane electromagnetic wave is the timeaverage of S over one or more cycles, which is called the wave intensity I. (We discussedthe intensity of sound waves in Chapter 17.) When this average is taken, we obtain anexpression involving the time average of cos2(kx ) %t), which equals . Hence, theaverage value of S (in other words, the intensity of the wave) is

(34.21)

Recall that the energy per unit volume, which is the instantaneous energy densityuE associated with an electric field, is given by Equation 26.13,

and that the instantaneous energy density uB associated with a magnetic field is givenby Equation 32.14:

uB "B

2

2&0

uE " 12 #0 E2

I " S av "E max B max

2&0"

E 2max

2&0c"

c2&0

B 2max

12

S "E2

&0c"

c&0

B 2

S "EB&0

S ' 1

&0 E ! B


! PITFALL PREVENTION 34.3 An Instantaneous

ValueThe Poynting vector given byEquation 34.19 is time-dependent. Its magnitude variesin time, reaching a maximumvalue at the same instant as themagnitudes of E and B do. Theaverage rate of energy transfer isgiven by Equation 34.21.

y

E

cBz x

S

Figure 34.7 The Poynting vector S for a plane electromag-netic wave is along the direction of wave propagation.

Poynting vector

! PITFALL PREVENTION 34.4 IrradianceIn this discussion, intensity isdefined in the same way as inChapter 17 (as power per unitarea). In the optics industry, how-ever, power per unit area is calledthe irradiance, and radiant inten-sity is defined as the power inwatts per solid angle (measuredin steradians).

Wave intensity

Because E and B vary with time for an electromagnetic wave, the energy densities also varywith time. When we use the relationships B " E/c and , the expression foruB becomes

Comparing this result with the expression for uE , we see that

That is, the instantaneous energy density associated with the magnetic field of anelectromagnetic wave equals the instantaneous energy density associated with theelectric field. Hence, in a given volume the energy is equally shared by the two fields.

The total instantaneous energy density u is equal to the sum of the energy den-sities associated with the electric and magnetic fields:

When this total instantaneous energy density is averaged over one or more cycles of anelectromagnetic wave, we again obtain a factor of . Hence, for any electromagneticwave, the total average energy per unit volume is

(34.22)

Comparing this result with Equation 34.21 for the average value of S, we see that

(34.23)

In other words, the intensity of an electromagnetic wave equals the averageenergy density multiplied by the speed of light.

I " S av " cu av

u av " #0 (E 2)av " 1

2 #0 E max2 "

B max2

2&0

12

u " uE ' uB " #0 E 2 "

B 2

&0

uB " uE " 12 #0 E

2 "B

2

2&0

uB "(E/c)2

2&0"

#0&0

2&0 E

2 " 12 #0 E

2

c " 1/√#0&0

S E C T I O N 3 4 . 3 • Energy Carried by Electromagnetic Waves 1075

Total instantaneous energydensity of an electromagneticwave

Average energy density of anelectromagnetic wave

Quick Quiz 34.2 An electromagnetic wave propagates in the ) y direction.The electric field at a point in space is momentarily oriented in the ' x direction. Themagnetic field at that point is momentarily oriented in the (a) ) x direction (b) ' ydirection (c) ' z direction (d) ) z direction.

Quick Quiz 34.3 Which of the following is constant for a plane electromag-netic wave? (a) magnitude of the Poynting vector (b) energy density uE (c) energy den-sity uB (d) wave intensity.

Example 34.2 Fields on the Page

the intensity of an electromagnetic wave is also given byEquation 34.21, we have

We must now make some assumptions about numbers toenter in this equation. If we have a 60-W lightbulb, itsoutput at 5% efficiency is approximately 3.0 W by visiblelight. (The remaining energy transfers out of the bulb byconduction and invisible radiation.) A reasonable distancefrom the bulb to the page might be 0.30 m. Thus, we have

I ""av

4,r 2 "

E 2max

2&0c

Estimate the maximum magnitudes of the electric andmagnetic fields of the light that is incident on this pagebecause of the visible light coming from your desk lamp.Treat the bulb as a point source of electromagnetic radia-tion that is 5% efficient at transforming energy comingin by electrical transmission to energy leaving by visiblelight.

Solution Recall from Equation 17.7 that the wave intensityI a distance r from a point source is I " "av/4,r 2, where "avis the average power output of the source and 4,r 2 is thearea of a sphere of radius r centered on the source. Because

34.4 Momentum and Radiation Pressure

Electromagnetic waves transport linear momentum as well as energy. It follows that, asthis momentum is absorbed by some surface, pressure is exerted on the surface. Weshall assume in this discussion that the electromagnetic wave strikes the surface atnormal incidence and transports a total energy U to the surface in a time interval .t.Maxwell showed that, if the surface absorbs all the incident energy U in this time inter-val (as does a black body, introduced in Section 20.7), the total momentum p trans-ported to the surface has a magnitude

(34.24)

The pressure exerted on the surface is defined as force per unit area F/A. Let uscombine this with Newton’s second law:

If we now replace p, the momentum transported to the surface by radiation, fromEquation 34.24, we have

We recognize (dU/dt)/A as the rate at which energy is arriving at the surface per unitarea, which is the magnitude of the Poynting vector. Thus, the radiation pressure Pexerted on the perfectly absorbing surface is

(34.25)

If the surface is a perfect reflector (such as a mirror) and incidence is normal, thenthe momentum transported to the surface in a time interval .t is twice that given byEquation 34.24. That is, the momentum transferred to the surface by the incominglight is p " U/c, and that transferred by the reflected light also is p " U/c. Therefore,

(34.26)

The momentum delivered to a surface having a reflectivity somewhere between thesetwo extremes has a value between U/c and 2U/c, depending on the properties of thesurface. Finally, the radiation pressure exerted on a perfectly reflecting surface fornormal incidence of the wave is3

(34.27)P "2Sc

p "2Uc

(complete reflection)

P "Sc

P "1A

dpdt

"1A

ddt

" Uc # "

1c

(dU/dt)

A

P "FA

"1A

dpdt

p "Uc (complete absorption)


3 For oblique incidence on a perfectly reflecting surface, the momentum transferred is (2U cos /)/cand the pressure is P " (2S cos2 /)/c where / is the angle between the normal to the surface and thedirection of wave propagation.

From Equation 34.14,

This value is two orders of magnitude smaller than theEarth’s magnetic field, which, unlike the magnetic field inthe light wave from your desk lamp, is not oscillating.

1.5 $ 10)7 TB max "E max

c"

45 V/m3.00 $ 108 m/s

"

45 V/m"

" √ (4, $ 10)7 T!m/A)(3.00 $ 108 m/s)(3.0 W)2,(0.30 m)2

E max " √ &0c "av

2,r 2

! PITFALL PREVENTION 34.5 So Many p’sWe have p for momentum and Pfor pressure, and these are bothrelated to " for power! Be sureyou keep these all straight.

Radiation pressure exerted on aperfectly absorbing surface

Momentum transported to aperfectly absorbing surface

Radiation pressure exerted on aperfectly reflecting surface

Although radiation pressures are very small (about 5 $ 10)6 N/m2 for directsunlight), they have been measured with torsion balances such as the one shown inFigure 34.8. A mirror (a perfect reflector) and a black disk (a perfect absorber) areconnected by a horizontal rod suspended from a fine fiber. Normal-incidence lightstriking the black disk is completely absorbed, so all of the momentum of the light istransferred to the disk. Normal-incidence light striking the mirror is totally reflected,and hence the momentum transferred to the mirror is twice as great as that transferredto the disk. The radiation pressure is determined by measuring the angle throughwhich the horizontal connecting rod rotates. The apparatus must be placed in a highvacuum to eliminate the effects of air currents.

NASA is exploring the possibility of solar sailing as a low-cost means of sendingspacecraft to the planets. Large sheets would experience radiation pressure from sunlightand would be used in much the way canvas sheets are used on earthbound sailboats. In1973 NASA engineers took advantage of the momentum of the sunlight striking the solarpanels of Mariner 10 (Fig. 34.9) to make small course corrections when the spacecraft’sfuel supply was running low. (This procedure was carried out when the spacecraft was inthe vicinity of the planet Mercury. Would it have worked as well near Pluto?)

S E C T I O N 3 4 . 4 • Momentum and Radiation Pressure 1077

Light

Blackdisk

Mirror

Figure 34.8 An apparatus for measuringthe pressure exerted by light. In practice,the system is contained in a high vacuum.

Figure 34.9 Mariner 10 used its solar panels to“sail on sunlight.”

Quick Quiz 34.4 To maximize the radiation pressure on the sails of aspacecraft using solar sailing, should the sheets be (a) very black to absorb as muchsunlight as possible or (b) very shiny, to reflect as much sunlight as possible?

Quick Quiz 34.5 In an apparatus such as that in Figure 34.8, the disks areilluminated uniformly over their areas. Suppose the black disk is replaced by one withhalf the radius. Which of the following are different after the disk is replaced? (a) radi-ation pressure on the disk, (b) radiation force on the disk, (c) radiation momentumdelivered to the disk in a given time interval.

Cour

tesy

of N

ASA

Conceptual Example 34.3 Sweeping the Solar System

cube of the radius of a spherical dust particle because it isproportional to the mass and therefore to the volume 4,r 3/3of the particle. The radiation pressure is proportional to thesquare of the radius because it depends on the planar crosssection of the particle. For large particles, the gravitationalforce is greater than the force from radiation pressure. Forparticles having radii less than about 0.2 &m, the radiation-pressure force is greater than the gravitational force, and as aresult these particles are swept out of the Solar System.

A great amount of dust exists in interplanetary space.Although in theory these dust particles can vary in size frommolecular size to much larger, very little of the dust in oursolar system is smaller than about 0.2 &m. Why?

Solution The dust particles are subject to two significantforces—the gravitational force that draws them toward theSun and the radiation-pressure force that pushes them awayfrom the Sun. The gravitational force is proportional to the


Example 34.4 Pressure from a Laser Pointer

Notice that if f " 1, which represents complete reflection,this equation reduces to Equation 34.27. For a beam that is70% reflected, the pressure is

To finalize the example, consider first the magnitude of thePoynting vector. This is about the same as the intensity ofsunlight at the Earth’s surface. (For this reason, it is not safeto shine the beam of a laser pointer into a person’s eyes;that may be more dangerous than looking directly at theSun.) Note also that the pressure has an extremely smallvalue, as expected. (Recall from Section 14.2 that atmos-pheric pressure is approximately 105 N/m2.)

What If? What if the laser pointer is moved twice as faraway from the screen? Does this affect the radiationpressure on the screen?

Answer Because a laser beam is popularly represented asa beam of light with constant cross section, one might betempted to claim that the intensity of radiation, andtherefore the radiation pressure, would be independent ofdistance from the screen. However, a laser beam does nothave a constant cross section at all distances from thesource—there is a small but measurable divergence of thebeam. If the laser is moved farther away from the screen,the area of illumination on the screen will increase,decreasing the intensity. In turn, this will reduce the radia-tion pressure.

In addition, the doubled distance from the screen willresult in more loss of energy from the beam due to scatter-ing from air molecules and dust particles as the light travelsfrom the laser to the screen. This will further reduce the ra-diation pressure.

5.4 $ 10)6 N/m2Pav " (1 ' 0.70) 955 W/m2

3.0 $ 108 m/s"

Many people giving presentations use a laser pointer todirect the attention of the audience to information on ascreen. If a 3.0-mW pointer creates a spot on a screen that is2.0 mm in diameter, determine the radiation pressure on ascreen that reflects 70% of the light that strikes it. Thepower 3.0 mW is a time-averaged value.

Solution In conceptualizing this situation, we do notexpect the pressure to be very large. We categorize this as acalculation of radiation pressure using something likeEquation 34.25 or Equation 34.27, but complicated by the70% reflection. To analyze the problem, we begin bydetermining the magnitude of the beam’s Poynting vector.We divide the time-averaged power delivered via theelectromagnetic wave by the cross-sectional area of thebeam:

Now we can determine the radiation pressure from thelaser beam. Equation 34.27 indicates that a completelyreflected beam would apply an average pressure ofPav " 2S av/c. We can model the actual reflection asfollows. Imagine that the surface absorbs the beam,resulting in pressure Pav " S av/c. Then the surface emitsthe beam, resulting in additional pressure Pav " S av/c. Ifthe surface emits only a fraction f of the beam (so that f isthe amount of the incident beam reflected), then thepressure due to the emitted beam is Pav " f S av/c. Thus,the total pressure on the surface due to absorption and re-emission (reflection) is

Pav "S av

c' f

S av

c" (1 ' f )

S av

c

S av ""av

A"

"av

,r 2 "

3.0 $ 10)3 W

, " 2.0 $ 10)3 m2 #2 " 955 W/m2

At the Interactive Worked Example link at http://www.pse6.com, you can investigate the pressure on the screen for variouslaser and screen parameters.

Example 34.5 Solar Energy

(B) Determine the radiation pressure and the radiationforce exerted on the roof, assuming that the roof covering isa perfect absorber.

Solution Using Equation 34.25 with Sav " 1 000 W/m2, wefind that the radiation pressure is

Because pressure equals force per unit area, this corre-sponds to a radiation force of

5.33 $ 10)4 N"

F " P av A " (3.33 $ 10)6 N/m2)(160 m2)

3.33 $ 10)6 N/m2P av "

S av

c"

1 000 W/m2

3.00 $ 108 m/s"

As noted in the preceding example, the Sun delivers about103 W/m2 of energy to the Earth’s surface via electromag-netic radiation.

(A) Calculate the total power that is incident on a roof ofdimensions 8.00 m $ 20.0 m.

Solution We assume that the average magnitude of thePoynting vector for solar radiation at the surface of theEarth is Sav " 1 000 W/m2; this represents the power perunit area, or the light intensity. Assuming that the radiationis incident normal to the roof, we obtain

1.60 $ 105 W"

"av " S av A " (1 000 W/m2)(8.00 $ 20.0 m2)

Interactive

34.5 Production of Electromagnetic Waves by an Antenna

Neither stationary charges nor steady currents can produce electromagnetic waves.Whenever the current in a wire changes with time, however, the wire emits electromag-netic radiation. The fundamental mechanism responsible for this radiation is theacceleration of a charged particle. Whenever a charged particle accelerates, itmust radiate energy.

Let us consider the production of electromagnetic waves by a half-wave antenna. Inthis arrangement, two conducting rods are connected to a source of alternating volt-age (such as an LC oscillator), as shown in Figure 34.10. The length of each rod isequal to one quarter of the wavelength of the radiation that will be emitted when theoscillator operates at frequency f. The oscillator forces charges to accelerate back andforth between the two rods. Figure 34.10 shows the configuration of the electric andmagnetic fields at some instant when the current is upward. The electric field lines,due to the separation of charges in the upper and lower portions of the antenna,resemble those of an electric dipole. (As a result, this type of antenna is sometimescalled a dipole antenna.) Because these charges are continuously oscillating betweenthe two rods, the antenna can be approximated by an oscillating electric dipole. Themagnetic field lines, due to the current representing the movement of chargesbetween the ends of the antenna, form concentric circles around the antenna and areperpendicular to the electric field lines at all points. The magnetic field is zero at allpoints along the axis of the antenna. Furthermore, E and B are 90° out of phase intime—for example, the current is zero when the charges at the outer ends of the rodsare at a maximum.

At the two points where the magnetic field is shown in Figure 34.10, the Poyntingvector S is directed radially outward. This indicates that energy is flowing away fromthe antenna at this instant. At later times, the fields and the Poynting vector reversedirection as the current alternates. Because E and B are 90° out of phase at pointsnear the dipole, the net energy flow is zero. From this, we might conclude (incor-rectly) that no energy is radiated by the dipole.

However, we find that energy is indeed radiated. Because the dipole fields fall off as1/r 3 (as shown in Example 23.6 for the electric field of a static dipole), they are negli-gible at great distances from the antenna. At these great distances, something elsecauses a type of radiation different from that close to the antenna. The source of thisradiation is the continuous induction of an electric field by the time-varying magneticfield and the induction of a magnetic field by the time-varying electric field, predictedby Equations 34.3 and 34.4. The electric and magnetic fields produced in this mannerare in phase with each other and vary as 1/r. The result is an outward flow of energy atall times.

S E C T I O N 3 4 . 5 • Production of Electromagnetic Waves by an Antenna 1079

photovoltaic cells, reducing the available power in part (A)by an order of magnitude. Other considerations reduce thepower even further. Depending on location, the radiationwill most likely not be incident normal to the roof and, evenif it is (in locations near the Equator), this situation existsfor only a short time near the middle of the day. No energyis available for about half of each day during the nighttimehours. Furthermore, cloudy days reduce the availableenergy. Finally, while energy is arriving at a large rate duringthe middle of the day, some of it must be stored for lateruse, requiring batteries or other storage devices. The resultof these considerations is that complete solar operation ofhomes is not presently cost-effective for most homes.

What If? Suppose the energy striking the roof could becaptured and used to operate electrical devices in the house.Could the home operate completely from this energy?

Answer The power in part (A) is large compared to thepower requirements of a typical home. If this power weremaintained for 24 hours per day and the energy could beabsorbed and made available to electrical devices, it wouldprovide more than enough energy for the average home.However, solar energy is not easily harnessed, and theprospects for large-scale conversion are not as bright asmay appear from this calculation. For example, the effi-ciency of conversion from solar energy is typically 10% for

EES S

Bout Bin

I

I

+

++

++

–

––

––

×

Figure 34.10 A half-wave antennaconsists of two metal rodsconnected to an alternating voltagesource. This diagram shows E andB at an arbitrary instant when thecurrent is upward. Note that theelectric field lines resemble thoseof a dipole (shown in Fig. 23.22).

The angular dependence of the radiation intensity produced by a dipole antennais shown in Figure 34.11. Note that the intensity and the power radiated are a maxi-mum in a plane that is perpendicular to the antenna and passing through itsmidpoint. Furthermore, the power radiated is zero along the antenna’s axis. Amathematical solution to Maxwell’s equations for the dipole antenna shows that theintensity of the radiation varies as (sin2 /)/r 2, where / is measured from the axis ofthe antenna.

Electromagnetic waves can also induce currents in a receiving antenna. The responseof a dipole receiving antenna at a given position is a maximum when the antenna axis isparallel to the electric field at that point and zero when the axis is perpendicular to theelectric field.


Quick Quiz 34.6 If the antenna in Figure 34.10 represents the source ofa distant radio station, rank the following points in terms of the intensity of theradiation, from greatest to least: (a) a distance d to the right of the antenna (b) adistance 2d to the left of the antenna (c) a distance 2d in front of the antenna (out ofthe page) (d) a distance d above the antenna (toward the top of the page).

Quick Quiz 34.7 If the antenna in Figure 34.10 represents the source of adistant radio station, what would be the best orientation for your portable radioantenna located to the right of the figure—(a) up–down along the page, (b) left–rightalong the page, or (c) perpendicular to the page?

34.6 The Spectrum of Electromagnetic Waves

The various types of electromagnetic waves are listed in Figure 34.12, which shows theelectromagnetic spectrum. Note the wide ranges of frequencies and wavelengths. Nosharp dividing point exists between one type of wave and the next. Remember that allforms of the various types of radiation are produced by the same phenome-non—accelerating charges. The names given to the types of waves are simply for con-venience in describing the region of the spectrum in which they lie.

Radio waves, whose wavelengths range from more than 104 m to about 0.1 m, arethe result of charges accelerating through conducting wires. They are generated bysuch electronic devices as LC oscillators and are used in radio and television communi-cation systems.

Microwaves have wavelengths ranging from approximately 0.3 m to 10)4 m andare also generated by electronic devices. Because of their short wavelengths, they arewell suited for radar systems and for studying the atomic and molecular properties ofmatter. Microwave ovens are an interesting domestic application of these waves. It hasbeen suggested that solar energy could be harnessed by beaming microwaves to theEarth from a solar collector in space.

Infrared waves have wavelengths ranging from approximately 10)3 m to thelongest wavelength of visible light, 7 $ 10)7 m. These waves, produced by moleculesand room-temperature objects, are readily absorbed by most materials. The infrared(IR) energy absorbed by a substance appears as internal energy because the energy agi-tates the atoms of the object, increasing their vibrational or translational motion,which results in a temperature increase. Infrared radiation has practical and scientificapplications in many areas, including physical therapy, IR photography, and vibrationalspectroscopy.

Visible light, the most familiar form of electromagnetic waves, is the part of theelectromagnetic spectrum that the human eye can detect. Light is produced by therearrangement of electrons in atoms and molecules. The various wavelengths of visible

y

x

Sθ

Figure 34.11 Angular depen-dence of the intensity of radiationproduced by an oscillating electricdipole. The distance from the ori-gin to a point on the edge of thegray shape is proportional to theintensity of radiation.

! PITFALL PREVENTION 34.6 “Heat Rays”Infrared rays are often called “heatrays.” This is a misnomer. While in-frared radiation is used to raise ormaintain temperature, as in thecase of keeping food warm with“heat lamps” at a fast-food restau-rant, all wavelengths of electro-magnetic radiation carry energythat can cause the temperature ofa system to increase. As an exam-ple, consider a potato baking inyour microwave oven.

light, which correspond to different colors, range from red (* $ 7 $ 10)7 m) to violet(* $ 4 $ 10)7 m). The sensitivity of the human eye is a function of wavelength, beinga maximum at a wavelength of about 5.5 $ 10)7 m. With this in mind, why do yousuppose tennis balls often have a yellow-green color?

Ultraviolet waves cover wavelengths ranging from approximately 4 $ 10)7 m to6 $ 10)10 m. The Sun is an important source of ultraviolet (UV) light, which is themain cause of sunburn. Sunscreen lotions are transparent to visible light but absorbmost UV light. The higher a sunscreen’s solar protection factor (SPF), the greater thepercentage of UV light absorbed. Ultraviolet rays have also been implicated in theformation of cataracts, a clouding of the lens inside the eye.

Most of the UV light from the Sun is absorbed by ozone (O3) molecules in theEarth’s upper atmosphere, in a layer called the stratosphere. This ozone shieldconverts lethal high-energy UV radiation to infrared radiation, which in turn warmsthe stratosphere. Recently, a great deal of controversy has arisen concerning the possi-ble depletion of the protective ozone layer as a result of the chemicals emitted fromaerosol spray cans and used as refrigerants.

X-rays have wavelengths in the range from approximately 10)8 m to 10)12 m. Themost common source of x-rays is the stopping of high-energy electrons upon bombard-ing a metal target. X-rays are used as a diagnostic tool in medicine and as a treatmentfor certain forms of cancer. Because x-rays damage or destroy living tissues and organ-isms, care must be taken to avoid unnecessary exposure or overexposure. X-rays arealso used in the study of crystal structure because x-ray wavelengths are comparable tothe atomic separation distances in solids (about 0.1 nm).

Gamma rays are electromagnetic waves emitted by radioactive nuclei (such as60Co and 137Cs) and during certain nuclear reactions. High-energy gamma rays are a

S E C T I O N 3 4 . 6 • The Spectrum of Electromagnetic Waves 1081

Wavelength

1 pm

1 nm

1 µm

1 cm

1 m

1 km

Long wave

AM

TV, FM

Microwaves

Infrared

Visible light

Ultraviolet

X-rays

Gamma rays

Frequency, Hz

1022

1021

1020

1019

1018

1017

1016

1015

1014

1013

1012

1011

1010

109

108

107

106

105

104

103

µ

Radio waves

1 mm

VioletBlueGreenYellow

Orange

Red

~400 nm

~700 nm

Figure 34.12 The electromagnetic spectrum. Note the overlap between adjacent wavetypes. The expanded view to the right shows details of the visible spectrum.

Wearing sunglasses that do notblock ultraviolet (UV) light is worsefor your eyes than wearing nosunglasses. The lenses of anysunglasses absorb some visiblelight, thus causing the wearer’spupils to dilate. If the glasses donot also block UV light, then moredamage may be done to the lens ofthe eye because of the dilatedpupils. If you wear no sunglasses atall, your pupils are contracted, yousquint, and much less UV lightenters your eyes. High-qualitysunglasses block nearly all the eye-damaging UV light.

Ron

Chap

ple/

Getty

Imag

es

component of cosmic rays that enter the Earth’s atmosphere from space. They havewavelengths ranging from approximately 10)10 m to less than 10)14 m. They arehighly penetrating and produce serious damage when absorbed by living tissues. Con-sequently, those working near such dangerous radiation must be protected with heavilyabsorbing materials, such as thick layers of lead.


Example 34.6 A Half-Wave Antenna

Thus, to operate most efficiently, the antenna should have alength of (3.16 m)/2 " 1.58 m. For practical reasons, carantennas are usually one-quarter wavelength in size.

3.16 m* "3.00 $ 108 m/s9.47 $ 107 Hz

"A half-wave antenna works on the principle that the optimumlength of the antenna is half the wavelength of the radiationbeing received. What is the optimum length of a car antennawhen it receives a signal of frequency 94.7 MHz?

Solution Equation 34.13 tells us that the wavelength of thesignal is

Quick Quiz 34.8 In many kitchens, a microwave oven is used to cookfood. The frequency of the microwaves is on the order of 1010 Hz. The wavelengthsof these microwaves are on the order of (a) kilometers (b) meters (c) centimeters(d) micrometers.

Quick Quiz 34.9 A radio wave of frequency on the order of 105 Hz isused to carry a sound wave with a frequency on the order of 103 Hz. The wavelengthof this radio wave is on the order of (a) kilometers (b) meters (c) centimeters(d) micrometers.

Electromagnetic waves, which are predicted by Maxwell’s equations, have the follow-ing properties:

• The electric field and the magnetic field each satisfy a wave equation. These two waveequations, which can be obtained from Maxwell’s third and fourth equations, are

(34.8)

(34.9)

• The waves travel through a vacuum with the speed of light c, where

(34.10)

• The electric and magnetic fields are perpendicular to each other and perpendicu-lar to the direction of wave propagation. (Hence, electromagnetic waves aretransverse waves.)

• The instantaneous magnitudes of E and B in an electromagnetic wave are relatedby the expression

(34.14)EB

" c

c "1

√&0#0" 3.00 $ 108 m/s

+2B+x

2 " &0#0 +2B+t2

+2E+x

2 " &0#0 +2E+t2

S U M M A RY

Take a practice test forthis chapter by clicking onthe Practice Test link athttp://www.pse6.com.

Questions 1083

• The waves carry energy. The rate of flow of energy crossing a unit area is describedby the Poynting vector S, where

(34.19)

• Electromagnetic waves carry momentum and hence exert pressure on surfaces. Ifan electromagnetic wave whose Poynting vector is S is completely absorbed by asurface upon which it is normally incident, the radiation pressure on that surface is

(34.25)

If the surface totally reflects a normally incident wave, the pressure is doubled.

The electric and magnetic fields of a sinusoidal plane electromagnetic wave propa-gating in the positive x direction can be written

(34.11)

(34.12)

where % is the angular frequency of the wave and k is the angular wave number. Theseequations represent special solutions to the wave equations for E and B. Thewavelength and frequency of electromagnetic waves are related by

(34.13)

The average value of the Poynting vector for a plane electromagnetic wave has amagnitude

(34.21)

The intensity of a sinusoidal plane electromagnetic wave equals the average valueof the Poynting vector taken over one or more cycles.

The electromagnetic spectrum includes waves covering a broad range of wavelengths, from long radio waves at more than 104 m to gamma rays at lessthan 10)14 m.

S av "E max B max

2&0"

E 2max

2&0c"

c2&0

B 2max

* "cf

"3.00 $ 108 m/s

f



P "Sc (complete absorption)

S ' 1

&0 E ! B

Radio stations often advertise “instant news.” If they meanthat you can hear the news the instant they speak it, istheir claim true? About how long would it take for amessage to travel across this country by radio waves,assuming that the waves could be detected at this range?

2. Light from the Sun takes approximately 8.3 min to reachthe Earth. During this time interval the Earth has contin-ued to rotate on its axis. How far is the actual direction ofthe Sun from its image in the sky?

3. When light (or other electromagnetic radiation) travelsacross a given region, what is it that oscillates? What is itthat is transported?

4. Do all current-carrying conductors emit electromagneticwaves? Explain.

5. What is the fundamental source of electromagneticradiation?

1. 6. If a high-frequency current is passed through a solenoidcontaining a metallic core, the core becomes warm due toinduction. Explain why the material rises in temperaturein this situation.

7. Does a wire connected to the terminals of a battery emitelectromagnetic waves? Explain.

If you charge a comb by running it through your hair andthen hold the comb next to a bar magnet, do the electricand magnetic fields produced constitute an electromag-netic wave?

9. List as many similarities and differences between soundwaves and light waves as you can.

10. Describe the physical significance of the Poynting vector.

11. For a given incident energy of an electromagnetic wave, whyis the radiation pressure on a perfectly reflecting surfacetwice as great as that on a perfect absorbing surface?

8.

Q U E S T I O N S


12. Before the advent of cable television and satellite dishes,city dwellers often used “rabbit ears” atop their sets (Fig.Q34.12). Certain orientations of the receiving antenna ona television set give better reception than others. Further-more, the best orientation varies from station to station.Explain.

13. Often when you touch the indoor antenna on a radio ortelevision receiver, the reception instantly improves. Why?

14. Explain how the (dipole) VHF antenna of a television setworks. (See Fig. Q34.12.)

15. Explain how the UHF (loop) antenna of a television setworks. (See Fig. Q34.12.)

16. Explain why the voltage induced in a UHF (loop) antennadepends on the frequency of the signal, while the voltagein a VHF (dipole) antenna does not. (See Fig. Q34.12.)

17. Electrical engineers often speak of the radiation resistanceof an antenna. What do you suppose they mean by thisphrase?

What does a radio wave do to the charges in the receivingantenna to provide a signal for your car radio?

19. An empty plastic or glass dish being removed from amicrowave oven is cool to the touch. How can this bepossible? (Assume that your electric bill has been paid.)

20. Why should an infrared photograph of a person look dif-ferent from a photograph taken with visible light?

Suppose that a creature from another planet had eyes thatwere sensitive to infrared radiation. Describe what the alienwould see if it looked around the room you are now in. Inparticular, what would be bright and what would be dim?

22. A welder must wear protective glasses and clothing toprevent eye damage and sunburn. What does this implyabout the nature of the light produced by the welding?

23. A home microwave oven uses electromagnetic waves with awavelength of about 12.2 cm. Some 2.4-GHz cordless tele-phones suffer noisy interference when a microwave oven isused nearby. Locate the waves used by both devices on theelectromagnetic spectrum. Do you expect them to inter-fere with each other?

21.

18.

Figure Q34.12 Questions 12, 14, 15, and 16, and Problem 49.The V-shaped pair of long rods is the VHF antenna and the loop isthe UHF antenna.

Geor

ge S

empl

e

Section 34.1 Maxwell’s Equations and Hertz’s Discoveries

1. A very long, thin rod carries electric charge with the lineardensity 35.0 nC/m. It lies along the x axis and movesin the x direction at a speed of 15.0 Mm/s. (a) Find theelectric field the rod creates at the point (0, 20.0 cm, 0).(b) Find the magnetic field it creates at the same point.

Note: Assume that the medium is vacuum unless specifiedotherwise.

(c) Find the force exerted on an electron at this point,moving with a velocity of (240 i) Mm/s.

Section 34.2 Plane Electromagnetic Waves

2. (a) The distance to the North Star, Polaris, is approxi-mately 6.44 $ 1018 m. If Polaris were to burn out today,in what year would we see it disappear? (b) How longdoes it take for sunlight to reach the Earth? (c) How longdoes it take for a microwave radar signal to travel fromthe Earth to the Moon and back? (d) How long does ittake for a radio wave to travel once around the Earth in agreat circle, close to the planet’s surface? (e) How long

1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide

= coached solution with hints available at http://www.pse6.com = computer useful in solving problem

= paired numerical and symbolic problems

P R O B L E M S

Problems 1085

does it take for light to reach you from a lightning stroke10.0 km away?

3. The speed of an electromagnetic wave traveling in a trans-parent nonmagnetic substance is , where 0 isthe dielectric constant of the substance. Determine thespeed of light in water, which has a dielectric constant atoptical frequencies of 1.78.

4. An electromagnetic wave in vacuum has an electric fieldamplitude of 220 V/m. Calculate the amplitude of thecorresponding magnetic field.

Figure 34.3 shows a plane electromagnetic sinu-soidal wave propagating in the x direction. Suppose thatthe wavelength is 50.0 m, and the electric field vibratesin the xy plane with an amplitude of 22.0 V/m. Calculate(a) the frequency of the wave and (b) the magnitudeand direction of B when the electric field has its maxi-mum value in the negative y direction. (c) Write anexpression for B with the correct unit vector, withnumerical values for B max, k, and %, and with its magni-tude in the form

6. Write down expressions for the electric and magneticfields of a sinusoidal plane electromagnetic wave having afrequency of 3.00 GHz and traveling in the positive x direc-tion. The amplitude of the electric field is 300 V/m.

In SI units, the electric field in an electromagnetic wave isdescribed by

Find (a) the amplitude of the corresponding magneticfield oscillations, (b) the wavelength *, and (c) the fre-quency f.

8. Verify by substitution that the following equations are solu-tions to Equations 34.8 and 34.9, respectively:

9. Review problem. A standing-wave interference pattern isset up by radio waves between two metal sheets 2.00 mapart. This is the shortest distance between the plates thatwill produce a standing-wave pattern. What is the funda-mental frequency?

10. A microwave oven is powered by an electron tube called amagnetron, which generates electromagnetic waves of fre-quency 2.45 GHz. The microwaves enter the oven and arereflected by the walls. The standing-wave pattern producedin the oven can cook food unevenly, with hot spots in thefood at antinodes and cool spots at nodes, so a turntable isoften used to rotate the food and distribute the energy. If amicrowave oven intended for use with a turntable isinstead used with a cooking dish in a fixed position, theantinodes can appear as burn marks on foods such ascarrot strips or cheese. The separation distance betweenthe burns is measured to be 6 cm 1 5%. From these data,calculate the speed of the microwaves.



Ey " 100 sin(1.00 $ 107x ) %t)

7.


5.

v " 1/√0&0#0

Section 34.3 Energy Carried by Electromagnetic Waves

11. How much electromagnetic energy per cubic meter is con-tained in sunlight, if the intensity of sunlight at the Earth’ssurface under a fairly clear sky is 1 000 W/m2?

12. An AM radio station broadcasts isotropically (equally in alldirections) with an average power of 4.00 kW. A dipolereceiving antenna 65.0 cm long is at a location 4.00 milesfrom the transmitter. Compute the amplitude of the emfthat is induced by this signal between the ends of thereceiving antenna.

What is the average magnitude of the Poynting vector5.00 miles from a radio transmitter broadcastingisotropically with an average power of 250 kW?

14. A monochromatic light source emits 100 W of electromag-netic power uniformly in all directions. (a) Calculate theaverage electric-field energy density 1.00 m from thesource. (b) Calculate the average magnetic-field energydensity at the same distance from the source. (c) Find thewave intensity at this location.

A community plans to build a facility to convert solarradiation to electrical power. They require 1.00 MW ofpower, and the system to be installed has an efficiency of30.0% (that is, 30.0% of the solar energy incident on thesurface is converted to useful energy that can power thecommunity). What must be the effective area of a perfectlyabsorbing surface used in such an installation, assumingsunlight has a constant intensity of 1 000 W/m2?

16. Assuming that the antenna of a 10.0-kW radio station radi-ates spherical electromagnetic waves, compute the maxi-mum value of the magnetic field 5.00 km from theantenna, and compare this value with the surface magneticfield of the Earth.

The filament of an incandescent lamp has a 150-2resistance and carries a direct current of 1.00 A. The fila-ment is 8.00 cm long and 0.900 mm in radius. (a) Calcu-late the Poynting vector at the surface of the filament,associated with the static electric field producing thecurrent and the current’s static magnetic field. (b) Findthe magnitude of the static electric and magnetic fields atthe surface of the filament.

18. One of the weapons being considered for the “Star Wars”antimissile system is a laser that could destroy ballisticmissiles. When a high-power laser is used in the Earth’satmosphere, the electric field can ionize the air, turning itinto a conducting plasma that reflects the laser light. Indry air at 0°C and 1 atm, electric breakdown occurs forfields with amplitudes above about 3.00 MV/m. (a) Whatlaser beam intensity will produce such a field? (b) At thismaximum intensity, what power can be delivered in acylindrical beam of diameter 5.00 mm?

19. In a region of free space the electric field at an instantof time is E " (80.0 i ' 32.0 j ) 64.0 k) N/C and themagnetic field is B " (0.200 i ' 0.080 0 j ' 0.290 k) &T.(a) Show that the two fields are perpendicular to eachother. (b) Determine the Poynting vector for these fields.

20. Let us model the electromagnetic wave in a microwaveoven as a plane traveling wave moving to the left, with anintensity of 25.0 kW/m2. An oven contains two cubical

17.

15.

13.


containers of small mass, each full of water. One has anedge length of 6.00 cm and the other, 12.0 cm. Energyfalls perpendicularly on one face of each container. Thewater in the smaller container absorbs 70.0% of the energythat falls on it. The water in the larger container absorbs91.0%. (That is, the fraction 0.3 of the incoming mi-crowave energy passes through a 6-cm thickness of water,and the fraction (0.3)(0.3) " 0.09 passes through a 12-cmthickness.) Find the temperature change of the water ineach container over a time interval of 480 s. Assume that anegligible amount of energy leaves either container byheat.

21. A lightbulb filament has a resistance of 110 2. The bulb isplugged into a standard 120-V (rms) outlet, and emits1.00% of the electric power delivered to it by electromag-netic radiation of frequency f. Assuming that the bulb iscovered with a filter that absorbs all other frequencies,find the amplitude of the magnetic field 1.00 m from thebulb.

22. A certain microwave oven contains a magnetron that hasan output of 700 W of microwave power for an electricalinput power of 1.40 kW. The microwaves are entirely transferred from the magnetron into the oven chamberthrough a waveguide, which is a metal tube of rectan-gular cross section with width 6.83 cm and height 3.81 cm. (a) What is the efficiency of the magnetron?(b) Assuming that the food is absorbing all the microwaves produced by the magnetron and that noenergy is reflected back into the waveguide, find thedirection and magnitude of the Poynting vector, averaged over time, in the waveguide near the entrance to the oven chamber. (c) What is the maximum electricfield at this point?

23. High-power lasers in factories are used to cut throughcloth and metal (Fig. P34.23). One such laser has a beamdiameter of 1.00 mm and generates an electric field

having an amplitude of 0.700 MV/m at the target. Find(a) the amplitude of the magnetic field produced, (b) theintensity of the laser, and (c) the power delivered by thelaser.

24. A 10.0-mW laser has a beam diameter of 1.60 mm. (a) Whatis the intensity of the light, assuming it is uniform across thecircular beam? (b) What is the average energy density of thebeam?

25. At one location on the Earth, the rms value of themagnetic field caused by solar radiation is 1.80 &T. Fromthis value calculate (a) the rms electric field due to solarradiation, (b) the average energy density of the solar component of electromagnetic radiation at this location,and (c) the average magnitude of the Poynting vector for the Sun’s radiation. (d) Compare the value found inpart (c) to the value of the solar intensity given in Example 34.5.

Section 34.4 Momentum and Radiation Pressure

26. A 100-mW laser beam is reflected back upon itself by amirror. Calculate the force on the mirror.

A radio wave transmits 25.0 W/m2 of power per unitarea. A flat surface of area A is perpendicular to thedirection of propagation of the wave. Calculate theradiation pressure on it, assuming the surface is aperfect absorber.

28. A possible means of space flight is to place a perfectlyreflecting aluminized sheet into orbit around the Earth andthen use the light from the Sun to push this “solar sail.”Suppose a sail of area 6.00 $ 105 m2 and mass 6 000 kg isplaced in orbit facing the Sun. (a) What force is exerted onthe sail? (b) What is the sail’s acceleration? (c) How longdoes it take the sail to reach the Moon, 3.84 $ 108 m away?Ignore all gravitational effects, assume that the accelerationcalculated in part (b) remains constant, and assume a solarintensity of 1 340 W/m2.

A 15.0-mW helium–neon laser (* " 632.8 nm) emits abeam of circular cross section with a diameter of 2.00 mm.(a) Find the maximum electric field in the beam.(b) What total energy is contained in a 1.00-m length ofthe beam? (c) Find the momentum carried by a 1.00-mlength of the beam.

30. Given that the intensity of solar radiation incident onthe upper atmosphere of the Earth is 1 340 W/m2, deter-mine (a) the intensity of solar radiation incident onMars, (b) the total power incident on Mars, and (c) theradiation force that acts on the planet if it absorbs nearlyall of the light. (d) Compare this force to the gravita-tional attraction between Mars and the Sun. (See Table13.2.)

31. A plane electromagnetic wave has an intensity of 750 W/m2. A flat, rectangular surface of dimensions50 cm $ 100 cm is placed perpendicular to the directionof the wave. The surface absorbs half of the energy andreflects half. Calculate (a) the total energy absorbed by thesurface in 1.00 min and (b) the momentum absorbed inthis time.

29.

27.

Figure P34.23 A laser cutting device mounted on a robot arm isbeing used to cut through a metallic plate.

Phili

ppe

Plai

lly/S

PL/P

hoto

Res

earc

hers

Problems 1087

32. A uniform circular disk of mass 24.0 g and radius 40.0 cmhangs vertically from a fixed, frictionless, horizontal hingeat a point on its circumference. A horizontal beam of elec-tromagnetic radiation with intensity 10.0 MW/m2 is inci-dent on the disk in a direction perpendicular to itssurface. The disk is perfectly absorbing, and the resultingradiation pressure makes the disk rotate. Find the anglethrough which the disk rotates as it reaches its new equilib-rium position. (Assume that the radiation is always perpen-dicular to the surface of the disk.)

Section 34.5 Production of Electromagnetic Wavesby an Antenna

33. Figure 34.10 shows a Hertz antenna (also known as a half-wave antenna, because its length is */2). The antenna islocated far enough from the ground that reflections donot significantly affect its radiation pattern. Most AM radiostations, however, use a Marconi antenna, which consists ofthe top half of a Hertz antenna. The lower end of this(quarter-wave) antenna is connected to Earth ground, andthe ground itself serves as the missing lower half. What arethe heights of the Marconi antennas for radio stationsbroadcasting at (a) 560 kHz and (b) 1 600 kHz?

34. Two hand-held radio transceivers with dipole antennas areseparated by a large fixed distance. If the transmittingantenna is vertical, what fraction of the maximum receivedpower will appear in the receiving antenna when it isinclined from the vertical by (a) 15.0°? (b) 45.0°? (c) 90.0°?

Two radio-transmitting antennas are separated by half thebroadcast wavelength and are driven in phase with eachother. In which directions are (a) the strongest and (b) theweakest signals radiated?

36. Review problem. Accelerating charges radiate electromag-netic waves. Calculate the wavelength of radiation pro-duced by a proton moving in a circle of radius Rperpendicular to a magnetic field of magnitude B.

35.

37. A very large flat sheet carries a uniformly distributedelectric current with current per unit width Js . Example30.6 demonstrated that the current creates a magneticfield on both sides of the sheet, parallel to the sheet andperpendicular to the current, with magnitude B " &0 Js . Ifthe current oscillates in time according to

Js " J max(cos %t) j " J max[cos()%t)] j ,

the sheet radiates an electromagnetic wave as shown inFigure P34.37. The magnetic field of the wave is describedby the wave function B " &0 J max[cos(kx ) %t)]k .(a) Find the wave function for the electric field in thewave. (b) Find the Poynting vector as a function of x and t.(c) Find the intensity of the wave. (d) What If? If the sheetis to emit radiation in each direction (normal to the planeof the sheet) with intensity 570 W/m2, what maximumvalue of sinusoidal current density is required?

Section 34.6 The Spectrum of ElectromagneticWaves

38. Classify waves with frequencies of 2 Hz, 2 kHz, 2 MHz,2 GHz, 2 THz, 2 PHz, 2 EHz, 2 ZHz, and 2 YHz onthe electromagnetic spectrum. Classify waves withwavelengths of 2 km, 2 m, 2 mm, 2 &m, 2 nm, 2 pm,2 fm, and 2 am.

39. The human eye is most sensitive to light having a wave-length of 5.50 $ 10)7 m, which is in the green-yellowregion of the visible electromagnetic spectrum. What is thefrequency of this light?

40. Compute an order-of-magnitude estimate for the fre-quency of an electromagnetic wave with wavelength equalto (a) your height; (b) the thickness of this sheet of paper.How is each wave classified on the electromagneticspectrum?

What are the wavelengths of electromagnetic waves in freespace that have frequencies of (a) 5.00 $ 1019 Hz and(b) 4.00 $ 109 Hz?

42. Suppose you are located 180 m from a radio transmitter.(a) How many wavelengths are you from the transmitterif the station calls itself 1 150 AM? (The AM bandfrequencies are in kilohertz.) (b) What If? What if thisstation is 98.1 FM? (The FM band frequencies are inmegahertz.)

43. A radar pulse returns to the receiver after a total traveltime of 4.00 $ 10)4 s. How far away is the object thatreflected the wave?

44. This just in! An important news announcement is transmit-ted by radio waves to people sitting next to their radios100 km from the station, and by sound waves to peoplesitting across the newsroom, 3.00 m from the newscaster.Who receives the news first? Explain. Take the speed ofsound in air to be 343 m/s.

45. The United States Navy has long proposed the construc-tion of extremely low-frequency (ELF) communicationsystems. Such waves could penetrate the oceans to reachdistant submarines. Calculate the length of a quarter-

41.

12

12

z

y

Js

E

B

x

c

Figure P34.37 Representation of the plane electromagnetic waveradiated by an infinite current sheet lying in the yz plane. Thevector B is in the z direction, the vector E is in the y direction, andthe direction of wave motion is along x. Both vectors havemagnitudes proportional to cos(kx ) %t).


wavelength antenna for a transmitter generating ELFwaves of frequency 75.0 Hz. How practical is this?

46. What are the wavelength ranges in (a) the AM radio band(540–1 600 kHz), and (b) the FM radio band (88.0–-108 MHz)?

Additional Problems

47. Assume that the intensity of solar radiation incident onthe cloudtops of the Earth is 1 340 W/m2. (a) Calculatethe total power radiated by the Sun, taking the averageEarth–Sun separation to be 1.496 $ 1011 m. (b) Deter-mine the maximum values of the electric and magneticfields in the sunlight at the Earth’s location.

48. The intensity of solar radiation at the top of the Earth’satmosphere is 1 340 W/m2. Assuming that 60% of theincoming solar energy reaches the Earth’s surface andassuming that you absorb 50% of the incident energy,make an order-of-magnitude estimate of the amount ofsolar energy you absorb in a 60-min sunbath.

Review problem. In the absence of cable input or asatellite dish, a television set can use a dipole-receivingantenna for VHF channels and a loop antenna forUHF channels (Fig. Q34.12). The UHF antennaproduces an emf from the changing magnetic fluxthrough the loop. The TV station broadcasts a signalwith a frequency f, and the signal has an electric-fieldamplitude E max and a magnetic-field amplitude B max atthe location of the receiving antenna. (a) UsingFaraday’s law, derive an expression for the amplitude ofthe emf that appears in a single-turn circular loopantenna with a radius r, which is small compared withthe wavelength of the wave. (b) If the electric field in thesignal points vertically, what orientation of the loop givesthe best reception?

50. Consider a small, spherical particle of radius r locatedin space a distance R from the Sun. (a) Show that theratio Frad/Fgrav is proportional to 1/r, where Frad isthe force exerted by solar radiation and Fgrav is the forceof gravitational attraction. (b) The result of part(a) means that, for a sufficiently small value of r, theforce exerted on the particle by solar radiation exceedsthe force of gravitational attraction. Calculate the valueof r for which the particle is in equilibrium underthe two forces. (Assume that the particle has a perfectlyabsorbing surface and a mass density of 1.50 g/cm3. Letthe particle be located 3.75 $ 1011 m from the Sun, anduse 214 W/m2 as the value of the solar intensity at thatpoint.)

A dish antenna having a diameter of 20.0 m receives(at normal incidence) a radio signal from a distantsource, as shown in Figure P34.51. The radio signalis a continuous sinusoidal wave with amplitudeE max " 0.200 &V/m. Assume the antenna absorbs allthe radiation that falls on the dish. (a) What is theamplitude of the magnetic field in this wave? (b) Whatis the intensity of the radiation received by thisantenna? (c) What is the power received by the antenna?(d) What force is exerted by the radio waves on theantenna?

51.

49.

Figure P34.51

Figure P34.54

52. One goal of the Russian space program is to illuminatedark northern cities with sunlight reflected to Earthfrom a 200-m diameter mirrored surface in orbit. Severalsmaller prototypes have already been constructed andput into orbit. (a) Assume that sunlight with intensity1 340 W/m2 falls on the mirror nearly perpendicularlyand that the atmosphere of the Earth allows 74.6% ofthe energy of sunlight to pass through it in clearweather. What is the power received by a city when thespace mirror is reflecting light to it? (b) The plan is forthe reflected sunlight to cover a circle of diameter 8.00 km. What is the intensity of light (the average mag-nitude of the Poynting vector) received by the city? (c) This intensity is what percentage of the vertical com-ponent of sunlight at Saint Petersburg in January, whenthe sun reaches an angle of 7.00° above the horizon atnoon?

In 1965, Arno Penzias and Robert Wilson discovered thecosmic microwave radiation left over from the Big Bangexpansion of the Universe. Suppose the energy density ofthis background radiation is 4.00 $ 10)14 J/m3. Deter-mine the corresponding electric field amplitude.

54. A hand-held cellular telephone operates in the 860- to900-MHz band and has a power output of 0.600 Wfrom an antenna 10.0 cm long (Fig. P34.54). (a) Findthe average magnitude of the Poynting vector 4.00 cmfrom the antenna, at the location of a typical person’s

53.

Amos

Mor

gan/

Getty

Imag

es

Problems 1089

head. Assume that the antenna emits energy withcylindrical wave fronts. (The actual radiation fromantennas follows a more complicated pattern.) (b) TheANSI/IEEE C95.1-1991 maximum exposure standard is0.57 mW/cm2 for persons living near cellular telephonebase stations, who would be continuously exposed to theradiation. Compare the answer to part (a) with thisstandard.

A linearly polarized microwave of wavelength 1.50 cm isdirected along the positive x axis. The electric fieldvector has a maximum value of 175 V/m and vibratesin the xy plane. (a) Assume that the magnetic fieldcomponent of the wave can be written in the formB " B max sin(kx ) %t) and give values for B max, k, and %.Also, determine in which plane the magnetic field vectorvibrates. (b) Calculate the average value of the Poyntingvector for this wave. (c) What radiation pressurewould this wave exert if it were directed at normalincidence onto a perfectly reflecting sheet? (d) Whatacceleration would be imparted to a 500-g sheet(perfectly reflecting and at normal incidence) withdimensions of 1.00 m $ 0.750 m?

56. The Earth reflects approximately 38.0% of the incidentsunlight from its clouds and surface. (a) Given that theintensity of solar radiation is 1 340 W/m2, what is the radi-ation pressure on the Earth, in pascals, at the locationwhere the Sun is straight overhead? (b) Compare this tonormal atmospheric pressure at the Earth’s surface, whichis 101 kPa.

An astronaut, stranded in space 10.0 m from hisspacecraft and at rest relative to it, has a mass (includingequipment) of 110 kg. Because he has a 100-W lightsource that forms a directed beam, he considers usingthe beam as a photon rocket to propel himself continu-ously toward the spacecraft. (a) Calculate how long ittakes him to reach the spacecraft by this method.(b) What If? Suppose, instead, that he decides to throwthe light source away in a direction opposite thespacecraft. If the mass of the light source is 3.00 kg and,after being thrown, it moves at 12.0 m/s relative to therecoiling astronaut, how long does it take for theastronaut to reach the spacecraft?

58. Review problem. A 1.00-m-diameter mirror focuses theSun’s rays onto an absorbing plate 2.00 cm in radius,which holds a can containing 1.00 L of water at 20.0°C.(a) If the solar intensity is 1.00 kW/m2, what is the inten-sity on the absorbing plate? (b) What are the maximummagnitudes of the fields E and B? (c) If 40.0% of theenergy is absorbed, how long does it take to bring thewater to its boiling point?

59. Lasers have been used to suspend spherical glass beads inthe Earth’s gravitational field. (a) A black bead has a massof 1.00 &g and a density of 0.200 g/cm3. Determine theradiation intensity needed to support the bead. (b) If thebeam has a radius of 0.200 cm, what is the power requiredfor this laser?

60. Lasers have been used to suspend spherical glass beads inthe Earth’s gravitational field. (a) A black bead has a mass

57.

55.

m and a density 3. Determine the radiation intensityneeded to support the bead. (b) If the beam has a radius r,what is the power required for this laser?

61. A microwave source produces pulses of 20.0-GHz radia-tion, with each pulse lasting 1.00 ns. A parabolic reflec-tor with a face area of radius 6.00 cm is used to focusthe microwaves into a parallel beam of radiation, asshown in Figure P34.61. The average power during eachpulse is 25.0 kW. (a) What is the wavelength of thesemicrowaves? (b) What is the total energy containedin each pulse? (c) Compute the average energy densityinside each pulse. (d) Determine the amplitude ofthe electric and magnetic fields in these microwaves.(e) Assuming this pulsed beam strikes an absorbingsurface, compute the force exerted on the surfaceduring the 1.00-ns duration of each pulse.

12.0 cm

Figure P34.61

62. The electromagnetic power radiated by a nonrelativisticmoving point charge q having an acceleration a is

where #0 is the permittivity of free space and c is the speedof light in vacuum. (a) Show that the right side of thisequation has units of watts. (b) An electron is placed in aconstant electric field of magnitude 100 N/C. Determinethe acceleration of the electron and the electromagneticpower radiated by this electron. (c) What If? If a proton isplaced in a cyclotron with a radius of 0.500 m and a mag-netic field of magnitude 0.350 T, what electromagneticpower does this proton radiate?

63. A thin tungsten filament of length 1.00 m radiates60.0 W of power in the form of electromagnetic waves. Aperfectly absorbing surface in the form of a hollow cylin-der of radius 5.00 cm and length 1.00 m is placed con-centrically with the filament. Calculate the radiationpressure acting on the cylinder. (Assume that the radia-tion is emitted in the radial direction, and ignore endeffects.)

64. The torsion balance shown in Figure 34.8 is used inan experiment to measure radiation pressure. The sus-pension fiber exerts an elastic restoring torque. Itstorque constant is 1.00 $ 10)11 N ! m/degree, and thelength of the horizontal rod is 6.00 cm. The beam froma 3.00-mW helium–neon laser is incident on the blackdisk, and the mirror disk is completely shielded.

" "q

2a 2

6,#0c3


Calculate the angle between the equilibrium positions ofthe horizontal bar when the beam is switched from “off”to “on.”

65. A “laser cannon” of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beamis E. The beam is aimed at an asteroid that is initiallymoving in the direction of the spacecraft. What is theacceleration of the asteroid relative to the spacecraft if thelaser beam strikes the asteroid perpendicular to its surface,and the surface is nonreflecting? The mass of the asteroidis m. Ignore the acceleration of the spacecraft.

66. A plane electromagnetic wave varies sinusoidally at90.0 MHz as it travels along the ' x direction. The peakvalue of the electric field is 2.00 mV/m, and it is directedalong the 1 y direction. (a) Find the wavelength, theperiod, and the maximum value of the magnetic field.(b) Write expressions in SI units for the space and timevariations of the electric field and of the magnetic field.Include numerical values and include subscripts to indi-cate coordinate directions. (c) Find the average powerper unit area that this wave carries through space.(d) Find the average energy density in the radiation (injoules per cubic meter). (e) What radiation pressurewould this wave exert upon a perfectly reflecting surfaceat normal incidence?

67. Eliza is a black cat with four black kittens: Penelope,Rosalita, Sasha, and Timothy. Eliza’s mass is 5.50 kg, andeach kitten has mass 0.800 kg. One cool night all fivesleep snuggled together on a mat, with their bodiesforming one hemisphere. (a) Assuming that the purringheap has uniform density 990 kg/m3, find the radius ofthe hemisphere. (b) Find the area of its curved surface.(c) Assume the surface temperature is uniformly 31.0°Cand the emissivity is 0.970. Find the intensity of radiationemitted by the cats at their curved surface, and (d) theradiated power from this surface. (e) You may think ofthe emitted electromagnetic wave as having a single pre-dominant frequency (of 31.2 THz). Find the amplitudeof the electric field just outside the surface of the cozypile, and (f) the amplitude of the magnetic field. (g) Arethe sleeping cats charged? Are they current-carrying?Are they magnetic? Are they a radiation source? Do theyglow in the dark? Give an explanation for your answersso that they do not seem contradictory. (h) What If? Thenext night the kittens all sleep alone, curling up intoseparate hemispheres like their mother. Find the totalradiated power of the family. (For simplicity, we ignorethroughout the cats’ absorption of radiation from theenvironment.)

68. Review problem. (a) An elderly couple has a solarwater heater installed on the roof of their house (Fig.P34.68). The heater consists of a flat closed box with

Note: Section 20.7 introduced electromagnetic radiationas a mode of energy transfer. The following threeproblems use ideas introduced both there and in thecurrent chapter.

extraordinarily good thermal insulation. Its interior ispainted black, and its front face is made of insulatingglass. Assume that its emissivity for visible light is 0.900and its emissivity for infrared light is 0.700. Assume thatlight from the noon Sun is incident perpendicular to theglass with an intensity of 1 000 W/m2, and that nowater enters or leaves the box. Find the steady-statetemperature of the interior of the box. (b) What If?The couple builds an identical box with no watertubes. It lies flat on the ground in front of the house.They use it as a cold frame, where they plant seeds inearly spring. Assuming the same noon Sun is at an eleva-tion angle of 50.0°, find the steady-state temperature ofthe interior of this box when its ventilation slots aretightly closed.

69. Review problem. The study of Creation suggests a Creatorwith an inordinate fondness for beetles and for small redstars. A small red star radiates electromagnetic waves withpower 6.00 $ 1023 W, which is only 0.159% of the lumi-nosity of the Sun. Consider a spherical planet in a circularorbit around this star. Assume the emissivity of the planetis equal for infrared and for visible light. Assume theplanet has a uniform surface temperature. Identify theprojected area over which the planet absorbs starlight andthe radiating area of the planet. If beetles thrive at a tem-perature of 310 K, what should be the radius of theplanet’s orbit?

Answers to Quick Quizzes

34.1 (c). Figure 34.3b shows that the B and E vectorsreach their maximum and minimum values at the sametime.

34.2 (c). The B field must be in the ' z direction in orderthat the Poynting vector be directed along the ) y direction.

34.3 (d). The first three choices are instantaneous values andvary in time. The wave intensity is an average over a fullcycle.

34.4 (b). To maximize the pressure on the sails, they should beperfectly reflective, so that the pressure is given by Equa-tion 34.27.

34.5(b), (c). The radiation pressure (a) does not changebecause pressure is force per unit area. In (b), thesmaller disk absorbs less radiation, resulting in a smaller

Figure P34.68

© B

ill B

anas

zew

ski/V

isual

s Unl

imite

d

Answers to Quick Quizzes 1091

force. For the same reason, the momentum in (c) isreduced.

34.6 (a), (b) " (c), (d). The closest point along the x axis inFigure 34.11 (choice a) will represent the highest inten-sity. Choices (b) and (c) correspond to points equidistantin different directions. Choice (d) is along the axis of theantenna and the intensity is zero.

34.7 (a). The best orientation is parallel to the transmittingantenna because that is the orientation of the electric

field. The electric field moves electrons in the receivingantenna, thus inducing a current that is detected andamplified.

34.8 (c). Either Equation 34.13 or Figure 34.12 can be used tofind the order of magnitude of the wavelengths.

34.9 (a). Either Equation 34.13 or Figure 34.12 can be used tofind the order of magnitude of the wavelength.

chapter 34 electromagnetic waves

Documents