701

CHAPTER 33 ELECTROMAGNETIC WAVES

Select odd-numbered solutions, marked with a dagger (), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.

33-1. (a) The potential difference is ,QVC

= so the electric field is .V QEd Cd

= = The capacitance is

0 ,ACd

= so

0

.QEA

=

(b)0

E EQEA

= =

(c) 0 Edisplacementd dQI

dt dt

= =

(d) dQdt

is the current charging the capacitor, so .displacementI I=

33-2. The current intercepted by the surface is I, thecurrent in the wire. The opening in the surface isA/4, where A the total surface area of one of thecapacitor plates. If we assume that thedisplacement current is evenly distributed acrossA, then the displacement current intercepted is3I/4 because the total displacement current mustbe equal to I. To find the total currentintercepted by the surface, we must recognizethat the current fluxes have opposite senses forthe wire and the platecurrent is entering thesurface where it intersects the wire anddisplacement current is exiting the surfacebetween the plates. Thus the net currentintercepted by the surface is I/4, which is justthe total displacement current contained withinthe loop of radius R/2 formed by the hole in thesurface between the plates.

33-3. (a) The displacement current between the plates is equal to the conduction current in the wires.So 4.0 A.displacementI =

(b) Idisplacement = 0 ;Eddt therefore,

0

displacementE Iddt

= =

1112 2 2

4.0 A 4.5 10 V m/s8.85 10 C /N m

= ii

CHAPTER 33

702

33-4. (a) dVdt

= 2 103 V/s. But E = V/d, where d is the plate separation of 0.2 cm. Then dEdt

=

61 1.0 10 V/m sdVd dt

= i

(b) 0 EdisplacementdI

dt

=

2 6 2 5 20 10 2.78 10E Edisplacement

d ddEr r I rdt dt dt

= = = =At r = 0.15 m, Idisplacement = 6.3 107 A;At r = 0.30 m, Idisplacement = 2.5 106 A.

(c) B = 0 02

r dEdt

= 132 8.33 10 T2r dEc dt

= at r = 0.15 m, and

121.67 10 TB = at r = 0.30 m

33-5. (a) Displacement current = current into capacitor = 0.10 A.(b) Uniform field between plates. Therefore, flux through a specified area is proportional to the

area. Area radius 5 cm as described takes up 520

2

2 = 1

16 of plate of capacitor. Since Id = 0 ;d

dt

Id Area; therefore, displacement current through that area = 30.10 A 6.3 10 A16

=

33-6. 0 0 .EdB ds

dt = &v For r < R,

( )

2 2

2

20 0

7 12 2 2 4 2 130 0

2

(2 ) 2

(4 10 T m/A)(8.85 10 C /N m )(5 10 V/m s ) (5.56 10 T/m s)

E

E

EA r Ctd r Ct

dtB r r Ct

B rCt rt rt

= =

=

=

= = = i i i i

At t = 0.50 s, r = 1.0 cm, B = 2.8 1015 TFor r > R,

( )

2 2

2

20 0

2

(2 ) 2

E

E

EA R Ctd R Ct

dtB r R Ct

= =

=

=

2 7 12 2 2 4 2 20 0

15

(4 10 T m/A)(8.85 10 C /N m )(5 10 V/m s )(0.05 m)

(1.39 10 T m/s)

R Ct tBr r

tr

= =

=

i i i

i

At t = 2.0 s, r = 6.0 cm, 14B 4.6 10 T=

CHAPTER 33

703

33-7. The displacement current between the capacitor plates must equal the current I coming in, i.e.,I = Id. The flux through the dielectric is given by

= free0

1(since );Q E E

=

therefore, ddt =

0

1 dQdt

= 0

;I

therefore, 0 .ddI Idt

= = Substituting this into Eq. 33.3 gives

0 0 0 0= ( + )ddd I I I dt

= +B si33-8. I = Id = 0 Ed

dt

= dQdt

= C ddt = Cmax cos t

= (2 1012 F)(0.5 V)(4 103 rad/s)cos (t) 94 10 cos( ) t=

33-9. (a) Id = 0 .ddt But = EA, where E is electric field through the space in the capacitor and A =

area of capacitor; therefore, Id = 0A .dEdt

But E = Vd

where V = voltage, d = distance across

capacitor; therefore, Id = 0 .A dVd dt

But C = 0 ;Ad

therefore, Id = C dVdt

= 2.0 106 F 1.0 103

V/s = 2.0 103 A.

(b) Real current I = .VR

We want V = IR = (2.0 103 A 5.0 105 )

= 1000 V. Since voltage rises at 1000 V/s, this occurs in 1.0 s.

(c) From (a), Id = 0A dV

d dt =

20 ( , ,r dV

d dt r = 0.20 m, dV

dt = 1 103 V/s

and I = 49

VR

= 4 ( / )9

V t tR

= 3(4)10

(9)R t.

R = 5 105 . Then B = 23

30 01 (4)10 (10 )2 (9) rt

r R d

+

B = 3

0(4)10(9)0.4

tR + 30 0 (0.2 10 ).

2

d We determine d through

C = 0 ,Ad

d = 0 AC

so that B = 3

0(4)10(9)0.4

tR + 0200 ,

2 CA

A = (.3)2.

at 0,t = B = 100200 8.9 10 T2

CA

=

at t = 1 s, B = 8.9 1010 + 8.9 1010 T 91.78 10 T=

at t = 2.0 s, B = (8.9 1010)2 + 8.9 1010 92.7 10 T=

CHAPTER 33

704

33-10. By Example 1, B = 0 02

rE0 cos t (since r < radius of capacitor).

Here, E0 = 35000 V

1.5 10 mVd

= = 3.33 106 V/m and = 2f

= 2 60 rad/s = 377 rad/s, and r = 20 cm = 0.20 m. Therefore,

B = 6 12 2

2

1.26 10 8.85 10 s2 m

0.20 m 377 1s

3.33 106 Vm

cos t

= 1.4 109 T cos t. Amplitude 91.4 10 T=

33-11. (a) 2 2 2 .I E I CtJ ER R R

= = = =(b) For r R, the Ampere-Maxwell law gives

0 0(2 ) EdB r I

dt = +

The current within radius r is 2 2

22 2( ) .

Ir CtrI J rR R

= = =

2 2

2 2

2

2

2 2

0 02 2

002

( )

(2 )

( )2

E

E

Ct r CtrEAR R

d Crdt R

r C CrB rR R

CrB tR

= = =

=

= +

= +

At t = 0, 0 0 22CrB

R

=

(c) For r R, the current within r is I = Ct, and the Ampere-Maxwell Law gives

( )

2

2

0 0

00

( )

(2 )

( )2

E

E

Ct REA CtR

d Cdt

B r Ct CCB tr

= = =

=

= +

= +

At t = 0, 0 02

CBr

=

As a check, note that the answers to (b) and (c) are the same at r = R.

33-12. The electric flux through one turn is .EEdJ I Ct CEA A A

A dt = = = = = An

Amperian loop around the coils encloses N turns, so the Ampere-Maxwell Law gives0 0 0

0 0 0 .NCCBl N I NC t B tl

= + = + = + At t = 0,

0 0 .NCBl

=

CHAPTER 33

705

33-13. For a medium with dielectric constant , Gauss Law becomes 0

= Qd E Aiv and

Maxwell-Amperes Law becomes

0 0B ( )Edd I

dt = + siv so that Maxwells equations become:

0

= 0Qd d

= E A B Ai iv v0 0= ( )E

ddd d Idt dt

= + E s B si iv v33-14. md q= Aiv (1), similar to Gauss Law for

electricity. Applying this to a point charge byconstructing a surface with radius r around the

charge gives 2 ,4mqBr

= as required. As with

electric charge, if we sent a flow of magneticcharge into a capacitor, there must be a magnetic

displacement current that is equal to the magnetic conduction current .mdqdt

The magnetic

displacement current will be proportional to the rate of change of the magnetic flux, by analogy

with electric displacement current. Then magnetic current Im must equal .Bd

dt This is verified

by (1): .m BB m mdq dd q Idt dt

= = = = Ai The version of Amperes Law containing

displacement current says 0 ( ),displacementd I I= + B siv so Faradays Law must be modified toinclude magnetic current: Therefore, B m

dd Idt

= + E siv (2). The other two laws remainthe same.

33-15. As implied in the text, an accelerated positive charge produces aradiation (transverse) electric field that points in the opposite directionfrom the component of the acceleration parallel to the radiation field.For points on the line perpendicular to the acceleration, the direction ofthe radiation field is exactly opposite the direction of the acceleration.For an accelerated electron, the radiation electric field on the lineperpendicular to the acceleration will point in the same direction as theacceleration because of the electrons negative charge. Suppose theelectron is initially moving to the right. When it stops, its accelerationwill point to the left, so the transverse electric will also point left. Inthe diagram, the pulse is propagating up in the top half of the figureand down in the bottom half. The electric component of the radiationfield along a line perpendicular to the acceleration will point lefteverywhere. To satisfy the right-hand rule relation between E, B, andthe propagation direction, the magnetic field must point out of the pagein the top half of the figure and into the page in the bottom half.

a

E B

E B

mdqdt

Bddt

CHAPTER 33

706

33-16. The diagram is a mirror image of Figure33.5. In the diagram for this problem, thecharge is initially stationary at point 1,which is at time t = 0. It undergoes aleftward acceleration a until time (indicated by the arrow pointing from 1to 2), which is at point 2, and then travelsto the left at constant speed v = a. Attime t it has reached point 3, which is a

distance 2

( )2

a v t + from point 1.

The large circle has radius ct and iscentered on point 1. The smaller circlehas radius c(t ) and is centered onpoint 2 to show how far the field hadreached after the disturbance ended. Thefield lines at the larger circle are directedfrom point 1. Lines are drawn from point3 to show the direction of the field at thatlocation. By extending those linesto the smaller circle centered on point 2, we can determine the change in the field caused by theacceleration. The kinks connecting the field lines from the smaller circle to the ones at thelarger circle represent the average change in the field caused by the acceleration. Note that thecomponent of the kink connecting the two parts of the field has a transverse component thatpoints opposite the component of the acceleration that is perpendicular to the Coulomb fielddirection (parallel to the radiation field direction).

33-17. To set this problem up, we use the nontechnicalinterpretation of the word acceleration to meanan increase in velocity because decelerationis a colloquial expression that means a decreasein velocity (referring to an acceleration thatpoints in the opposite direction from the initialvelocity). Assume there is a positive chargeinitially at point 1 in the diagram. A circle ofradius 7 shows how far the field from the chargehas reached from that position at time t. At t = 0,the charge moves to the right with constantacceleration a to the right for time to point 2. Itthen moves with constant acceleration a (thedeceleration) for time , finally coming to restagain at point 3. The largest circle has radius ctand is centered on point 1. The next smaller circlehas radius c(t ) and is centered on point 2. Thesmallest circle has radius c(t 2) and is centeredon point 3.

+++

3 2 1

CHAPTER 33

707

33-18. The diagram will look like Figure 33.8 except that if the electric field vectors point up as in 33.8,the magnetic field vectors will point in the z direction so that the direction of E B will point inthe x direction.

33-19. To satisfy the requirement that E B points up whenE points north, B must point west.

33-20. If the charge on the sheet is positive, there will be a Coulomb field pointing away from the sheeton each side. So on the x side, the Coulomb field will point in the x direction. The accelerationin the +y direction will cause a radiation field in the y direction. To satisfy the requirement thatE B points in the x direction, B must point in the +z direction.

33-21. Consider the small rectangle of length l shown. The gray arearepresents the region containing the electric field, which is pointing outof the paper. If the pulse has crossed a length x of the rectangle, theflux is E Elx = , and the rate of change of the flux is

( ) ,Ed d dxElx El Elcdt dt dt

= = = since the speed of the front edge of

the field is c. Then the displacement current/length along the loop is12 2 2 3 8

0

5

(8.85 10 C /N m )(4.0 10 V/m)(3.00 10 m/s)

1.1 10 A/m.

dI Ecl

= = =i

33-22. 1 ly = (3.00 108 m/s)(365 days)(24 h/day)(3600 s/h) = 9.46 1015 m

33-23. 1180.01 m 3.3 10 s

3.00 10 m/sxx c t t

c

= = = =

33-24.8

8

2(3.8 10 m) 2.5 s3.00 10 m/s

dtc

= = =

33-25.11

38

1.50 10 m 5.00 10 s (8.33 min)3.00 10 m/s

dtc

= = =

33-26.6

38

2.0 10 m 6.7 10 s3.00 10 m/s

dtc

= = =

33-27. 981.0 m 3.3 10 s

3.00 10 m/sxt

c

= = =

81 3.0 10 Hz (0.30 GHz)ft

= =

These limitations are based on fundamental properties of nature and cannot be overcome.

33-28. (a) vertical.(b) To maximize the flux (and thus the change in flux), the vector that defines the plane of thecircle must be parallel to the magnetic field. This means that the plane of the circle must includethe direction of propagation of the wave.

E l

c

E

B

CHAPTER 33

708

33-29. By right-hand rule,B points north.

90Magnitude 2.0 10E Tc

= = (direction ofpropagation)

33-30. If the original beam is unpolarized, the first polarizer transmits half of the incident intensity. Thesecond polarizer then transmits a fraction cos2 of that intensity. The final intensity is

20

2

0

1

cos2

cos 0.302

cos 0.60 39

final

final

II

II

= = =

= =

33-31. The incident polarization is at 45 to the preferential direction of the first polarizer, so theintensity after the first polarizer is

2 01 0 cos 45 2

II I= =

This light is now polarized at 45 to the preferred direction of the second polarizer, so2 01

2 1 cos 45 2 4III I= = =

33-32. The first Polaroid transmits half of the incident intensity. The second transmits a fraction cos2 of that, so

2

0

cos2

transmittedII

=

2

0

0

0

cos 3030 : 0.382

45 : 0.25

60 : 0.13

transmitted

transmitted

transmitted

II

II

II

= =

=

=

33-33. If the original beam is unpolarized, the first polarizer transmits half of the incident intensity. Thesecond polarizer then transmits a fraction cos2 of that intensity. The final intensity is

2 20 00cos cos 80 0.0152 2final

I II I = = = 1.5% of the incident light is transmitted.

33-34. If the original beam is unpolarized, the first polarizer transmits vertically polarized light with halfthe incident intensity. The second polarizer then transmits a fraction cos2 60 of that intensity:

2 01 12 cos 602 4 8

II II = = = The third polarizer is oriented 20 from the vertical, which is 40 from the polarization directionafter the second polarizer. The final intensity is

2 202 0cos 40 cos 40 0.0738final

II I I = = =

CHAPTER 33

709

33-35. The intensity transmitted by the first polarizer is 2 01 0 cos 45 .2II I= = The intensity transmitted

by the second polarizer is2 20

1

2

0

1

cos cos2

cos 0.302

cos 0.60 39

final

final

II I

II

= = = =

= =

33-36. (a) See Problem 31 for the solution to this part.(b) After the first polarizer, 21 0 cos 30 .I I= After the second polarizer,

( )22 22 1 0cos 30 cos 30 .I I I= = After the third polarizer,( )32 23 2 0 0cos 30 cos 30 0.42I I I I= = =

(c) After the first polarizer, 21 090cos .I IN

= After the second polarizer,

22 2

2 1 090 90cos cos .I I IN N

= = After N polarizers,

20

90cos .N

NI I N

= If N = 90,

( )90290 0 cos 1 0.97.I I= =As N , 2 290cos cos 0 1

N

= , so 0lim .NN I I =

33-37. (a) 2 20 sin cosx xE E t t

c c

= + 0 01E E= =

(b)At x = 0, ( ) ( )0( ) sin cost E t t = + E j k , so0

0

0

0

(0)

2

32

E

E

E

E

=

=

=

=

E k

E j

E k

E j

So the electric field rotates counterclockwise with angular frequency .

33-38. 0 0 cos cos .2E E = The intensity will be a maximum when the magnitude of 0E is amaximum.

00 cos cos 02

sin cos cos sin 02 2

dE dEd d

= = + =

t = 0

t = /(2)

t = /

t = 3/(2)

y

z

CHAPTER 33

710

2 2

2

cos sin ; sin cos2 2sin cos 0

2cos 11cos2

45

= =

+ = = = =

33-39. We will make the simplifyingassumption that the secondpolarizer acts as an idealpolarizing filter, i.e., thetransmitted intensity isdetermined by the projectedangle that the incoming wavesees when approaching thesecond Polaroid. The effect ofoblique incidence on thetransmitted intensity will beneglected. If I0 is the intensityof light incident on the secondpolarizer, then the transmittedintensity isI = I0 cos2

where 2 2 2

cos coscos sin cos

=

+

as can be seen from part (a). Parts (b) and (c) are views along the y-, and the z-axis respectively

shown for clarity. Thus 02 2 2cos

cos sin cosII

=

+(Note: Angle has been shown opposite to that shown in the book, for clarity.)

33-40.8

6

3.00 10 m/s 1.85 m162.6 10 Hz

cf

= = =

33-41.8

113

3.00 10 m/s 3.0 10 Hz10 m

cf = = =

CHAPTER 33

711

33-42.8 8

6

3.00 10 m/s 3.00 10 m/s 120 m2.5 10 Hz

cf f

= = = =

Use a spreadsheet to make a table of all the wavelengths.

f (MHz) (m)2.5 1205.0 6010 3015 2020 15

33-43. By cf

= we have:

AM 530 kHz 8

3

3.00 10 m/s 566 m;530 10 Hz

= = 1600 kHz 8

3

3.00 10 m/s 187 m1600 10 Hz

= =

FM 88 MHz 8

6

3.00 10 m/s 3.41 m;88 10 Hz

= = 108 MHz 8

6

3.00 10 m/s 2.78 m108 10 Hz

= =

33-44. The cyclotron frequency is Eq. 30.7 .2qBf

m= For protons,

198

27

8

8

(1.6 10 C)(8.0 T) 1.22 10 Hz2 (1.67 10 kg)

3.00 10 m/s 2.5 m1.22 10 Hz

f

cf

= =

= = =

33-45. (a)8 8

1018

3.00 10 m/s 3.00 10 m/s 1.5 10 m (X-ray)2.0 10 Hz

cf f

= = = =

(b)8 8

210

3.00 10 m/s 3.00 10 m/s 1.0 10 m, or 1.0 cm (microwave radio)3.0 10 Hz

cf f

= = = =

(c)8 8

63.00 10 m/s 3.00 10 m/s 5.0 10 m (electric wave)60 Hz

cf f

= = = =

33-46.8

92

3.00 10 m/s 1.4 10 Hz21 10 m

cf = = =

33-47.8 8

10

3.00 10 m/s 3.00 10 m/s 0.027 m1.1 10 Hz

cf f

= = = =

33-48. max = 550 nm, which is yellow-green, according to the caption in Figure 33-24.1/2 510 nm (aqua) and 610 nm (yellow-orange). 1/4 490 nm (blue) and 640 nm (orange).

33-49.2 12 2 2 2

8 30 (8.85 10 C /N m )(80 V/m) 2.8 10 J/m2 2Eu

= = =

i

33-50.2 12 2 2 2

14 30 0 (8.85 10 C /N m )(0.12 V/m) 6.4 10 J/m2 2E BEu u

= = = =

i

33-51.2 2 2 8 10 2

6 27

0 0 0

( ) (3.00 10 m/s)(2.0 10 T) 9.5 10 W/m4 10 T m/A

E cB cBSc c

= = = = =i

CHAPTER 33

712

33-52.20

02ES

c=

7 8 3 2 30 0

360

0 8

2 2(4 10 T m/A)(3.00 10 m/s)(1.4 10 W/m ) 1.03 10 V/m

1.03 10 V/m 3.42 10 T3.00 10 m/s

E cS

EBc

= = == = =

i

33-53. B = 0Ec

= 80.50

3 10T 91.7 10 .T=

By right-hand rule, B pointsvertically downwards.Direction of Poynting vector =direction of propagation NORTH.=

Magnitude S = 0

1

EB

= 61

1.26 10 0.50 1.7 109

4 26.6 10 W/m=

33-54.20

02ES

c=

7 8 9 2 50 0

530

0 8

2 2(4 10 T m/A)(3.00 10 m/s)(1.0 10 W/m ) 8.68 10 V/m

8.68 10 V/m 2.89 10 T3.00 10 m/s

E cS

EBc

= = == = =

i

33-55. The average intensity S at a distance r from a source emitting power P equally in all directions is

2 .4PSr

= The power is2 17 2 8 2 284 (4 )(4.3 10 m) (1.2 10 W/m ) 2.8 10 W.P r S = = =

33-56. 5 23 21.2 W 2.44 10 W/m

(1.25 10 m)PSA

= = =

27 8 5 20

0 00

4

2 2(4 10 T m/A)(3.00 10 m/s)(2.44 10 W/m )2

1.4 10 V/m

ES E cSc

= = ==

i

33-57. 2 11 2 3 2 264 (4 )(1.5 10 m) (1.4 10 W/m ) 4.0 10 W.P r S = = =

33-58. ( )3

8 223

Power 6.0 10 W 4.8 10 W/mArea 2.0 10 m

S

= = =

2 22 6

63

8

/ 2 2 1.3 10 V/m2

1.332 10 V/m 4.4 10 T3.00 10 m/s

o rmsrms o o rms o

o o

rmsrms

E EE E E E S E cSc c

EBc

= = = = = =

= = =

CHAPTER 33

713

33-59. Magnitude of average energy flux at 5 km is given by2 2

5 207 8

0

(0.22 V/m) 6.42 10 W/m .2 2(4 10 T m/A)(3.00 10 m/s)ES

c

= = =iPower of transmitter = flux area = S 4r2 = 6.4 105 W/m2 4(5000 m)2 = 2.02 104 W

33-60. Power incident on cell = 0.10 2W

cm 13 cm2 = 1.3 W.

Power out of cell = VI = 0.45 V 0.20 A = 0.09 W. Therefore,

Efficiency = Power outPower in

= 0.09 0.069 ( 7%)1.3

=

33-61. The electric field from the transmitter is produced by accelerated charges, so the magnitude is

proportional to 1/r. Since E 1 ,r

we have E1 = 001

;rEr

therefore, at 12.0 km,

E1 = (0.13 V/m)(6/12) = 0.065 V/m. At 18.0 km, E1 = (0.13 V/m)(6/18) = 0.043 V/m.

33-62. 2 .4PSr

= 20

202 4

E PSc r

= =

7 80

0

1 1 (4 10 T m/A)(3.00 10 m/s)(100 W) 39 m2 2.0V/m 2

cPrE

= = =i

33-63.3

3 24 2

5 10 W 6.37 10 W/m(5 10 m)

PSA

= = =

28 3 2 30

0 0 00

2 2( )(3.00 10 m/s)(6.37 10 W/m ) 2.19 10 V/m2ES E cS

c

= = = =

The total energy contained within a volume V of the field is 2

0 0 .2EW uV V= = The volume of a

cylinder of length l and radius r is r2l, so12 2 2 3 4

10(8.85 10 C /N m )(2.19 10 V/m)( )(5 10 m)(0.1 m) 1.67 10 J2

W

= =

i

33-64. 780.50 J 2.5 10 W

2.0 10 sWPt

= = =

3

12 23 2

28 12 2 70

0 0 00

5 10 W 7.96 10 W/m(1.0 10 m)

2 2( )(3.00 10 m/s)(7.96 10 W/m ) 7.75 10 V/m2

PSAES E cS

c

= = =

= = = =33-65. The average intensity S at a distance r from a source emitting power P equally in all directions is

22 2

75 W 1.49 W/m .4 4 (2.0 m)

PSr

= = = The radiation pressure is 2[ ] ,Spressurec

= since the

mirror reflects the light. The magnitude of the force on the mirror is2

2 211

8

2 ( )[ ] [ ]

2(1.49 W/m )( )(0.05 m) 7.81 10 N3.00 10 m/s

mirrorS RF pressure area of mirrorc

= =

= =

CHAPTER 33

714

33-66.3

6 22 3 2

50 10 W 1.59 10 W/m4 4 (50 10 m)

PSr

= = =

28 6 20

0 0 00

2 2( )(3.00 10 m/s)(1.59 10 W/m ) 0.0346 V/m2ES E cS

c

= = = =33-67. The total power output is ,P eSA= where e is the efficiency. Then

63 21.0 10 8.0 10 m .

(0.25)(500)PAeS

= = =

33-68. (a) Energy through lens = 0.10 2W

cm 2(10 cm) ;

4 therefore, Energy flux at spot = energy

area =

2 22

2 2

0.10 W/cm ( /4)(10 cm) 40 W/cm( /4)(0.50) cm

=

(b) Yes!=33-69. Let S be flux at the binoculars. With naked eye, energy through eye per second = 7.02 S. Energy

through binoculars = 502 S. All this goes into the eye. Therefore, Factor increase =2

2

50 51 times7.0

SS

=

33-70. 2 .4P PSA r

= = By Eq. 33.24, 20

0

.2ES

c= Therefore,

20

20

.2 4E P

c r = Then

7 8 47 8 40

2 4 2 40

6

(2 10 )(3.00 10 )(10 )(4 10 T m/A)(3.00 10 m/s)(10 W)2 2 (2 10 V/m) 2 10

m 3.9 10 m

cPrE

= = =

=

i

33-71. The radiation pressure is [ ] Spressurec

= , since the radiation is absorbed. The magnitude of the

radiation force is2 2 6 2

8

8

( ) (1400 W/m )( )(6.38 10 m)[ ] [ ]3.00 10 m/s

6.0 10 N.

earthr

S RF pressure cross sectional areac

= = =

=

The magnitude of the gravitational force exerted by the sun is11 2 2 24 30

222 11 2

(6.67 10 N m /kg )(5.98 10 kg)(1.99 10 kg) 3.5 10 N.(1.50 10 m)

earth sung

Gm mFr

= = =

i

The effect of the radiation pressure is negligible.33-72. The potential difference from one end of the wire to the other end is V = IR, so the magnitude of

the electric field within the wire is E = V / l. The field points in the direction of I. At the surface

of the wire, the magnitude of the magnetic field is 0 .2

IBr

= The field lines form a circular

pattern around the wire with the direction at any point given by the right hand rule. The Poynting

vector is given by 0

.

=

E BS E and B are perpendicular to each other, so the magnitude of S is

0

.2

EB VISrl

= = The direction of the Poynting vector is radially inward toward the center of the

CHAPTER 33

715

wire. Note that the magnitude of the Poynting vector is equal to the electrical power dissipated bythe wire divided by its surface area. This means the power dissipated comes from the electric andmagnetic fields around and within the wire.

33-73. (a)2 12 2 2 3

17 30 (8.85 10 C /N m )(2.0 10 V/m) 1.77 10 J/m .2 2electricEu

= = =

i For each pulse,uelectric = umagnetic = 1.77 1017 J/m3.(b) Suppose the waves are propagating along the x direction (one to the right, the other to theleft), and at some instant both electric fields point in the +y direction. Then, since they aretraveling

in opposite directions, their magnetic fields must cancel as shown in the diagram. If at some otherinstant, their electric fields point in opposite directions along the y axis, then their magnetic fieldspoint in the same direction but the electric fields cancel. Thus at every instant the superposition ofthe two waves has one of the net field components equal to zero while the other component hastwice the magnitude of the individual pulses. For the specific case shown, 34.0 10 V/m,totalE

=

and total 0.B = If the electric fields cancel, then3

single 11total 8 2

2.0 10 V/m2 2 1.33 10 T3.0 10 m/s

EB

c

= = =

(c) The energy density for each field is proportional to the square of the amplitude of that field.Since one of the net fields in the superposition is always zero and the nonzero field has twice themagnitude of the field in each individual pulse, the nonzero field will have four times the energydensity of one of the individual pulses. To give a concrete example, suppose that the electricfields are aligned as shown in the diagram. Then -17 37.08 10 J/m ,electricu = and 0.magneticu =

The other possibility is 17 3 = 7.08 10 J/m ,magneticu and 0.electricu = Note that in either case the

energy density is equal to the total energy density in the individual fields as calculated in (a).

33-74. Eq. 33.33 says 2 2

0 02 2 .E E

x t =

If 0 sin( )yE E kx t= , then

22

0 02

22

0 02

cos( ), sin( )

cos( ), sin( )

y

y

E EkE kx t k E kx tx x

E EE kx t E kx tt t

= =

= =

Substituting into the wave equation gives

2 20 0 0 0sin( ) sin( ).k E kx t E kx t =

B

E

B

E

CHAPTER 33

716

The sine factors cancel and were left with2

0 0

1 ,k

= which means the wave is propagating with speed c. So the equation given does describe anelectromagnetic wave.The wave is moving in the +x direction because of the minus sign in the argument of the sinefunction. By the right hand rule, B must point in the +z direction when E points in the +y

direction. So the equation for B is 0 sin( ).zEB kx tc

=

33-75.8

6

3.00 10 m/s 80.0 m3.75 10 Hz

cf

= = =

1

7 1

2 0.0785 m

2 2.56 10 s

k

f

= =

= =

The wave is traveling in the z direction because of the plus sign in the argument of the cosinefunction. Since the directions of the fields and propagation directions are related by the right handrule, the electric field must point in the +y direction when the magnetic field points in the +xdirection to give the correct propagation direction. Thus the equation for the electric field is

0 cos( )yE cB kx t= +

33-76. Define a new variable u = x ct. Then 2 2

2 2;f f f fx u x u

= =

and

2 22

2 2; .f f u f f fc ct u t u x u

= = =

Substituting into the wave equation gives

2 2 2 22

0 0 0 02 2 2 20 0f f f fc

x t u u = =

because 2 0

0

1 .c

= Thus f(x ct) satisfies the

wave equation.For the specific case of 2 2 2 20 0exp[ ( ) ] exp[ ]E E k x ct E k u= = , we find

( )2 2 2

0

2 22 2 2 2 2 2 2 2 2 2 2 2

0 0 02 2

2 2 20

2 22 2 2 2 2 2 2 2 2 2 2 2

0 0 02 2

2 exp[ ]

2 exp[ ] (2 ) exp[ ] 2 exp[ ] 2 1

2 exp[ ]

2 exp[ ] (2 ) exp[ ] 2 exp[

E E k uE k ux uE E k E k u k u E k u k E k u k u

x uE Ec ck uE k ut uE Ec c k E k u ck u E k u c k E k u

t u

= =

= = + =

= =

= = =

( )2 2 2] 2 1k u

Substituting this into 2 2

0 02 2 0E E

x t =

2 2

20 02 2 0

f fcu u

=

completes the proof because

2 0

0

1 ,c

= causing all the terms to cancel to give 0 = 0.

CHAPTER 33

717

33-77. Applying the line integral in the Ampere-Maxwell Law around the rectangle in the xz plane inFigure 33.27 gives ( ) .B ds Bdz B dB dz dB dz= + = &v The electric flux through the rectangleis ,Ed Edx dz = which gives .E

E dxdzt t

=

Then the Ampere-Maxwell Law gives

0 0EdB dz dx dzt

=

, or

0 0 ,B Ex t

=

which is Eq. 33.32.

33-78. 20 0 0 ( )Edisplacementd dE dEI A r

dt dt dt

= = =

12 2 2 2 13(8.85 10 C /N m )( )(0.15 m) (3.8 10 V/m s) 24 A= =i i33-79. (a) The current in the thin wire is given by I = 0 .V sin tV

R R

=

(b) Displacement current

0 0 0 0 0 0( ) ( / )E

dd d EA d V d A dV AI A V cos t

dt dt dt d dt d

= = = = =

(c) Current arriving at the outside terminal must equal current through the thin wire plus

displacement current: 0 0 0outsideV sin t AI V cos t

R d

= +

(d) 0 ( ).dd I I= + B si As in Example 1, we have2rB = 0(I + Id)in, 0 in

( ) .2

d I IBr

+= Here, I = 0 .V sin t

R

But Id is not the entire displacement current, since r < radius of plate.

Since Id Ed

dt

area, this gives Idin = dI r

2 =

20 rd

V0 cos t; therefore,

B = 2

0 0 0 0 0 00 0

12 2 V sin t r V sin t rV cos t V cos t

r R d rR d

+ = +

CHAPTER 33

718

33-80. Assume a positively charged particle isinitially moving to the right. Let t = 0 bethe instant the particle comes to rest. At it begins to slow down with a constantacceleration pointing to the left as shown.At some time t before it stopped it wasat the point shown. The large circlecentered on the point labeled with t hasradius ct and shows how far the fieldproduced at that instant has propagatedby the time the particle stops. Thesmaller circle is centered on the pointlabeled and has radius c. Field linesare drawn from the point at t = 0 to showthe kinks produced by the accelerationfrom t = to t = 0.

33-81.6

8

3.141 6.4 10 0.067 s3 10

rtc

= =

33-82. 20

costransmittedII

=

0

0

0

20 : 0.88

40 : 0.59

60 : 0.25

transmitted

transmitted

transmitted

II

II

II

=

=

=

33-83. After the first polarizer, 21 0 cos 60.I I= The second polarizer is inclined 60 to the first, so2 4 2 4 2

2 1 0cos 60 cos 60 (0.50 W/m ) cos 60 0.31 W/mI I I= = = =

33-84. (a) In minus z-direction froms plus sign in ,ztc

+ since

waves in form f(z + vt) travel in minus direction.

(b) From diagram, E makes angle = tan1 21

63 with direction,x=

27 with axis, andy 90 with axis.z

CHAPTER 33

719

(c) Since B = 0Ec

in magnitude, and B-field is related to

E by right-hand rule, and since wave is propagating inminus-z-direction, we have

B = 0 Ec

j sin ztc

+ + 20 E

ci sin zt

c

+

33-85. E = 140 V/m east, and the propagation direction is up.By the right-hand rule, B must point north so thedirection of E B is up. The magnitude of B is

78

150 V/m 5.00 10 T.3.00 10 m/s

EBc

= = = Thus the

magnetic field vector is B = 5.00 107 T north.

33-86. Assuming the wavelength is specified in air, 8

6

3.00 10 m/s 75 Hz4 10 m

cf = = =

33-87. 5 23 21.2 W 2.44 10 W/m

(1.25 10 m)PSA

= = =

2 20 rms

0 0

7 8 5 2 3rms 0

5rmsrms

2

(4 10 T m/A)(3.00 10 m/s)(2.44 10 W/m ) 9.60 10 V/m

3.2 10 T

E ESc c

E cS

EBc

= =

= = == =

33-88. (a) S = Energy flux = PowerArea

= 9

24 2

5 10 W 50 W/m(10 m)

=

(b) E0 = 02 cS = 7 8 22(4 10 T m/A)(3.00 10 m/s)(50 W/m ) 194 V/m =i V/m

195 V/m= and B = 70 6.5 10 TEc

=

33-89. Flux at 10 km = Power .Area

Area = 4r2 = 4(10,000 m)2 = 1.26 109 m2; therefore, Flux = \f(5000

W, 1.26 109 m2) = 4.0 106 W/m2 = S. By (36), E0 = 02 0.055 V/m c =S and B = 0Ec

=

108 1

0.055 V/m 1.8 103 10 ms

T

=

33-90. (a)( ) ( )

( )8

223

12 A 1.7 10 /m3.8 10 V/m

1.3 10 m

IEA

= = =

(b)( ) ( )

( )7

303

4 10 T m/A 12 A1.8 10 T

2 2 1.3 10 mIBr

= = =

(c)( ) ( )

( )2 3

27

0

3.8 10 V/m 1.8 10 T56.5 W/m

4 10 T m/AEBS

= = =i

B

E

CHAPTER 33

720

(d) ( ) ( ) ( ) ( )2 32 56.5 W/m 2 1.3 10 m 1 m 0.46 WP S rl = = =(e)

( ) ( ) ( )( )

2 822

23

12 A 1.7 10 /m 1 m0.46 W

1.3 10 m

I lP I RA

= = = =

33-91. Power received by the Arecibo telescope = Flux Area= 4.1 1025 (150)2 202.9 10 W=

Total power emitted by the quasar = Flux 4R2 (R is the earth-quasar distance) =4.1 1025 4(2.8 109 9.46 1015)2 273.6 10 W=

33-92. (a)2

3 42 2 2

mW W cm W10 10 10 100 .cm mW m m

= Then

( ) ( ) ( )2 7 8 20 0 00

2 2 4 10 T m/A 3 10 m/s 100 W/m 270 V/m2ES E cS c

= = = =i70

0 8

274.5 V/m 9.2 10 T3 10 m/s

EBc

= = =

(b) ( ) ( )2 2100 W/m 1 m (100 J/s)/(4.186 J/cal) 24 cal/sP SA= = = =33-93. (a) Power = flux area. Area = r2 = (3.5 103)2 m2 = 3.85 105 m2. Therefore, P = 8.8

1011 W/m2 3.85 105 m2 153.4 10 W=

(b) Energy flux = PowerArea

, and at a distance of r, area = 4r2. Therefore,

Flux = 2 .4Pr

Minimum flux = 8.8 1011 W/m2. Therefore, max distance

r = 4 flux

P

= r = 26

1711

3.9 10 5.9 10 m,4 (8.8 10 )

= which is the distance of faintest possible

star. Volume that distance covers around us = 43r3

= 43(5.9 1017 m)3 = 8.8 1053 m3. One cubic light year = (9.47 1015 m)3

= 8.5 1047 m3, contains 3.5 103 stars. Therefore, number of visible stars

= 8.8 1053 m3 3

47 3

3.5 10 stars 3600 stars.8.5 10 m

=

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