Chapter 34 Solutions
34.1
Since the light from this star travels at 3.00 6.44 × 1018 m
×
108 m/ s, the last bit bit of light light w ill ill hit the Earth i n
= 2.15 × 1010 s = 680 680 years. Therefore, it w ill disap pear from th e sky in the year
8
3.00 × 10 m / s
1999 + 680 = 2.68 × 10 3 A.D.
v
34.2
1
=
k µ µ 0 e0
1
=
1.78
c = 0.750c = 2.25 × 108 m / s
34.3
E =c B
34.4
E max max = v is the generalized version of Equation 34.13. Bma x
=
Bm ax
34.5
220 = 3.00 × 108; B
or
E m ax v
=
7.60 × 10 −3 V / m 8
(2/ 3)(3 3)(3.00 .00 × 10 m /
N ⋅ m T ⋅ C ⋅ m = 3.80 × 10–11 T = s ) V ⋅ C N ⋅ s
× 108 m / s
(a) f λ = c
or
f (50.0 m) = 3.00
(b)
E =c B
or
22.0 = 3.00 × 108 Bma x
(c)
k =
2π
λ
=
B = 7.33 × 10–7 T = 733 nT
so
2π = 0.126 m –1 50.0
an d
38.0 p T
so
f = 6.00 × 106 Hz = 6.00 6.00 MH z
so
Bma x = (73.3 n T)(–k )
ω = 2π f = 2π (6.00
×
10 6 s –1) = 3.77
rad / s t ) = (73.3 n T) cos (0.126 x – 3.77 × 107 t) (–k ) B = Bm ax cos(kx – ω t
34.6
ω = 2π f = 6.00π × 109 s –1 = 1.88 × 1010 s -1 k =
2π
λ
=
E = 300
ω c
=
6.00π × 109 s –1 3.00 × 10 m / s 8
= 20.0π = 62.8 m –1
V cos 62.8 x − 1.88 × 1010 t m
(
)
Bma x =
E 300 300 V/ V/ m = = 1.00 µ T c 3.00 × 108 m / s
B = (1.00 µ T) cos (62.8 (62.8 x – 1.88 × 1010 t )
×
10 7
Chapter 34 Solutions 299
34.7
(a) B =
(b)
2π 2π = = 0.628 µ m k 1.00 × 10 7 m – 1
λ =
c
(c) f =
34.8
100 100 V/ V/ m E = = 3.33 × 10–7 T = 0.333 µ T c 3.00 × 10 8 m / s
λ
=
3.00 × 108 m / s 6.28
× 10
–7
= 4.77 × 1014 H z
m
E = E m ax cos(kx – ω t t )
∂ E ∂ x
∂ 2 E ∂ x
2
∂ E
t )( = – E m ax sin(kx – ω t )( k )
∂ t t 2
t )( = – E m ax cos(kx – ω t )(k )
t )(– = – E m ax sin(kx – ω t )(–ω )
∂ 2 E 2
∂ t t
t )(– = – E m ax cos(kx – ω t )(–ω )2
We must show:
∂ E ∂ 2 E µ = e 0 0 ∂ x 2 ∂ t t 2
That is,
2 −( k 2 ) E E m ax cos ( kx − ω t ) = − µ 0e0 ( − ω ) E ma x cos ( kx − ω t ) 2
But this is true, because
k
ω 2
2
1 1 = = 2 = µ 0 e0 λ c f λ
The proof for the wave of magnetic field is precisely similar.
*34.9
In the fundamental mode, there is a single loop in the standing wave between the plates. Therefore, Therefore, the distance between th e plates is equal to half a w avelength.
λ = 2 L = 2(2.00 m) = 4.00 m Thus, f =
*34.10
d A to A
c
λ
3.00 × 108 m / s = = 7.50 × 10 7 Hz = 75.0 75.0 MH z 4.00 m
= 6 cm ±
λ = 12 cm
±
5%
5%
=
λ 2
300 Chapter 34 Solutions
v
34.11
f = (0.12 m ± = λ f
S = I =
(
5%) 2.45 × 10 9 s
−1
)=
2. 9 × 10 8 m s
±
5%
U Uc = = uc A t V
I Energy 1000 1000 W/ m 2 =u= = = 3.33 µ J/ m 3 8 c U n i t V o lu lu m e 3.00 × 10 m / s
34.12
S av =
E m ax
P
4 π r 2
=
=
4π (4.00 × 1609 m) 2
2µ 0 cSav
∆V m ax = E m ax
34.13
4.00 × 10 3 W
= 0.07 .0761 V /
= 7.68 µ W / m 2
m
· L = (76.1 m V/ V/ m )( )(0.650 m ) = 49 49 .5 .5 m V (a m p l it it u d e )
r = ( 5.00 mi )(1609 m / m i ) = 8.04 × 10 3 m
S=
P
2
4 π r
=
250 × 10 3 W 4π (8.04 × 10 m ) 3
2
= 307 µ W / m 2
or
35.0 m V (r m s) s)
Chapter 34 Solutions 301
Goal Solution What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropicall isotropically y w ith an average p ower of 250 250 kW? G : As the distance from from the source is increase increased, d, the pow er p er un it area w ill ill d ecrease, ecrease, so so at a distance of 5 miles from the source, the power per unit area will be a small fraction of the Poynting vector near the source. O : The Poynting vector is the power per unit area, where A is the surface area of a sphere with a 5-mile radius. A : The Poynting vector is
r = ( 5.00 mi )(1609 m / m i) = 8045 m
In meters,
and the magnitude is S
L:
=
250 × 10 3 W 2
(4 π )(8045) )(8045)
= 3.07 × 10 −4
W/ m2 2
The magnitu de of the Poynting vector vector ten m eters from the sou rce is is 199 199 W/ m , on the order of a million times larger than it is 5 miles away! It is surprising to realize how little little pow er is actually actually received by a radio (at the 5-mile distance, the signal would only be about 30 nW, assuming a 2
receiving area of about 1 cm ).
I =
34.14
100 W 4π (1.00 m)
u=
2
= 7.9 7.96 6 W/ m 2
I = 2.65 × 10 – 8 J/ m 3 = 26.5 26.5 n J/ m 3 c 1
(a)
13.3 n J/ m 3 u E = 2 u = 13.3
(b)
u B = 2 u = 13.3 13.3 n J/ m 3
1
(c) I = 7.96 7.96 W/ m 2
34.15
Power outp ut = (pow er inpu t)(ef t)(effi fici ciency) ency)
Thus,
an d
power out 1.00 × 106 W Power input = = = 3.33 × 106 W eff 0.300
A =
P
I
=
3.33 × 10 6 W 1.00 × 10
3
W/ m
2
= 3.33 × 10 3 m 2
302 Chapter 34 Solutions
2
Bma x c
I =
*34.16
=
2µ 0
P
4π r 2
P 2µ 0 = 4 π r 2 c
Bma x =
(10.0 × 103 )(2)(4π × 10−7 ) 2 4 π ( 5.00 × 10 3 ) ( 3.00 × 108 )
= 5.16 × 10 –10 T
Since the magnetic field of the Earth is approximately 5 100,00 100,000 0 times stron ger.
34.17
(a)
P
= I 2 R = 150 W; A = 2π rL = 2 π (0.900
S=
P A
(b) B = µ 0
E =
I µ 0(1.00) = = 222 µ T 2π r 2π (0.900 × 10 –3 )
=
Note: S =
150 V IR = = 1.88 1.88 kV/ m L 0.0800 m
EB = 332 332 kW/ m 2 µ 0
10 –5 T, the Earth's field is some
× 10–3 m)(0.0800 m) = 4.52 × 10 –4
= 332 332 kW/ m 2 (points radially radially inward )
∆V ∆ x
×
m2
Chapter 34 Solutions 303
34.18
(a)
)(N / C) · (0.20 (0.200 0 i + 0.0800 j + 0.290 k ) µ T E · B = (80.0 i + 32.0 j – 64.0 k )(N E · B= (16.0 + 2.56 – 18.56)N
(b)
S=
S=
· s/ C 2 · m = 0
1 (80.0 i + 32.0 j – 64.0 k )(N/ C) × (0.200 i + 0.0800 j + 0.290 k ) µ T E × B= µ 0 4π × 10 –7 T · m / A (6.40 k – 23.2 j – 6.40 k + 9.28 i – 12.8 j + 5.12 i )10 – 6 W / m 2 4π × 10 –7
S = (11.5 i – 28.6 j) W / m 2
34.19
2
= 30. 30.9 9 W/ m 2 at – 68.2° from the + x axis
We call the current I rms and the intensity I . The power radiated at this frequency frequency is
P
0.0100(∆V rm s)2 = (0.0100)(∆V rm s ) I = 1.31 W I rm rm s = R
If it is isotropic, the intensity one meter away is I =
P A
Bm ax
*34.20
=
1.31 W 4π (1.00 m)2 2µ 0 I
=
c
=
(a)
efficiency
(b)
Sav
=
S av
=
Sav
= E m ax
P
A
=
=
= 0.10 0.104 4 W/ m 2 = S av =
(
2 4 π × 10
c 2 Bmax 2 µ 0
−7 T ⋅ m / A 0. 10 104 W / m 2
)(
3.00 × 10 8 m / s
useful useful power output total pow er inpu inpu t 700 W
(0.0683 m )(0.0381 m )
)=
29.5 n Τ
× 100% = 700 W × 100% = 1400 W
= 2.69 × 105
50.0%
W m2
269 kW kW m 2 toward the oven chamber chamber 2
(c)
E m ax
2µ 0c
=
2 4 π × 10 −7
T ⋅ m m W V 3.00 × 10 8 2.69 × 10 5 = 1.42 × 10 4 = 14.2 kV m 2 A s m m
304 Chapter 34 Solutions
34.21
(a) Bma x =
7.00 × 105 N / C E max max = = 2.33 m T c 3.00 × 108 m / s
2
E max (7.00 × 10 5 )2 (b) I = = = 650 650 MW/ m 2 2 µ 0 c 2(4π × 10 –7 )(3.00 × 108)
(c) I =
34.24
(a) I =
P
= I A = (6.50
2 E m ax
2µ 0c
π (0.80 800 × 10
= I =
m)
m 2)
π 4
(1.00 × 10 – 3 m )
(4 π r 2 ); solving for r ,
(10. 0 × 10 − 3 )W −3
× 108 W/
2
=
r =
2
= 510 W
Pµ 0 c
2 E ma x 2 π
=
(100 W)µ 0 c 2 π (15. 0 V / m )2
=
5.16 m
4.97 4.97 kW/ m 2
4.97 × 10 3 J / m 2 ⋅ s
=
16.6 µ J/ m 3
(b)
u av
(a)
E = cB = (3.00 × 10 8 m / s) s )(1.80 × 10 − 6 T) = 540 540 V/ V/ m
(b)
u av
=
(c)
Sav
= cuav = (3.00 × 108 )(2.58 × 10− 6 ) =
(d)
This This is 77.3 77.3% % of of the flux flux in in Examp Examp le 34.5 34.5 . It may be clo cloudy, udy, or the Sun m ay be setti setting. ng.
3.00 × 10 8 m / s
c
B
2
µ 0
=
(1.80 × 10 − 6 )2 − 4 π × 10 7
=
2.58 µ J/ m 3
For complete absorption, P =
34.25
*34.26
A
:
Power = SA =
34.22
34.23
P
(a)
P
= (Sav )( A ) = (6. 00 00 W /
)(
S c
=
m 2 40.0 × 10
7 73 73 W / m 2
25.0 3.00 × 108
−4 m 2
= 83.3 nP a
) = 2.40 × 10−2 J/
s
In one second, the total energy U impinging on the mirror is is 2.40 2.40 transferred each second for total reflection is 2U 2(2.40 × 10–2 J) p = = c 3.00 × 108 m / s (b)
F =
= 1.60 × 10 –10
kg · m s
dp 1.60 × 10–10 k g · m / s = = 1.60 × 10–10 N d t 1s
(each second)
× 10 –2 J.
The momentum p
Chapter 34 Solutions 305
2
34.27
(a )
Th e ra ra d ia ia ti tio n p r es es su su r e i s
(2) (2)(1340 1340 W / m ) 8
3.00 × 10 m / s
× 105 m 2
Multiplying Multiplying by the total area,
A = 6.00
(b)
The acc accelera elerati tion on is: is:
a=
(c)
It w il ill t ak ak e a ti tim e t where:
d = 2 at 2
34.28
(a)
(b)
N / m2
gives: F = 5.36 N
F 5.36 N = = 8.93 × 10– 4 m / s 2 6000 kg m
2 d a
=
(
2 3. 84 84 × 10 8 m
(
)
8.93 × 10 −4 m / s 2
)
= 9.27 × 105 s =
The pressure P upon the mirror is is
P=
2Sav
where A is the cross-sectional area of the beam and
Sav
=
The force on the mirror is then
F = PA
Therefore,
F =
I =
34.29
= 8.93 × 10 − 6
1
t =
or
2
P
π r 2
E m ax
=
=
2 E m ax
2µ 0 c
P
( 2µ 0c) π r 2
15 × 10–3 J/ s 3.00 × 108 m / s
(c) p =
= 1.90 1.90 kN/ C
(1.00 m ) = 50.0 p J
U 5 × 10 –11 –19 = kg · m/ s 8 = 1.67 × 10 c 3.00 × 10
10.7 days
c P
A
2 P 2P A A = = c A c
2(100 × 10 −3 ) 8
(3 × 10 )
=
6.67
× 10–10 N
306 Chapter 34 Solutions
34.30
(a )
If PS is the total power radiated by the Sun, and r E an d r M are the radii of the orbits of the planets Earth and Mars, then the intensities of the solar radiation at these planets are: I E =
PS
(b)
2 R M = I M (π R ) = (577
) (
W m 2 π 3.37 × 106 m
=
I M c
(
2 π R R M
)=
P M
=
GM S M M 2 r M
6.67 × 10 −11 ( =
5 77 77 W / m 2
)
2
=
2.06 × 10 16 W
c
=
2.06 × 1016 W 3.00 × 10
8
m s
=
S M c
=
I M c
6.87 × 10 7 N
)(
)(
N ⋅ m 2 kg 2 1.991 × 10 30 kg 6.42 × 10 23 kg
wh ich ich is ~1013 times stronger
(a) (a)
1.496 × 1011 m ) 2.28 × 1011 m =
The attrac attractive tive gravit gravitati ational onal forc forcee exerte exerted d on Mars by the Sun Sun is
F g
34.31
W m
2
If Mars behaves behaves as a perfect perfect absorber, it feels feels pressure P =
and force force F = PA (d)
2 4 π r M
Mars interc intercepts epts the power falli falling ng on its circ circular ular face face:: P M
(c) (c)
PS
2
2
r I M = I E E = (1340 r M
Thus,
I M =
an d
2 4 π r E
(2.28 × 1011 m )
2
) = 1.64 × 1021 N
than the repulsive force force of (c). (c).
The total total energy energy absorbe absorbed d by the surfac surfacee is is W 0.500 × 1.00 m 2 (60.0 60.0 s ) = 11.3 kJ U = 21 I A A t = 21 750 2 m
(
( )
(b)
)
The total energy incident incident on the surface surface in this time time is 2 U = 22.5 kJ, with U = 11.3 kJ being absorbed absorbed an d U = 11.3 11.3 kJ being being reflected. reflected. The total momentu m tra nsferred to the su rface rface is p = (momentu m from absorption absorption
) + (momentum from reflection )
3 U 2U 3U 3(11.3 × 10 J) + = = = p = c c c 3.00 × 108 m s
34.32
Sav
=
µ 0 J m2 ax c 8
or
570 =
(4 π × 10
1.13 × 10 − 4 kg ⋅ m s
−7 ) J 2
m ax (3.00
8
× 108 )
so
J m ax
= 3.4 3.48 8 A/ m 2
Chapter 34 Solutions 307
34.33
(a)
P
µ J 2 c = Sav A = 0 ma x A 8
4 π × 10−7 (10.0)2 (3.00 × 108 ) (1.20 × 0.400) = P= 8 (b)
*34.34
=
Sav
P
=
µ 0 J m2 ax c 8
( ∆V )
=
(4 π × 10 −7 (10.0)2 (3.00 × 10 8 )
2
or
R
P
8
=
2.26 kW
4.71 4.71 kW/ m 2
∝ (∆V )2 y
θ
E ∆V = (−) E y ⋅ ∆ y = E y ⋅ l cos θ
∆V ∝ cos θ
34.35
so
P
∝ cos 2 θ
(a)
θ = 15.0° :
P
= P m ax cos 2 (15.0°) = 0.933P m ax = 93.3 93.3% %
(b)
θ = 45.0° :
P
= P ma x cos ( 45.0°) = 0.500P ma x = 50.0 50.0% %
(c)
θ = 90.0° :
P
= P ma x cos 2 ( 90.0°) = 0
(a)
Construct Constructiv ivee interf interfere erence nce oc occurs when d cos θ = n λ fo r some integer n . cos θ = n
∴ (b)
2
λ = n = 2n λ / / 2 d
λ
n
= 0, ± 1, ± 2 , . . .
strong signal @ θ = cos –1 0 = 90°, 270°
Destruc Destructi tive ve interf interfere erence nce oc occurs when d cos θ =
∴
2n + 1 λ : 2
cos θ = 2 n + 1
weak signal @ θ = cos –1 (±1) = 0°, 180°
l
eceiving ntenna
308 Chapter 34 Solutions
Goal Solution Two radio-transmitting antennas are separated by half the broadcast wavelength and are driven in phase with each other. In wh ich ich d irections irections are (a) (a) the strongest and (b) the weakest signals signals radiated? G : The strength of the radiated signal will be a function of the location around the two antennas and will depend on the interference of the waves.
helps to to visualize this this situation. The two antenna s are driven in p hase, wh ich ich m ea ns O : A diagram helps that they both create maximu m electric electric field field strength at the same time, time, as show show n in the diagram. Th e radio EM waves travel radially outwards from the antennas, and the received signal will be the vector sum of the two w aves. (a) Along Along the perpendicular perpendicular bisec bisector tor of the line line joining the antennas, the d istanc istancee is is the same to A : (a) both transm itting itting anten nas. The transmitters oscil oscillate late in in ph ase, so along this line line the two signals w ill be received in phase, constructively interfering to produce a maximum signal strength that is twice the amp litude litude of one transmitter. transmitter. (b) Along the the ext extended ended line line joini joining ng the sources, sources, the wave from from the more d istant istant antenna m u st travel one-half wa velength farther, so so the w aves are received received 180° 180° out of phase. They interfere destructively to produce the weakest signal with zero amplitude. L:
Radio stations may use an antenna array to direct the radiated signal toward a highly-populated region and reduce the signal strength delivered to a sparsely-populated area.
34.36
λ = λ =
34.37
c f c f
= 536 m
so
h=
λ
= 188 m
so
h=
λ
4
4
ΣF = ma ⇒
For the proton:
=
134 m
=
46.9 m
2
sin 90.0 90.0° = mv R qvB sin
The period and frequency of the proton’s circular motion are therefore: T =
R 2π v
=
2 π m qB
=
2 π 1.67 × 10 −27 kg
(
)
1.60 × 10 −19 C (0.350 T )
(
)
= 1.87 × 10−7
s
The charge w ill ill radiate at this same frequency, with
34.38
For the proton,
f = 5.34 × 106 H z .
λ =
c f
=
ΣF = ma yields
3.00 × 10
8
m s
=
6
5.34 × 10 H z
qvB sin sin 90.0° = R 2 π
The period of the proton’s circular motion is therefore:
T =
The frequency of the proton's motion is
f = 1/ T
The charge w ill ill rad iate elect electromagn romagn etic waves at th is frequency, with
λ =
v
c f
=
56.2 m
mv
2
R
2 π m qB
= cT =
2π mc qB
Chapter 34 Solutions 309
From the electromagnetic spectrum chart and accompanying text discussion, the following identifications are made:
*34.39
Frequency f
Wavelength, λ = c f
Classification
2 H z = 2 × 100 H z 3 2 kHz = 2 × 10 H z 2 MHz = 2 × 106 H z 2 GHz = 2 × 109 H z 2 THz = 2 × 1012 H z 15 2 PHz = 2 × 10 H z 2 EHz = 2 × 1018 H z 2 ZHz = 2 × 10 21 H z 2 YHz = 2 × 10 24 H z
150 Mm 150 k m 150 m 15 cm 150 µ m 150 n m 150 p m 150 fm 150 a m
Radio Ra d i o Ra d i o M icr o w a v e In fr a r e d U lt r a v io le t x-r a y Gam m a ray G a m m a Ra y
Wavelength,
Frequency f = c λ
Classification
1. 5 × 10 5 H z 1. 5 × 108 H z 1. 5 × 1011 H z 1. 5 × 1014 H z
Radio Radio Microwave Infrared
1. 5 × 1017 H z 1. 5 × 10 20 H z
Ultraviole Ultraviolet/ t/ x-ray x-ray x-ray/ x-ray/ Gamma ray ray
1. 5 × 10 23 H z 1. 5 × 10 26 H z
Gamma ray Gamma ray
λ 2 k m = 2 × 10 3 m 2 m = 2 × 10 0 m 2 m m = 2 × 10 −3 m − 2 µ m = 2 × 10 6 m − 2 n m = 2 × 10 9 m − 2 p m = 2 × 10 12 m − 2 fm = 2 × 10 15 m − 2 a m = 2 × 10 18 m
*34.40
(a) f =
(b)
34.42
λ
3 × 108 m / s = ~ 10 8 H z 1.7 m
r ad a d io io w a ve ve
1000 1000 pages, 500 500 sheets, sheets, is is about 3 cm thick thick so one sheet sheet is is about 6
f =
*34.41
c
f =
3 × 108 m / s 6
c
λ
× 10 –5 m
=
~ 10 1 013 H z
3.00 × 108 m / s 5.50
× 10
–7
m
× 10 – 5
m thick
in fr fr a re re d
= 5.45 × 1014 H z
(a)
λ =
c 3.00 × 108 m / s = = 261 m f 1150 × 103 / s
so
180 m = 0.690 0.690 w avelength s 261 m
(b)
3.00 × 108 m / s c = = 3.06 m λ = f 98.1 × 106 / s
so
180 m = 58.9 58.9 w avelength s 3.06 m
310 Chapter 34 Solutions
34.43
× 1019 H z) λ = 3.00 × 108 m / s:
λ = 6.00 × 10 –12 m = 6.00 pm
(a) f λ = c
g iv e s
(5.00
(b) f λ = c
g iv e s
(4.00 × 109 H z) λ = 3.00 × 108 m / s: λ = 0.075 m = 7.50 cm
1
1
Time to r each object object = 2 (total time of flight) = 2 (4.00 × 10 – 4 s) = 2.00 × 10 – 4 s
*34.44
Thus, d = vt = (3.00
34.45
× 10 8 m/ s)(2 s)(2.0 .00 0 × 10 – 4 s) = 6.00 × 10 4 m
= 60.0 km
The time for the radio signal to travel 100 km is:
t r r =
The sound wave to travel 3.00 m across the room in:
t s =
Therefore, listeners listeners 100 100 km aw ay
100 × 10 3 m 8
3.00 × 10 m / s
= 3.33 × 10 – 4 s
3.00 m = 8.75 × 10– 3 s 343 m/ s
will receive receive the news before before the people in in the ne w sro om
by a total time difference of
∆t = 8.75 × 10– 3 s – 3.33 × 10– 4 s = 8.41 × 10– 3 s
*34.46
c
λ =
The length of a quarter-wavelength antenna would be
L = 1.00 × 106 m
f
=
3.00 × 108 m s
The wavelength of an ELF wave of frequency 75.0 Hz is
75.0 75.0 H z
=
Thus, while the project may be theoretically possible, it is not very practical.
(a )
(b)
Fo r th th e A M b an an d ,
Fo r t he he FM b an an d ,
λ ma x
=
λ mi n
=
λ ma x
=
λ mi n
=
c f mi n c f m ax c f mi n c f m ax
= = = =
3.00 × 108 m / s 540 × 10 3 H z 3.00 × 108 m / s 1600 × 10 3 H z 3.00 × 108 m / s 88.0 × 106 H z 3.00 × 108 m / s 108 × 106 H z
=
556 m
=
187 m
=
3.41 m
=
2.78 m
m
1.00 × 10 3 km
0.621 mi = 1.00 km
L = (1000 km )
or
34.47
= 4.00 × 106
621 mi
Chapter 34 Solutions 311
CH 4 : f mi mi n = 66 MH z
34.48
λ max = 4.55 m λ min = 4.17 m
f m ax = 72 MHz
CH 6 : f mi mi n = 82 MH z
λ max = 3.66 m λ min = 3.41 m
f m ax = 88 MHz
CH 8 : f mi mi n = 180 MH z
λ max = 1.67 m λ min = 1.61 m
f m ax = 186 MH z
34.49
(a)
(b)
P
= S A = (134 (1340 0 W/ m 2)4π (1.496
S=
S=
2 cBm ax
2µ 0 2 E m ax
2µ 0 c
so
Bma x =
so
E m ax
=
2µ 0S c
2µ 0 cS
=
=
3.77
× 10 26 W
2(4 π × 10 −7 N / A 2 )(1340 W / m 2 ) 8
3.00 × 10 m / s
(
2 4 π × 10
−7 3.00 × 108 1340 ( )
)(
)
=
= 3.35 µ T
1.01 1.01 kV/ m
Supp ose you cover a 1.7 m-by-0.3 m section section of beach blanket. Supp ose the eleva tion ang le of the Sun is 60°. 60°. Then th e target area you fill fill in in th e Sun Sun 's field field of view is
*34.50
(1.7 m
)( 0.3 m )( cos 30° ) = 0.4 m 2
N ow I =
34.51
× 1011 m )2 =
(a)
P A
= E : A t
E = IAt = 1340
ε = − d Φ B = − d ( BA cos θ ) = − A dt
dt
W m2
(0.6)(0.5)(0.4 m 2) 3600 s ~ 106 J
d
( Bma x cos ω t cos θ ) = A Bm ax ω (sin (sin ω t cos θ ) dt
ε (t ) = 2π f f Bma x A sin 2 π f f t cos θ = 2π 2 r 2 f Bm ax cos θ sin 2 π f f t Thus,
ε m ax = 2π 2r 2 f Bm ax cos θ
, where θ is the angle between the magnetic field and the
normal to the loop. (b)
If E is vertical, then B is horizontal horizontal,, so the plane of the loop loop should be vertical vertical plane should contain the line line of sight sight to the transmitter .
and the
312 Chapter 34 Solutions
34.52
(a)
F grav
s ρ ( 4 / 3) π r 3 = GM 2sm = GM 2 R R
where M s = mass of Sun, r = rad ius of particle particle and R = distance from Sun to particle.
Since F ra ra d
(b)
*34.54
F rad F grav
1 3SR 2 1 = ∝ 4cGM ρ r r s
From From the res result ult found found in part (a) (a),, when F grav g rav = F ra ra d , we have
r =
34.53
S π r 2 = , c
(
)(
3 214 W / m 2 3.75 × 1011 m
)
r =
2
(
)(
)(
)(
)
=
3.78
× 10–7 m
E max max = 6.67 × 10 –16 T c
2 E ma x
× 10–17 W/
m2
(b)
S av =
(c)
P
(d)
F = PA
(a) (a)
The elec electri tricc fiel field d between the plates is E = ∆V l, directed dow nw ard in the the figure. figure. The magnetic field field between the p late's edges is B = µ 0 i 2 π r counterclockwise.
2 µ 0 c
= 5.31
= S av A = 1.67
× 10–14 W
= S av A = c
5.56
The Poynting vector is:
× 10–23 N
S
=
1
µ 0
(≈ weigh t of 3000 3000 H atom s!) s!)
E× B =
( ∆V ) i 2 π r l
i B S
(radially (radially outward )
+ + + ++ + + E - - - - --
+
+ + + + + - - - +
i
The lateral lateral surface surface area surround surround ing the ele elect ctric ric fiel field d volume is
A
(c) (c)
4cGM s ρ
− 4 6. 67 67 × 10 11 N ⋅ m 2 kg 2 1.991 × 10 30 kg 1500 k g m 3 3.00 × 10 8 m s
(a) Bmax =
(b)
3S R 2
= 2π r l,
so the power output is
P
∆V i = SA = ( ) (2π r l) = (∆ V )i 2π r l
As the capacitor capacitor charges, the polarity of the p lates and hence the direction direction of the electric electric field field is un changed . Reversing Reversing the current reverses the direction direction of the magn etic field, field, and therefore the Poynting vector. The Poynting vector is now directed radially inward.
Chapter 34 Solutions 313
*34.55
(a) (a)
The magnetic magnetic fiel field d in in the encl enclose osed d volum e is directed directed upward , dB di with magnitude B = µ 0 ni and increasing at the rate = µ 0n . dt dt The changing magnetic field induces an electric field around any circle circle of rad ius r , accordin accordin g to Farad ay’s ay’s Law:
B
E
E ( 2 π r ) = − µ 0n
di
dt ( π r ) 2
E = − E=
or
Then,
S=
1
µ 0
(b)
E× B =
(c)
(clockwise) i
µ 0nr di µ ni ( 0 ) µ 0 2 dt 1
µ 0 n 2 r i di
The power flo flowing wing into into the volume volume is is S . Therefore, P
dt
2
S
dt
2
µ 0 nr di
S=
or the Poyn ting vector is is
µ 0nr di
2 P
dt
= SA lat
inward,
(radially (radially inward )
where A lat is the lateral area perpendicular to
µ n 2r i di = 0 (2 π r l) = 2 dt
µ 0 π n 2 r 2li
di dt
Ta k in g A cross to be the cross-sectional area perpendicular to B , the induced voltage between the ends of the inductor, which has N = n l turn s, is is dB di di ∆V = ε = N A cross = n l µ 0n ( π r 2 ) = µ 0 π n 2r 2l dt dt dt
and it is observed that
*34.56
(a)
P
= ( ∆V ) i
= IA = 1340 W2 [π (100 m )2 ] = 4.21 × 107 m
The power inci incident dent on on the mirror mirror is: is:
P I
The power reflected through the atmosphere is
P R
=
P R
=
3.14 × 107 W
=
= 0.746 .746( 4.21 4.21 × 107
W
)=
3.14 × 107 W
0.62 0.625 5 W/ m 2
(b)
S
(c)
Noon sunshine in Saint Saint Peter Petersburg sburg produ ces ces this this power-per-area power-per-area on a horizontal horizontal surface: surface: P N
A
(
π 4.00 × 103 m
= 0. 74 746(1340 W /
)
2
)
m 2 sin 7.00 7.00° = 122 W / m 2
The radiation intensity received from the mirror is
0. 62 625 W m 2 100% = 122 W m 2
0.51 0.513% 3%
of that from the the noon Sun Sun in Janu Janu ary.
W
314 Chapter 34 Solutions
1
34.57
2
u=2
e0 E m ax
*34.58
2u
=
E m ax
e0
(Equation 34.21)
= 95.1 95.1 mV/ m
The area over which we model the antenna as radiating is the lateral surface of a cylinder, A
= 2 π r l = 2 π ( 4.00 × 10−2 =
P
(a )
Th e in in t en en s it it y is th th e n :
S
(b)
Th e st aan n d a rd rd is is:
0.570
A
=
0.600 W −2 2 2.51 × 10 m
mW cm
2
= 0.570
=
)
m (0.100 m ) = 2.51 × 10
23. 9 W m
(a) Bmax =
k =
2
m W 1.00 × 10 −3 W 1.00 × 10 4 cm 2 cm
2
1.00 m W
While it is on, the telephone is over the standard by
34.59
−2 m 2
23. 9 W m 2 5. 70 70 W m 2
1.00 m
=
2
= 5.70
4.19 time s
E max 175 175 V/ V/ m max = = 5.83 × 10–7 T 8 c 3.00 × 10 m / s
2π
λ
=
2π = 419 419 rad/ m (0.0150 (0.0150 m )
ω = kc = 1.26 × 1011 r a d / s Since S is along x , and E is along y , B must be be in the z direction .
(b)
S av =
(c)
Pr =
(d)
a=
E ma ma x Bma x = 40.6 40.6 W/ m 2 2 µ 0
2S = 2.71 × 10 –7 N / m 2 c
ΣF m
=
PA m
− 2.71 × 10 7 N / m 2 )(0.750 m 2 ) ( = = 0.500 kg
4.06 × 10
−7 m / s 2
(That is S
∝ E × B.)
W m2
Chapter 34 Solutions 315
*34.60
(a )
(b)
A t s te te ad ad y -s -s ta ta te te ,
P in
= Pou t
W
0.90 0.900 0 1000 1000
or
900 W m 2 T = −8 0.700 .700( 5.67 × 10 W
m
m 2 ⋅ K4 )
W
P=
F =
W m
2
⋅K
4
= eσ A T 4 .
A T 4
14
=
388 K = 115° 115°C C
A sin sin 50.0 50.0° = (0.700) 5.67 × 10 −8
W m
2
⋅K
4
A T 4
14
=
363 K = 90.0 90.0 °C
F I = A c IA c
=
P
c
=
100 J / s 8
3.00 × 10 m / s
a = 3.03 × 10–9 m / s 2
t =
m
2
(900 W m 2 ) sin sin 50.0 50.0° T = 0.700 .700( 5.67 × 10 −8 W m 2 ⋅ K 4 )
or
(b)
2
P ou t
The box box of horizontal horizontal area area A , presents projected area A sin 50.0° perpendicular to the sunlight. Then by the same same reasoning,
(a)
A = (0.700) 5.67 × 10−8
Thus,
0.900 .900 100 1000
34.61
and the p ower radiated radiated ou t is
2 x a
= 8.12 × 10 4 s =
= 3.33 × 10−7
an d
N
= (110 kg )a 1
x = 2 at 2
22.6 h
0 = (1 (107 k g) g)v – (3.00 (3.00 kg)(12.0 kg)(12.0 m/ s – v ) = (107 kg)v – 36.0 36.0 kg · m/ s + (3.00 (3.00 kg)v
v=
36.0 = 0.32 0.327 7 m/ s 110
t = 30.6 s
316 Chapter 34 Solutions
Goal Solution An astronaut, stranded in space 10.0 10.0 m from his spacecraf spacecraftt and at rest relative to it, has a m ass (includ ing equ ipm ent) of 110 kg. Since Since he has a 100100-W W light light sou rce that forms a d irected beam, h e d ecides ecides to u se th e beam as a photon rocket to prop el himself continu ously toward th e spacecraft spacecraft.. (a) Calculate Calculate how long it takes him to reach reach the spacecraf spacecraftt by this method . (b) Sup pose, instead, he decides decides to throw the light sour ce aw ay in a direction opp osite the spacecraft. spacecraft. If the ma ss of the light source has a mass of 3.00 3.00 kg an d , after after being being throw n, m oves at 12.0 12.0 m/ s relative to the recoiling astronaut , how long does it take for the astronaut to reach the spacecraft? spacecraft? G : Based on our everyday experience, the force exerted by photons is too small to feel, so it may take a very long time time (maybe days!) days!) for the astronaut to travel 10 m with his “p hoton rocket.” rocket.” Using th e mom entum of the throw n light seems like a better better solution, but it will still still take a w hile (maybe a few minu tes) for the astronaut to reach the spacecraf spacecraftt because because his mass is so so mu ch larger than the mass of the light source. O : In part (a), the radiation pressure can be used to find the force that accelerates the astronaut toward the spacecraft. spacecraft. In part (b), (b), the principle of conservation conservation of momentum can be app lied lied to find find the tim e required to travel the 10 m.
Light exerts exerts on the astronau astronau t a pressure A : (a) Light
By N ewton’s 2nd 2nd law,
P=F A F =
SA
a=
F
=
c
m
=S
=
=
2 x
8
3.00 × 10 m / s
−7 N 110 kg
= 21 at 2 ,
=
a
100 J / s
3.33 × 10
This acceleration is constant, so the distance traveled is x t =
force of c , and a force
= 3.33 × 10−7
= 3.03 × 10−9
N
m / s2
and the amoun t of time it travels is
2(10.0 m ) 3.03 × 10 −9 m / s 2
= 8.12 × 10 4 s = 22.6 22.6 h
(b) Because ecause there are no external external forc forces, es, the mom entu m of the astronaut astronaut before before throwing the light is the the same as afterward afterward s when the now 107107-kg kg astronaut is mo ving at speed v towards the spacecraft and the light light is is moving moving away from from the spacec spacecraf raftt at (12.0 12.0 m / s − v ) . Thus, p i = p f gives 0 = (107 kg )v − ( 3.00 kg )(12. 0 m / s − v ) 0 = (107 kg )v − ( 36.0 kg ⋅ m / s ) + ( 3.00 kg )v v
=
t = L:
36.0 110
x v
=
= 0. 32 327 m / 10.0 10.0 m
0. 32 327 m / s
s
= 30.6 30.6 s
Throwing the light away is certainly a more expedient way to reach the spacecraft, but there is not mu ch chance chance of retrieving retrieving the lamp lamp u nless it has a very long cord. cord. How long w ould the cord need to be, and does its length dep end on how hard th e astronaut throw s the lamp? (You (You shou ld v erify erify th at the minimum cord length is 367 m, independent of the speed that the lamp is thrown.)
Chapter 34 Solutions 317
The 38.0% of the intensity S
34.62
= 1340
W m
2
that is reflected exerts a pressure
The absorbed light exerts pressure
2Sr
2 0.380 .380)S = (
P1
=
P2
= Sa = (
c
c
0.620)S
c
Altogether the pressure at the subsolar point on Earth is
(a)
(b)
1.38S 1.38(1 1.38(1340 340 W/ m 2) Ptot = P1 + P2 = = = 6.16 × 10– 6 Pa 8 c 3.00 × 10 m / s Pa Ptot
=
1.01 × 10 5 N / m 2 6.16 × 10 −6 N / m 2
=
1.64
× 1010 times smaller smaller than atm ospheric pressure
2
Think of light light going up and being absorbed absorbed by the bead w hich hich p resents a face face area π r b .
34.63
The light pressure is P =
S I = . c c
2
(a)
I π π r b 4 3 F l = = m g = ρ π r b g c 3
(b)
P
I =
an d
= IA = (8.32 × 10 7 W/ m 2 )π (2.00
× 10–3 m )2 =
1/ 3
4ρ gc 3m
4 πρ
3
=
8.32 × 10 7 W / m 2
1.05 kW
2
Think of light light going up and being absorbed absorbed by the bead wh ich ich p resents face face area area π r b .
34.64
If we take the bead to be perfectly absorbing, the light pressure is P =
(a)
F l = F g
so
I =
F lc A
=
F g c A
From the definition of density,
=
(b)
P
= IA =
π r 24ρ gc gc 3 m 3
π
1/ 3 4π ρ
3m
F l
c
A
π r b2 m V
2/ 3
Substituting for r b , I =
c
I
= =
m gc
ρ =
m gc 4 πρ
S av
=
m 4 3
π r b3 2/ 3
1
so
r b
1/ 3
4ρ m = gc 3 π
=
1/ 3
= ( 43 π ρ m )
4ρ gc gc 3 m 3
1/ 3
4 π ρ
c
318 Chapter 34 Solutions
The mirror intercepts power
34.65
P
= I 1 A 1 = (1.00 × 10 3 W/ m 2)π (0.500 m)2 = 785 W
In the image,
=
P
=
(a)
I 2
(b)
E max I 2 = 2 µ 0 c
A 2
785 W
π (0.0200 m )
=
2
6 25 25 k W / m 2
2
Bmax =
(c)
1/ 2 = [2(4π × 10–7)(3.00 E ma ma x = (2 µ 0c I 2 )
so
× 108)(6.25 × 105)]1/ 2=
21.7 21.7 kN/ C
E max max = 72.4 µ T c
0.400 P t = m c ∆ T 0.400(785 W)t = (1.00 kg) 4186
J (100°C – 20.0°C) kg · C°
3.35 × 105 J = 1.07 × 103 s = 17.8 m in t = 314 W
34.66
c
3.00 × 10 8 m s 20.0 × 10 9 s −1
(a)
λ =
(b)
U = P ( ∆t ) = 25.0 × 10 3
(c)
u av
= U =
u av
= 7.37 × 10− 3
(d)
(e)
f
=
U
(π r 2 )
V
2u av
E m ax
=
Bm ax
= E m ax =
F = PA
e0
c
=
l
=
=
1.50 cm
J
− − 1.00 × 10 9 s ) = 25.0 × 10 6 J = ( s U
(π r 2 )c(∆t )
J m3
=
=
25.0 × 10 −6 J
π (0.0600 m ) 3.00 × 108 m s 1.00 × 10 −9 s 2
(
)
4.08 × 10 4 V m 3.00 × 10
(
)(
)
7. 37 37 m J m 3
− 2 7. 37 37 × 10 3 J m 3 − 8.85 × 10 12 C 2 N ⋅ m 2
8
25.0 µ J
m s
= 4.08 × 10 4
= 1.36 × 10− 4
V m
=
40.8 40.8 kV/ m
T = 136 µ T
S cu av A = u A = 7.37 × 10− 3 A = = av c c
J m
3
π (0.0600 m )2 = 8.33 × 10− 5
N
=
83.3 µ N
Chapter 34 Solutions 319
34.67
(a)
(b)
On the right right side side of the equatio equation, n,
F = ma = qE
(C 2
a=
or
The radiated power is then:
(c)
(
C2 m s2
v 2 = qvB r
F = mar = m
P
qE m
)
2
)
N ⋅ m 2 (m s )
3
9.11 × 10 − 31 kg 2
v
so
34.69
F = PA
c
A c
= SA = (
2 π r lc
P / A
c
Therefore,
*34.70
P
A ) A
c
=
P
=
6.00 × 1023 R 2 4 π R
R
=
W
2
8
2π (0.0500 (0.0500 m )(1 )(1.00 m )(3.00 )(3.00 × 10 m \ s )
l = Pl , 2 2c
τ = F
16 π σ T 4
m s2
1.80 × 10
−24 W
(
e I in π r 2
2
2
=
6.37 × 10–7 Pa
τ = κ θ
and
perp end icular icular to the starlight. starlight.
=
1. 75 75 × 10 − 27 W
2
− 1.60 × 10 19 ) ( 5.62 × 1014 ) ( q a = = = 6 π 0 c3 6 π 8.85 × 10 −12 3.00 × 108 3 ( )( )
3.00 × 10– 2 d eg
The plan et, of rad ius r , presents a
It rad iates over area 4 π r 2 .
) = eσ (4π r 2 )T 4
(π r ) = eσ (4π r )T
6.00 × 10 23 W
1.76 × 1013 m s 2
2
2 2
We take R to be the plan et’s et’s distance from its star.
e
s
2
60.0 60.0 W
= P,
= P out :
s
J
= =W
m
3.00 × 10 −3 )(0.0600) ( = = θ = 2 c κ 2( 3. 00 00 × 108 )(1.00 × 10 −11 )
P in
N ⋅m
qBr
Pl
At steady-state,
=
e
c
pr ojected ojected area π r 2
⋅s ⋅m
3
2 −19 q B r (1.60 × 10 ) (0.350) (0.500) v a= = 2 = = 5.62 × 1014 2 r m (1.67 × 10−27 )
The proton then radiates
= Power =
4
1.60 × 10 −19 ) (1.76 × 1013 ) ( = = = 6 π 0 c3 6 π 8.85 × 10 −12 3.00 × 108 3 ( )( ) q2 a2
2 2
S
C
2
e
The p roton accele accelerates rates at
P=
N ⋅ m 2 ⋅ C2 ⋅ m 2 ⋅ s3
1.60 × 10 −19 C )(100 N C ) ( = =
2
34.68
=
4
R 2 T 4 6.00 × 10 23 W = 1 6 π σ R
so that
6.00 × 10 23 W = 4.77 × 10 9 m 4 − 8 2 4 16 π 5.67 × 10 W m ⋅ K ( 310 K )
(
)
= 4.77 Gm
320 Chapter 34 Solutions
2
The light intensity is
34.71
I = Sav =
S
2 µ 0c 2
E
The light pressure is
P=
For the asteroid,
PA = m a
c
=
E
2 µ 0 c
2
= 21 e0 E 2 2
34.72
f = 90.0 MHz,
(a)
λ =
T =
c f
1 f
Bm ax
(b)
=
E ma x
= 2.00 × 10− 3 V /
E m ax c
I =
(d)
I = cuav
(e)
P=
2µ 0 c
2 I
= 6.67 × 10−12 T =
6.67 pT
c
=
u av = 1.77 × 10
(2)(5.31 × 10 − 9 ) 3.00 × 10
8
=
3.54 × 10
− 17 Pa
x − t 3.33 m 11.1 ns
6.67 p T)k cos2 π B = (6.67
− (2.00 × 10 3 )2 − = 5.31 × 10 9 W / m 2 − 7 8 2(4 π × 10 )(3.00 × 10 ) so
=
2m
s = 11.1 11.1 ns
x − t j 11.1 ns 3.33 m 11.1
(c)
e0 E A
m = 20 200 m V/ V/ m
00 m V / m ) co s 2 π E = (2. 00
2 E m ax
a=
3.33 m
= 1.11 × 10− 8
=
an d
− 17 J / m 3