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Unformatted text preview: traveled by the
electromagnetic wave divided by the speed of light,
t = s/c. The distance s/2 from one city to the satellite is
given by the Pythagorean theorem as (see the drawing) s
=
2 3.5 (3.6 × 10 7 m) 2 + × 10 6 m 2 2 1289 Satellite = 3.6 × 10 7 m The time for the wave to travel from one city up to the
satellite and back to the other is s 2(3.6 × 10 7 m)
= 0.24 s
=
c 3.0 × 10 8 m/s
______________________________________________________________________________
t= 22. REASONING AND SOLUTION According to Equation 16.8, we have
P
P
1.2 × 10−3 W
S=
=
=
= 3.8 × 102 W/m 2
2
2
−3
A πr
π (1.0 × 10 m )
______________________________________________________________________________
23. REASONING AND SOLUTION Using Equations 24.5b and 24.3, we find the following
results: S
=
cε 0 1.23 × 109 W/m 2
=
( 3.00 × 108 m/s ) 8.85 × 10−12 C2 / ( N ⋅ m2 ) a. =
Erms b. Brms = Erms/c = 2.27 × 10 –3 6.81 × 10 N/C
5 T ______________________________________________________________________________
24. REASONING The relationship between the intensity S of an electromagnetic wave and its
electric field E is given by Equation 24.5b as S = cε 0 E 2 . For example, if the magnitude of
the electric field triples, the intensity increases by a factor of 32 = 9.
The magnitude of the magnetic field is given by Equation 24.3 as B = E / c . Even though
the magnitude of the magnetic field is much smaller than that of the electric field, tripling 1290 ELECTROMAGNETIC WAVES the magnetic field also causes the intensity to increase by a factor of 32 = 9. This can be
c2
B.
seen by examining Equation 24.5c, S = µ0 SOLUTION
a. When the magnitude of the electric field is 315 N/C, the intensity of the electromagnetic
wave is
2
S = cε 0 E (24.5b) ( ) ( ) 2
= 108 m/s 8.85 × 10−12 C2 / N ⋅ m 2 ( 315 N/C ) = 2
3.00 ×
263 W/m When the magnitude of the electric field is 945 N/C, the intensity of the electromagnetic
wave is ( ) ( ) 2
2
8
12
2
2
2
S == 10 m/s 8.85 × 10− C / N ⋅ m ( 945 N/C ) =W/m
cε 0 E
3.00 ×
2370 b. The magnitudes of the magnetic fields associated with each electric field are B
= E
315 N/C
−6
=
= 1.05 × 10 T
8
c 3.00 × 10 m/s B
= E
945 N/C
=
= 3.15 × 10−6 T
8
c 3.00 × 10 m/s (24.3) c. The intensities of the waves associated with each value of the magnetic field are
8
c2
3.00 × 10 m/s
S= B =
1.05 × 10−6 T
−7
µ0
4π × 10 T ⋅ m/A ( ) 2 263
= W/m 2 (24.5c) 2
c2
3.00 × 108 m/s
S= B =
3.15 × 10−6 T =
2370 W/m 2
−7
µ0
4π × 10 T ⋅ m/A
______________________________________________________________________________ ( 25. ) SSM REASONING AND SOLUTION
2
a. According to Equation 24.5b, the average intensity is S = cε 0 Erms . In addition, the average intensity is the average power P divided by the area A. Therefore,
=
Erms S
=
cε 0 P
=
cε 0 A 1.20 × 104 W
=
(3.00 × 108 m/s)[8.85 × 10−12C2 /(N ⋅ m 2 )](135 m 2 ) 183 N/C Chapter 24 Problems 1291 b. Then, from Erms = cBrms (Equation 24.3), we have Erms 183 N/C
= 6.10 ×10 –7 T
8
c
3.00×10 m/s
______________________________________________________________________________
Brms = = 26. REASONING The average intensity of a wave is the average power per unit area that
passes perpendicularly through a surface. Thus, the average power P of the wave is the
product of the average intensity S and the area A. The average intensity is related to the
2
rmsvalue Erms of the electric field by S = cε 0 Erms .
2
SOLUTION The average power is P = S A . Since S = cε 0 Erms (Equation 24.5b), we have P
A
= S= 2
( cε 0 Erms ) A ( 3.00
= ×108 m/s ) 8.85 ×10−12 C2 / ( N ⋅ m 2 ) ( 2.0 ×109 N/C ) (1.6 ×10−5 m 2 ) 2 = 1.7 ×1011 W
______________________________________________________________________________
27. SSM REASONING AND SOLUTION The energy is equal to the power P multiplied by
the time t. The power, on the other hand, is equal to product of the intensity S of the wave
and the area A through which the wave passes.
9 Energy = P t = (SA)t = (1390 W/m2)(25 m × 45 m)(3600 s) = 5.6 × 10 J
______________________________________________________________________________
28. REASONING The average intensity S is related to the distance r from the bulb by S = P / ( 4π r 2 ) (Equation 16.9), where P is the average power radiated by the bulb. The intensity of an electromagnetic wave is related to the magnitude E of its electric field by
S = cε 0 E 2 (Equation 24.5b). According to the discussion in Section 24.4, if the intensity is
an average intensity, then the value for the electric field must be an rms value, not a peak
value. The peak value E0 and the rms value Erms are related by Erms = E0 / 2 .
SOLUTION
a. The average intensity of the wave is S
= P
150.0 W
=
= 0.477 W/m 2
2
2
4π r
4π ( 5.00 m ) (16.9) 1292 ELECTROMAGNETIC WAVES b. The average intensity S is related to the rms value Erms of the electric field by
2
S = cε 0 Erms (Equation 24.5b). Solving for the electric field gives Erms
= S
=
cε 0 2 ( 0.477 W/m
= 13.4 N/C
8
3.00 × 10 m/s 8...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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