30. Eq. 33-27 suggests that the slope in an intensity versus inverse-square-distance graph
(I plotted versus r 2 ) is P/4. We estimate the slope to be about 20 (in SI units) which
means the power is P = 4(30) 2.5 102 W.
29. (a) The upward force supplied by radiation pressure in this case (Eq. 33-32) must be
equal to the magnitude of the pull of gravity (mg). For a sphere, the projected area
(which is a factor in Eq. 33-32) is that of a circle A = r2 (not the entire surfa
28. The mass of the cylinder is m = ( D 2 / 4) H , where D is the diameter of the cylinder.
Since it is in equilibrium
Fnet = mg Fr =
HD 2 g D 2 2 I
= 0.
4
4 c
We solve for H:
H=
=
2I 2P 1
=
gc D 2 / 4 gc
2(4.60W)
[(2.6010 m) / 4](9.8m/s 2 )(3.0 108
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
26. We require Fgrav = Fr or G and solve for the area A: cGmM s (6.67 1011 N m2 / kg 2 )(1500 kg)(1.99 1030 kg)(2.998 108 m / s) A= = 2 . . 2 Id es 2(140 103 W / m2 )(150 1011 m) 2 = 9.5 105 m2 = 0.95 km2 . mM s 2 IA = , 2 d es c
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
24. (a) We note that the cross section area of the beam is d 2/4, where d is the diameter
of the spot (d = 2.00). The beam intensity is
I=
P
d 2 / 4
=
5.00 103 W
b gc
9
h
2.00 633 10 m
2
= 3.97 109 W / m2 .
/4
(b) The radiation pressure is
pr =
I 3.97 10
23. Let f be the fraction of the incident beam intensity that is reflected. The fraction
absorbed is 1 f. The reflected portion exerts a radiation pressure of
pr =
2 f I0
c
and the absorbed portion exerts a radiation pressure of
pa =
(1 f ) I 0
,
c
where
22. (a) The radiation pressure produces a force equal to
(1.4103 W/m 2 ) ( 6.37 106 m )
I
2
= 6.0 108 N.
Fr = pr ( R ) = ( Re ) =
8
2.998 10 m/s
c
2
2
e
(b) The gravitational pull of the Sun on Earth is
GM s M e ( 6.67 10
=
Fgrav =
2
d es
= 3.6 1022 N,
w
21. The plasma completely reflects all the energy incident on it, so the radiation pressure
is given by pr = 2I/c, where I is the intensity. The intensity is I = P/A, where P is the
power and A is the area intercepted by the radiation. Thus
2 (1.5 109 W )
18. From the equation immediately preceding Eq. 33-12, we see that the maximum value
of B/t is Bm . We can relate Bm to the intensity:
Bm =
2c 0 I
Em
=
,
c
c
and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we
relate to w
19. Since the surface is perfectly absorbing, the radiation pressure is given by pr = I/c,
where I is the intensity. Since the bulb radiates uniformly in all directions, the intensity a
distance r from it is given by I = P/4r2, where P is the power of the
18. From the equation immediately preceding Eq. 33-12, we see that the maximum value
of B/t is Bm . We can relate Bm to the intensity:
Bm =
2c 0 I
Em
=
,
c
c
and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we
relate to w
17. (a) The average rate of energy flow per unit area, or intensity, is related to the electric
2
field amplitude Em by I = Em / 2 0c , so
c
hc
hc
Em = 2 0cI = 2 4 107 H / m 2.998 108 m / s 10 106 W / m2
h
= 8.7 102 V / m.
(b) The amplitude of the magneti
16. (a) The expression Ey = Em sin(kx t) it fits the requirement at point P [it] is
decreasing with time if we imagine P is just to the right (x > 0) of the coordinate origin
(but at a value of x less than /2k = /4 which is where there would be a maximum,
2
15. (a) We use I = Em /20c to calculate Em:
c
hc
hc
h
Em = 2 0 I c = 2 4 107 T m / A 140 103 W / m2 2.998 108 m / s
.
= 103 103 V / m.
.
(b) The magnetic field amplitude is therefore
Bm =
Em 103 104 V / m
.
=
= 3.43 106 T.
8
c
2.998 10 m / s
14. (a) The power received is
Pr = (1.0 10
12
W)
(300 m )2 / 4
4 ( 6.37 10 m )
6
(b) The power of the source would be
c
hc
P = 4 r 2 I = 4 2.2 104 ly 9.46 1015 m / ly
2
= 1.4 1022 W.
LM 10 10 W OP
.
h M 4 .6.37 10 m P = 11 10
hQ
Nc
2
12
15
6
2
W.
13. (a) The magnetic field amplitude of the wave is
Bm =
2.0 V / m
Em
=
= 6.7 109 T.
2.998 108 m / s
c
(b) The intensity is
b
g
2
2.0 V / m
E2
I= m =
= 5.3 103 W / m2 .
8
7
2 0c 2 4 10 T m / A 2.998 10 m / s
c
hc
h
(c) The power of the source is
P = 4r 2
11. The intensity is the average of the Poynting vector:
I = Savg
c
hc
2
3.0 108 m / s 10 104 T
.
cBm
=
=
2
2 0
2 126 106 H / m
.
c
h
h
2
= 12 106 W / m2 .
.
10. (a) The amplitude of the magnetic field in the wave is
Bm =
5.00 V / m
Em
.
=
= 167 108 T.
2.998 108 m / s
c
(b) The intensity is the average of the Poynting vector:
I = Savg
b
g
2
5.00 V / m
E2
= m=
= 3.31 102 W / m2 .
7
8
2 0c 2 4 10 T m / A 2.998 1
27. If the beam carries energy U away from the spaceship, then it also carries momentum
p = U/c away. Since the total momentum of the spaceship and light is conserved, this is
the magnitude of the momentum acquired by the spaceship. If P is the power of t
9. (a) The amplitude of the magnetic field is
Bm =
Em
2.0V/m
=
= 6.67 109 T 6.7 109 T.
8
c 2.998 10 m/s
(b) Since the E -wave oscillates in the z direction and travels in the x direction, we have Bx
= Bz = 0. So, the oscillation of the magnetic field is p
4. (a) The frequency of the radiation is 3.0 108 m / s c f= = = 4.7 103 Hz. 5 6 (10 10 )(6.4 10 m) . (b) The period of the radiation is
T=
1 1 = = 212 s = 3 min 32 s. f 4.7 103 Hz