Download chapter 24: electromagnetic waves

College Physics Student Solutions Manual Chapter 24 CHAPTER 24: ELECTROMAGNETIC WAVES 24.1 MAXWELL’S EQUATIONS: ELECTROMAGNETIC WAVES PREDICTED AND OBSERVED 1. Verify that the correct value for the speed of light c is obtained when numerical values for the permeability and permittivity of free space ( µ 0 and ε 0 ) are entered into 1
the equation c =
. µ0ε 0
Solution We know that µ 0 = 4π × 10 −7 T ⋅ m / A, and from Section 19.1, we know that ε 0 = 8.85 × 10 -12 F/m, so that the equation becomes: 1
c=
(4π ×10
−7
)
T ⋅ m / A (8.8542 ×10
-12
F/m)
= 2.998 ×10 8 m/s = 3.00 ×10 8 m/s The units work as follows: [c] =
1
T ⋅ F/A
=
A
=
T⋅F
C/s
=
(N ⋅ s/C ⋅ m )× C 2 /J
(
)
J⋅m
=
N ⋅ s2
(N ⋅ m )m
N ⋅ s2
=
m2
= m/s s2
24.3 THE ELECTROMAGNETIC SPECTRUM 8. A radio station utilizes frequencies between commercial AM and FM. What is the frequency of a 11.12-­‐m-­‐wavelength channel? Solution Using the equation c = fλ , we can solve for the frequency since we know the speed of light and are given the wavelength; f=
17. c 2.998 ×108 m/s
=
= 2.696 ×10 7 s−1 = 26.96 MHz λ
11.12 m
If the Sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the Sun is 1.50 × 1011 m away? 183 €
College Physics Solution 23. Student Solutions Manual Chapter 24 d
, and since we know the speed of light and the distance from the t
d 1.50 ×1011 m
sun to the earth, we can calculate the time: t = =
= 500 s c 3.00 ×108 m/s
We know that v =
(a) What is the frequency of the 193-­‐nm ultraviolet radiation used in laser eye surgery? (b) Assuming the accuracy with which this EM radiation can ablate the cornea is directly proportional to wavelength, how much more accurate can this UV be than the shortest visible wavelength of light? Solution (a) Using the equation c = fλ we can calculate the frequency given the speed of light and the wavelength of the radiation: f =
c
λ
=
3.00 ×10 8 m/s
= 1.55 ×1015 s −1 = 1.55 ×1015 Hz -7
1.93 ×10 m
λvisible 380 nm
=
= 1.97. λUV
193 nm
In other words, the UV radiation is 97% more accurate than the shortest wavelength of visible light, or almost twice as accurate. (b) The shortest wavelength of visible light is 380 nm, so that: 24.4 ENERGY IN ELECTROMAGNETIC WAVES 31. Solution Find the intensity of an electromagnetic wave having a peak magnetic field strength of 4.00 × 10 −9 T . 2
Using the equation I ave =
(
2
I ave
cB 0
we see that: 2µ 0
)(
cB
3.00 × 10 8 m/s 4.00 × 10 −9 T
= 0 =
2 µ0
2 4π × 10 −7 T ⋅ m/A
(
)
2
)
= 1.91× 10 −3 W/m 2 The units work as follows: [I ] = (m/s)T
2
T ⋅ m/A
=
T ⋅ A (N/A ⋅ m)(A ) N
J/m W
=
=
=
=
s
s
s.m s ⋅ m m 2
184 College Physics 36. Solution Student Solutions Manual Chapter 24 Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1.00 × 1011 V / m for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on a 1.00 - mm2 area? E
= c , we can determine the maximum magnetic field B
strength given the maximum electric field strength: (a) Using the equation E0 1.00 ×1011 N/C
B0 =
=
= 333 T, recalling that 1 V/m = 1 N/C. c
3.00 ×108 m/s
(b) Using the equation I ave =
result from part (a): cε 0 E02
, we can calculate the intensity without using the 2
cε 0 E 02
I=
2
(3.00 × 10
=
8
)(
)(
m/s 8.85 × 10 −12 C 2 /N ⋅ m 1.00 × 1011 N/C
2
2
)
19
= 1.33 × 10 W/m
2
(c) We can get an expression for the power in terms of the intensity: P = IA, and from the equation E = Pt , we can express the energy in terms of the power provided. Since we are told the time of the laser pulse, we can calculate the energy delivered to a 1.00 mm 2 area per pulse: E = PΔt = IAΔt
2
⎛ 1 m ⎞
−9
= 1.328 × 10 W/m 1.00 mm ⎜
⎟ 1.00 × 10 s 1000
mm
⎝
⎠
4
= 1.33 × 10 J = 13.3 kJ
(
40. Solution 19
2
)(
2
)
(
)
Integrated Concepts What capacitance is needed in series with an 800 - µH inductor to form a circuit that radiates a wavelength of 196 m? Using the equation f 0 =
1
2π LC
, we can find the capacitance in terms of the 185 College Physics Student Solutions Manual resonant frequency: C =
Chapter 24 1
. Substituting for the frequency, using the equation 4π Lf 02
2
2
(
λ2
196 m )
= 1.35 ×10 −11 F = 13.5 pF c = fλ gives: C = 2 2 2
−4
8
4π Lc 4π (8.00 ×10 H )(3.00 ×10 m/s)
The units work as follows: [C ] =
44. €
Solution m2
2
H(m/s)
=
s2
s2
s
s
A⋅s C
=
= =
=
= = F H Ω ⋅ s Ω V/A
V
V
Integrated Concepts Electromagnetic radiation from a 5.00-­‐mW laser is concentrated on a 1.00 - mm2 area. (a) What is the intensity in W/m2 ? (b) Suppose a 2.00-­‐nC static charge is in the beam. What is the maximum electric force it experiences? (c) If the static charge moves at 400 m/s, what maximum magnetic force can it feel? €
(a) From the equation I =
P 5.00 × 10 −3 W
P
, we know: I = =
= 5.00 × 10 3 W/m 2 −6
2
A
A 1.00 × 10 m
2
(b) Using the equation I ave =
cε0 E0
, we can solve for the maximum electric field: 2
2I
2(5.00 × 10 3 W/m 2 )
E0 =
=
= 1.94 × 10 3 N/C. 8
−12
2
2
cε 0
(3.00 × 10 m/s)(8.85 × 10 C /N ⋅ m )
So, using the equation E =
(
F
we can calculate the force on a 2.00 nC charges: q
)(
)
F = qE0 = 2.00 × 10 −9 C 1.94 × 10 3 N/C = 3.88 × 10 −6 N E
= c , we can write the maximum B
magnetic force in terms of the electric field, since the electric and magnetic fields are related for electromagnetic radiation: (c) Using the equations F = qvB sin θ and FB ,max
(
)
(
)
qvE0
2.00 × 10 −9 C (400 m/s) 1.94 × 10 3 N/C
= qvB0 =
=
= 5.18 × 10 −12 N 8
c
3.00 × 10 m/s
So the electric force is approximately 6 orders of magnitude stronger than the magnetic force. 186 College Physics 50. Solution Student Solutions Manual Unreasonable Results An LC circuit containing a 1.00-­‐pF capacitor oscillates at such a frequency that it radiates at a 300-­‐nm wavelength. (a) What is the inductance of the circuit? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent? (a) Using the equations f 0 =
f =
L=
c
λ
=
1
2π LC
λ2
2
4π Cc
1
2π LC
and c = fλ , we can solve for the inductance: , so that −7
2
2
(3.00 ×10 m)
=
4π (1.00 ×10 F)(3.00 ×10 m/s)
2
−12
8
(b) This inductance is unreasonably small. (c) The wavelength is too small. Chapter 24 187 2
= 2.53 ×10 −20 H