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Chapter 32: Problem 10

The electric field of a sinusoidal electromagnetic wave obeys the equation \(E = (375 V/m)\) cos [(1.99 \(\times\) 107 rad/m)x + (5.97 \(\times\) 10\(^{15}\) rad/s)\(t\)]. (a) What is the speed of the wave? (b) What are the amplitudes of the electric and magnetic fields of this wave? (c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?

Short Answer

Expert verified
Wave speed: 3.00x10^8 m/s. Amplitudes: 375 V/m (electric), 1.25x10^-6 T (magnetic). Frequency: 9.51x10^14 Hz. Wavelength: 316 nm. Period: 1.05x10^-15 s. Not visible.

Step by step solution

01

Identify the Wave Properties

The electric field equation is given by:\[E = (375 \text{ V/m}) \cos [(1.99 \times 10^7 \text{ rad/m}) x + (5.97 \times 10^{15} \text{ rad/s}) t].\]This follows the general form \( E = E_0 \cos (kx - \omega t) \), where \( k = 1.99 \times 10^7 \text{ rad/m} \) and \( \omega = 5.97 \times 10^{15} \text{ rad/s} \), representing the wave number and the angular frequency.
02

Calculate the Wave Speed

The speed \( v \) of the wave is given by \( v = \frac{\omega}{k} \). Substituting the given values:\[v = \frac{5.97 \times 10^{15} \text{ rad/s}}{1.99 \times 10^7 \text{ rad/m}} = 3.00 \times 10^8 \text{ m/s}.\]This is the speed of light, confirming the wave is an electromagnetic wave.
03

Determine Amplitudes of Electric and Magnetic Fields

The amplitude of the electric field is directly given as \( E_0 = 375 \text{ V/m} \). The amplitude of the magnetic field \( B_0 \) can be calculated using the relation \( c = \frac{E_0}{B_0} \), where \( c \) is the speed of light:\[B_0 = \frac{E_0}{c} = \frac{375 \text{ V/m}}{3.00 \times 10^8 \text{ m/s}} = 1.25 \times 10^{-6} \text{ T}.\]
04

Calculate Frequency, Wavelength, and Period

The frequency \( f \) is related to the angular frequency by \( \omega = 2\pi f \):\[f = \frac{\omega}{2\pi} = \frac{5.97 \times 10^{15} \text{ rad/s}}{2\pi} \approx 9.51 \times 10^{14} \text{ Hz}.\]The wavelength \( \lambda \) is given by \( \lambda = \frac{2\pi}{k} \):\[\lambda = \frac{2\pi}{1.99 \times 10^7 \text{ rad/m}} \approx 3.16 \times 10^{-7} \text{ m} \text{ or } 316 \text{ nm}.\]The period \( T \) is the reciprocal of frequency \( T = \frac{1}{f} \):\[T \approx \frac{1}{9.51 \times 10^{14} \text{ Hz}} \approx 1.05 \times 10^{-15} \text{ s}.\]
05

Determine If the Light is Visible

Visible light typically ranges from about 400 nm to 700 nm. Since the wavelength of our wave is 316 nm, it falls in the ultraviolet range and is not visible to the human eye.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed
The speed of a wave is determined by the relationship between its angular frequency and wave number. For electromagnetic waves, like the one described, the wave speed is crucial for understanding how the wave propagates through space. It is represented by the formula \( v = \frac{\omega}{k} \), where \( \omega \) is the angular frequency in radians per second and \( k \) is the wave number in radians per meter.
In the example given, substituting \( \omega = 5.97 \times 10^{15} \text{ rad/s} \) and \( k = 1.99 \times 10^7 \text{ rad/m} \) into the formula gives a wave speed \( v = 3.00 \times 10^8 \text{ m/s} \).
  • This calculated speed is exactly the speed of light, which confirms that the wave is an electromagnetic wave.
  • Understanding wave speed is fundamental, as it indicates how quickly energy or information carried by the wave travels across space.
Electric and Magnetic Fields
Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. The electric field's amplitude is given as 375 V/m for the wave under consideration. Understanding the magnitude of these fields is essential for applications in technologies like telecommunications and medical imaging.
The amplitude of the magnetic field, denoted as \( B_0 \), can be found using the relationship between the speed of light \( c \), the electric field \( E_0 \), and magnetic field \( B_0 \) with the equation \( c = \frac{E_0}{B_0} \).
  • Substituting the speed of light and the given electric field into the equation gives the magnetic field amplitude \( B_0 = 1.25 \times 10^{-6} \text{ T} \).
  • This shows the interdependence of electric and magnetic fields in an electromagnetic wave.
Frequency and Wavelength
The frequency and wavelength of a wave are two fundamental characteristics that determine its behavior and how it interacts with other matter and waves. For electromagnetic waves, frequency \( f \) refers to the number of oscillations that occur per second. It can be calculated from the angular frequency \( \omega \) using the formula \( f = \frac{\omega}{2\pi} \).
In this scenario, the frequency is approximately \( 9.51 \times 10^{14} \text{ Hz} \). The wavelength \( \lambda \), which represents the distance over which the wave's shape repeats, is determined from the wave number \( k \) by \( \lambda = \frac{2\pi}{k} \). The calculated wavelength is around 316 nm.
  • High frequency and short wavelength waves carry more energy, which is evident in this wave due to its extremely high frequency and short wavelength.
  • Such properties are essential in understanding light's interactions, such as absorption and scattering.
Visible Light Spectrum
The visible light spectrum is a portion of the electromagnetic spectrum that is visible to the human eye, ranging from about 400 nm to 700 nm. This range is critical in various technologies like cameras and displays, where perception of light comes into play.
In this exercise, the calculated wavelength of the wave is 316 nm, classifying it as ultraviolet (UV) light, which lies outside the visible spectrum.
  • UV light, although not visible, is essential in processes like sterilization and fluorescence testing.
  • Understanding the spectra of light is crucial in fields ranging from physics to biology, as it affects how organisms perceive their environment and how scientific instruments are designed.

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Most popular questions from this chapter

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electricfield vector \(\vec{E}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (\(Hint\): Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portions back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 \(\mu\)m in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34. (a) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? (b) What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0 cm longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Consider each of the following electric- and magneticfield orientations. In each case, what is the direction of propagation of the wave? (a) \(\vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{\jmath}\); (b) \(\vec{E} = E\hat{\jmath}\), \(\vec{B} = B\hat{\imath}\); (c) \(\vec{E} = -E\hat{k}\) , \(\vec{B} = -B\hat{\imath}\); (d) \(vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{k}\).

An electromagnetic wave with frequency 5.70 \(\times\) 10\(^{14}\) Hz propagates with a speed of 2.17 \(\times\) 10\(^8\) m/s in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.
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