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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 sinusoidal electromagnetic wave is propagating in vacuum in the +\(z\)-direction. If at a particular instant and at a certain point in space the electric field is in the +\(x\)-direction and has magnitude 4.00 V/m, what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

The energy flow to the earth from sunlight is about 1.4 kW/m\(^2\). (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about 1.5 \(\times\) 10\(^{11}\) m. Find the total power radiated by the sun.

Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is 4.82 \(\times 10^{-11}\) T. Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electricfield amplitude.

The sun emits energy in the form of electromagnetic waves at a rate of 3.9 \(\times\) 10\(^{26}\) W. This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r = R = 6.96 \times 10^5\) km) and at \(r = R/\)2, in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about 1.0 \(\times\) 10\(^4\) Pa; at \(r = R/\)2, the gas pressure is calculated from solar models to be about 4.7 \(\times\) 10$^{13} Pa. Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate \(E_{max}\) and \(B_{max}\) for the 700-nm light at a distance of 5.00 m from the source.

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