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
Step by step solution
Identify the Wave Properties
Calculate the Wave Speed
Determine Amplitudes of Electric and Magnetic Fields
Calculate Frequency, Wavelength, and Period
Determine If the Light is Visible
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wave Speed
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
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
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
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.