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Chapter 32: Problem 8

The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=-(375 \mathrm{V} / \mathrm{m}) \sin \left[\left(5.97 \times 10^{15} \mathrm{rad} / \mathrm{s}\right) t+\right.\) \(\left(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x ]\) . (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?(c) What is the speed of the wave?

Short Answer

Expert verified
(a) 375 V/m, 1.25 渭T; (b) 9.51x10^14 Hz, 316 nm, not visible, 1.05 fs; (c) 3.0x10^8 m/s.

Step by step solution

01

Identify the electric field amplitude

The amplitude of the electric field, denoted as \( E_0 \), is the coefficient in front of the sine function in the given equation. From the equation \( E=-(375 \mathrm{V} / \mathrm{m}) \sin \left[(5.97 \times 10^{15} \mathrm{rad}/\mathrm{s})t + (1.99 \times 10^{7} \mathrm{rad}/\mathrm{m}) x \right] \), the amplitude \( E_0 \) is \( 375 \mathrm{V} / \mathrm{m} \).
02

Calculate the magnetic field amplitude

The amplitudes of the electric and magnetic fields in an electromagnetic wave are related by the speed of light \( c \) using the formula: \( B_0 = \frac{E_0}{c} \). Given \( E_0 = 375 \mathrm{V} / \mathrm{m} \) and \( c = 3 \times 10^8 \mathrm{m/s} \), the magnetic field amplitude \( B_0 \) is calculated as follows: \( B_0 = \frac{375}{3 \times 10^8} = 1.25 \times 10^{-6} \mathrm{T} \).
03

Find the frequency of the wave

The angular frequency \( \omega \) is given as \( 5.97 \times 10^{15} \mathrm{rad/s} \). The frequency \( f \) can be found using the relation \( f = \frac{\omega}{2\pi} \): \( f = \frac{5.97 \times 10^{15}}{2\pi} \approx 9.51 \times 10^{14} \mathrm{Hz} \).
04

Calculate the wavelength

The wave number \( k \) given is \( 1.99 \times 10^7 \mathrm{rad/m} \), and wavelength \( \lambda \) is calculated using the relation \( \lambda = \frac{2\pi}{k} \): \( \lambda = \frac{2\pi}{1.99 \times 10^7} \approx 3.16 \times 10^{-7} \mathrm{m} \).
05

Determine the period of the wave

The period \( T \) of the wave can be found using the relation \( T = \frac{1}{f} \). Using the previously calculated frequency \( f = 9.51 \times 10^{14} \mathrm{Hz} \), we find \( T = \frac{1}{9.51 \times 10^{14}} \approx 1.05 \times 10^{-15} \mathrm{s} \).
06

Assess visibility status of the light

The wavelength \( 3.16 \times 10^{-7} \mathrm{m} \) is equivalent to \( 316 \mathrm{nm} \), which falls in the ultraviolet range. Ultraviolet light is not visible to the human eye.
07

Calculate the speed of the wave

The speed of a wave is given by \( v = \lambda \times f \). Using \( \lambda \approx 3.16 \times 10^{-7} \mathrm{m} \) and \( f \approx 9.51 \times 10^{14} \mathrm{Hz} \), we find \( v = \lambda \times f = \approx 3.00 \times 10^8 \mathrm{m/s} \), which is the speed of light.

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude, denoted by \( E_0 \), is a crucial parameter in describing electromagnetic waves. It represents the maximum strength of the electric field and directly influences how the wave interacts with materials. In the equation provided, \( E = -(375 \, \text{V/m}) \sin[(5.97 \times 10^{15} \, \text{rad/s}) t + (1.99 \times 10^7 \, \text{rad/m}) x] \), the amplitude \( E_0 \) can be easily identified as the coefficient in front of the sine function, which is \( 375 \, \text{V/m} \). This value indicates the peak electric field strength of the wave.
Magnetic Field Amplitude
Electromagnetic waves have both electric and magnetic field components, which oscillate perpendicularly to each other and to the direction of wave propagation. The magnetic field amplitude \( B_0 \) in electromagnetic waves is related to the electric field amplitude \( E_0 \) and the speed of light \( c \) through the equation \( B_0 = \frac{E_0}{c} \). Given \( E_0 = 375 \, \text{V/m} \) and the speed of light \( c = 3 \times 10^8 \, \text{m/s} \), we calculate \( B_0 \) as \( 1.25 \times 10^{-6} \, \text{T} \). This value signifies the maximum strength of the magnetic field associated with the wave.
Wavelength Calculation
The wavelength \( \lambda \) of an electromagnetic wave is the distance over which the wave's shape repeats, which is inversely related to its wave number \( k \). The relationship between wavelength and wave number is given by \( \lambda = \frac{2\pi}{k} \). From the equation we know \( k = 1.99 \times 10^7 \, \text{rad/m} \), thus \( \lambda \) is calculated as \( \frac{2\pi}{1.99 \times 10^7} \), which approximates to \( 3.16 \times 10^{-7} \, \text{m} \) or \( 316 \, \text{nm} \). This wavelength falls within the ultraviolet range, which is not visible to the human eye.
Frequency and Period of Waves
The frequency \( f \) of a wave refers to how many cycles it completes per second, while the period \( T \) is the time taken for one complete cycle. These are linked by the relationship \( f = \frac{1}{T} \). The angular frequency \( \omega \) from the equation is \( 5.97 \times 10^{15} \, \text{rad/s} \), and frequency \( f \) can be found using \( f = \frac{\omega}{2\pi} \). This results in \( f \approx 9.51 \times 10^{14} \, \text{Hz} \). The wave's period \( T \) then is \( \frac{1}{f} \approx 1.05 \times 10^{-15} \, \text{s} \). Such high frequencies and short periods are typical of electromagnetic waves in the ultraviolet spectrum.
Speed of Light
The speed of electromagnetic waves in a vacuum is constant at approximately \( 3.00 \times 10^8 \, \text{m/s} \), often referred to simply as the speed of light \( c \). This speed is a fundamental constant in physics, crucial for calculations involving electromagnetic waves. Using the relation \( v = \lambda \times f \), where \( \lambda \approx 3.16 \times 10^{-7} \, \text{m} \) and \( f \approx 9.51 \times 10^{14} \, \text{Hz} \), the wave speed is confirmed to be \( 3.00 \times 10^8 \, \text{m/s} \). This consistency demonstrates the reliability of using these calculations in understanding electromagnetic wave behavior.

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Most popular questions from this chapter

A simusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric-field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in pliysics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is 2.5 \(\mathrm{km}\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090 \(\mathrm{V} / \mathrm{m}\) and that the frequency of the waves is 244 \(\mathrm{MHz}\) . Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

The intensity of a cylindrical laser beam is 0.800 \(\mathrm{W} / \mathrm{m}^{2}\) The cross-sectional area of the beam is \(3.0 \times 10^{-4} \mathrm{m}^{2}\) and the intensity is uniform across the cross section of the beam. (a) What is the average power output of the laser? (b) What is the rms value of the electric field in the beam?

An electromagnetic wave of wavelength \(435 \mathrm{~nm}\) is traveling in vacuum in the \(-z\) -direction. The electric field has amplitude \(2.70 \times 10^{-3} \mathrm{~V} / \mathrm{m}\) and is parallel to the \(x\) -axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t)\).

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{m} / \mathrm{s}\) 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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