Problem 8 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \( 375 \mathrm{V} / \mathrm{m} \).

Step 2: 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: \( B_0 = \frac{375}{3 \times 10^8} = 1.25 \times 10^{-6} \mathrm{T} \).

Step 3: 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 \( f = \frac{\omega}{2\pi} \): \( f = \frac{5.97 \times 10^{15}}{2\pi} \approx 9.51 \times 10^{14} \mathrm{Hz} \).

Step 4: 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 \( \lambda = \frac{2\pi}{k} \): \( \lambda = \frac{2\pi}{1.99 \times 10^7} \approx 3.16 \times 10^{-7} \mathrm{m} \).

Step 5: 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 \( T = \frac{1}{9.51 \times 10^{14}} \approx 1.05 \times 10^{-15} \mathrm{s} \).

Step 6: 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.

Step 7: 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.

Key Concepts

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.