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Chapter 23: Problem 8

A sinusoidal electromagnetic wave having a magnetic field of amplitude \(1.25 \mu \mathrm{T}\) and a wavelength of \(432 \mathrm{nm}\) is traveling in the \(+x\) direction through empty space. (a) What is the frequency of this wave? (b) What is the amplitude of the associated electric field? (c) Write the equations for the electric and magnetic fields as functions of \(x\) and \(t\) in the form of Equations 23.3 .

Short Answer

Expert verified
(a) Frequency is approximately \(6.94 \times 10^{14}\) Hz. (b) Electric field amplitude is 375 V/m. (c) Equations are: \(E(x, t) = 375 \cos(1.45 \times 10^7 x - 4.36 \times 10^{15} t)\) V/m and \(B(x, t) = 1.25 \times 10^{-6} \cos(1.45 \times 10^7 x - 4.36 \times 10^{15} t)\) T.

Step by step solution

01

Determine the Frequency of the Wave

The frequency of a wave, \( f \) can be found using the relation between speed of light \( c \), wavelength \( \lambda \), and frequency:\[ f = \frac{c}{\lambda} \]where \( c = 3 \times 10^8 \, \text{m/s} \) and \( \lambda = 432 \, \text{nm} = 432 \times 10^{-9} \, \text{m} \). Calculate \( f \):\[ f = \frac{3 \times 10^8}{432 \times 10^{-9}} = \frac{3 \times 10^8}{4.32 \times 10^{-7}} \]\[ f \approx 6.94 \times 10^{14} \, \text{Hz} \]
02

Find the Amplitude of the Electric Field

Using the relation between the electric field \( E_0 \) and the magnetic field \( B_0 \) amplitudes in an electromagnetic wave:\[ E_0 = cB_0 \]where \( B_0 = 1.25 \, \mu\text{T} = 1.25 \times 10^{-6} \, \text{T} \). Calculate \( E_0 \):\[ E_0 = (3 \times 10^8) \times (1.25 \times 10^{-6}) \] \[ E_0 = 375 \, \text{V/m} \]
03

Write the Equations for the Fields

The general forms of the electric and magnetic fields in a sinusoidal wave traveling in the +x direction are given by:\[ E(x, t) = E_0 \cos(kx - \omega t) \] \[ B(x, t) = B_0 \cos(kx - \omega t) \]where:\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{432 \times 10^{-9}} \approx 1.45 \times 10^7 \, \text{m}^{-1} \]and \( \omega = 2\pi f \approx 2\pi (6.94 \times 10^{14}) \approx 4.36 \times 10^{15} \, \text{rad/s} \). Thus, the equations are:\[ E(x, t) = 375 \cos(1.45 \times 10^7 x - 4.36 \times 10^{15} t) \, \text{V/m} \] \[ B(x, t) = 1.25 \times 10^{-6} \cos(1.45 \times 10^7 x - 4.36 \times 10^{15} t) \, \text{T} \]

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Key Concepts

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

Wave Frequency
Understanding wave frequency is crucial when studying electromagnetic waves. It tells us how many wave crests pass a point per second. Frequency, denoted as \( f \), is directly related to the wave's speed and its wavelength. In mathematical terms, frequency is expressed as:\[ f = \frac{c}{\lambda} \]where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \), and \( \lambda \) represents wavelength.
  • If the wavelength is small, the frequency is high, meaning the wave crests are closer together.
  • A larger wavelength means a lower frequency, with wave crests farther apart.
For instance, in our exercise, the wavelength was \( 432 \, \text{nm} \), which resulted in a high frequency of approximately \( 6.94 \times 10^{14} \, \text{Hz} \). This tells us the wave is oscillating very quickly.
Electric Field Amplitude
The electric field amplitude is a significant feature of electromagnetic waves. It quantifies the wave's ability to exert force on charges. In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of wave propagation.To find the electric field amplitude \( E_0 \), we can use the relationship:\[ E_0 = cB_0 \]where \( B_0 \) is the magnetic field amplitude and \( c \) is again the speed of light.For the given magnetic field in our problem, \( B_0 = 1.25 \, \mu\text{T} \), the corresponding electric field amplitude \( E_0 \) is calculated to be \( 375 \, \text{V/m} \).
  • Increased magnetic field amplitude leads to a proportionally larger electric field amplitude.
  • This relationship shows the interconnected nature of separate components within an electromagnetic wave.
Magnetic Field
The magnetic field component of an electromagnetic wave represents the part that responds to magnetic charges or reconfigurations. It鈥檚 usually weaker than the electric field but still critical in wave mechanics. In our exercise, the magnetic field has an amplitude value of \( 1.25 \, \mu\text{T} \). This small value highlights the subtle but essential role of magnetic fields in electromagnetic waves.
  • Even small magnetic fields can significantly impact when combined with their counterpart electric fields.
  • The interplay between electric and magnetic fields leads to energy transmission through space.
Electromagnetic waves, such as light, shine by their ability to interleave magnetic and electric fields that support each other perfectly.
Wavelength
Wavelength, represented by \( \lambda \), is the physical length of one complete wave cycle. From crest to crest or trough to trough, it dictates the wave's spatial frequency.In our original problem, the wavelength was given as \( 432 \, \text{nm} \), a measure often used in the context of light waves.
  • A key insight from wavelength is its inverse relationship with frequency: shorter wavelengths mean higher frequency, and vice versa.
  • This relationship explains why light waves can have a visible spectrum, where different wavelengths contribute to different colors.
Wavelength is a fundamental property influencing how waves spread, interact with materials, and are absorbed or transmitted.
Wave Equations
Wave equations capture the dynamic nature of electromagnetic waves as they traverse space and time. These equations use trigonometric functions to model how the fields vary over position \( x \) and time \( t \).The general form of wave equations are:\[ E(x, t) = E_0 \cos(kx - \omega t) \]\[ B(x, t) = B_0 \cos(kx - \omega t) \]where:
  • \( k \) is the wave number, calculated by \( k = \frac{2\pi}{\lambda} \).
  • \( \omega \) is the angular frequency, \( \omega = 2\pi f \).
For this specific problem:
  • \( k \approx 1.45 \times 10^7 \, \text{m}^{-1} \)
  • \( \omega \approx 4.36 \times 10^{15} \, \text{rad/s} \)
These equations reveal how field strength changes with both time and position in space, depicting the beautiful interplay of forces in an electromagnetic wave.

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Most popular questions from this chapter

In a physics lab, light with wavelength \(490 \mathrm{nm}\) travels in air from a laser to a photocell in \(17.0 \mathrm{~ns}\). When a slab of glass \(0.840 \mathrm{~m}\) thick is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light \(21.2 \mathrm{~ns}\) to travel from the laser to the photocell. What is the wavelength of the light in the glass?

The speed of light with a wavelength of \(656 \mathrm{nm}\) in heavy flint glass is \(1.82 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is the index of refraction of the glass at this wavelength?

A beaker with a mirrored bottom is filled with a liquid whose index of refraction is 1.63. A light beam strikes the top surface of the liquid at an angle of \(42.5^{\circ}\) from the normal. At what angle from the normal will the beam exit from the liquid after traveling down through it, reflecting from the mirrored bottom, and returning to the surface?

Plane-polarized light passes through two polarizers whose axes are oriented at \(35.0^{\circ}\) to each other. If the intensity of the original beam is reduced to \(15.0 \%,\) what was the polarization direction of the original beam, relative to the first polarizer?

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