/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A sinusoidal electromagnetic wav... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 32: Problem 7

A sinusoidal electromagnetic wave of frequency \(6.10 \times 10^{14} \mathrm{Hz}\) travels in vacuum in the \(+z\) -direction. The \(\overrightarrow{\boldsymbol{B}}\) -field is parallel to the \(y\) -axis and has amplitude \(5.80 \times 10^{-4}\) T. Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t) .\)

Short Answer

Expert verified
The electric field is \(\overrightarrow{\boldsymbol{E}}(z, t) = (1.74 \times 10^{5}) \sin(1.28 \times 10^{7}z - 3.83 \times 10^{15}t) \hat{i}\) and the magnetic field is \(\overrightarrow{\boldsymbol{B}}(z, t) = (5.80 \times 10^{-4}) \sin(1.28 \times 10^{7}z - 3.83 \times 10^{15}t) \hat{j}\)."

Step by step solution

01

Identify the Wave Characteristics

The electromagnetic wave is sinusoidal and traveling in the +z-direction. We know that the magnetic field (\(\overrightarrow{\boldsymbol{B}}\)) is parallel to the y-axis, and it has an amplitude of \(5.80 \times 10^{-4}\) T. The frequency of the wave is given as \(6.10 \times 10^{14}\, \mathrm{Hz}.\) From this, we need to find the expressions for both the electric field (\(\overrightarrow{\boldsymbol{E}}(z, t)\)) and the magnetic field (\(\overrightarrow{\boldsymbol{B}}(z, t)\)).
02

Find the Angular Frequency and Wavelength

The angular frequency \(\omega\) is calculated using the formula \(\omega = 2\pi f\), where \(f = 6.10 \times 10^{14}\, \mathrm{Hz}.\) Thus, \(\omega = 2\pi \times 6.10 \times 10^{14}\, \mathrm{Hz} = 3.83 \times 10^{15}\, \mathrm{rad/s}.\) The wavelength \(\lambda\) is found using the speed of light in vacuum, \(c = 3.00 \times 10^{8}\, \mathrm{m/s}.\) The relation is \(\lambda = \frac{c}{f}\). Substituting the values, \(\lambda = \frac{3.00 \times 10^{8}\, \mathrm{m/s}}{6.10 \times 10^{14}\, \mathrm{Hz}} = 4.92 \times 10^{-7}\, \mathrm{m}.\)
03

Write the Magnetic Field Equation

The magnetic field vector \(\overrightarrow{\boldsymbol{B}}(z, t)\) is parallel to the y-axis and can be expressed as: \[\overrightarrow{\boldsymbol{B}}(z, t) = B_0 \sin(kz - \omega t + \phi_B) \hat{j}\]where \(B_0 = 5.80 \times 10^{-4}\, \mathrm{T}\) is the amplitude, \(k = \frac{2\pi}{\lambda} = \frac{2\pi}{4.92 \times 10^{-7}\, \mathrm{m}} = 1.28 \times 10^{7}\, \mathrm{m}^{-1}\) is the wave number, and assuming zero phase constant \(\phi_B = 0\). Thus, \[\overrightarrow{\boldsymbol{B}}(z, t) = (5.80 \times 10^{-4}) \sin(1.28 \times 10^{7}z - 3.83 \times 10^{15}t) \hat{j}\]
04

Use the Relationship Between E and B to Find the Electric Field

In a vacuum, the magnitudes of \(\overrightarrow{\boldsymbol{E}}\) and \(\overrightarrow{\boldsymbol{B}}\) are related by \(c = \frac{E_0}{B_0}\). Therefore, \[E_0 = c \cdot B_0 = (3.00 \times 10^{8}\, \mathrm{m/s}) \cdot (5.80 \times 10^{-4}\, \mathrm{T}) = 1.74 \times 10^{5}\, \mathrm{V/m}.\] The direction of \(\overrightarrow{\boldsymbol{E}}\) is perpendicular to the direction of \(\overrightarrow{\boldsymbol{B}}\) and the direction of wave propagation. Thus, \(\overrightarrow{\boldsymbol{E}}\) is parallel to the x-axis.
05

Write the Electric Field Equation

The electric field vector \(\overrightarrow{\boldsymbol{E}}(z, t)\) can be expressed as: \[\overrightarrow{\boldsymbol{E}}(z, t) = E_0 \sin(kz - \omega t + \phi_E) \hat{i}\]where \(E_0 = 1.74 \times 10^{5}\, \mathrm{V/m}\) is the amplitude, the wave number \(k\) and angular frequency \(\omega\) are the same as those for \(\overrightarrow{\boldsymbol{B}}(z, t)\). Assuming zero phase constant \(\phi_E = 0\), \[\overrightarrow{\boldsymbol{E}}(z, t) = (1.74 \times 10^{5}) \sin(1.28 \times 10^{7}z - 3.83 \times 10^{15}t) \hat{i}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Sinusoidal Waves
A sinusoidal wave is a type of wave that oscillates in intensity and is represented by the mathematical function sinus or cosine. In the electromagnetic spectrum, these waves can be found in light, radio waves, and more.
Sinusoidal waves are characterized by their smooth, repetitive oscillations, which are similar to the motion of a sine curve. These waves have peaks and troughs, where the peak is the maximum displacement from its equilibrium position and the trough is the lowest.
  • Think of a sinusoidal wave in terms of its amplitude, frequency, and wavelength, each critical in defining its behavior.
  • The frequency refers to how many cycles occur in a second, usually measured in Hertz (Hz).
  • The wavelength is the distance between consecutive peaks or troughs.
  • Amplitude is the wave's height from the center line to a peak. In electromagnetic waves, amplitude is related to the strength of the electric or magnetic field.
Sinusoidal waves are central in understanding the propagation of electromagnetic waves because they neatly illustrate how fields vary in space and time.
Wave Number
The wave number is an essential concept when studying waves, particularly electromagnetic ones. Denoted as \( k \), the wave number is a measure of how many wave cycles fit into a unit distance. It describes the spatial frequency of a wave, basically how `crammed' the waves are into a given space.
Wave number \( k \) is calculated using the formula \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength of the wave. The wave number gives insight into the wave's spatial variation and is expressed in radians per meter (rad/m).
  • Smaller wavelengths result in larger wave numbers, meaning more cycles fit into a given distance.
  • The wave number helps in determining how rapidly the wave oscillates in space鈥攚hich is opposite to frequency, that describes time oscillations.
  • In electromagnetic waves, knowing the wave number is crucial to understand how the wave interacts with different materials.
The wave number serves an important role, especially when creating equations for wave propagation through different mediums.
Angular Frequency
Angular frequency is another vital aspect of wave characteristics, denoted as \( \omega \). It describes how quickly the wave oscillates with time.
Given in radians per second, it provides an understanding of how fast these oscillations occur. To calculate angular frequency, use the formula \( \omega = 2\pi f \), where \( f \) is the frequency in Hertz.
  • Higher angular frequencies indicate that the wave oscillates more rapidly.
  • It represents the rate of rotation in a circular motion if that circular path was the wave.
  • Angular frequency is particularly important in electromagnetism because it also relates to the energy of photons in a given wave.
Understanding angular frequency is key to grasping wave dynamics and energy distribution in electromagnetic waves.
Electric and Magnetic Fields Equations
In electromagnetic waves, electric and magnetic fields are interlinked and form sinusoidal wave patterns.
The electromagnetic equations for these fields are used to predict wave behavior, and they reflect how fields vary over time and distance.
  • The electric field \( \overrightarrow{E}(z, t) \) and magnetic field \( \overrightarrow{B}(z, t) \) are expressed in sinusoidal forms such as \[ \overrightarrow{E}(z, t) = E_0 \sin(kz - \omega t) \hat{i} \]
  • Correspondingly, the magnetic field is \[ \overrightarrow{B}(z, t) = B_0 \sin(kz - \omega t) \hat{j} \]
  • Here, \( E_0 \) and \( B_0 \) are the amplitudes, and \( kz - \omega t \) describes wave displacement.
In these equations, the wave number \( k \) and angular frequency \( \omega \) identify how these fields oscillate in space and time. The electric and magnetic fields are always perpendicular to each other and the direction of wave propagation, following the right-hand rule.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An intense light source radiates uniformly in all directions. At a distance of 5.0 \(\mathrm{m}\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10^{-6} \mathrm{Pa}\) . What is the total average power output of the source?

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\) (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M )\) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r .\) The relevant area is the cross-sectional area of the particle, \(n o t\) the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000 \(\mathrm{kg} / \mathrm{m}^{3}\) . Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{W}\) . Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

In the \(25-\) fil Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 \(\mathrm{W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

A space probe \(2.0 \times 10^{10} \mathrm{m}\) from a star measures the total intensity of electrumagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{W} / \mathrm{m}^{2} .\) If the star radiates uniformly in all directions, what is its total average power output?

In a TV picture, ghost images are formed when the signal from the transmitter travels to the receiver both directly and indirectly after reflection from a building or other large metallic mass. In a 25 -inch set, the ghost is about 1.0 \(\mathrm{cm}\) to the right of the principal image if the reflected signal arrives 0.60\(\mu\) s after the principal signal. In this case, what is the difference in path lengths for the two signals?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Translational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Rotational Dynamics

Read Explanation

Force

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.