/*! 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 20 (II) An electromagnetic wave has... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 31: Problem 20

(II) An electromagnetic wave has an electric field given by \(\overrightarrow{\mathbf{E}}=\hat{\mathbf{i}}(225 \mathrm{~V} / \mathrm{m}) \sin \left[\left(0.077 \mathrm{~m}^{-1}\right) z-\left(2.3 \times 10^{7} \mathrm{rad} / \mathrm{s}\right) t\right]\) (a) What are the wavelength and frequency of the wave? (b) Write down an expression for the magnetic field.

Short Answer

Expert verified
(a) Wavelength: 81.6 m, Frequency: 3.66 MHz; (b) Magnetic field: \( \overrightarrow{\mathbf{B}} = \hat{\mathbf{j}} (7.5 \times 10^{-7}) \sin(0.077z - 2.3 \times 10^7 t) \).

Step by step solution

01

Convert Wavenumber to Wavelength

The given wave number is \( k = 0.077 \, \text{m}^{-1} \). The relationship between wavenumber \( k \) and wavelength \( \lambda \) is \( k = \frac{2\pi}{\lambda} \). Solve for \( \lambda \) using \( \lambda = \frac{2\pi}{k} \).\( \lambda = \frac{2\pi}{0.077} \approx 81.6 \, \text{m} \).
02

Calculate Frequency from Angular Frequency

The angular frequency is given by \( \omega = 2.3 \times 10^7 \, \text{rad/s} \). Use the relationship between angular frequency \( \omega \) and frequency \( f \), which is \( \omega = 2\pi f \). Solve for \( f \) using \( f = \frac{\omega}{2\pi} \).\( f = \frac{2.3 \times 10^7}{2\pi} \approx 3.66 \times 10^6 \, \text{Hz} \).
03

Determine Magnetic Field Amplitude

The amplitude of the magnetic field \( B_0 \) can be found using the relationship \( E_0 = c B_0 \), where \( c \) is the speed of light \( 3 \times 10^8 \, \text{m/s} \). Solve for \( B_0 \) using \( B_0 = \frac{E_0}{c} \).\( B_0 = \frac{225}{3 \times 10^8} = 7.5 \times 10^{-7} \, \text{T} \).
04

Compute Magnetic Field

The expression for the magnetic field \( \overrightarrow{\mathbf{B}} \) is similar to that of the electric field, with the direction noted to be perpendicular to \( \overrightarrow{\mathbf{E}} \). For a wave traveling in the positive \( z \)-direction and \( \overrightarrow{\mathbf{E}} \) along \( \hat{\mathbf{i}} \), \( \overrightarrow{\mathbf{B}} \) should point along \( \hat{\mathbf{j}} \) (perpendicular to both \( \hat{\mathbf{i}} \) and direction of travel).\( \overrightarrow{\mathbf{B}} = \hat{\mathbf{j}} \left(7.5 \times 10^{-7} \right) \sin \left( 0.077 z - 2.3 \times 10^7 t \right) \).

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.

Wavelength
Wavelength refers to the distance over which the shape of the wave repeats itself, such as from crest to crest or trough to trough. In a mathematical sense, it is represented by the symbol \( \lambda \). For the given electromagnetic wave with a wave number \( k = 0.077 \, \text{m}^{-1} \), we find the wavelength using the formula:
  • The relationship between wave number \( k \) and wavelength \( \lambda \) is \( k = \frac{2\pi}{\lambda} \).
  • Solving for \( \lambda \), we get \( \lambda = \frac{2\pi}{k} \).
  • Substituting the given value, \( \lambda = \frac{2\pi}{0.077} \approx 81.6 \, \text{m} \).
Wavelength is an essential characteristic of electromagnetic waves, influencing how they interact with different materials and environments.
Materials may absorb or reflect certain wavelengths differently, affecting how we perceive them in daily life.
Frequency
Frequency measures how often the wave repeats itself over time and is denoted by the symbol \( f \). It is related directly to how fast the electromagnetic wave oscillates.
For the given problem, the wave is described using the angular frequency \( \omega = 2.3 \times 10^7 \, \text{rad/s} \).
  • The relationship between angular frequency \( \omega \) and frequency \( f \) is \( \omega = 2\pi f \).
  • To find \( f \), solve \( f = \frac{\omega}{2\pi} \).
  • Using the provided \( \omega \), we get \( f = \frac{2.3 \times 10^7}{2\pi} \approx 3.66 \times 10^6 \, \text{Hz} \).
Frequency is critical as it determines the energy of the photons in the wave.
Higher frequency waves, like X-rays, have more energy and can penetrate materials more effectively than lower frequency waves like radio waves.
Magnetic Field
Electromagnetic waves consist of oscillating electric and magnetic fields at right angles to each other and to the direction of wave propagation.
The magnetic field part of an electromagnetic wave is not always as obvious but is just as crucial as the electric field.
  • The relationship between the electric field amplitude \( E_0 \) and magnetic field amplitude \( B_0 \) is given by \( E_0 = c B_0 \), where \( c \) is the speed of light.
  • Solving for \( B_0 \), we get \( B_0 = \frac{E_0}{c} \).
  • With \( E_0 = 225 \, \text{V/m} \), calculate \( B_0 = \frac{225}{3 \times 10^8} = 7.5 \times 10^{-7} \, \text{T} \).
  • The direction of the magnetic field \( \overrightarrow{\mathbf{B}} \) is perpendicular to both the direction of electric field \( \overrightarrow{\mathbf{E}} \) and the direction in which the wave travels.
In this exercise, the magnetic field expression follows a similar sinusoidal form as the electric field, but points along a different axis, indicating the multi-dimensional nature of electromagnetic waves.

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

(III) The Arecibo radio telescope in Puerto Rico can detect a radio wave with an intensity as low as \(1 \times 10^{-23} \mathrm{~W} / \mathrm{m}^{2}\) As a "best-case" scenario for communication with extraterrestrials, consider the following: suppose an advanced civilization located at point \(A\), a distance \(x\) away from Earth, is somehow able to harness the entire power output of a Sun-like star, converting that power completely into a radio-wave signal which is transmitted uniformly in all directions from A. ( \(a\) ) In order for Arecibo to detect this radio signal, what is the maximum value for \(x\) in light-years \(\left(11 \mathrm{y} \approx 10^{16} \mathrm{~m}\right) ?\) (b) How does this maximum value compare with the 100,000 -ly size of our Milky Way galaxy? The intensity of sunlight at Earth's orbital distance from the Sun is \(1350 \mathrm{~W} / \mathrm{m}^{2}\).

(II) The powerful laser used in a laser light show provides a 3-mm diameter beam of green light with a power of \(3 \mathrm{~W}\). When a space-walking astronaut is outside the Space Shuttle, her colleague inside the Shuttle playfully aims such a laser beam at the astronaut's space suit. The masses of the suited astronaut and the Space Shuttle are \(120 \mathrm{~kg}\) and \(103,000 \mathrm{~kg},\) respectively. \((a)\) Assuming the suit is perfectly reflecting, determine the "radiation-pressure" force exerted on the astronaut by the laser beam. ( \(b\) ) Assuming the astronaut is separated from the Shuttle's center of mass by \(20 \mathrm{~m}\), model the Shuttle as a sphere in order to estimate the gravitation force it exerts on the astronaut. (c) Which of the two forces is larger, and by what factor?

(1) If the magnetic field in a traveling EM wave has a peak magnitude of 12.5 \(\mathrm{nT}\) , what is the peak magnitude of the electric field?

At a given instant in time, a traveling EM wave is noted to have its maximum magnetic field pointing west and its maximum electric field pointing south. In which direction is the wave traveling? If the rate of energy flow is \(560 \mathrm{~W} / \mathrm{m}^{2}\), what are the maximum values for the two fields?

(11) A high-energy pulsed laser emits a 1.0 -ns-long pulse of average power \(1.8 \times 10^{11}\) W. The beam is \(2.2 \times 10^{-3}\) m in radius. Determine \((a)\) the energy delivered in each pulse, and \((b)\) the rms value of the electric field.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Quantum Physics

Read Explanation

Magnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Translational Dynamics

Read Explanation

Astrophysics

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.