/*! 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 44 The electric-field component of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 44

The electric-field component of a sinusoidal electromagnetic wave traveling through a plastic cylinder is given by the equation \(E=(5.35 \mathrm{~V} / \mathrm{m}) \cos \left[\left(1.39 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x-(3.02 \times\right.\) \(\left.\left.10^{15} \mathrm{rad} / \mathrm{s}\right) t\right] .\) (a) Find the frequency, wavelength, and speed of this wave in the plastic. (b) What is the index of refraction of the plastic? (c) Assuming that the amplitude of the electric field does not change, write a comparable equation for the electric field if the light is traveling in air instead of in plastic. Fn. AE

Short Answer

Expert verified
(a) Frequency: \(4.81 \times 10^{14}\) Hz, Wavelength: \(4.52 \times 10^{-7}\) m, Speed: \(2.17 \times 10^{8}\) m/s. (b) Index: 1.38. (c) New equation: \(E=(5.35\text{ V/m}) \cos[(1.01 \times 10^{7}\text{ rad/m}) x - (3.02 \times 10^{15}\text{ rad/s}) t]\).

Step by step solution

01

Understand the Equation

The electric-field component of the wave is given by the equation \(E=(5.35 \text{ V/m}) \cos\left[(1.39 \times 10^{7} \text{ rad/m}) x - (3.02 \times 10^{15} \text{ rad/s}) t\right]\). Here, \((1.39 \times 10^{7} \text{ rad/m})\) is the wave number \(k\), and \((3.02 \times 10^{15} \text{ rad/s})\) is the angular frequency \(\omega\).
02

Determine Frequency

The relationship between angular frequency \(\omega\) and frequency \(f\) is given by the formula \(\omega = 2\pi f\). Rearrange to find \(f\): \[f = \frac{\omega}{2\pi} = \frac{3.02 \times 10^{15} \text{ rad/s}}{2\pi} \approx 4.81 \times 10^{14} \text{ Hz}\].
03

Calculate Wavelength in Plastic

The wavelength \(\lambda\) in a medium is related to the wave number \(k\) by \(k = \frac{2\pi}{\lambda}\). Rearrange this to \(\lambda = \frac{2\pi}{k}\): \[\lambda = \frac{2\pi}{1.39 \times 10^{7} \text{ rad/m}} \approx 4.52 \times 10^{-7} \text{ m}\].
04

Determine Speed in Plastic

The speed \(v\) of the wave in the medium is given by \(v = f\lambda\). Using \(f = 4.81 \times 10^{14} \text{ Hz}\) and \(\lambda = 4.52 \times 10^{-7} \text{ m}\), we find: \[v = (4.81 \times 10^{14} \text{ Hz})(4.52 \times 10^{-7} \text{ m}) \approx 2.17 \times 10^{8} \text{ m/s}\].
05

Compute Index of Refraction

The index of refraction \(n\) is the ratio of the speed of light in vacuum \(c\) (approximately \(3.00 \times 10^{8} \text{ m/s}\)) to the speed in the medium \(v\). So, \(n = \frac{c}{v}\): \[n = \frac{3.00 \times 10^{8} \text{ m/s}}{2.17 \times 10^{8} \text{ m/s}} \approx 1.38\].
06

Write Equation for Air

In air, the speed of light is approximately the same as in a vacuum. So, the speed \(v\) is \(3.00 \times 10^{8} \text{ m/s}\). Calculate the new wave number \(k_{\text{air}}\) with the same frequency \(f\): \(k_{\text{air}} = \frac{2\pi f}{c} = \frac{2\pi (4.81 \times 10^{14} \text{ Hz})}{3.00 \times 10^{8} \text{ m/s}} \approx 1.01 \times 10^{7} \text{ rad/m}\). The equation for the electric field in air is \[E = (5.35 \text{ V/m}) \cos \left[(1.01 \times 10^{7} \text{ rad/m}) x - (3.02 \times 10^{15} \text{ rad/s}) t\right]\].

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.

Wave Speed
Wave speed is a fundamental concept when studying electromagnetic waves. It describes how fast the wave travels through a medium. To find the wave speed, you need to consider both its frequency and wavelength. The formula to determine the wave speed (\(v\)) is simple: the product of frequency (\(f\)) and wavelength (\(\lambda\)). So, \[v = f \times \lambda\].In vacuum and air, light travels at \(3.00 \times 10^8 \text{ m/s}\), which is considered its maximum speed. In the original problem, the speed in plastic is reduced. This is because different materials can slow down electromagnetic waves in varying degrees.
Wavelength
Wavelength is the distance between successive peaks (or troughs) of a wave. It's crucial in defining wave properties and behaviors. In mathematical terms, it relates to the wave number (\(k\)), which is given as \[k = \frac{2\pi}{\lambda}\]. Rearranging this gives the formula for wavelength: \[\lambda = \frac{2\pi}{k}\].In the exercise, the wavelength of the wave in the plastic is smaller than in the air. A smaller wavelength in a medium implies that the wave travels slower compared to its speed in air or vacuum.
Index of Refraction
The index of refraction (\(n\)) is a measure of how much a material can slow down light. It is a dimensionless number that tells us the ratio of the speed of light in vacuum to the speed in the medium. Calculated as \[n = \frac{c}{v}\], where \(c\) is the speed of light in vacuum and \(v\) is the speed in the medium.In this problem, the index of refraction for the plastic is around 1.38. This means the wave travels slower in the plastic compared to in air, which allows us to understand how light interacts with different materials, resulting in phenomena like bending, or refraction.
Angular Frequency
Angular frequency (\(\omega\)) represents how many radians a wave cycles through in one second. It is closely related to the ordinary frequency, and we can convert between them using \(\omega = 2\pi f\).In terms of behavior, a higher angular frequency indicates quicker oscillations of the wave. For the electromagnetic wave in the exercise, the high angular frequency (\(3.02 \times 10^{15} \text{ rad/s}\)) indicates rapid oscillation of the wave within very short time intervals. Understanding this concept is essential in physics, particularly when analyzing wave patterns and interactions.

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

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic 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?

A circular loop of wire has radius \(7.50 \mathrm{~cm}\). A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is \(0.0195 \mathrm{~W} / \mathrm{m}^{2}\), and the wavelength of the wave is \(6.90 \mathrm{~m}\). What is the maximum emf induced in the loop?

An electromagnetic standing wave in air of frequency \(750 \mathrm{MHz}\) is set up between two conducting planes \(80.0 \mathrm{~cm}\) apart. At which positions between the planes could a point charge be placed at rest so that it would remain at rest? Explain.

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, not 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 mag nitude. 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 (d) Explain why dust particles with a radius less than or why not? that found in part (c) are unlikely to be found in the solar system.

If the eye receives an average intensity greater than \(1.0 \times 10^{2} \mathrm{~W} / \mathrm{m}^{2}\), damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in \(\mathrm{mW}\) ) that a laser beam \(1.5 \mathrm{~mm}\) in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in \(\mathrm{W} / \mathrm{cm}^{2} .\)
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Measurements

Read Explanation

Space Physics

Read Explanation

Turning Points in Physics

Read Explanation

Quantum 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.