/*! 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 27 A plane electromagnetic wave, wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A plane electromagnetic wave, with wavelength \(3.0 \mathrm{~m}\), travels in vacuum in the positive direction of an \(x\) axis. The electric field, of amplitude \(300 \mathrm{~V} / \mathrm{m},\) oscillates parallel to the \(y\) axis. What are the (a) frequency, (b) angular frequency, and (c) angular wave number of the wave? (d) What is the amplitude of the magnetic field component? (e) Parallel to which axis does the magnetic field oscillate? (f) What is the timeaveraged rate of energy flow in watts per square meter associated with this wave? The wave uniformly illuminates a surface of area \(2.0 \mathrm{~m}^{2} .\) If the surface totally absorbs the wave, what are \((\mathrm{g})\) the rate at which momentum is transferred to the surface and (h) the radiation pressure on the surface?

Short Answer

Expert verified
(a) 1.0 x 10^8 Hz, (b) 2Ï€ x 10^8 rad/s, (c) 2Ï€/3 rad/m, (d) 1.0 x 10^-6 T, (e) z-axis, (f) 119 W/m^2, (g) 238 W, (h) 3.97 x 10^-7 N/m^2.

Step by step solution

01

Calculating Frequency

The frequency \( f \) of a wave is related to its wavelength \( \lambda \) and the speed of light \( c \) in a vacuum by the equation \( f = \frac{c}{\lambda} \). Given \( \lambda = 3.0 \; \text{m} \) and \( c = 3.0 \times 10^8 \; \text{m/s} \), we find: \[ f = \frac{3.0 \times 10^8 \; \text{m/s}}{3.0 \; \text{m}} = 1.0 \times 10^8 \; \text{Hz} \]
02

Calculating Angular Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by the formula \( \omega = 2\pi f \). Using \( f = 1.0 \times 10^8 \; \text{Hz} \), we get: \[ \omega = 2\pi \times 1.0 \times 10^8 = 2\pi \times 10^8 \; \text{rad/s} \]
03

Calculating Angular Wave Number

The angular wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \). With a wavelength \( \lambda = 3.0 \; \text{m} \), we have: \[ k = \frac{2\pi}{3.0} = \frac{2\pi}{3} \; \text{rad/m} \]
04

Amplitude of Magnetic Field

The amplitude of the magnetic field \( B_0 \) related to the electric field \( E_0 \) is given by \( B_0 = \frac{E_0}{c} \). With \( E_0 = 300 \; \text{V/m} \) and \( c = 3.0 \times 10^8 \; \text{m/s} \): \[ B_0 = \frac{300}{3.0 \times 10^8} = 1.0 \times 10^{-6} \; \text{T} \]
05

Direction of Magnetic Field Oscillation

In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of wave propagation. Since the electric field oscillates parallel to the \( y \)-axis and the wave travels in the positive \( x \)-direction, the magnetic field oscillates parallel to the \( z \)-axis.
06

Calculating Time-Averaged Energy Flow

The time-averaged energy flow per unit area, or the intensity \( S \), is given by \( S = \frac{1}{2}c\varepsilon_0 E_0^2 \). Using \( \varepsilon_0 = 8.85 \times 10^{-12} \; \text{F/m} \) and \( E_0 = 300 \; \text{V/m} \): \[ S = \frac{1}{2} \times 3.0 \times 10^8 \times 8.85 \times 10^{-12} \times 300^2 \approx 119 \; \text{W/m}^2 \]
07

Rate of Momentum Transfer

The rate of momentum transfer per unit area is equal to the intensity \( S \), thus the total rate is \( S \times 2.0 \). \[ \text{Rate} = 119 \times 2.0 = 238 \; \text{W (or Nm/s)} \]
08

Calculating Radiation Pressure

The radiation pressure \( P \) for complete absorption is given by \( P = \frac{S}{c} \). Using \( S = 119 \; \text{W/m}^2 \) and \( c = 3.0 \times 10^8 \; \text{m/s} \): \[ P = \frac{119}{3.0 \times 10^8} = 3.97 \times 10^{-7} \; \text{N/m}^2 \]

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 91Ó°ÊÓ!

Key Concepts

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

Frequency Calculation
The frequency of an electromagnetic wave is a key parameter that defines how many oscillations occur in one second. It's calculated using the equation: \[ f = \frac{c}{\lambda} \]where:
  • \( f \) is the frequency in Hertz (Hz),
  • \( c \) is the speed of light in a vacuum, approximately \( 3.0 \times 10^8 \; \text{m/s} \), and
  • \( \lambda \) is the wavelength of the wave.
To calculate the frequency for a wave with a wavelength of \( 3.0 \; \text{m} \), plug in the values:
  • \( f = \frac{3.0 \times 10^8}{3.0} = 1.0 \times 10^8 \; \text{Hz} \)
This means the wave oscillates \( 100 \) million times per second as it travels through a vacuum.
Angular Frequency
Angular frequency describes how fast the wave oscillates and is expressed in radians per second. It is closely related to the frequency and is given by:\[ \omega = 2\pi f \]
  • \( \omega \) is the angular frequency in radians per second,
  • \( f \) is the regular frequency in Hertz.
Using the previously calculated frequency \( f = 1.0 \times 10^8 \; \text{Hz} \), we find:
  • \( \omega = 2\pi \times 1.0 \times 10^8 = 2\pi \times 10^8 \; \text{rad/s} \)
Angular frequency provides a deeper understanding of the wave's oscillation speed on a circular scale.
Radiation Pressure
Radiation pressure is the pressure exerted by an electromagnetic wave when it strikes a surface. This occurs because the wave carries momentum, which can be transferred to the surface. For complete absorption, the radiation pressure \( P \) can be calculated using the formula:\[ P = \frac{S}{c} \]where:
  • \( S \) is the intensity of the wave (time-averaged energy flow per unit area),
  • \( c \) is the speed of light.
From the step-by-step solution, we derived that \( S = 119 \; \text{W/m}^2 \). Substituting in the values gives:
  • \( P = \frac{119}{3.0 \times 10^8} \approx 3.97 \times 10^{-7} \; \text{N/m}^2 \)
Although appearing minuscule, radiation pressure plays significant roles in areas such as solar sails for spacecraft propulsion.
Magnetic Field Amplitude
In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of wave propagation. They oscillate in harmony but with different magnitudes. The amplitude of the magnetic field \( B_0 \) is related to the electric field amplitude \( E_0 \) by the equation:\[ B_0 = \frac{E_0}{c} \]where:
  • \( E_0 \) is the electric field amplitude,
  • \( c \) is the speed of light.
Given \( E_0 = 300 \; \text{V/m} \) and \( c = 3.0 \times 10^8 \; \text{m/s} \), substitute these values to find:
  • \( B_0 = \frac{300}{3.0 \times 10^8} = 1.0 \times 10^{-6} \; \text{T} \)
This value indicates the strength of the magnetic field component of the wave, allowing understanding of its electromagnetic structure.

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

Prove, for a plane electromagnetic wave that is normally incident on a flat surface, that the radiation pressure on the surface is equal to the energy density in the incident beam. (This relation between pressure and energy density holds no matter what fraction of the incident energy is reflected.)

An electromagnetic wave with frequency \(4.00 \times 10^{14} \mathrm{~Hz}\) travels through vacuum in the positive direction of an \(x\) axis. The wave has its electric field oscillating parallel to the \(y\) axis, with an amplitude \(E_{m^{*}}\) At time \(t=0,\) the electric field at point \(P\) on the \(x\) axis has a value of \(+E_{m} / 4\) and is decreasing with time. What is the distance along the \(x\) axis from point \(P\) to the first point with \(E=0\) if we search in (a) the negative direction and (b) the positive direction of the \(x\) axis?

The magnetic component of a polarized wave of light is given by \(B_{x}=(4.00 \mu \mathrm{T}) \sin \left[k y+\left(2.00 \times 10^{15} \mathrm{~s}^{-1}\right) t\right] .\) (a) In which direction does the wave travel, (b) parallel to which axis is it polarized, and (c) what is its intensity? (d) Write an expression for the electric field of the wave, including a value for the angular wave number. (e) What is the wavelength? (f) In which region of the electromagnetic spectrum is this electromagnetic wave?

What is the radiation pressure \(1.5 \mathrm{~m}\) away from a \(500 \mathrm{~W}\) lightbulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.

A beam of intensity \(I\) reflects from a long. totally reflecting cylinder of radius \(R ;\) the beam is perpendicular to the central axis of the cylinder and has a diamcter larger than \(2 R .\) What is the beam's force per unit length on the cylinder?

See all solutions

Recommended explanations on Physics Textbooks

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.