/*! 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 25 \(\bullet$$\bullet\) A sinusoida... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate: (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

Short Answer

Expert verified
(a) \(8.47 \times 10^8 \; \mathrm{Hz}\); (b) \(1.80 \times 10^{-10} \; \mathrm{T}\); (c) \(0.38 \; \mathrm{W/m^2}\).

Step by step solution

01

Calculate the Frequency of the Wave

The relation between the speed of light \( c \), frequency \( f \), and wavelength \( \lambda \) is given by the formula: \[ c = f \cdot \lambda \]Where \( c = 3 \times 10^8 \; \mathrm{m/s} \) is the speed of light. Substitute \( \lambda = 0.354 \; \mathrm{m} \): \[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{0.354} \approx 8.47 \times 10^8 \; \mathrm{Hz} \]
02

Calculate the Magnetic-Field Amplitude

The magnetic field amplitude \( B_0 \) is related to the electric field amplitude \( E_0 \) and the speed of light \( c \) by the equation: \[ B_0 = \frac{E_0}{c} \]Substitute \( E_0 = 5.40 \times 10^{-2} \; \mathrm{V/m} \) and \( c = 3 \times 10^8 \; \mathrm{m/s} \): \[ B_0 = \frac{5.40 \times 10^{-2}}{3 \times 10^8} \approx 1.80 \times 10^{-10} \; \mathrm{T} \]
03

Calculate the Intensity of the Wave

The intensity \( I \) of the wave is calculated using the electric field amplitude \( E_0 \), the magnetic permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} \), and the speed of light \( c \): \[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]Substitute the given values: \[ I = \frac{1}{2} \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times (5.40 \times 10^{-2})^2 \approx 0.38 \; \mathrm{W/m^2} \]

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.

Wavelength Calculation: Understanding the Basics
Wavelength is a critical concept when discussing electromagnetic waves, as it determines properties like frequency and energy. Wavelength (\(\lambda\)) is the distance between consecutive peaks of an electromagnetic wave. For this exercise, the wavelength of the wave is provided as 35.4 cm, which can be converted to meters as 0.354 m. The relationship between the speed of light (\(c\)), frequency (\(f\)), and wavelength is described by the equation \[ c = f \cdot \lambda \] From this equation, knowing any two of these values lets you calculate the third. In our exercise, after substituting the known values, we found the frequency to be approximately \(8.47 \times 10^8\) Hz. Understanding this relationship is crucial for solving many wave-related problems across physics.
Electric Field Amplitude: Significance in Electromagnetic Waves
The electric field amplitude (\(E_0\)) is a measure of the strength of the electric component of an electromagnetic wave. At a distance of 250 meters from the antenna in our scenario, the electric field amplitude is given as \(5.40 \times 10^{-2}\) V/m.The amplitude of the electric field determines the wave’s energy content and is a key quantity in calculating other features of electromagnetic waves, like intensity and magnetic field amplitude. A larger amplitude signifies a more powerful wave, having implications in applications ranging from communications to medical technologies.
Magnetic Field Amplitude: Relation to Electric Field
The magnetic field amplitude (\(B_0\)) reflects the strength of the magnetic part of the electromagnetic wave. This concept is often discussed alongside the electric field amplitude.In electromagnetic waves, particularly those traveling in vacuum or free space, the electric field amplitude (\(E_0\)) and the magnetic field amplitude are related by the speed of light (\(c\)):\[ B_0 = \frac{E_0}{c} \] Using the given electric field amplitude of \(5.40 \times 10^{-2}\) V/m, we calculated the magnetic field amplitude to be approximately \(1.80 \times 10^{-10}\) T. This relationship is fundamental in understanding the interplay between electric and magnetic components in waves.
Wave Intensity Calculation: Energy Flow Through a Surface
Intensity (\(I\)) is a measure of power per unit area carried by a wave. It's a crucial quantity in understanding how much energy a wave transfers through a certain area over time.For electromagnetic waves, intensity is computed using the formula:\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \] Here, \(\varepsilon_0\) is the permittivity of free space, with a value of \(8.85 \times 10^{-12} \mathrm{F/m}\). Using our electric field amplitude and speed of light, the wave intensity was calculated as approximately \(0.38 \mathrm{W/m^2}\). Understanding intensity helps us grasp how waves can affect materials they encounter, making it an important concept across various scientific and practical applications.

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

\(\bullet\) Light with a frequency of \(5.80 \times 10^{14}\) Hz travels in a block of glass that has an index of refraction of \(1.52 .\) What is the wavelength of the light (a) in vacuum and (b) in the glass?

\(\bullet\) The intensity at a certain distance from a bright light source is 6.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

\(\bullet\) A sinusoidal electromagnetic wave having a magnetic field of amplitude 1.25 \(\mu\) and a wavelength of 432 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 \((23.3) .\)

\(\bullet$$\bullet\) You (height of your eyes above the water, 1.75 \(\mathrm{m}\) ) are standing 2.00 \(\mathrm{m}\) from the edge of a 2.50 -m-deep swimming pool. You notice that you can barely see your cell phone, which went missing a few minutes before, on the bottom of the pool. How far from the side of the pool is your cell phone?

\(\bullet$$\bullet\) A thin beam of light in air is incident on the surface of a lanthanum flint glass plate having a refractive index of 1.80 . What is the angle of incidence, \(\theta_{a}\) of the beam with this plate, for which the angle of refraction is \(\theta_{a} / 2 ?\) Both angles are measured relative to the normal.

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.