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Chapter 34: Problem 18

An electromagnetic wave with a wavelength of \(450 \mathrm{~nm}\) travels through vacuum in the negative direction of a \(y\) axis with its electric component directed parallel to the \(x\) axis. The \(\mathrm{rms}\) value of the electric component is \(5.31 \times\) \(10^{-6} \mathrm{~V} / \mathrm{m} .\) Write an equation for the magnetic component in the form of Eq. \(34-3\), but complete with numbers.

Short Answer

Expert verified
The magnetic component of the wave is \( B(z, t) = 2.50 \times 10^{-14} \ \mathrm{T} \cos \left(\frac{2\pi}{450 \times 10^{-9}} z - 2\pi f t \right) \.

Step by step solution

01

Understand the Given Parameters

An electromagnetic wave has a wavelength \( \lambda = 450 \ \mathrm{nm} = 450 \times 10^{-9} \ \mathrm{m} \) and it travels through a vacuum with an rms value of the electric component given by \( E_{rms} = 5.31 \times 10^{-6} \ \mathrm{V/m} \).
02

Relation Between Electric and Magnetic Fields

The magnetic component \(B\) of an electromagnetic wave can be related to the electric component \(E\) using the speed of light \(c\). The relationship is given by \(B = \frac{E}{c} \), where \(c = 3 \times 10^8 \ \mathrm{m/s} \).
03

Calculate the Amplitude of the Magnetic Component

First, calculate the amplitude of the electric field using \( E_{\text{max}} = \sqrt{2} \times E_{rms} \): \[ E_{\text{max}} = \sqrt{2} \times 5.31 \times 10^{-6} \ \mathrm{V/m} \approx 7.51 \times 10^{-6} \ \mathrm{V/m} \].
04

Compute the Maximum Magnetic Component

Using \(B_{\text{max}} = \frac{E_{\text{max}}}{c} \), calculate the magnetic component: \[ B_{\text{max}} = \frac{7.51 \times 10^{-6} \ \mathrm{V/m}}{3 \times 10^8 \ \mathrm{m/s}} \approx 2.50 \times 10^{-14} \ \mathrm{T} \].
05

Formulate the Magnetic Component Equation

An electromagnetic wave in the negative direction of the \(y\) axis with its magnetic component traveling along the \(z\) axis can be written in the form \( B(z, t) = B_{\text{max}} \cos(kz - \omega t) \), where \(k = \frac{2\pi}{\lambda} \) and \omega = 2\pi \times f \. For this wave, the equation becomes: \[ B(z, t) = (2.50 \times 10^{-14} \ \mathrm{T}) \cos \left(\frac{2\pi}{450 \times 10^{-9}}z - 2\pi f t \right) \]

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Key Concepts

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

Electromagnetic Wave Equation
Electromagnetic waves are solutions to Maxwell's equations. They describe how electric and magnetic fields (E and B fields) propagate through space. The general form of an electromagnetic wave traveling in the z-direction can be described by the electric field:

\( E(z, t) = E_{\text{max}} \times \text{cos}(kz - \theta t) \)

and the magnetic field:

\( B(z, t) = B_{\text{max}} \times \text{cos}(kz - \theta t) \)

where \( k = \frac{2\text{蟺}}{\text{位}} \) is the wave number and \( \theta = 2\text{蟺}f \) is the angular frequency. These equations show how the fields oscillate over time and space. The amplitude represents the peak values of E and B fields.
Electric and Magnetic Fields Relationship
In electromagnetic waves, the electric (E) and magnetic (B) fields are perpendicular to each other. They also move together in phase, meaning they reach their maximum and minimum values simultaneously. One crucial relationship between these fields is given by:

\( B = \frac{E}{c} \)

where \( c \) is the speed of light in a vacuum (\( c \text{ = } 3 \times 10^8 \text{ m/s} \)). This relationship means that as the electric field oscillates, the magnetic field oscillates proportionally. The magnitude of the magnetic field can be calculated if the electric field is known.
Wave Propagation in Vacuum
Electromagnetic waves can travel through a vacuum without needing a medium, like air or water. In a vacuum, the waves propagate at the speed of light (\( c = 3 \times 10^8 \text{ m/s} \)). This property makes light and other electromagnetic waves incredibly important for space communication. The absence of a medium means there is no loss of energy due to absorption or scattering, ensuring efficient transmission over vast distances.
RMS Value Calculation
Root Mean Square (RMS) values are used to quantify the effective amplitude of oscillating quantities like the electric field in electromagnetic waves. The RMS value gives a measure of the average power of the oscillation. For an electric field:

\( E_{\text{rms}} = \frac{E_{\text{max}}}{\text{鈭�2}} \)

Given the RMS value \( E_{\text{rms}} = 5.31 \times 10^{-6} \text{ V/m} \), the peak or maximum value is:

\( E_{\text{max}} = \text{鈭�2} \times E_{\text{rms}} \)

This makes it easier to relate RMS values to peak values in practical scenarios, like solving problems involving electromagnetic waves.
Wavelength and Frequency Relation
The wavelength \( \text{位} \) and frequency \( f \) of an electromagnetic wave are closely related through the speed of light. The relationship is given by:

\( c = \text{位} \times f \)

where \( c \) is the speed of light (\( c = 3 \times 10^8 \text{ m/s} \)). This equation means that if you know the wavelength of a wave, you can determine its frequency, and vice versa. For the given wave with wavelength \( \text{位} = 450 \times 10^{-9} \text{ m} \), the frequency can be found by:

\( f = \frac{c}{\text{位}} \)

Understanding this relationship is key in many fields, such as electronics, optics, and telecommunications.

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Most popular questions from this chapter

An unpolarized beam of light is sent through a stack of four polarizing sheets, oriented so that the angle between the polarizing directions of adjacent sheets is \(30^{\circ} .\) What fraction of the incident intensity is transmitted by the system?

Someone plans to float a small, totally absorbing sphere \(0.500 \mathrm{~m}\) above an isotropic point source of light, so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere's density is \(19.0 \mathrm{~g} / \mathrm{cm}^{3}\) and its radius is \(2.00 \mathrm{~mm}\). (a) What power would be required of the light source? (b) Even if such a source were made, why would the support of the sphere be unstable?

Project Seafarer was an ambitious proposal to construct an enormous antenna, buried underground on a site about \(10000 \mathrm{~km}^{2}\) in area. Its purpose was to transmit signals to submarines while they were deeply submerged. If the effective wavelength were \(1.0 \times 10^{4}\) Earth radii, what would be (a) the frequency and (b) the period of the radiations emitted? Ordinarily, electromagnetic radiations do not penetrate very far into conductors such as seawater.

A particle in the solar system in under the combined influence of the Sun's gravitational attraction and the radiation force due to the Sun's rays. Assume that the particle is a sphere of density \(1.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and that all the incident light is absorbed. (a) Show that, if its radius is less than some critical radius \(R\), the particle will be blown out of the solar system. (b) Calculate the critical radius.

Although light appears to travel at a speed that is for all practical purposes infinite, for some modern purposes the time delay due to light travel time is of great importance. The Global Positioning System (GPS) allows you to determine your position from comparison of the time delays between radio signals from 4 satellites at a height of \(20,000 \mathrm{~km}\) above the surface of the earth. (There are actually 24 of these satellites. Your GPS picks out the closest 4 to your current position.) In order to get some idea of how important the speed of light is in establishing your position with one of these gadgets, make some simple assumptions. Assume that a satellite is almost directly overhead. Then figure out how far the satellite will move in the time it takes light (the radio signal) to get from the satellite to your GPS receiver. This estimates how far off the reading of your position would be if your device didn't include the speed of light in its calculations. To do this: (a) Figure out what speed the satellite must be traveling to be in a circular orbit. (b) Estimate the time it would take for a radio signal to get from the satellite to your receiver. (c) Estimate how far the satellite would move in that time. If you ignore light travel time, this tells about how wrong you would get the satellite's position (and therefore how wrong you would get your position).
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