/*! 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 19 Find the wave \(*\) Write out ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the wave \(*\) Write out formulas for \(\mathbf{E}\) and \(\mathbf{B}\) that specify a plane electromagnetic sinusoidal wave with the following characteristics. The wave is traveling in the direction \(-\hat{\mathbf{x}}\); its frequency is 100 megahertz (MHz), or \(10^{8}\) cycles per second; the electric field is perpendicular to the \(\hat{\mathbf{z}}\) direction.

Short Answer

Expert verified
\(\mathbf{E} = E_0\cos(kx - \omega t + \phi)\hat{\mathbf{y}}\) and \(\mathbf{B} = \frac{E_0}{c}\cos(kx - \omega t + \phi)\hat{\mathbf{z}}\)

Step by step solution

01

Understanding physical quantities

Given that the wave is traveling in the direction \(-\hat{\mathbf{x}}\), its frequency is \(100 MHz\) or \(10^8 Hz\), and the electric field is perpendicular to the \(\hat{\mathbf{z}}\) direction.
02

Write out the formula for the electric field \(\mathbf{E}\)

We firstly construct the electric field. Since the electric field is perpendicular to the \(\hat{\mathbf{z}}\) direction, it can only have a \(\mathbf{y}\) component. Thus, the electric field can be written as follows: \(\mathbf{E} = E_0\cos(kx - \omega t + \phi)\hat{\mathbf{y}}\)
03

Relate the electric and magnetic fields

The magnetic field \(\mathbf{B}\) of an electromagnetic wave is always perpendicular to the electric field and goes as \(\mathbf{B} = \frac{\mathbf{E}}{c}\) where \(c\) is the speed of light in vacuum.
04

Write out the formula for the magnetic field \(\mathbf{B}\)

Given \(\mathbf{B} = \frac{\mathbf{E}}{c}\) and because \(\mathbf{B}\) should be orthogonal to both \(\mathbf{E}\) and the direction of wave propagation (which is \(-\hat{\mathbf{x}}\)), \(\mathbf{B}\) should point in the \(\hat{\mathbf{z}}\) direction. Hence \(\mathbf{B} = \frac{E_0}{c}\cos(kx - \omega t + \phi)\hat{\mathbf{z}}\).

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Key Concepts

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

Electric Field
The electric field (\( \mathbf{E} \)) is a crucial part of understanding electromagnetic waves. It is a vector field that represents the force a positive charge would experience at any point in space. In the context of electromagnetic waves like light, the electric field is time-varying and spatially oscillating. This means it changes with time and position, moving through space in the form of waves.

In the given exercise, the electric field of the wave is specified to be perpendicular to the \( \hat{\mathbf{z}} \) direction. This implies that the field can only have components in the \( \mathbf{y} \) direction if we consider a standard Cartesian coordinate system. Therefore, it can be mathematically represented as:
  • \( \mathbf{E} = E_0\cos(kx - \omega t + \phi)\hat{\mathbf{y}} \)
Where \( E_0 \) is the maximum amplitude of the electric field, \( k \) is the wave number, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.
Magnetic Field
The magnetic field (\( \mathbf{B} \)) is another essential component of electromagnetic waves. Just like the electric field, it oscillates in a wave form. However, it is always perpendicular to both the electric field and the direction of wave propagation. This configuration helps maintain the integrity of the wave's predefined energy patterns.

In our exercise, the magnetic field must also follow these perpendicularity rules. It is expressed in terms of the electric field over the speed of light because of the connection between their magnitudes in electromagnetic waves:
  • \( \mathbf{B} = \frac{E_0}{c}\cos(kx - \omega t + \phi)\hat{\mathbf{z}} \)
Here, \( c \) is the speed of light in a vacuum, making this relation crucial for calculating \( \mathbf{B} \) given \( \mathbf{E} \). This formula shows that the peak amplitudes of \( \mathbf{B} \) and \( \mathbf{E} \) are proportionately connected.
Wave Propagation
Wave propagation is the phenomenon by which waves travel through different mediums. For electromagnetic waves, such propagation occurs even in a vacuum. These waves are characterized by their dynamic oscillations of electric and magnetic fields at right angles to one another and also orthogonal to their direction of travel.

In this exercise, the wave travels in the negative \( \hat{\mathbf{x}} \) direction. This directionality influences how the fields are oriented, as both \( \mathbf{E} \) and \( \mathbf{B} \) must sustain perpendicularity both mutually and in relation to the wave's travel path. Understanding this angular relation is essential for grasping how energy transmits across space.
Frequency
The frequency of an electromagnetic wave is fundamentally how many oscillations it completes per second. It is usually measured in Hertz (Hz). Frequency is a pivotal property because it determines the wave's energy and the spectrum it belongs to, such as radio waves, microwaves, visible light, etc.

In our scenario, the wave's frequency is given as \( 100 \) megahertz (MHz), equivalent to \( 10^8 \) Hz. High-frequency waves like this possess higher energy and shorter wavelengths, which influence their behavior in various mediums. Understanding frequency is key because it's directly related to the wave's angular frequency \( \omega \), calculated as:
  • \( \omega = 2 \pi \times \text{frequency} \)
This connection helps determine the oscillatory motions of the electric and magnetic fields.
Speed of Light
The speed of light (\( c \)) is a fundamental constant of nature, representing how fast electromagnetic waves move through a vacuum. It’s an incredibly high speed, approximately \( 3 \times 10^8 \) meters per second. Recognizing this speed is crucial for determining the relationship between the electric and magnetic fields in electromagnetic waves.

In your exercise, the speed of light is instrumental when relating field magnitudes as noted in \( \mathbf{B} = \frac{\mathbf{E}}{c} \). This shows that a change in the electric field is reciprocated proportionately in the magnetic field due to their intrinsic linkage through \( c \). Understanding the speed of light allows you to make accurate predictions and computations about wave behavior and properties across different scenarios.

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Most popular questions from this chapter

An electromagnetic wave ** Here is a particular electromagnetic field in free space: $$ \begin{array}{ll} E_{x}=0, \quad E_{y}=E_{0} \sin (k x+\omega t), & E_{z}=0 \\ B_{x}=0, \quad B_{y}=0, & B_{z}=-\left(E_{0} / c\right) \sin (k x+\omega t) \end{array} $$ (a) Show that this field can satisfy Maxwell's equations if \(\omega\) and \(k\) are related in a certain way. (b) Suppose \(\omega=10^{10} \mathrm{~s}^{-1}\) and \(E_{0}=1 \mathrm{kV} / \mathrm{m}\). What is the wavelength? What is the energy density in joules per cubic meter, averaged over a large region? From this calculate the power density, the energy flow in joules per square meter per second.

Energy flow in a capacitor ** A capacitor is charged by having current flow in a thin straight wire from the middle of one circular plate to the middle of the other (as opposed to wires coming in from infinity, as in the example in Section 9.6.2). The electric field inside the capacitor increases, so the energy density also increases. This implies that there must be a flow of energy from somewhere. As in Problem 9.10, this "somewhere" is the wire. Verify that the flux of the Poynting vector away from the wire equals the rate of change of the energy stored in the field. (Of course, we would need to place a battery somewhere along the wire to produce the current flow, and this battery is where the energy flow originates. See Galili and Goihbarg (2005).)

Spherically symmetric current A spherically symmetric (and constant) current density flows radially inward to a spherical shell, causing the charge on the shell to increase at the constant rate \(d Q / d t\). Verify that Maxwell's equation, \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t\), is satisfied at points outside the shell.

\(B\) in a discharging capacitor, via conduction current \(*\) As mentioned in Exercise \(9.15\), the magnetic field inside a discharging capacitor can be calculated by summing the contributions from all elements of conduction current. This calculation is extremely tedious. We can, however, get a handle on the contribution from the plates' conduction current in a much easier way that doesn't involve a nasty integral. If we make the usual assumption that the distance \(s\) between the plates is small compared with their radius \(b\), then any point \(P\) inside the capacitor is close enough to the plates so that they look essentially like infinite planes, with a surface current density equal to the density at the nearest point. (a) Determine the current that crosses a circle of radius \(r\) in the capacitor plates, and then use this to find the surface current density. Hint: The charge on each plate is essentially uniformly distributed at all times. (b) Combine the field contributions from the wires and the plates to show that the field at a point \(P\) inside the capacitor, a distance \(r\) from the axis of symmetry, equals \(B=\mu_{0} I r / 2 \pi b^{2}\) (Assume \(s \ll r\), so that you can approximate the two wires as a complete infinite wire.)

Reflected wave ** A sinusoidal wave is reflected at the surface of a medium whose properties are such that half the incident energy is absorbed. Consider the field that results from the superposition of the incident and the reflected wave. An observer stationed somewhere in this field finds the local electric field oscillating with a certain amplitude \(E\). What is the ratio of the largest such amplitude noted by any observer to the smallest amplitude noted by any observer? (This is called the voltage standing wave ratio, or, in laboratory jargon, VSWR.)

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