/*! 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 4 Consider each of the following e... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider each of the following electric- and magnetic-field orientations. In each case, what is the direction of propagation of the wave? (a) \(\overrightarrow{\boldsymbol{E}}=E \hat{\imath}, \overrightarrow{\boldsymbol{B}}=-B \hat{\jmath} ;\) (b) \(\overrightarrow{\boldsymbol{E}}=E \hat{\jmath}, \overrightarrow{\boldsymbol{B}}=B \hat{\imath} ;(\mathrm{c}) \overrightarrow{\boldsymbol{E}}=-E \hat{\boldsymbol{k}}, \overrightarrow{\boldsymbol{B}}=-B \hat{\imath}\) (d) \(\overrightarrow{\boldsymbol{E}}=E \hat{\imath}, \overrightarrow{\boldsymbol{B}}=-B \hat{\boldsymbol{k}}\)

Short Answer

Expert verified
The direction of propagation of the waves is: (a) in the negative z direction i.e. \(- \hat{k}\), (b) in the negative z direction i.e. \(- \hat{k}\), (c) in the positive y direction i.e. \( \hat{j}\), and (d) in the positive y direction i.e. \( \hat{j}\).

Step by step solution

01

Determine the Direction of Propagation for Case A

For case (a), the electric field is \(E \hat{\imath}\) and the magnetic field is \(-B \hat{\jmath}\). Using the rule of cross product, the direction of propagation \( \overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B} = E \hat{\imath} \times (-B \hat{\jmath}) = - EB \hat{k} \). Therefore, the direction of propagation is along \(- \hat{k}\) or in the negative z direction.
02

Determine the Direction of Propagation for Case B

For case (b), the electric field is \(E \hat{\jmath}\) and the magnetic field is \(B \hat{\imath}\). The direction of propagation is \( \overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B} = E \hat{\jmath} \times B \hat{\imath} = - EB \hat{k} \). Therefore, the direction of propagation is along \(- \hat{k}\) or in the negative z direction.
03

Determine the Direction of Propagation for Case C

For case (c), the electric field is \(-E \hat{k}\) and the magnetic field is \(-B \hat{\imath}\). The direction of propagation is \( \overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B} = (-E \hat{k}) \times (-B \hat{\imath}) = EB \hat{j} \). Therefore, the direction of propagation is along \( \hat{j}\) or in the positive y direction.
04

Determine the Direction of Propagation for Case D

For case (d), the electric field is \(E \hat{\imath}\) and the magnetic field is \(-B \hat{k}\). The direction of propagation is \( \overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B} = E \hat{\imath} \times (-B \hat{k}) = EB \hat{j} \). Therefore, the direction of propagation is along \( \hat{j}\) or in the positive y direction.

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Key Concepts

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

Cross Product in Physics
The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. It has extensive applications in physics, particularly in the study of electromagnetic waves. The cross product of two vectors results in a third vector that is perpendicular to the plane containing the original vectors.

Mathematically, if we have two vectors, \( \overrightarrow{A} \text{ and } \overrightarrow{B} \), their cross product \( \overrightarrow{A} \times \overrightarrow{B} \) is a vector not aligned with either \( \overrightarrow{A} \text{ or } \overrightarrow{B} \), but orthogonal to both. The magnitude of this product is given by \( |\overrightarrow{A} \times \overrightarrow{B}| = |\overrightarrow{A}||\overrightarrow{B}|\sin(\theta) \), where \( \theta \) is the smaller angle between the two vectors. Importantly, the direction of the resultant vector follows the right-hand rule, where your right hand's thumb points in the direction of the cross product when your fingers curl from the first vector to the second in the shortest way.

This operation is crucial for understanding electromagnetic wave propagation, as it helps determine the direction of wave travel using the electric and magnetic fields.
Electric Field Direction
In electromagnetic waves, the electric field direction represents the oscillation direction of the electric field strength. It is a vector quantity with both magnitude and direction, usually represented by the symbol \( \overrightarrow{E} \).

The direction of the electric field is critical in determining the behavior and propagation direction of an electromagnetic wave. According to Maxwell's equations, a time-varying electric field creates a magnetic field, and vice versa. The changing electric field induces the magnetic field to orient perpendicularly to the electric field. Thus, understanding the direction of the electric field helps in visualizing the orientation and subsequent dynamics of the whole electromagnetic wave.

In the exercises provided, we use unit vector notation to indicate the direction of the electric field vector, such as \( \hat{\imath}, \hat{\jmath}, \text{and } \hat{k} \) for the x, y, and z coordinates, respectively. This precise notation is essential for applying the cross product correctly.
Magnetic Field Direction
Similar to the electric field, the magnetic field in an electromagnetic wave also has a specific direction, denoted by \( \overrightarrow{B} \). The magnetic field is always perpendicular to the electric field and, by the nature of electromagnetic waves, also perpendicular to the direction of wave propagation.

In the context of electromagnetic waves, the direction of the magnetic field is crucial for using the right-hand rule in cross product calculations, where it is the second vector after the electric field. For instance, knowing the orientation of both \( \overrightarrow{E} \text{ and } \overrightarrow{B} \) allows us to compute the direction of energy flow in the wave.

When analyzing problems involving electromagnetic waves, it's essential to correctly interpret the direction of the magnetic field indicated by the unit vectors. Precise understanding of this helps in visualizing and determining the wave's propagation direction, which is demonstrated in the given step-by-step solutions.
Poynting Vector
The Poynting vector, symbolized by \( \overrightarrow{S} \), represents the energy flux (the rate of energy transfer per unit area) of an electromagnetic wave. Mathematically, it is defined as the cross product of the electric and magnetic field vectors: \( \overrightarrow{S} = \overrightarrow{E} \times \overrightarrow{B} \). This vector points in the direction in which energy is flowing, and its magnitude is proportional to the rate at which energy is transmitted through a given area perpendicular to that direction.

In the exercise solutions, we use the Poynting vector to identify the direction of propagation of electromagnetic waves. The Poynting vector is essential because it not only informs us about the direction of wave travel but also the distribution of energy in space, which is key to understanding phenomena such as radiation pressure and light intensity.

By calculating the Poynting vector, we gain insights into the intrinsic relationship between electric and magnetic fields and how energy is transported in electromagnetic waves. Thus, the concept of the Poynting vector is a fundamental aspect of the physics governing electromagnetic radiation and is crucial for students to master in order to comprehend the propagation and dynamics of such waves.

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Most popular questions from this chapter

Medical x rays are taken with electromagnetic waves having a wavelength of around \(0.10 \mathrm{nm}\) in air. What are the frequency, period, and wave number of such waves?

A sinusoidal electromagnetic wave emitted by a mobile 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 phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

An electromagnetic standing wave in air has frequency \(75.0 \mathrm{MHz}\). (a) What is the distance between nodal planes of the \(\vec{E}\) field? (b) What is the distance between a nodal plane of \(\overrightarrow{\boldsymbol{E}}\) and the closest nodal plane of \(\overrightarrow{\boldsymbol{B}} ?\)

The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a \(50.0 \mathrm{~W}\) (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is \(1575.42 \mathrm{MHz}\). Assume that a satellite transmits half of its power at each frequency and that the waves travel uniformly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton's laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel \(1.50 \mathrm{~cm}\) on a side that absorbs all of the beam, what average pressure does the signal exert on it? (d) What wavelength must the receiver be tuned to?

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 magnitude. 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 or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

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