/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 32: Problem 4

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

Short Answer

Expert verified
The wave propagates in directions: (a) +z, (b) -z, (c) +y, (d) -y.

Step by step solution

01

Recall the Right-Hand Rule for Electromagnetic Waves

To determine the direction of wave propagation, we use the right-hand rule for electromagnetic waves. For a wave traveling in a certain direction, the electric field vector \( \vec{E} \), the magnetic field vector \( \vec{B} \), and the wave propagation vector \( \vec{k} \) are mutually perpendicular and follow the right-hand rule: if you point the fingers of your right hand in the direction of \( \vec{E} \) and curl them towards \( \vec{B} \), your thumb points in the direction of \( \vec{k} \), the wave propagation.
02

Analyze Case (a)

For case (a), \( \vec{E} = E \hat{\imath} \) and \( \vec{B} = -B \hat{\jmath} \). Point your fingers in the direction of \( \hat{\imath} \) (positive x-axis), then curl them towards the negative y-axis. Your thumb points in the positive z-direction (\( \hat{k} \)). Therefore, the wave propagates in the positive z-direction.
03

Analyze Case (b)

For case (b), \( \vec{E} = E \hat{\jmath} \) and \( \vec{B} = B \hat{\imath} \). Point your fingers in the direction of \( \hat{\jmath} \) (positive y-axis), then curl them towards the positive x-axis. Your thumb points in the negative z-direction (\(-\hat{k}\)). Therefore, the wave propagates in the negative z-direction.
04

Analyze Case (c)

For case (c), \( \vec{E} = -E \hat{k} \) and \( \vec{B} = -B \hat{\imath} \). Point your fingers in the negative z-direction, then curl them towards the negative x-direction. Your thumb points in the positive y-direction (\( \hat{\jmath} \)). Therefore, the wave propagates in the positive y-direction.
05

Analyze Case (d)

For case (d), \( \vec{E} = E \hat{\imath} \) and \( \vec{B} = -B \hat{k} \). Point your fingers in the direction of \( \hat{\imath} \) (positive x-axis), then curl them towards the negative z-axis. Your thumb points in the negative y-direction (\(-\hat{\jmath}\)). Therefore, the wave propagates in the negative y-direction.

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 91影视!

Key Concepts

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

Right-Hand Rule
Understanding the right-hand rule is vital when dealing with electromagnetic waves. It is a simple yet powerful tool that helps determine the direction of wave propagation. Imagine your right hand in action: **pointing your fingers in the direction of the electric field vector** (\( \vec{E} \)), and then **curling them towards the magnetic field vector** (\( \vec{B} \)). Your thumb will naturally point towards the direction of the wave vector (\( \vec{k} \)), indicating where the wave is traveling. This mutual orthogonality of electric field, magnetic field, and propagation direction forms the essence of electromagnetic waves. The right-hand rule is easily applied to different scenarios by adjusting the orientation of your hand:
  • Point fingers in the **electric field direction**.
  • Curl them towards the **magnetic field direction**.
  • Notice where your thumb points for the ** wave propagation direction**.
Remember, the key is to keep everything perpendicular - fingers, curl, and thumb - reflecting the perfection of nature's 90-degree efficiency in wave dynamics.
Electric Field
An electric field (\( \vec{E} \)) is more than just a force; it is a vector field around charged particles. This invisible influence nudges charges and is vital in electrical processes. In electromagnetic waves, the electric field is one of the two perpendicular vectors that travel through space, carrying energy from one place to another. Whenever we talk about the **direction of an electric field**, we refer to the direction that a positive charge would move if placed in the field. Usually expressed as a vector, the electric field helps define the path for wave propagation when your right hand is involved. For instance, a directional change of the electric field shifts the whole orientation, influencing the wave's path. The characteristics of the electric field in scenarios:
  • **Determines charge motion:** Positive charges move with the field; negative against it.
  • **Acts perpendicular:** In waves, it always maintains a perpendicular relationship with the magnetic field vector.
  • **Provider of energy:** Acts as one-half of the duo transmitting energy in the form of waves.
Remember that the electric field is crucial to determining how and where electromagnetic energy shifts.
Magnetic Field
The magnetic field (\( \vec{B} \)) is the silent partner in the dance of electromagnetic waves. Just like the electric field, it is a vector field navigating around magnetic materials and moving charges. But it does not act alone. Instead, it synchronizes perfectly with the electric field to create electromagnetic waves. When considering electromagnetic propagation, the magnetic field is **perpendicular to both the electric field and the propagation direction**. This perfect 90-degree relationship is necessary for the correct application of the right-hand rule and for wave efficiency.Key facts about magnetic fields:
  • **Vector by nature:** Always has a direction, just like the electric field.
  • **Aligned orthogonally:** In the wave construct, it remains perpendicular to the electric field vector.
  • **Completes the energy pair:** Alongside the electric field, it ferries energy through space.
The harmonious interaction between electric and magnetic fields makes the movement of energy possible, shifting space in unison to create coherent electromagnetic waves that smoothly carry power from one point to another.

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

An electromagnetic wave with frequency 65.0 Hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude \(7.20 \times 10 ^ { - 3 } \mathrm { V } / \mathrm { m }\) . (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

Laser Safety. If the eye receives an average intensity greater than \(1.0 \times 10 ^ { 2 } \mathrm { W } / \mathrm { m } ^ { 2 }\) , damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in mW) that a laser beam 1.5 \(\mathrm { mm }\) in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in W/cm?

The sun emits energy in the form of electromagnetic waves at a rate of \(3.9 \times 10 ^ { 26 } \mathrm { W }\) . This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r = R = 6.96 \times 10 ^ { 5 } \mathrm { km }\) ) and at \(r = R / 2 ,\) in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about \(1.0 \times 10 ^ { 4 }\) Pa; at \(r = R / 2 ,\) the gas pressure is calculated from solar models to be about \(4.7 \times 10 ^ { 13 }\) Pa. Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

Microwave Oven. The microwaves in a certain microwave oven have a wavelength of 12.2\(\mathrm { cm }\) (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing- wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0\(\mathrm { cm }\) longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) Verify this by substituting \(E _ { y } ( x , t )\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) m; permeability \(\mu = \mu _ { 0 } )\) . since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Astrophysics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electricity

Read Explanation

Medical Physics

Read Explanation
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.