/*! 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 2 Consider each of the electric- a... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec { \boldsymbol { E } }\) in the \(+ x\) -direction, \(\vec { \boldsymbol { B } }\) in the \(+ y\) - direction; (b) \(\vec { \boldsymbol { E } }\) in the \(- y\) - direction, \(\vec { \boldsymbol { B } }\) in the \(+ x\) -direction; \(( \mathrm { c } ) \vec { \boldsymbol { E } }\) in the \(+ z\) -direction, \(\vec { \boldsymbol { B } }\) in the \(- x\) -direction; (d) \(\vec { \boldsymbol { E } }\) in the \(+ y\) -direction, \(\vec { \boldsymbol { B } }\) in the \(- z\) -direction.

Short Answer

Expert verified
(a) +z; (b) +z; (c) +y; (d) +x.

Step by step solution

01

Understanding Wave Propagation

To determine the direction of wave propagation, remember that the electric field vector \( \vec{\boldsymbol{E}} \), the magnetic field vector \( \vec{\boldsymbol{B}} \), and the direction of propagation \( \vec{\boldsymbol{k}} \) are mutually perpendicular in an electromagnetic wave. The direction of wave propagation is given by the cross product \( \vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} \).
02

Calculate Direction for Case (a)

Given: \( \vec{\boldsymbol{E}} \) in the \(+x\)-direction, \( \vec{\boldsymbol{B}} \) in the \(+y\)-direction. Calculate \( \vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} \). The cross product \( (\hat{i} \times \hat{j}) \) is \( \hat{k} \), hence the wave propagates in the \(+z\)-direction.
03

Calculate Direction for Case (b)

Given: \( \vec{\boldsymbol{E}} \) in the \(-y\)-direction, \( \vec{\boldsymbol{B}} \) in the \(+x\)-direction. Calculate \( \vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} \). The cross product \( (-\hat{j} \times \hat{i}) \) is \( \hat{k} \), thus the wave propagates in the \(+z\)-direction.
04

Calculate Direction for Case (c)

Given: \( \vec{\boldsymbol{E}} \) in the \(+z\)-direction, \( \vec{\boldsymbol{B}} \) in the \(-x\)-direction. Calculate \( \vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} \). The cross product \( (\hat{k} \times -\hat{i}) \) is \( \hat{j} \), indicating the wave propagates in the \(+y\)-direction.
05

Calculate Direction for Case (d)

Given: \( \vec{\boldsymbol{E}} \) in the \(+y\)-direction, \( \vec{\boldsymbol{B}} \) in the \(-z\)-direction. Calculate \( \vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} \). The cross product \( (\hat{j} \times -\hat{k}) \) is \( \hat{i} \), so the wave propagates in the \(+x\)-direction.

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are fascinating phenomena that consist of oscillating electric and magnetic fields. These fields are perpendicular to one another as well as to the direction in which the wave is propagating. This trio forms a right-handed directional system and is a fundamental concept in physics.
  • The electric field ( \(\vec{\boldsymbol{E}}\)
  • The magnetic field ( \(\vec{\boldsymbol{B}}\)
  • The direction of propagation (\(\vec{\boldsymbol{k}}\)
This structure is required for understanding any scenario that involves electromagnetic waves.
Example: Consider an example where the electric field is pointing in one direction (say the x-axis) and the magnetic field in another (like the y-axis). The resulting wave will propagate in a direction that is perpendicular to both, which would be the z-axis, completing a three-dimensional pathway.
Electric Field
For an electromagnetic wave, the electric field ( \(\vec{\boldsymbol{E}}\) ) is a vector quantity that represents the ability of the wave to exert a force on charged particles. It determines the direction and magnitude of the force that a charged particle will experience.
  • The electric field vector oscillates perpendicular to the direction of wave propagation.
  • It influences how rapidly the wave can affect or accelerate charges.
It's important to understand that in a typical electromagnetic wave scenario, like light, the electric component is perpendicular not only to the direction of travel but also to the magnetic field. This orientation provides a method to visually and intuitively determine the path of energy transmission.
Magnetic Field
The magnetic field ( \(\vec{\boldsymbol{B}}\) ) component of an electromagnetic wave is also a vector. Just like the electric field, it is always perpendicular to the electric field as well as the wave's direction of travel.
  • It works hand in hand with the electric field to propagate the wave.
  • Unlike electric fields, magnetic fields only act on moving charges.
In practice, this field is essential for the self-sustainability of the wave. The mutual induction between the electric field and magnetic field that changes with time allows for the wave to propagate through space without requiring a medium.
Cross Product
In physics, the cross product is a vital algebraic operation that helps to determine the directional aspect when dealing with vector quantities like electric and magnetic fields. It is employed to deduce the propagation direction for electromagnetic waves.
  • The cross product of two vectors, such as \(\vec{\boldsymbol{E}}\) and \(\vec{\boldsymbol{B}}\), results in a new vector that is perpendicular to both.
  • Its magnitude is proportional to the sine of the angle between the original vectors, making it useful for determining not only direction but also the strength of interaction between the fields.
Knowing how to calculate and interpret the cross product can unlock a deeper understanding of 3D vector relations, which is critical for mastering wave propagation concepts using \(\vec{\boldsymbol{E}} \times \vec{\boldsymbol{B}} = \vec{\boldsymbol{k}}\) as a guide.
Vector Analysis
Vector analysis involves using mathematical tools and methods to study vectors in fields such as physics and engineering. It is especially useful for understanding concepts like electromagnetic waves.
  • The orientation of vectors (like \(\vec{\boldsymbol{E}}\) and \(\vec{\boldsymbol{B}}\)), and their respective cross product, provides insights into wave motion.
  • By analyzing the vectors' directions and performing operations like cross products and dot products, a wealth of information is revealed about wave strength, direction, and velocity.
Vector analysis equips you with the skills to navigate between the microscopic explanation of phenomena like electromagnetic waves and their macroscopic implications, assisting in problem-solving and the prediction of real-world phenomena.

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Most popular questions from this chapter

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 electric-field component of a sinusoidal electromagnetic wave traveling through a plastic cylinder is given by the equation \(E = ( 5.35 \mathrm { V } / \mathrm { m } ) \cos \left[ \left( 1.39 \times 10 ^ { 7 } \mathrm { rad } / \mathrm { m } \right) x - ( 3.02 \times \right.\) \(10 ^ { 15 } \mathrm { rad } / \mathrm { s } ) t ]\) . (a) Find the frequency, wavelength, and speed of this wave in the plastic. (b) What is the index of refraction of the plastic? (c) Assuming that the amplitude of the electric field does not change, write a comparable equation for the electric field if the light is traveling in air instead of in plastic.

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

Radio station WCCO in Minneapolis broadcasts at a frequency of 830\(\mathrm { kHz }\) . At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is \(4.82 \times 10 ^ { - 111 } \mathrm { T.Calculate }\) (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

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.

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