/*! 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 magneticfield orientations. In each case, what is the direction of propagation of the wave? (a) \(\vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{\jmath}\); (b) \(\vec{E} = E\hat{\jmath}\), \(\vec{B} = B\hat{\imath}\); (c) \(\vec{E} = -E\hat{k}\) , \(\vec{B} = -B\hat{\imath}\); (d) \(vec{E} = E\hat{\imath}\), \(\vec{B} = -B\hat{k}\).

Short Answer

Expert verified
(a) -\hat{k}, (b) -\hat{k}, (c) \hat{\jmath}, (d) \hat{\jmath}.

Step by step solution

01

Understand the Wave Propagation

For an electromagnetic wave in free space, the electric field \(\vec{E}\), the magnetic field \(\vec{B}\), and the direction of propagation \(\vec{k}\) are all perpendicular to each other. This relationship can be expressed as \(\vec{k} = \vec{E} \times \vec{B}\), where \(\times\) denotes the cross product.
02

Determine Direction for Part (a)

Given \(\vec{E} = E\hat{\imath}\) and \(\vec{B} = -B\hat{\jmath}\), compute \(\vec{k}\) using the cross product. \(\vec{k} = (E\hat{\imath}) \times (-B\hat{\jmath}) = -EB(\hat{\imath} \times \hat{\jmath}) = -EB\hat{k}\). Thus, the wave propagates in the \(-\hat{k}\) direction.
03

Determine Direction for Part (b)

Given \(\vec{E} = E\hat{\jmath}\) and \(\vec{B} = B\hat{\imath}\), compute \(\vec{k}\). \(\vec{k} = (E\hat{\jmath}) \times (B\hat{\imath}) = EB(\hat{\jmath} \times \hat{\imath}) = -EB\hat{k}\). Thus, the wave propagates in the \(-\hat{k}\) direction.
04

Determine Direction for Part (c)

Given \(\vec{E} = -E\hat{k}\) and \(\vec{B} = -B\hat{\imath}\), compute \(\vec{k}\). \(\vec{k} = (-E\hat{k}) \times (-B\hat{\imath}) = EB(\hat{k} \times \hat{\imath}) = EB\hat{\jmath}\). Thus, the wave propagates in the \(\hat{\jmath}\) direction.
05

Determine Direction for Part (d)

Given \(\vec{E} = E\hat{\imath}\) and \(\vec{B} = -B\hat{k}\), compute \(\vec{k}\). \(\vec{k} = (E\hat{\imath}) \times (-B\hat{k}) = -EB(\hat{\imath} \times \hat{k}) = EB\hat{\jmath}\). Thus, the wave propagates in the \(\hat{\jmath}\) direction.

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Key Concepts

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

Wave Propagation Direction
Understanding the direction of wave propagation is crucial in electromagnetism. When dealing with electromagnetic waves, such as light, they propagate through space in a specific direction. This direction is denoted by the vector \( \vec{k} \). In simpler terms, \( \vec{k} \) shows where the wave is headed.

In the electromagnetic spectrum, both the electric field \( \vec{E} \) and the magnetic field \( \vec{B} \) are always perpendicular to each other and to the wave propagation direction. This means if you know the orientations of \( \vec{E} \) and \( \vec{B} \), you can determine the direction of \( \vec{k} \).

To find \( \vec{k} \), you use the formula \( \vec{k} = \vec{E} \times \vec{B} \). This equation is a mathematical representation of the direction of the wave by using the cross product. By calculating \( \vec{k} \), you determine the wave's travel path.
Electric Field Orientation
The electric field \( \vec{E} \) is a vector field that represents the electric force per unit charge at different points in space. In an electromagnetic wave, the direction of \( \vec{E} \) is one of the key factors influencing the wave's propagation.

Electric fields are typically oriented along one axis in a three-dimensional coordinate system, such as \( \hat{i} \), \( \hat{j} \), or \( \hat{k} \). The chosen axis depends on the situation or the problem at hand.

Orientation of \( \vec{E} \) is significant because it is directly perpendicular to where the wave will travel and also perpendicular to \( \vec{B} \), the magnetic field. In exercises involving electromagnetic waves, identifying \( \vec{E} \) is the first step in figuring out the direction of \( \vec{k} \), which ultimately determines the propagation direction.
Magnetic Field Orientation
Just like the electric field, the magnetic field \( \vec{B} \) is a vector. It represents the magnetic influence on moving charges, current loops, and magnetized materials. In electromagnetic waves, \( \vec{B} \) interacts with the electric field to dictate the wave’s direction.

Magnetic fields in these scenarios are also oriented along an axis in the three-dimensional system, such as \( \hat{i} \), \( \hat{j} \), or \( \hat{k} \). A crucial point to remember is \( \vec{B} \) is perpendicular to both the electric field \( \vec{E} \) and the wave propagation direction \( \vec{k} \).

This perpendicular relationship forms the right-handed coordinate system, where \( \vec{E} \), \( \vec{B} \), and \( \vec{k} \) interact according to the right-hand rule. Thus, knowing the orientation of \( \vec{B} \) is vital in determining \( \vec{k} \).
Cross Product in Physics
The cross product, denoted by \( \times \), is a fundamental operation in vector algebra used in physics to find a vector perpendicular to two given vectors. In the context of electromagnetic waves, it is employed to calculate the direction of wave propagation \( \vec{k} \).

In mathematical terms, the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is another vector \( \vec{C} \), which is orthogonal to both \( \vec{A} \) and \( \vec{B} \). The magnitude of \( \vec{C} \) is given by \( |\vec{C}| = |\vec{A}| |\vec{B}| \sin(\theta) \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \). The direction of \( \vec{C} \) follows the right-hand rule.

In electromagnetic problems, \( \vec{k} = \vec{E} \times \vec{B} \), helps us find the wave's direction. Using the cross product ensures that \( \vec{k} \) is perpendicular to both the electric and magnetic fields, adhering to the fundamental properties of wave propagation in free space.

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Most popular questions from this chapter

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the -\(z\)-direction. The electric field has amplitude 2.70 \(\times\) 10\(^{-3}\) V/m and is parallel to the \(x\)-axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\vec{E} (z, t)\) and \(\vec{B} (z, t)\).

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portions back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 \(\mu\)m in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34. (a) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? (b) What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

Fields from a Light Bulb. We can reasonably model a 75-W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5\(\%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in W/m\(^2\)) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N/C. What is the electric-field amplitude 20.0 cm from the source?

If the eye receives an average intensity greater than 1.0 \(\times\) 10\(^2\) W/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 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\(^2\).

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