/*! 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 10 A plane electromagnetic wave has... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A plane electromagnetic wave has a maximum electric field magnitude of \(3.20 \times 10^{-4} \mathrm{~V} / \mathrm{m}\). Find the magnetic field amplitude.

Short Answer

Expert verified
The magnetic field amplitude is \( 1.07 \times 10^{-12} \text{ T} \).

Step by step solution

01

Introduction to the Problem

We need to determine the amplitude of the magnetic field for an electromagnetic wave, given the maximum electric field amplitude. We will apply the relationship between electric field and magnetic field in an electromagnetic wave.
02

Understand the Relationship

In a plane electromagnetic wave, the electric field (E) and magnetic field (B) are related by the equation: \[E = cB\] where E is the amplitude of the electric field, B is the amplitude of the magnetic field, and c denotes the speed of light in vacuum, approximately equal to 3.00 \times 10^8 m/s.
03

Arrange the Formula

To find the magnetic field amplitude B, rearrange the formula:\[B = \frac{E}{c}\]This gives us the expression for determining the amplitude of the magnetic field when the electric field amplitude is known.
04

Insert Known Values

Insert the given values into the rearranged formula. The electric field magnitude E is given as 3.20 \times 10^{-4} V/m, and c equals 3.00 \times 10^8 m/s:\[B = \frac{3.20 \times 10^{-4}}{3.00 \times 10^8}\]
05

Calculate the Magnetic Field Amplitude

Perform the division to find B:\[B = 1.07 \times 10^{-12} \text{ T}\](The unit 'T' stands for Tesla, the standard unit for magnetic field amplitude.)

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 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field Amplitude
The magnetic field amplitude is an essential property of electromagnetic waves, determining the maximum strength of the magnetic field component. In an electromagnetic wave, such as light, the magnetic field interacts with the electric field to propagate the wave through space.

To find the magnetic field amplitude from a given electric field amplitude, we can use the relationship:
  • Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation.
  • The electric field (\(E\)) and magnetic field (\(B\)) amplitudes are connected through the speed of light (\(c\)).
Given the formula \(E = cB\), one can rearrange to solve for the magnetic field amplitude:\[B = \frac{E}{c}\]By plugging in the known values for the electric field amplitude and the speed of light, we find the specific value for the magnetic field amplitude, expressed in Tesla (T). This process exemplifies how understanding one parameter of a wave helps infer others, emphasizing the intrinsic relationship between electric and magnetic fields in electromagnetic waves.
Electric Field Amplitude
Electric field amplitude refers to the maximum strength of the electric field component in an electromagnetic wave. This amplitude plays a critical role in determining how strongly the wave can exert force on electric charges it encounters.

In the context of electromagnetic waves, such as light or radio waves, the electric field amplitude is one of the primary characteristics:
  • The electric field (\(E\)) defines how charged particles at a given location are affected by the wave.
  • In most equations and problems involving electromagnetic waves, electric field amplitude is given or used to find other values, like magnetic field amplitude.
Understanding electric field amplitude helps illuminate how electromagnetic waves interact with materials and devices. For instance, the biological impact of electromagnetic radiation is partly governed by the electric field intensity. Estimating the amplitude allows us to predict and measure interactions in various contexts, from communication technologies to medical imaging.
Speed of Light
The speed of light (\(c\)), a fundamental constant of nature, plays a critical role in the propagation of electromagnetic waves, linking the electric and magnetic field amplitudes. It is universally defined as approximately 3.00 \(\times 10^8\) meters per second in vacuum.

Key points about the speed of light include:
  • It represents the maximum speed at which information and matter can travel.
  • The constancy of the speed of light underpins many physical laws and concepts, including Einstein's theory of relativity.
  • In electromagnetic wave equations, it allows us to find relationships between electric (\(E\)) and magnetic (\(B\)) field amplitudes.
Using the formula \(E = cB\), we see how the speed of light can be used to determine one field from the other in electromagnetic phenomena. This interdependence emphasizes the integral nature of light's speed in understanding how electromagnetic waves traverse space. Its signification traces into everyday applications like satellite communication, where precise timing hinges on light's speed. Thus, grasping the speed of light's role sheds light on a multitude of technological and scientific endeavors.

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

Figure 33-61 depicts a simplistic optical fiber: a plastic core \(\left(n_{1}=\right.\) \(1.58\) ) is surrounded by a plastic sheath \(\left(n_{2}=1.53\right) .\) A light ray is incident on one end of the fiber at angle \(\theta\). The ray is to undergo total internal reflection at point \(A\), where it encounters the core-sheath boundary. (Thus there is no loss of light through that boundary.) What is the maximum value of \(\theta\) that allows total internal reflection at \(A\) ?

A beam of initially unpolarized light is sent through two polarizing sheets placed one on top of the other. What must be the angle between the polarizing directions of the sheets if the intensity of the transmitted light is to be one-third the incident intensity?

A plane electromagnetic wave traveling in the positive direction of an \(x\) axis in vacuum has components \(E_{x}=E_{y}=0\) and \(E_{z}=(2.0 \mathrm{~V} / \mathrm{m}) \cos \left[\left(\pi \times 10^{15} \mathrm{~s}^{-1}\right)(t-x / c)\right] .(\mathrm{a}) \mathrm{What}\) is the amplitude of the magnetic field component? (b) Parallel to which axis does the magnetic field oscillate? (c) When the electric field component is in the positive direction of the \(z\) axis at a certain point \(P\), what is the direction of the magnetic field component there?

Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at \(1.0 \mathrm{MW}\). What is the intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system, \(4.3\) ly away? (An alien civilization at that distance might be able to watch \(X\) Files.) A light-year (ly) is the distance light travels in one year.

A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through \(360^{\circ}\) while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of \(5.0\) during the rotation, what fraction of the intensity of the original beam is associated with the beam's polarized light?

See all solutions

Recommended explanations on Physics Textbooks

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.