/*! 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 8 (I) If the electric field in an ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(I) If the electric field in an EM wave has a peak magnitude of \(0.57 \times 10^{-4} \mathrm{~V} / \mathrm{m}\), what is the peak magnitude of the magnetic field strength?

Short Answer

Expert verified
The peak magnetic field strength is approximately \( 1.9 \times 10^{-13} \text{ T} \).

Step by step solution

01

Understand the relationship

In an electromagnetic wave, the electric field (E) and magnetic field (B) are related by the speed of light (c). The relationship is given by the formula: \[ c = \frac{E}{B} \]where \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \), \( E \) is the peak electric field magnitude, and \( B \) is the peak magnetic field magnitude.
02

Rearrange the formula to solve for B

Rearrange the formula to solve for the peak magnetic field (B). The equation becomes:\[ B = \frac{E}{c} \]
03

Insert the given values

We are given the peak electric field magnitude \( E = 0.57 \times 10^{-4} \text{ V/m} \). Insert this value and the speed of light \( c = 3 \times 10^8 \text{ m/s} \) into the equation:\[ B = \frac{0.57 \times 10^{-4}}{3 \times 10^8} \]
04

Perform the calculation

Calculate the value of \( B \):\[ B = \frac{0.57 \times 10^{-4}}{3 \times 10^8} = 1.9 \times 10^{-13} \text{ T} \]So, the peak magnetic field magnitude is approximately \( 1.9 \times 10^{-13} \) Tesla.

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.

Electric Field
The electric field is a crucial part of any electromagnetic wave. It's a region where charged particles experience a force. In electromagnetic waves, the electric field alternates direction and magnitude as the wave travels through space. It's measured in volts per meter (V/m), which provides a sense of the force experienced by a charged particle in the field.

One important aspect of electric fields in electromagnetic waves is that they are always perpendicular to the magnetic field. This means that if the electric field travels in one direction, the magnetic field will always be at a right angle to it. The strength or the peak magnitude of this electric field tells us how strong the force would be at its maximum. In the given exercise, this value is given as \(0.57 \times 10^{-4}\mathrm{~V} / \mathrm{m}\).

Understanding this concept is vital for computing the characteristics of electromagnetic waves and correlates directly to understanding how different components of the waves interact.
Magnetic Field
The magnetic field forms another essential component of electromagnetic waves. Just like the electric field, the magnetic field also propagates perpendicular to the direction of wave travel and to the electric field. Magnetic fields are measured in Tesla (T), which gives us an idea of their strength.

When we talk about electromagnetic waves, the magnetic field and electric field work together, oscillating at right angles to each other. The peak magnetic field, which we determined through a calculation using the speed of light, indicates the highest field strength reached during the oscillation.

In the original exercise, by rearranging and applying the formula \(c = \frac{E}{B}\) where \(c\) is the speed of light, the peak magnetic field strength was computed to be approximately \( 1.9 \times 10^{-13} \text{ T} \). This formula highlights the interdependence of electric and magnetic fields in electromagnetic waves and how one can determine the peak of the magnetic field given the electric field and the speed of light.
Speed of Light
The speed of light is a fundamental constant in physics, denoted by \(c\). It is the maximum speed at which all light and other forms of electromagnetic radiation travel in a vacuum. This value is approximately \(3 \times 10^8 \text{ m/s}\), an essential constant for various calculations in physics.

In the context of electromagnetic waves, the speed of light is crucial as it provides the relationship between the electric field \(E\) and the magnetic field \(B\). By understanding this relationship, we can calculate one field's magnitude if we know the other. In the exercise, it helped derive the magnetic field's peak magnitude using the equation \(B = \frac{E}{c}\).

Remember, the speed of light is not just a number - it is a facet of how our universe operates. Its invariance underpins many theories and makes it a cornerstone in the study and understanding of all electromagnetic phenomena.

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

The electric and magnetic fields of a certain EM wave in free space are given by \(\overrightarrow{\mathbf{E}}=E_{0} \sin (k x-\omega t) \hat{\mathbf{j}}+E_{0} \cos (k x-\omega t) \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}=B_{0} \cos (k x-\omega t) \hat{\mathbf{j}}-B_{0} \sin (k x-\omega t) \hat{\mathbf{k}} .\) (a) Show that \(\overrightarrow{\mathbf{E}}\) and \(\overrightarrow{\mathbf{B}}\) are perpendicular to each other at all times \((b)\) For this wave, \(\overrightarrow{\mathbf{E}}\) and \(\overrightarrow{\mathbf{B}}\) are in a plane parallel to the \(y z\) plane. Show that the wave moves in a direction perpendicular to both \(\overrightarrow{\mathbf{E}}\) and \(\overrightarrow{\mathbf{B}}\). (c) At any arbitrary choice of position \(x\) and time \(t,\) show that the magnitudes of \(\overrightarrow{\mathbf{E}}\) and \(\overrightarrow{\mathbf{B}}\) always equal \(E_{0}\) and \(B_{0},\) respectively. \((d)\) At \(x=0,\) draw the orientation of \(\overrightarrow{\mathbf{E}}\) and \(\overrightarrow{\mathbf{B}}\) in the \(y z\) plane at \(t=0 .\) Then qualitatively describe the motion of these vectors in the \(y z\) plane as time increases. [Note: The EM wave in this Problem is "circularly polarized."]

Suppose that a right-moving EM wave overlaps with a leftmoving EM wave so that, in a certain region of space, the total electric field in the \(y\) direction and magnetic field in the \(z\) direction are given by \(E_{y}=E_{0} \sin (k x-\omega t)+E_{0} \sin (k x+\omega t)\) and \(B_{z}=B_{0} \sin (k x-\omega t)-B_{0} \sin (k x+\omega t),(a)\) Find the mathematical expression that represents the standing electric and magnetic waves in the \(y\) and \(z\) directions, respectively. (b) Determine the Poynting vector and find the \(x\) locations at which it is zero at all times.

(I) If the magnetic field in a traveling EM wave has a peak magnitude of \(12.5 \mathrm{nT}\), what is the peak magnitude of the electric field?

Light is emitted from an ordinary lightbulb filament in wave-train bursts about \(10^{-8} \mathrm{~s}\) in duration. What is the length in space of such wave trains?

The metal walls of a microwave oven form a cavity of dimensions \(37 \mathrm{~cm} \times 37 \mathrm{~cm} \times 20 \mathrm{~cm}\). When 2.45 -GHz microwaves are continuously introduced into this cavity, reflection of incident waves from the walls set up standing waves with nodes at the walls. Along the 37 -cm dimension of the oven, how many nodes exist (excluding the nodes at the wall) and what is the distance between adjacent nodes? [Because no heating occurs at these nodes, most microwaves rotate food while operating.

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.