/*! 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 28 The magnetic field in a microwav... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 28

The magnetic field in a microwave traveling through vacuum has amplitude $4.00 \times 10^{-11} \mathrm{T}\( and frequency \)120 \mathrm{GHz}$. Find the amplitude and frequency of the electric field.

Short Answer

Expert verified
Question: Given the amplitude and frequency of the magnetic field of a microwave traveling through a vacuum, calculate the amplitude and frequency of the electric field. Answer: The amplitude of the electric field is 1.20 x 10^-2 N/C and its frequency is 120 x 10^9 Hz.

Step by step solution

01

Recall the relationship between the electric field and the magnetic field in an electromagnetic wave

In a vacuum, the relationship between the electric field amplitude (\(E_0\)) and the magnetic field amplitude (\(B_0\)) is given by the equation: $$E_0 = cB_0$$ where \(c\) is the speed of light in a vacuum, approximately equal to \(3.00 \times 10^8\ \text{m/s}\).
02

Calculate the electric field amplitude

Given the magnetic field amplitude \(B_0 = 4.00 \times 10^{-11}\ \mathrm{T}\), we can find the electric field amplitude using the equation from Step 1: $$E_0 = cB_0$$ $$E_0 = (3.00 \times 10^8\ \mathrm{m/s})(4.00 \times 10^{-11}\ \mathrm{T})$$ $$E_0 = 1.20 \times 10^{-2}\ \mathrm{N/C}$$ Therefore, the amplitude of the electric field in the microwave is \(1.20 \times 10^{-2}\ \mathrm{N/C}\).
03

Identify the frequency of the electric field

Since the frequency of the electric field is the same as the frequency of the magnetic field, we can directly say that the frequency of the electric field is \(120\ \mathrm{GHz}\). It's important to note that, in the context of electromagnetic waves, frequency usually should be expressed in Hz. Converting GHz to Hz, we have: $$120\ \mathrm{GHz} = 120 \times 10^9\ \mathrm{Hz}$$ Thus, the frequency of the electric field is \(120 \times 10^9\ \mathrm{Hz}\). In summary, the amplitude of the electric field is \(1.20 \times 10^{-2}\ \mathrm{N/C}\) and its frequency is \(120 \times 10^9\ \mathrm{Hz}\).

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!

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

Light of wavelength 692 nm in air passes into window glass with an index of refraction of \(1.52 .\) (a) What is the wavelength of the light inside the glass? (b) What is the frequency of the light inside the glass?

The intensity of the sunlight that reaches Earth's upper atmosphere is approximately \(1400 \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the average energy density? (b) Find the rms values of the electric and magnetic fields.
Vertically polarized light with intensity \(I_{0}\) is normally incident on an ideal polarizer. As the polarizer is rotated about a horizontal axis, the intensity \(I\) of light transmitted through the polarizer varies with the orientation of the polarizer \((\theta),\) where \(\theta=0\) corresponds to a vertical transmission axis. Sketch a graph of \(I\) as a function of \(\theta\) for one complete rotation of the polarizer $\left(0 \leq \theta \leq 360^{\circ}\right)$.(W) tutorial: polarized light).
How long does it take sunlight to travel from the Sun to Earth?
A police car's radar gun emits microwaves with a frequency of $f_{1}=36.0 \mathrm{GHz}$. The beam reflects from a speeding car, which is moving away at \(43.0 \mathrm{m} / \mathrm{s}\) with respect to the police car. The frequency of the reflected microwave as observed by the police is \(f_{2} .\) (a) Which is larger, \(f_{1}\) or \(f_{2} ?\) (b) Find the frequency difference \(f_{2}-f_{1}\) [Hint: There are two Doppler shifts. First think of the police as source and the speeder as observer. The speeding car "retransmits" a reflected wave at the same frequency at which it observes the incident wave. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Space Physics

Read Explanation

Medical Physics

Read Explanation

Circular Motion and Gravitation

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.