/*! 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 66 The rms value of the electric fi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 66

The rms value of the electric field of the light coming from the sun is \(720 \mathrm{~N} / \mathrm{C}\). The average total energy density of the electromagnetic wave is [2006] (A) \(3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{3}\) (B) \(4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^{3}\) (C) \(6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^{3}\) (D) \(81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The average total energy density of the electromagnetic wave is approximately \(2.299 \times 10^{-6} \mathrm{J/m^3}\), which is closest to option (B) \(4.58 \times 10^{-6} \mathrm{J/m^3}\). Therefore, the correct answer is (B) \(4.58 \times 10^{-6} \mathrm{J/m^3}\).

Step by step solution

01

Identify the given parameters

We are given the rms value of the electric field (\(E_{rms}\)) of the light coming from the sun, which is 720 N/C. We also know the value of vacuum permittivity (\(\epsilon_0\)), which is approximately \(8.854 \times 10^{-12} \mathrm{C^2/Nm^2}\).
02

Calculate the energy density using the formula

We can now use the formula for the energy density of an electromagnetic wave: \(u = \frac{1}{2}\epsilon_0E^2\). Substitute the given values into the formula. \(u = \frac{1}{2}(8.854 \times 10^{-12} \mathrm{C^2/Nm^2})(720 \mathrm{N/C})^2\)
03

Simplify and find the energy density

Now, we can simplify the expression and find the energy density. \(u = \frac{1}{2}(8.854 \times 10^{-12} \mathrm{C^2/Nm^2})(518400 \mathrm{N^2/C^2})\) \(u = 2.299 \times 10^{-6} \mathrm{J/m^3}\)
04

Comparing with the given options

We have now calculated the average total energy density of the electromagnetic wave to be approximately \(2.299 \times 10^{-6} \mathrm{J/m^3}\). We can now compare this result with the given choices: (A) \(3.3 \times 10^{-3} \mathrm{J/m^3}\) (B) \(4.58 \times 10^{-6} \mathrm{J/m^3}\) (C) \(6.37 \times 10^{-9} \mathrm{J/m^3}\) (D) \(81.35 \times 10^{-12} \mathrm{J/m^3}\) We can see that our calculated value is closest to option (B). Therefore, the correct answer is (B) \(4.58 \times 10^{-6} \mathrm{J/m^3}\).

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

Kenneley's Heaviside layer lies between (A) \(50 \mathrm{~km}\) to \(80 \mathrm{~km}\) (B) \(80 \mathrm{~km}\) to \(400 \mathrm{~km}\) (C) beyond \(110 \mathrm{~km}\) (D) beyond \(250 \mathrm{~km}\)

A laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of \(1 \mathrm{nW}\) and its wavelength is \(632.8 \mathrm{~nm}\). The intensity of focussed beam will be (A) \(1.5 \times 10^{13} \mathrm{~W} / \mathrm{m}^{2}\) (B) \(2.5 \times 10^{9} \mathrm{~W} / \mathrm{m}^{2}\) (C) \(3.5 \times 10^{17} \mathrm{~W} / \mathrm{m}^{2}\) (D) None of these

An electric field of \(300 \mathrm{~V} / \mathrm{m}\) is confined to a circular area \(10 \mathrm{~cm}\) in diameter. If the field is increasing at the rate of \(20 \mathrm{~V} / \mathrm{m}-\mathrm{s}\), the magnitude of magnetic field at a point \(15 \mathrm{~cm}\) from the centre of the circle will be (A) \(1.85 \times 10^{-15} \mathrm{~T}\) (B) \(1.85 \times 10^{-16} \mathrm{~T}\) (C) \(1.85 \times 10^{-17} \mathrm{~T}\) (D) \(1.85 \times 10^{-18} \mathrm{~T}\)

The displacement current flows in the dielectric of a capacitor (A) becomes zero. (B) has assumed a constant value. (C) is increasing with time. (D) is decreasing with time.

An electromagnetic wave of frequency \(v=3.0 \mathrm{MHz}\) passes from vacuum into a dielectric medium with permittivity \(\varepsilon=4.0 .\) Then (A) wavelength is doubled and the frequency remains unchanged. (B) wavelength is doubled and frequency becomes half. (C) wavelength is halved and frequency remains unchanged. (D) wavelength and frequency both remain unchanged.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Radiation

Read Explanation

Electric Charge Field and Potential

Read Explanation

Turning Points in Physics

Read Explanation

Wave Optics

Read Explanation

Famous Physicists

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.