Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 31: Problem 35

A monochromatic point source of light emits \(1.5 \mathrm{~W}\) of electromagnetic power uniformly in all directions. Find the Poynting vector at a point situated at each of the following locations: a) \(0.30 \mathrm{~m}\) from the source b) \(0.32 \mathrm{~m}\) from the source c) \(1.00 \mathrm{~m}\) from the source

Short Answer

Expert verified
Question: Calculate the Poynting vector (energy flux) at points with different distances from a point source of light that emits 1.5 W of electromagnetic power uniformly in all directions: a) 0.30m, b) 0.32m, and c) 1.00m. Answer: a) At 0.30m, the Poynting vector is 1.326 W/m². b) At 0.32m, the Poynting vector is 1.161 W/m². c) At 1.00m, the Poynting vector is 0.119 W/m².

Step by step solution

01

Find the area at 0.30m distance

First, we need to find the surface area for a sphere whose radius is 0.30m. We'll use the formula for the surface area of a sphere: $$A = 4\pi r^2$$ With \(r = 0.30 \mathrm{~m}\): $$A_{0.30m} = 4\pi (0.30)^2 = 4\pi (0.09) = 1.131\mathrm{~m²}$$
02

Calculate the Poynting vector at 0.30m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.30m distance from the source: $$\vec{S}=\frac{P}{A}$$ With \(P = 1.5 \mathrm{~W}\) and \(A = 1.131\mathrm{~m²}\): $$\vec{S}_{0.30m}=\frac{1.5}{1.131} = 1.326\mathrm{~W/m²}$$
03

Find the area at 0.32m distance

As we did in Step 1, we need to find the surface area for a sphere whose radius is 0.32m: $$A_{0.32m} = 4\pi (0.32)^2 = 4\pi (0.1024) = 1.292\mathrm{~m²}$$
04

Calculate the Poynting vector at 0.32m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.32m distance from the source: $$\vec{S}_{0.32m}=\frac{P}{A}$$ With \(A = 1.292\mathrm{~m²}\): $$\vec{S}_{0.32m}=\frac{1.5}{1.292} = 1.161\mathrm{~W/m²}$$
05

Find the area at 1.00m distance

Finally, we need to find the surface area for a sphere whose radius is 1.00m: $$A_{1.00m} = 4\pi (1.00)^2 = 4\pi = 12.566\mathrm{~m²}$$
06

Calculate the Poynting vector at 1.00m distance

We'll use the formula for the Poynting vector to find the energy flux at 1.00m distance from the source: $$\vec{S}_{1.00m}=\frac{P}{A}$$ With \(A = 12.566\mathrm{~m²}\): $$\vec{S}_{1.00m}=\frac{1.5}{12.566} = 0.119\mathrm{~W/m²}$$ To summarize, the Poynting vector for the given distances is: a) \(0.30\mathrm{~m}\) : \(\vec{S}_{0.30m}=1.326\mathrm{~W/m²}\) b) \(0.32\mathrm{~m}\) : \(\vec{S}_{0.32m}=1.161\mathrm{~W/m²}\) c) \(1.00\mathrm{~m}\) : \(\vec{S}_{1.00m}=0.119\mathrm{~W/m²}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Silica aerogel, an extremely porous, thermally insulating material made of silica, has a density of \(1.00 \mathrm{mg} / \mathrm{cm}^{3}\). A thin circular slice of aerogel has a diameter of \(2.00 \mathrm{~mm}\) and a thickness of \(0.10 \mathrm{~mm}\). a) What is the weight of the aerogel slice (in newtons)? b) What is the intensity and radiation pressure of a \(5.00-\mathrm{mW}\) laser beam of diameter \(2.00 \mathrm{~mm}\) on the sample? c) How many \(5.00-\mathrm{mW}\) lasers with a beam diameter of \(2.00 \mathrm{~mm}\) would be needed to make the slice float in the Earth's gravitational field? Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

What is the distance between successive heating antinodes in a microwave oven's cavity? A microwave oven typically operates at a frequency of \(2.4 \mathrm{GHz}\).

The current flowing in a solenoid that is \(20.0 \mathrm{~cm}\) long and has a radius of \(2.00 \mathrm{~cm}\) and 500 turns decreases from \(3.00 \mathrm{~A}\) to \(1.00 \mathrm{~A}\) in \(0.1 .00 \mathrm{~s} .\) Determine the magnitude of the induced electric field inside the solenoid \(1.00 \mathrm{~cm}\) from its center.

Which of the following statements concerning electromagnetic waves are incorrect? (Select all that apply.) a) Electromagnetic waves in vacuum travel at the speed of light. b) The magnitudes of the electric field and the magnetic field are equal. c) Only the electric field vector is perpendicular to the direction of the wave's propagation. d) Both the electric field vector and the magnetic field vector are perpendicular to the direction of propagation. e) An electromagnetic wave carries energy only when \(E=B\).
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Torque and Rotational Motion

Read Explanation

Atoms and Radioactivity

Read Explanation

Electricity and Magnetism

Read Explanation

Wave Optics

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.

Sign-up for free