/*! 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 13 Sunlight just outside Earth's at... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sunlight just outside Earth's atmosphere has an intensity of \(1.40 \mathrm{~kW} / \mathrm{m}^{2} .\) Calculate (a) \(E_{m}\) and (b) \(B_{m}\) for sunlight there, assuming it to be a plane wave.

Short Answer

Expert verified
\(E_m \approx 1026 \, \text{V/m}\), \(B_m \approx 3.42 \times 10^{-6} \, \text{T}\).

Step by step solution

01

Understand the Problem

We are given the intensity of sunlight just outside Earth's atmosphere as \(1.40 \, \text{kW/m}^2\). We need to find the electric field amplitude \(E_m\) and the magnetic field amplitude \(B_m\) for sunlight, assuming it is a plane wave.
02

Recall the Relationship with Intensity

For a plane electromagnetic wave, the intensity \(I\) is given by the equation: \[I = \frac{c \varepsilon_0}{2} E_m^2\]where \(c\) is the speed of light and \(\varepsilon_0\) is the permittivity of free space. This will help us calculate the electric field amplitude \(E_m\).
03

Solve for Electric Field Amplitude \(E_m\)

Rearrange the equation from Step 1 to solve for \(E_m\): \[E_m = \sqrt{\frac{2I}{c \varepsilon_0}}\]Substitute \(I = 1.40 \, \text{kW/m}^2\), \(c = 3.00 \times 10^8 \, \text{m/s}\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\):\[E_m = \sqrt{\frac{2 \times 1400}{3 \times 10^8 \times 8.85 \times 10^{-12}}}\]
04

Calculate Electric Field Amplitude \(E_m\)

Compute the expression from Step 2:\[E_m \approx \sqrt{\frac{2800}{2.655 \times 10^{-3}}} \]\[E_m \approx \sqrt{1055.76 \times 10^3} = 1026 \, \text{V/m}\]
05

Recall the Relationship between \(E_m\) and \(B_m\)

The electric and magnetic fields of a plane wave are related by the speed of light \(c\): \[E_m = cB_m\]This gives the relationship to find \(B_m\) once \(E_m\) is known.
06

Solve for Magnetic Field Amplitude \(B_m\)

Rearrange the equation from Step 4 to solve for \(B_m\):\[B_m = \frac{E_m}{c}\]Use \(E_m = 1026 \, \text{V/m}\) from Step 3 and \(c = 3.00 \times 10^8 \, \text{m/s}\):\[B_m = \frac{1026}{3.00 \times 10^8}\]\[B_m = 3.42 \times 10^{-6} \, \text{T}\]
07

Solution

The electric field amplitude \(E_m\) is approximately 1026 V/m and the magnetic field amplitude \(B_m\) is approximately \(3.42 \times 10^{-6} \, \text{T}\).

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.

intensity of sunlight
The intensity of sunlight, especially just outside the Earth's atmosphere, is a measure of how much energy from sunlight hits a specific area in a given time. This intensity is quantified as energy per unit area and is measured in units like kilowatts per square meter (kW/m²). In the context of electromagnetic waves, intensity helps us understand how powerful the sunlight is.

In our example, the intensity is given as \(1.40 \text{ kW/m}^2\). What this means is that for every square meter in space just outside the Earth, sunlight delivers 1400 watts of energy. This value is crucial as it allows us to determine other properties of the electromagnetic wave, such as electric and magnetic field amplitudes.

It's interesting to note that this intensity is just an average. The actual intensity on Earth can vary due to factors like the angle of the sunlight, atmospheric conditions, and the time of year. Understanding intensity is the first step in delving deeper into the behavior of sunlight as a plane electromagnetic wave.
electric field amplitude
Electric field amplitude, symbolized as \(E_m\), is a key characteristic of electromagnetic waves like sunlight. It represents the strength of the electric field component of the wave.

In the context of light as an electromagnetic wave, the electric field amplitude determines how strongly the wave can exert a force on electric charges it encounters. To find the electric field amplitude from the sunlight intensity, we use the formula:
  • \(E_m = \sqrt{\frac{2I}{c \varepsilon_0}}\)
Where \(c\) is the speed of light, and \(\varepsilon_0\) is the permittivity of free space. These constants help relate the macroscopic property of intensity to the microscopic field strength.

In the given exercise, substituting the known values helps us calculate \(E_m\) as approximately 1026 V/m. This indicates a significant ability to interact with charges, reflecting the strength of sunlight's electric field just outside Earth's atmosphere.
magnetic field amplitude
Magnetic field amplitude, or \(B_m\), complements the electric field amplitude in describing an electromagnetic wave. It quantifies the strength of the magnetic component of the wave. Electromagnetic waves consist of these interconnected electric and magnetic fields that oscillate perpendicularly to each other and the direction of wave travel.

Once we have the electric field amplitude, the magnetic field amplitude can be found using the relationship:
  • \(B_m = \frac{E_m}{c}\)
This equation arises from Maxwell's equations and reflects the constant speed with which electromagnetic waves travel through a vacuum.

With an \(E_m\) of 1026 V/m, substituting into the equation for \(B_m\) gives a value of approximately \(3.42 \times 10^{-6}\) T. This small yet significant value highlights how magnetic fields accompany electric fields in the propagation of electromagnetic waves like sunlight.
speed of light
The speed of light, denoted by \(c\), is a fundamental constant in physics, signifying how fast electromagnetic waves travel in a vacuum. Its value is \(3.00 \times 10^8 \text{ m/s}\). This speed is not just a measure of how fast light moves but also essential in linking electric and magnetic fields within electromagnetic waves.

Understanding the speed of light is crucial for calculations in electromagnetism. For example, it helps calculate the electric and magnetic field amplitudes from the intensity of the wave. The relationship between \(E_m\), \(B_m\), and \(c\) is part of how electromagnetic radiation is characterized as having both electric and magnetic identities moving in unison.

Moreover, \(c\) is important beyond just calculations. It forms the backbone of our understanding that nothing in the universe travels faster than light in a vacuum, providing a universal speed limit. This concept unifies electromagnetism and special relativity, reinforcing our comprehensive grasp of the universe's physical laws.

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

A small spaceship with a mass of only \(1.5 \times 10^{3} \mathrm{~kg}\) (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a \(10 \mathrm{~kW}\) laser beam, what speed will the ship attain in \(1.0\) day because of the momentum carried away by the beam?

A beam of intensity \(I\) reflects from a long, totally reflecting cylinder of radius \(R ;\) the beam is perpendicular to the central axis of the cylinder and has a diameter larger than \(2 R\). What is the beam's force per unit length on the cylinder?

What is the radiation pressure \(1.5 \mathrm{~m}\) away from a \(500 \mathrm{~W}\) lightbulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.

Two polarizing sheets, one directly above the other, transmit \(p \%\) of the initially unpolarized light that is perpendicularly incident on the top sheet. What is the angle between the polarizing directions of the two sheets?

During a test, a NATO surveillance radar system, operating at \(12 \mathrm{GHz}\) at \(180 \mathrm{~kW}\) of power, attempts to detect an incoming stealth aircraft at \(90 \mathrm{~km}\). Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft's location? The aircraft reflects radar waves as though it has a cross-sectional area of only \(0.22 \mathrm{~m}^{2} .\) (b) What is the power of the aircraft's reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value of the magnetic field of the reflected radar beam?

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.