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Chapter 33: Problem 7

What is the intensity of a traveling plane electromagnetic wave if \(B_{m}\) is \(1.0 \times 10^{-4} \mathrm{~T} ?\)

Short Answer

Expert verified
The intensity is approximately 1193.25 W/m虏.

Step by step solution

01

Identify the formula for intensity

The intensity of a traveling electromagnetic wave is given by the formula \( I = \frac{1}{2}c\varepsilon_0 E_{m}^2 \), where \( c \) is the speed of light, \( \varepsilon_0 \) is the permittivity of free space, and \( E_m \) is the maximum electric field. We also know that \( E_m = cB_m \), where \( B_m \) is the maximum magnetic field.
02

Substitute known values

Given that \( B_m = 1.0 \times 10^{-4} \mathrm{~T} \), we can find the maximum electric field using \( E_m = cB_m \). Here, \( c = 3.0 \times 10^{8} \mathrm{~m/s} \). So, \( E_m = 3.0 \times 10^{8} \times 1.0 \times 10^{-4} = 3.0 \times 10^{4} \mathrm{~V/m} \).
03

Calculate intensity using given parameters

Substitute \( E_m = 3.0 \times 10^{4} \mathrm{~V/m} \), \( c = 3.0 \times 10^{8} \mathrm{~m/s}\), and \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m} \) into the intensity formula: \[ I = \frac{1}{2} \times 3.0 \times 10^{8} \times 8.85 \times 10^{-12} \times (3.0 \times 10^{4})^2 \].
04

Perform the calculation

Calculate step by step: First, compute \( (3.0 \times 10^{4})^2 = 9.0 \times 10^{8} \). Then multiply \( 3.0 \times 10^{8} \times 8.85 \times 10^{-12} = 2.655 \times 10^{-3} \). Finally, calculate the intensity: \( I = \frac{1}{2} \times 2.655 \times 10^{-3} \times 9.0 \times 10^{8} \), which gives \( I = 1193.25 \mathrm{~W/m^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Wave
Electromagnetic waves are fascinating phenomena that travel through space and carry energy. They are made up of oscillating electric and magnetic fields that propagate perpendicular to each other.
Electromagnetic waves include a wide spectrum, ranging from radio waves with long wavelengths to extremely short gamma rays. This means they play integral roles in various technologies and natural processes.
The speed of an electromagnetic wave in a vacuum is constant and equal to the speed of light, denoted by \( c = 3.0 \times 10^{8} \ m/s \). It is this consistent speed that allows us to determine other variables like wave intensity.
Magnetic Field
The magnetic field within an electromagnetic wave is crucial as it helps define the wave's characteristics. Magnetic fields are regions where magnetic forces can be observed, typically described by the magnetic induction \( B \). For electromagnetic waves, this is commonly captured in the concept of the maximum magnetic field \( B_{m} \).
When referring to electromagnetic wave intensity, \( B_{m} \) is critical for calculating the maximum electric field \( E_m \) using the relationship \( E_{m} = cB_{m} \). Knowing \( B_m \) enables the full characterization of the wave's intensity and its capacity to transmit energy.
For the scenario \( B_{m} = 1.0 \times 10^{-4} \ T \), this magnetic field guides the formulation of the corresponding electric field and its influence on wave intensity.
Electric Field
Electric fields are essential components of electromagnetic waves, representing the force a charge would experience at any point in space. In the context of electromagnetic waves, the electric field vector is one that oscillates perpendicular to the direction of the magnetic field vector.
The maximum electric field \( E_m \) is related to the maximum magnetic field \( B_m \) through the equation \( E_{m} = cB_{m} \). In this context, knowing \( B_m \) allows you to compute \( E_m \), an essential step to finding the wave's intensity.
For example, with a given \( B_{m} = 1.0 \times 10^{-4} \ T \), we find \( E_m \) to be \( 3.0 \times 10^{4} \ V/m \), confirming that electric fields are pivotal for calculating energy transmission in waves.
Permittivity of Free Space
Permittivity of free space, represented as \( \varepsilon_0 \), is a fundamental constant that represents how electric fields interact with the vacuum. Its value is about \( 8.85 \times 10^{-12} \ F/m \) and it plays a vital role in electric and electromagnetic theory.
This constant influences the intensity of electromagnetic waves through the formula \( I = \frac{1}{2}c\varepsilon_0 E_{m}^2 \). It indicates the ability of the vacuum to permit electric field lines, which is crucial for deriving the intensity of the wave.
Understanding \( \varepsilon_0 \) aids in several applications, from telecommunications involving wave propagation to intricate electric field calculations in physics.

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Most popular questions from this chapter

We want to rotate the direction of polarization of a beam of polarized light through \(90^{\circ}\) by sending the beam through one or more polarizing sheets. (a) What is the minimum number of sheets required? (b) What is the minimum number of sheets required if the transmitted intensity is to be more than \(60 \%\) of the original intensity?

Prove, for a plane electromagnetic wave that is normally incident on a flat surface, that the radiation pressure on the surface is equal to the energy density in the incident beam. (This relation between pressure and energy density holds no matter what fraction of the incident energy is reflected.)

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) Assuming that Earth (and its atmosphere) behaves like a flat disk perpendicular to the Sun's rays and that all the incident energy is absorbed, calculate the force on Earth due to radiation pressure. (b) For comparison, calculate the force due to the Sun's gravitational attraction.

In Fig. \(33-41,\) a beam of unpolarized light, with intensity \(43 \mathrm{~W} / \mathrm{m}^{2}\), is sent into a system of two polarizing sheets with polarizing dircctions at angles \(\theta_{1}=70^{\circ}\) and \(\theta_{2}=90^{\circ}\) to the \(y\) axis. What is the intensity of the light transmitted by the system?

A ray of white light traveling through fused quartz is incident at a quartz- air interface at angle \(\theta_{1}\). Assume that the index of refraction of quartz is \(n=1.456\) at the red end of the visible range and \(n=1.470\) at the blue end. If \(\theta_{1}\) is (a) \(42.00^{\circ},\) (b) \(43.10^{\circ}\), and (c) \(44.00^{\circ},\) is the refracted light white, white dominated by the red end of the visible range, or white dominated by the blue end of the visible range, or is there no refracted light?
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