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\(\bullet$$\bullet \mathrm{A}\) sinusoidal electromagnetic wave of frequency \(6.10 \times 10^{14} \mathrm{Hz}\) travels in vacuum in the \(+x\) -direction. The magnetic field is parallel to the \(y\) -axis and has amplitude \(5.80 \times 10^{-4} \mathrm{T} .\) (a) Find the magnitude and direction of the electric field. (b) Write the wave functions for the electric and magnetic fields in the form of Equations \((23.3).\)

Short Answer

Expert verified
(a) Electric field magnitude is \(1.74 \times 10^{5} \text{ V/m}\) in the \(z\)-direction. (b) Wave functions: \(E(x, t)=(1.74 \times 10^{5}) \sin(kx-\omega t)\) and \(B(x, t)=(5.80 \times 10^{-4}) \sin(kx-\omega t)\).

Step by step solution

01

Relationship between Electric and Magnetic Fields

Electromagnetic waves consist of mutually perpendicular electric and magnetic fields propagating in a direction perpendicular to both. The magnitudes of these fields are related by the speed of light in a vacuum, \( c \), as given by the equation: \( E = cB \), where \( E \) is the magnitude of the electric field and \( B \) is the magnitude of the magnetic field.
02

Calculate the Electric Field Magnitude

Given: \( B = 5.80 \times 10^{-4} \text{ T} \) and \( c = 3.00 \times 10^{8} \text{ m/s} \). Calculate \( E \) using \( E = cB \): \[ E = (3.00 \times 10^{8} )(5.80 \times 10^{-4}) = 1.74 \times 10^{5} \text{ V/m} \]. Thus, the magnitude of the electric field is \( 1.74 \times 10^{5} \text{ V/m} \).
03

Determine the Direction of the Electric Field

The direction of the electric field is perpendicular both to the direction of wave propagation (\( +x \) direction) and to the magnetic field (\( y \)-axis). Therefore, the electric field must be along the \( z \)-axis.
04

Determine the Wave Functions

In the form of equations \( 23.3 \), the wave function for the electric field is given by: \( E(z, t) = E_0 \sin(kx - \omega t) \). The angular frequency \( \omega = 2\pi f = 2 \pi \times 6.10 \times 10^{14} \), and the wave number \( k = \frac{\omega}{c} \). Substitute the values to get: \( E(x, t) = (1.74 \times 10^{5}) \sin(kx - \omega t) \) (along \( z \)-direction).Similarly, for the magnetic field, \( B(x, t) = B_0 \sin(kx - \omega t) \), where \( B_0 = 5.80 \times 10^{-4} \text{ T} \) (along \( y \)-direction).

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Key Concepts

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

Electric Field Calculation
Electromagnetic waves are fascinating and can be characterized by both electric and magnetic fields that oscillate in harmony. In any electromagnetic wave traveling through a vacuum, the electric field (\( E \)) is interlinked with the magnetic field (\( B \)) through the simple and fundamental relationship: \[ E = cB \]This equation tells us that the magnitude of the electric field is equal to the speed of light (\( c = 3.00 \times 10^8 \text{ m/s} \)) multiplied by the magnitude of the magnetic field.

In a given problem, if we know the magnetic field amplitude is \( 5.80 \times 10^{-4} \text{ T} \), the electric field's magnitude can be calculated by inputting these values into the formula:
  • Given: \( B = 5.80 \times 10^{-4} \text{ T} \)
  • Use: \( E = cB \)
  • Calculation: \( E = (3.00 \times 10^8 \text{ m/s})(5.80 \times 10^{-4} \text{ T}) = 1.74 \times 10^5 \text{ V/m} \)
We conclude that the electric field's magnitude is \( 1.74 \times 10^5 \text{ V/m} \). This formula is incredibly useful for connecting the electric and magnetic aspects of electromagnetic waves.
Magnetic Field
Magnetic fields in electromagnetic waves are just as crucial as electric fields, having their own amplitude and direction. Such fields are oriented perpendicular to both the electric field and the direction of wave propagation.

In our specific problem, the magnetic field is oriented parallel to the y-axis while the wave propagates in the +x direction. This orientation is essential, as it dictates the direction of the electric field to ensure that all components maintain perpendicular alignment:
  • The magnetic field's amplitude is given as \( 5.80 \times 10^{-4} \text{ T} \)
  • It follows the equation for wave functions: \( B(x, t) = B_0 \sin(kx - \omega t) \)
This sin function allows us to describe the oscillating nature of the magnetic field as the wave moves through space and time, with \( B_0 \) representing the wave’s maximum magnetic field strength. By understanding these core principles, we can comprehend the dynamic and interconnected relationship between electric and magnetic fields in electromagnetic waves.
Wave Function
The wave function is a mathematical representation that describes how electromagnetic waves vary in space and time. Both the electric and magnetic components of an electromagnetic wave can be expressed through wave functions, typically presented in sinusoidal forms.

For an electromagnetic wave traveling in the +x direction with a frequency of \( 6.10 \times 10^{14} \text{ Hz} \), we use the wave function formula to describe the electric field component. This is noted as:\[ E(x, t) = E_0 \sin(kx - \omega t) \]where:
  • \( E_0 \) is the amplitude of the electric field, \( 1.74 \times 10^5 \text{ V/m} \)
  • \( k \) is the wave number, calculated as \( k = \frac{\omega}{c} \) with \( \omega \) being the angular frequency \( 2\pi f \)
  • \( \omega \) is \( 2\pi \times 6.10 \times 10^{14} \)
Similarly for the magnetic field:\[ B(x, t) = B_0 \sin(kx - \omega t) \]where \( B_0 \) is the magnetic field amplitude. These equations mirror the sinusoidal nature of wave propagation and provide valuable insights into how waves behave over time and across distances, presenting a full picture of the oscillating fields within an electromagnetic wave.

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Most popular questions from this chapter

\(\bullet\) Visible light. The wavelength of visible light ranges from 400 nm to 700 nm. Find the corresponding ranges of this light's (a) frequency, (b) angular frequency, (c) wave number.

\(\bullet$$\bullet \mathrm{A} 45^{\circ}-45^{\circ}-90^{\circ}\) prism is immersed in water. A ray of light is incident normally on one of the prism's shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism?

\(\bullet\) Light with a frequency of \(5.80 \times 10^{14}\) Hz travels in a block of glass that has an index of refraction of \(1.52 .\) What is the wavelength of the light (a) in vacuum and (b) in the glass?

\(\bullet\) The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=-(375 \mathrm{V} / \mathrm{m}) \sin [(5.97 \times\) \(10^{15} \operatorname{rad} / \mathrm{s} ) t+\left(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x ] .\) (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave?

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 \(\mathrm{m}^{2} .\) At the window, the electric field of the wave has rea value 0.0200 \(\mathrm{V} / \mathrm{m} .\) How much energy does this wave carry through the window during a 30.0 s commercial?

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