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An electromagnetic wave of wavelength \(435 \mathrm{~nm}\) is traveling in vacuum in the \(-z\) -direction. The electric field has amplitude \(2.70 \times 10^{-3} \mathrm{~V} / \mathrm{m}\) and is parallel to the \(x\) -axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t)\).

Short Answer

Expert verified
(a) Frequency is approximately \(6.90 \times 10^{14} \text{Hz}\). (b) Magnetic-field amplitude is \(9.00 \times 10^{-12} \text{T}\). (c) Vector equations: \(\overrightarrow{E}(z, t) = 2.70 \times 10^{-3} \cos(1.45\times10^7 z + 4.33\times10^{15} t) \hat{i}\), \(\overrightarrow{B}(z, t) = 9.00 \times 10^{-12} \cos(1.45\times10^7 z + 4.33\times10^{15} t) \hat{j}\).

Step by step solution

01

Convert Wavelength to Frequency

The frequency of an electromagnetic wave, \( f \), is related to its wavelength, \( \lambda \), by the speed of light in vacuum. The relationship is given by:\[f = \frac{c}{\lambda}\]where \( c = 3 \times 10^8 \ \text{m/s} \). Given \( \lambda = 435 \ \text{nm} \), first convert this to meters: \( \lambda = 435 \times 10^{-9} \ \text{m} \).Now plug into the formula:\[f = \frac{3 \times 10^8}{435 \times 10^{-9}} \approx 6.90 \times 10^{14} \ \text{Hz}\]
02

Find Magnetic-Field Amplitude

The amplitude of the magnetic field, \( B_0 \), can be found using the relation between the electric field amplitude, \( E_0 \), and \( B_0 \):\[E_0 = c \cdot B_0\]Rearrange the formula to solve for \( B_0 \):\[B_0 = \frac{E_0}{c} = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \ \text{T}\]
03

Write Electric Field Equation

For a wave traveling in the \(-z\) direction with the electric field in the \(x\)-axis, we write:\[\overrightarrow{E}(z, t) = E_0 \cos(kz + \omega t) \hat{i}\]where \( E_0 = 2.70 \times 10^{-3} \ \text{V/m} \).Since \( k = \frac{2\pi}{\lambda} \approx 1.45 \times 10^7 \ \text{m}^{-1} \), and the angular frequency \( \omega = 2\pi f \approx 4.33 \times 10^{15} \ \text{rad/s} \):\[\overrightarrow{E}(z, t) = (2.70 \times 10^{-3}) \cos(1.45 \times 10^7 z + 4.33 \times 10^{15} t) \hat{i}\]
04

Write Magnetic Field Equation

For the magnetic field of a wave traveling in the \(-z\) direction, perpendicular to both \( \overrightarrow{E} \) and the direction of travel, we write:\[\overrightarrow{B}(z, t) = B_0 \cos(kz + \omega t) \hat{j}\]where \( B_0 = 9.00 \times 10^{-12} \ \text{T} \).Substituting the values from Step 2 and Step 3:\[\overrightarrow{B}(z, t) = (9.00 \times 10^{-12}) \cos(1.45 \times 10^7 z + 4.33 \times 10^{15} t) \hat{j}\]

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Key Concepts

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

Wavelength to Frequency Conversion
Electromagnetic waves move at the speed of light in a vacuum. The speed of light, denoted as \( c \), is a well-known constant value of \( 3 \times 10^8 \ ext{m/s} \). This speed is crucial when you're converting between wavelength and frequency, two essential properties of waves. The formula connecting wavelength \( \lambda \) and frequency \( f \) is:
  • \( f = \frac{c}{\lambda} \)
In this equation, if you know the wavelength of the wave, you can easily find its frequency by dividing the speed of light by the wavelength. Remember, the units are important here. Wavelength often comes in nanometers (nm) for these equations, so be sure to convert to meters (m) by multiplying by \( 10^{-9} \).
In our example, with a wavelength of 435 nm, the conversion goes as follows:
  • \( \lambda = 435 \times 10^{-9} \ ext{m} \)
  • \( f = \frac{3 \times 10^8}{435 \times 10^{-9}} \approx 6.90 \times 10^{14} \ ext{Hz} \)
This relationship helps us understand that as the wavelength decreases, the frequency increases—a fundamental concept in wave physics.
Electric Field and Magnetic Field Relationship
Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. The relationship between the electric field \( E \) and the magnetic field \( B \) is direct and can be given by the formula:
  • \( E = c \cdot B \)
Here, \( c \) is again the speed of light. The electric field's amplitude \( E_0 \) can be used to find the magnetic field's amplitude \( B_0 \) using:
  • \( B_0 = \frac{E_0}{c} \)
In practice, if you know the amplitude of the electric field, finding the magnetic field's amplitude is straightforward. For our scenario, with an electric field amplitude of \( 2.70 \times 10^{-3} \ ext{V/m} \):
  • \( B_0 = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \ ext{T} \)
This illustrates how intimately tied these two fields are in electromagnetic waves. They sustain each other as the wave propagates through space.
Wave Equation Derivation
Understanding wave equations is essential for describing electromagnetic wave behavior. The vector wave equations for the electric field \( \overrightarrow{E} \) and the magnetic field \( \overrightarrow{B} \) involve amplitude, wave number \( k \), and angular frequency \( \omega \).For an electric field propagating in the \(-z\)-direction and oscillating along the \(x\)-axis, the equation is:
  • \( \overrightarrow{E}(z, t) = E_0 \cos(kz + \omega t) \hat{i} \)
Here:
  • \( E_0 = 2.70 \times 10^{-3} \ ext{V/m} \)
  • \( k = \frac{2\pi}{\lambda} \approx 1.45 \times 10^7 \ ext{m}^{-1} \)
  • \( \omega = 2\pi f \approx 4.33 \times 10^{15} \ ext{rad/s} \)
The corresponding magnetic field, perpendicular and propagating in the same direction, can be expressed as:
  • \( \overrightarrow{B}(z, t) = B_0 \cos(kz + \omega t) \hat{j} \)
  • \( B_0 = 9.00 \times 10^{-12} \ ext{T} \)
Wave equations show the spatial and temporal dependence of the fields, depicting how they vary over time and space. These relationships govern how electromagnetic waves travel through a vacuum, maintaining the interconnected nature of their fields.

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Most popular questions from this chapter

An electromagnetic wave with frequency 65.0 \(\mathrm{Hz}\) travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude \(7.20 \times 10^{-3} \mathrm{V} / \mathrm{m}\) . (a) What is the speed of propagation of the wave? (b) what is the wavelength of the wave? (c) What is the amplitude of the magnetic field? (d) What is the intensity of the wave?

A sinusoidal electromagnetic wave of frequency \(6.10 \times 10^{14} \mathrm{Hz}\) travels in vacuum in the \(+z\) -direction. The \(\overrightarrow{\boldsymbol{B}}\) -field is parallel to the \(y\) -axis and has amplitude \(5.80 \times 10^{-4}\) T. Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t) .\)

We can reasonably model a 75-W incandescent light-bulb as a sphere 6.0 \(\mathrm{cm}\) in diameter. Typically, only about 5\(\%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\mathrm{W} / \mathrm{m}^{2} )\) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in pliysics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is 2.5 \(\mathrm{km}\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090 \(\mathrm{V} / \mathrm{m}\) and that the frequency of the waves is 244 \(\mathrm{MHz}\) . Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

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