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Chapter 32: Problem 8

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 \(\vec { E } ( z , t )\) and \(\vec { \boldsymbol { B } } ( z , t )\)

Short Answer

Expert verified
(a) Frequency is approximately \(6.90 \times 10^{14} Hz\). (b) Magnetic field amplitude is about \(9.00 \times 10^{-12} T\). (c) Vector equations are \( \vec{E}(z,t) = 2.70 \times 10^{-3} \cos(kz + \omega t) \hat{i}\) and \( \vec{B}(z,t) = 9.00 \times 10^{-12} \cos(kz + \omega t) \hat{j}\).

Step by step solution

01

Calculate the Frequency

The speed of light in vacuum is given by the equation \( c = \lambda f \), where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \), \( \lambda \) is the wavelength, and \( f \) is the frequency. First, convert the given wavelength from nanometers to meters: \( 435 \, \text{nm} = 435 \times 10^{-9} \, \text{m} \). Next, solve for the frequency \( f \):\[f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{435 \times 10^{-9} \, \text{m}}\]Calculate the result to find \( f \).
02

Calculate the Magnetic Field Amplitude

Using the relation between the electric field \( E \) and the magnetic field \( B \) amplitude in an electromagnetic wave in vacuum, given by \( c = \frac{E}{B} \), rearrange to find \( B \):\[ B = \frac{E}{c} \]Substitute \( E = 2.70 \times 10^{-3} \, \text{V/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \) to find \( B \).
03

Write the Vector Equations

The electric field vector for a wave traveling in the negative z-direction can be written as:\[ \vec{E}(z,t) = E_0 \cos(kz + \omega t) \hat{i} \]where \( E_0 = 2.70 \times 10^{-3} \, \text{V/m} \), the wave number \( k = \frac{2\pi}{\lambda} \), and the angular frequency \( \omega = 2\pi f \). Write \( k \) and \( \omega \) using the previously calculated frequency and wavelength.Similarly, the magnetic field vector is:\[ \vec{B}(z,t) = B_0 \cos(kz + \omega t) \hat{j} \]Use the previously calculated \( B_0 \) value to express the full magnetic field equation.

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Key Concepts

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

Frequency Calculation
To determine the frequency of an electromagnetic wave, we start by recognizing the relationship between the speed of light, wavelength, and frequency. The equation is given by:
  • Speed of light: \(c = \lambda f\), where \(c = 3 \times 10^8 \, \text{m/s}\).
  • Wavelength \(\lambda = 435 \, \text{nm}\), which converts to meters as \(435 \times 10^{-9} \, \text{m}\).
  • Solve for frequency \(f\) using: \[f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{435 \times 10^{-9} \, \text{m}}\]
Calculating this gives us the frequency of the electromagnetic wave. Remember, frequency is how many times a wave passes a point per second, measured in Hertz (Hz). It's foundational as it determines the energy and type of wave, such as radio, visible light, or gamma rays.
Wavelength Conversion
When working with electromagnetic waves, it's often necessary to convert the wavelength to a standard unit of meters. This is because the speed of light, often used in calculations, is provided in meters per second.
  • Given wavelength in nanometers: \(\lambda = 435 \, \text{nm}\).
  • Conversion factor: 1 nm = \(10^{-9}\) meters.
  • Converted wavelength: \(435 \, \text{nm} = 435 \times 10^{-9} \, \text{m}\).
Converting these units ensures consistency across calculations and makes the integration of the values with constants like the speed of light seamless. Always keep track of the units to avoid errors.
Magnetic Field Amplitude
In electromagnetic waves, there is a direct relationship between the electric field amplitude and the magnetic field amplitude, given by:
  • Relationship: \(c = \frac{E}{B}\)
  • Where \(E\) is the electric field amplitude, \(2.70 \times 10^{-3} \, \text{V/m}\).
  • Solve for \(B\) using: \[ B = \frac{E}{c} \]
  • Substitute \(c = 3 \times 10^8 \, \text{m/s}\) to find \(B\).
Calculating this provides the amplitude of the magnetic field. This demonstrates one of Maxwell's equations principles indicating that electric and magnetic fields are directly proportional in a vacuum, traveling at the speed of light.
Vector Equations
For electromagnetic waves, understanding the vector nature of the fields is crucial as they describe the wave's propagation and orientation.
The general formula for the electric field vector in the negative z-direction is:
  • \( \vec{E}(z,t) = E_0 \cos(kz + \omega t) \hat{i} \), where:
    • \(E_0 = 2.70 \times 10^{-3} \, \text{V/m}\)
    • Wave number \(k = \frac{2\pi}{\lambda}\)
    • Angular frequency \(\omega = 2\pi f\)
Similarly, the formula for the magnetic field vector is:
  • \( \vec{B}(z,t) = B_0 \cos(kz + \omega t) \hat{j} \)
These vector equations illustrate how electromagnetic fields oscillate perpendicular to the direction of wave propagation and to each other, maintaining the transverse wave property of light.

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Most popular questions from this chapter

A sinusoidal electromagnetic wave is propagating in vacuum in the \(+ z\) -direction. If at a particular instant and at a certain point in space the electric field is in the \(+ x - x -\) -direction and has mag- nitude \(4.00 \mathrm { V } / \mathrm { m } ,\) what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, 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\(\mathrm { MH } z\) . 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?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) Verify this by substituting \(E _ { y } ( x , t )\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) m; permeability \(\mu = \mu _ { 0 } )\) . since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

An electromagnetic standing wave in a certain material has frequency \(1.20 \times 10 ^ { 10 } \mathrm { Hz }\) and speed of propagation \(2.10 \times\) \(10 ^ { 8 } \mathrm { m } / \mathrm { s } .\) (a) What is the distance between a nodal plane of \(\vec { \boldsymbol { B } }\) and the closest antinodal plane of \(\vec { \boldsymbol { B } } ?\) (b) What is the distance between an antinodal plane of \(\vec { \boldsymbol { E } }\) and the closest antinodal plane of \(\vec { \boldsymbol { B } }\) ? (c) What is the distance between a nodal plane of \(\vec { \boldsymbol { E } }\) and the closest nodal plane of \(\vec { \boldsymbol { B } }\) ?

Laser Surgery. Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portions back into place. In one such procedure, a laser beam has a wavelength of 810\(\mathrm { nm }\) and delivers 250\(\mathrm { mW }\) of power spread over a circular spot 510\(\mu \mathrm { m }\) in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34 (a) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?
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