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

Short Answer

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

Step by step solution

01

Calculate the Frequency

The wave frequency can be found using the speed of light \(c\) and the wavelength \(\lambda\). The formula is \(f = \frac{c}{\lambda}\). Where \(c = 3 \times 10^8 \text{ m/s}\) and \(\lambda = 435 \text{ nm} = 435 \times 10^{-9} \text{ m}\). Substituting these values, \[ f = \frac{3 \times 10^8}{435 \times 10^{-9}} = 6.90 \times 10^{14} \text{ Hz} \].
02

Calculate the Magnetic Field Amplitude

The electric field \( E \) and the magnetic field \( B \) in an electromagnetic wave are related by the equation \( B = \frac{E}{c} \). Given \( E = 2.70 \times 10^{-3} \text{ V/m} \), substituting in the equation gives \[ B = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \text{ T} \].
03

Write the Electric Field Equation

The electric field vector can be written as \( \vec{E}(z, t) = E_0 \hat{i} \cos(kz + \omega t) \). Here, \(E_0 = 2.70 \times 10^{-3} \text{ V/m} \), \(k = \frac{2\pi}{\lambda}\), and the angular frequency \(\omega = 2\pi f\). Substitute \(\lambda\) and \(f\) to find \[ \vec{E}(z, t) = 2.70 \times 10^{-3} \hat{i} \cos\left(\frac{2\pi}{435 \times 10^{-9}} z + 2\pi \times 6.90 \times 10^{14} t\right) \].
04

Write the Magnetic Field Equation

The magnetic field vector is \( \vec{B}(z, t) = B_0 \hat{j} \cos(kz + \omega t) \). Here, \(B_0 = 9.00 \times 10^{-12} \text{ T}\), with the same wave number \(k\) and frequency \(\omega\) as before. Thus, \[ \vec{B}(z, t) = 9.00 \times 10^{-12} \hat{j} \cos\left(\frac{2\pi}{435 \times 10^{-9}} z + 2\pi \times 6.90 \times 10^{14} t\right) \].

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude in an electromagnetic wave is a measure of the maximum strength of the electric field at a given point. It's an important parameter since it tells us how strong the electric force would be on a charged particle in the wave. In our example, the electric field amplitude is given as \(2.70 \times 10^{-3} \text{ V/m}\). This value indicates the peak electric field strength in the wave as it propagates.

It's essential to note that the amplitude of the electric field affects the energy carried by the wave. This energy is proportional to the square of the amplitude, meaning a more extensive amplitude implies a more energetic wave. Understanding electric field amplitudes helps in various applications including communication systems and understanding light behavior.
Magnetic Field Amplitude
In electromagnetic waves, the magnetic field component is perpendicular to the electric field component, and its amplitude can be derived from the electric field and speed of the wave. The magnetic field amplitude is crucial as it helps determine the overall energy and behavior of the wave.

The relationship between the electric field \(E\) and the magnetic field \(B\) is given by the equation \(B = \frac{E}{c}\), where \(c\) is the speed of light. For example, with an electric field of \(2.70 \times 10^{-3} \text{ V/m}\), we find:
  • \( B = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \text{ T} \)
This demonstrates that the magnetic field, though small in amplitude, plays a significant role in the wave's dynamics, interacting with the electric field to propagate the wave through space.
Frequency Calculation
Frequency is a critical property of electromagnetic waves, representing how many wave cycles pass a point per unit time. The formula \(f = \frac{c}{\lambda}\) is used to calculate frequency, where \(c\) is the speed of light, and \(\lambda\) is the wavelength.

In our case, the wavelength \(\lambda\) is given as \(435 \text{ nm}\) (converted to meters, \(435 \times 10^{-9} \text{ m} \)), and using the speed of light \(c = 3 \times 10^8 \text{ m/s}\), we find:
  • \( f = \frac{3 \times 10^8}{435 \times 10^{-9}} = 6.90 \times 10^{14} \text{ Hz} \)
Knowing the frequency is vital for applications like tuning antennas and understanding color in optics, as frequency relates to wave energy and its various manifestations.
Wave Equation
The wave equation describes the electric and magnetic fields in terms of space and time variably. For example, for the electric field, it is represented as \(\vec{E}(z, t) = E_0 \hat{i} \cos(kz + \omega t)\). Here, \(E_0\) is the electric field amplitude, \(k\) is the wave number \(\left(\frac{2\pi}{\lambda}\right)\), and \(\omega\) is the angular frequency \(\left(2\pi f\right)\), ensuring that the fields change sinusoidally.

For the given wave, the electric field vector becomes:
  • \( \vec{E}(z, t) = 2.70 \times 10^{-3} \hat{i} \cos\left(\frac{2\pi}{435 \times 10^{-9}} z + 2\pi \times 6.90 \times 10^{14} t\right) \)
The magnetic field vector follows similarly:
  • \( \vec{B}(z, t) = 9.00 \times 10^{-12} \hat{j} \cos\left(\frac{2\pi}{435 \times 10^{-9}} z + 2\pi \times 6.90 \times 10^{14} t\right) \)
With these equations, one can graph the wave or use it to solve problems involving electromagnetic waves in various settings.
Vector Fields
Vector fields represent how vectors, like electric and magnetic fields, are distributed in space and may change over time. They use vectors to indicate both the magnitude and direction of the field at each point. In electromagnetic wave propagation, this helps in visualizing the fields moving through space.

The electric field vector in our case, \(\vec{E}(z, t)\), is directed along the \(x\)-axis, showing all the wave’s electric field influences are parallel along this axis, while the magnetic field \(\vec{B}(z, t)\) is along the \(y\)-axis. These perpendicular orientations are crucial because:
  • Electromagnetic waves in a vacuum always have perpendicular electric and magnetic fields.
  • The fields evolve as cosinusoidal variations in time and space.
Understanding vector fields assists in solving real-world problems, like determining how waves propagate along communication lines or how they interact with materials.

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Most popular questions from this chapter

A satellite \(575 \mathrm{~km}\) above the earth's surface transmits sinusoidal electromagnetic waves of frequency \(92.4 \mathrm{MHz}\) uniformly in all directions, with a power of \(25.0 \mathrm{~kW}\). (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring \(15.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm}\) oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

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 \(\overrightarrow{\boldsymbol{B}}\) and the closest antinodal plane of \(\overrightarrow{\boldsymbol{B}}\) ? (b) What is the distance between an antinodal plane of \(\overrightarrow{\boldsymbol{E}}\) and the closest antinodal plane of \(\overrightarrow{\boldsymbol{B}}\) ? (c) What is the distance between a nodal plane of \(\overrightarrow{\boldsymbol{E}}\) and the closest nodal plane of \(\overrightarrow{\boldsymbol{B}}\) ?

An intense light source radiates uniformly in all directions. At a distance of \(5.0 \mathrm{~m}\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10^{-6} \mathrm{~Pa}\). What is the total average power output of the source?

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, if 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 \(\overrightarrow{\boldsymbol{E}}(x, t)=E_{y}(x, t) \hat{\boldsymbol{j}}\) propagating in the \(+x\)-direction within a conductor is $$ \frac{\delta^{2} E_{y}(x, t)}{\delta x^{2}}=\frac{\mu}{\rho} \frac{\delta E_{y}(x, t)}{\delta t} $$ 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 \(K_{C}=\sqrt{\omega \mu / 2 \rho}\). 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?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu}\), and calculate this distance for a radio wave with frequency \(f=\) \(1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m}\); 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 cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from WCCO is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

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