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Chapter 12: Problem 2

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the \(+x\)-direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(+y_{-\text {direction; }}\) (b) \(\overrightarrow{\boldsymbol{E}}\) in the \(-y\)-direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(+x\)-direction; (c) \(\overrightarrow{\boldsymbol{E}}\) in the \(+z\) direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-x\)-direction; \((\) d) \(\overrightarrow{\boldsymbol{E}}\) in the \(+y\)-direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-z\)-direction.

Short Answer

Expert verified
(a) +z; (b) -z; (c) -y; (d) +x.

Step by step solution

01

Understand the Right-Hand Rule

The direction of wave propagation is perpendicular to both the electric field (\( \vec{E} \)) and the magnetic field (\( \overrightarrow{\boldsymbol{B}} \)). Using the right-hand rule, point your fingers in the direction of \( \vec{E} \), curl them towards \( \overrightarrow{\boldsymbol{B}} \), and your thumb will point in the direction of wave propagation.
02

Analyze Situation (a)

For (a), \( \vec{E} \) is in the \(+x\)-direction and \( \overrightarrow{\boldsymbol{B}} \) is in the \(+y\)-direction. Point your right hand in the \(+x\)-direction and curl your fingers towards the \(+y\)-direction. Your thumb points in the \(+z\)-direction, which is the direction of wave propagation.
03

Analyze Situation (b)

For (b), \( \vec{E} \) is in the \(-y\)-direction and \( \overrightarrow{\boldsymbol{B}} \) is in the \(+x\)-direction. Point your right hand in the \(-y\)-direction and curl your fingers towards the \(+x\)-direction. Your thumb points in the \(-z\)-direction, which indicates the direction of wave propagation.
04

Analyze Situation (c)

For (c), \( \vec{E} \) is in the \(+z\)-direction and \( \overrightarrow{\boldsymbol{B}} \) is in the \(-x\)-direction. Point your right hand in the \(+z\)-direction and curl your fingers towards the \(-x\)-direction. Your thumb will point in the \(-y\)-direction, showing the direction of wave propagation.
05

Analyze Situation (d)

For (d), \( \vec{E} \) is in the \(+y\)-direction and \( \overrightarrow{\boldsymbol{B}} \) is in the \(-z\)-direction. Point your right hand in the \(+y\)-direction and curl your fingers towards the \(-z\)-direction. Your thumb will point in the \(+x\)-direction, which signifies the direction of wave propagation.

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Key Concepts

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

Right-Hand Rule
The right-hand rule is a useful mnemonic in electromagnetism to determine the direction of a vector, especially in relation to cross products. When dealing with electromagnetic waves, it's applied to deduce the wave propagation direction. To use the right-hand rule:
  • Extend your fingers in the direction of the electric field ( \( \vec{E} \)), which represents one dimension of the wave.
  • Curl your fingers towards the direction of the magnetic field ( \( \overrightarrow{\boldsymbol{B}} \)). This involves a motion perpendicular to the electric field.
  • Your extended thumb will naturally point in the direction of the wave propagation, which is perpendicular to both the electric and magnetic fields.
The right-hand rule helps visualize and confirm these perpendicular relationships in a three-dimensional space, which can often be challenging to grasp initially.
Electric Field Orientation
The electric field orientation dictates a critical axis for the wave's behavior. When you think about the electric field in electromagnetic waves, it is always perpendicular to the direction the wave moves. For scenario (a), the electric field is oriented in the \( +x \)-direction, meaning it travels along the positive x-axis. Similarly, in situations (b), (c), and (d), the orientations change:
  • (b) Electric Field ( \( \vec{E} \)) is directed in the \( -y \)-direction.
  • (c) Electric Field ( \( \vec{E} \)) is directed in the \( +z \)-direction.
  • (d) Electric Field ( \( \vec{E} \)) is directed in the \( +y \)-direction.
This orientation significantly influences how we use the right-hand rule to determine the direction of wave propagation.
Magnetic Field Orientation
The magnetic field's orientation provides another axis of the electromagnetic wave and completes the cross product necessary for determining wave direction. It, too, remains perpendicular to the electric field. Magnetic field orientations for given cases are:
  • (a) Magnetic Field ( \( \overrightarrow{\boldsymbol{B}} \)) is in the \( +y \)-direction.
  • (b) Magnetic Field ( \( \overrightarrow{\boldsymbol{B}} \)) is in the \( +x \)-direction.
  • (c) Magnetic Field ( \( \overrightarrow{\boldsymbol{B}} \)) is in the \( -x \)-direction.
  • (d) Magnetic Field ( \( \overrightarrow{\boldsymbol{B}} \)) is in the \( -z \)-direction.
These orientations work alongside their corresponding electric field orientations, forming a cross product that guides our understanding of the wave's overall path.
Wave Propagation Direction
Using the insights from the right-hand rule, and understanding the orientations for both electric and magnetic fields, we can determine the propagation direction of each wave case:
  • For (a), with \( \vec{E} \) in the \( +x \)-direction and \( \overrightarrow{\boldsymbol{B}} \) in the \( +y \)-direction, the wave propagates along the \( +z \)-direction.
  • For (b), \( \vec{E} \) is in the \( -y \)-direction and \( \overrightarrow{\boldsymbol{B}} \) is in the \( +x \)-direction, resulting in a propagation direction of \( -z \).
  • For (c), \( \vec{E} \) is in the \( +z \)-direction with \( \overrightarrow{\boldsymbol{B}} \) in the \( -x \)-direction, and the wave travels in the \( -y \)-direction.
  • For (d), the \( \vec{E} \) is aligned \( +y \), and \( \overrightarrow{\boldsymbol{B}} \) is \( -z \), guiding the wave in the \( +x \)-direction.
Understanding these directional interrelationships allows for greater comprehension in wave dynamics and is crucial for analyzing electromagnetic phenomena.

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Most popular questions from this chapter

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.

A plane sinusoidal electromagnetic wave in air has a wavelength of \(3.84 \mathrm{~cm}\) and an \(\vec{E}\)-field amplitude of \(1.35 \mathrm{~V} / \mathrm{m}\). (a) What is the frequency? (b) What is the \(\overrightarrow{\boldsymbol{B}}\)-field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area \(0.240 \mathrm{~m}^{2}\) perpendicular to the direction of propagation?

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in mag nitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why (d) Explain why dust particles with a radius less than or why not? that found in part (c) are unlikely to be found in the solar system.

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.

Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of \(316 \mathrm{~kW}\). Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home \(5.00 \mathrm{~km}\) away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?
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