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A plane electromagnetic wave, with wavelength \(3.0 \mathrm{~m}\), travels in vacuum in the positive direction of an \(x\) axis. The electric field, of amplitude \(300 \mathrm{~V} / \mathrm{m}\), oscillates parallel to the \(y\) axis. What are the (a) frequency, (b) angular frequency, and (c) angular wave number of the wave? (d) What is the amplitude of the magnetic field component? (e) Parallel to which axis does the magnetic field oscillate? (f) What is the timeaveraged rate of energy flow in watts per square meter associated with this wave? The wave uniformly illuminates a surface of area \(2.0 \mathrm{~m}^{2}\). If the surface totally absorbs the wave, what are \((\mathrm{g})\) the rate at which momentum is transferred to the surface and (h) the radiation pressure on the surface?

Step by step solution

01

Calculate the frequency of the wave

The speed of light in a vacuum is given by \( c = 3.0 \times 10^8 \text{ m/s} \). The frequency, \( f \), of a wave is related to the speed of light and wavelength \( \lambda \) by the formula:\[f = \frac{c}{\lambda}\]Given \( \lambda = 3.0 \text{ m} \), we have:\[f = \frac{3.0 \times 10^8 \text{ m/s}}{3.0 \text{ m}} = 1.0 \times 10^8 \text{ Hz}\]

02

Calculate the angular frequency

Angular frequency \( \omega \) is related to frequency \( f \) by the equation:\[\omega = 2\pi f\]Using \( f = 1.0 \times 10^8 \text{ Hz} \), we get:\[\omega = 2\pi \times 1.0 \times 10^8 = 2\pi \times 10^8 \text{ rad/s}\]

03

Determine the angular wave number

The angular wave number \( k \) relates to the wavelength \( \lambda \) by the equation:\[k = \frac{2\pi}{\lambda}\]Using \( \lambda = 3.0 \text{ m} \), we find:\[k = \frac{2\pi}{3.0} \approx 2.09 \text{ rad/m}\]

04

Calculate the amplitude of the magnetic field component

The amplitude of the magnetic field \( B_0 \) can be derived from the relation between the electric field and the speed of light:\[B_0 = \frac{E_0}{c}\]Where \( E_0 = 300 \text{ V/m} \). Calculate \( B_0 \):\[B_0 = \frac{300}{3.0 \times 10^8} = 1.0 \times 10^{-6} \text{ T}\]

05

Determine the oscillation direction of the magnetic field

As the electric field oscillates parallel to the \( y \) axis and the wave propagates along the \( x \) axis, the magnetic field oscillates perpendicular to both, which is parallel to the \( z \) axis.

06

Calculate the time-averaged rate of energy flow (Poynting vector)

The time-averaged rate of energy flow per unit area, also known as the Poynting vector \( S \), is given by:\[S = \frac{E_0 B_0}{2\mu_0} = \frac{E_0^2}{2c\mu_0}\]Where \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \). Substituting \( E_0 = 300 \text{ V/m} \):\[S = \frac{300^2}{2 \times 3.0 \times 10^8 \times 4\pi \times 10^{-7}} \approx 119.4 \text{ W/m}^2 \]

07

Determine the rate of momentum transfer to the surface

The rate of momentum transfer is the power absorbed divided by the speed of light. If the surface area is absorbed by \( A = 2.0 \text{ m}^2 \), the total power absorbed is \( P = S \times A \) and the momentum transfer rate \( F \) is:\[F = \frac{P}{c} = \frac{119.4 \times 2.0}{3.0 \times 10^8} \approx 7.96 \times 10^{-7} \text{ N}\]

08

Calculate the radiation pressure on the surface

The radiation pressure \( P_{rad} \) for a totally absorbing surface is given by:\[P_{rad} = \frac{S}{c}\]Thus:\[P_{rad} = \frac{119.4}{3.0 \times 10^8} \approx 3.98 \times 10^{-7} \text{ Pa}\]

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Key Concepts

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

Frequency Calculation

To understand electromagnetic waves, it's essential to calculate their frequency, which tells us how many cycles occur in one second. This is a critical property for describing wave behavior. The relationship between the speed of light in a vacuum, denoted as \(c\), and the wavelength \(\lambda\) helps us determine frequency \(f\) using the formula \(f = \frac{c}{\lambda}\). For a wavelength of 3 meters, the wave's frequency is \(1.0 \times 10^8\) Hz. This concept is pivotal as the frequency is what allows us to differentiate between different types of electromagnetic radiation, from radio waves to gamma rays. Calculating frequency helps in many fields like telecommunications, enabling engineers to design systems for efficient information transmission.

Angular Frequency

Once you have figured out the frequency of an electromagnetic wave, finding the angular frequency \(\omega\) is the next step, which describes how quickly the wave oscillates in radians per second rather than cycles. The relationship is given by \(\omega = 2\pi f\). Given the frequency of \(1.0 \times 10^8\) Hz, the angular frequency is \(2\pi \times 10^8\) rad/s. This concept is useful because angular frequency helps in analyzing oscillatory systems, including wave mechanics, and bridges the gap between linear frequency concepts and rotational motion analysis.

Angular Wave Number

The angular wave number \(k\) represents the wave's spatial frequency, indicating how many radians the wave phase changes per meter. To find this, use the formula \(k = \frac{2\pi}{\lambda}\). For a 3-meter wavelength, \(k\) calculates to approximately \(2.09\) rad/m. Understanding the angular wave number is vital for determining how waves propagate through different media, influencing how engineers and scientists analyze wave interference, diffraction, and other phenomena on a spatial scale.

Magnetic Field Amplitude

In electromagnetic waves, calculating the magnetic field's amplitude \(B_0\) is crucial as it determines the wave's magnetic strength. It is derived from the electric field amplitude \(E_0\), using the relation \(B_0 = \frac{E_0}{c}\). With an electric field amplitude of 300 V/m, the magnetic field amplitude is \(1.0 \times 10^{-6}\) T. This information is essential for understanding the full electromagnetic spectrum's behavior and interactions with materials, as it gives insight into both electric and magnetic components of the wave.

Poynting Vector

The Poynting vector \(S\) represents the energy flow rate per unit area for electromagnetic waves, giving us insight into the wave's power. It can be calculated using the formula \(S = \frac{E_0^2}{2c\mu_0}\), where \(\mu_0\) is the magnetic permeability of free space. Substituting the electric field amplitude of 300 V/m gives \(S \approx 119.4 \text{ W/m}^2\). Understanding the Poynting vector is important for assessing the power delivered by electromagnetic waves, with applications in everything from solar energy absorption to radio communication.

Momentum Transfer

Momentum transfer in electromagnetic waves is a consequence of their wave-particle duality, meaning the waves can exert force on objects. For a wave fully absorbed by a surface, the momentum transfer rate \(F\) is calculated as the power absorbed \(P\) divided by the speed of light \(c\). Here, for a surface absorbing \(S = 119.4 \text{ W/m}^2\) over \(2 \text{ m}^2\), \(F \approx 7.96 \times 10^{-7} \text{ N}\). This explains how radiation can apply pressure, useful in space engineering for propulsion systems like solar sails.

Radiation Pressure

The concept of radiation pressure \(P_{rad}\) explains why electromagnetic waves exert pressure upon interacting with surfaces. For a perfectly absorbing surface, use \(P_{rad} = \frac{S}{c}\). For an energy density of \(119.4 \text{ W/m}^2\), the radiation pressure is \(3.98 \times 10^{-7} \text{ Pa}\). Understanding this pressure is essential in various technological applications, such as optical trapping in physics experiments, assisting in manipulating small particles and molecules with laser beams.