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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?

Short Answer

Expert verified
(a) 6.84 x 10^7 m/s (b) 1.05 x 10^6 m (c) 1.05 x 10^-10 T (d) 4.90 x 10^-10 W/m^2.

Step by step solution

01

Determine the speed of propagation

The speed of propagation \( v \) of an electromagnetic wave in a material is given by the equation \[ v = \frac{c}{\sqrt{\varepsilon_r \cdot \mu_r}} \]where \( c = 3 \times 10^8 \) m/s is the speed of light in a vacuum, \( \varepsilon_r = 3.64 \) is the dielectric constant, and \( \mu_r = 5.18 \) is the relative permeability. Calculate \( v \):\[ v = \frac{3 \times 10^8}{\sqrt{3.64 \cdot 5.18}} \approx 6.84 \times 10^7 \text{ m/s} \].
02

Calculate the wavelength of the wave

The wavelength \( \lambda \) of a wave is given by\[ \lambda = \frac{v}{f} \]where \( v = 6.84 \times 10^7 \) m/s is the speed of the wave, and \( f = 65 \) Hz is the frequency. Substitute these values to find the wavelength:\[ \lambda = \frac{6.84 \times 10^7}{65} \approx 1.05 \times 10^6 \text{ m} \].
03

Find the amplitude of the magnetic field

The amplitude of the magnetic field \( B_0 \) can be determined from the relationship\[ E_0 = vB_0 \]where \( E_0 = 7.20 \times 10^{-3} \) V/m is the electric field amplitude and \( v = 6.84 \times 10^7 \) m/s is the speed of the wave. Solve for \( B_0 \):\[ B_0 = \frac{E_0}{v} = \frac{7.20 \times 10^{-3}}{6.84 \times 10^7} \approx 1.05 \times 10^{-10} \text{ T} \].
04

Compute the intensity of the wave

The intensity \( I \) of an electromagnetic wave is given by\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 n \]where \( n = \sqrt{\varepsilon_r \cdot \mu_r} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m is the permittivity of free space, and \( E_0 = 7.20 \times 10^{-3} \) V/m. First, calculate \( n \):\[ n = \sqrt{3.64 \times 5.18} \approx 4.38 \].Then calculate \( I \):\[ I = \frac{1}{2} \cdot 3 \times 10^8 \cdot 8.85 \times 10^{-12} \cdot (7.20 \times 10^{-3})^2 \cdot 4.38 \approx 4.90 \times 10^{-10} \text{ W/m}^2 \].

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Key Concepts

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

Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a measure that reflects how well a material can store electrical energy in an electric field. In essence, it determines how much a material can reduce the electric field compared to a vacuum. This is crucial for understanding how electromagnetic waves behave in different media.
The dielectric constant, represented by \( \varepsilon_r \), is dimensionless. It indicates the amount of electric charge a material can store compared to the capacity of a vacuum. For example, a dielectric constant of 3.64 means that the material can store 3.64 times the electric charge a vacuum can.
  • This value directly affects the speed of electromagnetic wave propagation through the material. A higher dielectric constant tends to slow down the wave.
  • It is also an important parameter in calculating the wave intensity and other characteristics of electromagnetic waves in the material.
Understanding the dielectric constant helps predict how electromagnetic fields interact with different materials, which is vital in designing materials for various electronic and telecommunication applications.
Relative Permeability
Relative permeability is a measure of a material's ability to support the formation of a magnetic field within itself. It is compared to the permeability of free space (or vacuum). In simpler terms, it describes how much easier or harder it is for a magnetic field to pass through a material compared to going through a vacuum.
Represented by \( \mu_r \), this dimensionless quantity indicates the effectiveness of a material in conducting magnetic lines of force.
  • A relative permeability greater than 1 means the material is more permeable than a vacuum. The given value of 5.18 suggests that the material is significantly more capable of conducting magnetic fields.
  • This property directly influences the wave's speed. The magnetic property, along with the dielectric constant, dictates how electromagnetic waves propagate through the medium.
Gaining insights into relative permeability helps in various applications such as designing magnetic materials and understanding how materials influence electromagnetic wave propagation, especially in magnetic and insulating contexts.
Speed of Light
The speed of light in a vacuum is a fundamental constant of nature, approximately \(3 \times 10^8\) meters per second. However, when light passes through a material, its speed decreases, depending on the material's dielectric constant and relative permeability.
The speed at which an electromagnetic wave travels through a material is given by \[ v = \frac{c}{\sqrt{\varepsilon_r \cdot \mu_r}} \]where \( c \) is the speed of light in a vacuum, \( \varepsilon_r \) is the dielectric constant, and \( \mu_r \) is the relative permeability. This formula demonstrates that both the dielectric and magnetic properties of a material influence light's speed.
  • In the given material with \( \varepsilon_r = 3.64 \) and \( \mu_r = 5.18 \), light speed becomes approximately \(6.84 \times 10^7\) meters per second, showing a substantial reduction from its speed in a vacuum.
  • This knowledge is crucial for applications like fiber optic communications and material science.
Understanding how light travels through different substances reveals the interactions between electromagnetic waves and materials, which is vital for the development of optical technologies.
Magnetic Field Amplitude
The magnetic field amplitude is a key parameter in describing electromagnetic waves. It represents the maximum strength of the magnetic component in the wave and is closely tied to the electric field amplitude.
Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other. The relationship between them is established by\[ E_0 = vB_0 \]where \( E_0 \) is the electric field amplitude, \( v \) is the speed of the wave, and \(B_0 \) is the magnetic field amplitude. Using this equation, one can find the magnetic field amplitude if the other values are known.
  • In the given scenario, \(B_0\) is determined to be approximately \(1.05 \times 10^{-10}\) Tesla, showcasing the relative contribution of the magnetic field.
  • The magnetic field amplitude helps in calculating wave intensity and thereby the energy transmission capability of the wave.
Understanding magnetic field amplitude is essential for applications in wave transmission and electromagnetic field interaction studies, providing a deeper insight into the behavior of waves in various media.

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Most popular questions from this chapter

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0 \(\mathrm{MHz}\) in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing wave pattern is determined to be in its eighth harmonic, how long is the cavity?

A space probe \(2.0 \times 10^{10} \mathrm{m}\) from a star measures the total intensity of electrumagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{W} / \mathrm{m}^{2} .\) If the star radiates uniformly in all directions, what is its total average power output?

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?

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 aver- age 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?

The electric-field amplitude near a certain radio transmitter is \(3.85 \times 10^{-3} \mathrm{V} / \mathrm{m}\) What is the amplitude of \(\overrightarrow{\boldsymbol{B}} ?\) How does this compare in magnitude with the earth's field?

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