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Chapter 32: Problem 14

An electromagnetic wave with frequency 65.0 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} V/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?

Short Answer

Expert verified
(a) 6.91 x 10^7 m/s; (b) 1.06 x 10^6 m; (c) 1.04 x 10^{-10} T.

Step by step solution

01

Calculate the Speed of Propagation

The speed of propagation of an electromagnetic wave in a medium is given by the formula \( v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} \), where \( c \) is the speed of light in vacuum \( (3.00 \times 10^8 \text{ m/s}) \), \( \varepsilon_r \) is the dielectric constant, and \( \mu_r \) is the relative permeability. Substituting the given values, \( \varepsilon_r = 3.64 \) and \( \mu_r = 5.18 \), we calculate: \[ v = \frac{3.00 \times 10^8}{\sqrt{3.64 \times 5.18}} = \frac{3.00 \times 10^8}{\sqrt{18.8352}} = \frac{3.00 \times 10^8}{4.34} = 6.91 \times 10^7 \text{ m/s}. \]
02

Calculate the Wavelength

The wavelength \( \lambda \) of the wave can be found using the relationship \( \lambda = \frac{v}{f} \), where \( v \) is the speed of the wave and \( f \) is the frequency. With \( v = 6.91 \times 10^7 \text{ m/s} \) and \( f = 65.0 \text{ Hz} \), we have: \[ \lambda = \frac{6.91 \times 10^7}{65.0} = 1.06 \times 10^6 \text{ m}. \]
03

Calculate the Amplitude of the Magnetic Field

To find the amplitude of the magnetic field \( B_0 \), use the formula \( B_0 = \frac{E_0}{v} \), where \( E_0 = 7.20 \times 10^{-3} \text{ V/m} \) is the amplitude of the electric field. Substituting the values, we get: \[ B_0 = \frac{7.20 \times 10^{-3}}{6.91 \times 10^7} = 1.04 \times 10^{-10} \text{ T}. \]

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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, often represented as \( \varepsilon_r \), is a measure of a material's ability to store electrical energy in an electric field. It is a relative value, highlighting how much easier electrical energy is stored in a material compared to a vacuum.
Using the dielectric constant, we can understand how a medium interacts with an electromagnetic wave as it propagates.
Some characteristics to note include:
  • It affects the speed of light in the medium.
  • A higher dielectric constant usually reduces the speed of light through the material.
  • It also influences the capacitance of capacitors in electronics.
In our exercise, the dielectric constant value is 3.64, showing the relative effect of the medium on electromagnetic waves compared to a vacuum.
Relative Permeability
Relative permeability, symbolized by \( \mu_r \), is crucial in understanding how a material responds to magnetic fields. It's a factor by which the permeability of a medium is greater than that of free space.
This quantity illustrates how a medium impacts the propagation of electromagnetic waves, particularly magnetic fields, and can impact the wave's speed.
Key points about relative permeability:
  • A higher value suggests the medium allows a stronger magnetic field to pass through.
  • It, along with the dielectric constant, is important for calculating the refractive index.
  • Materials with high permeability are typically used in inductors and transformers.
In the problem, the relative permeability of 5.18 indicates a significant enhancement of magnetic interactions within the medium.
Speed of Light
When discussing electromagnetic waves, the speed of light, \( c \), is a critical value. Defined as approximately \( 3.00 \times 10^8 \text{ m/s} \) in a vacuum, it represents the maximum possible speed for any form of communication across the universe.
However, this speed can be reduced in various media depending on factors such as dielectric constant and relative permeability.
Things to remember about the speed of light in mediums:
  • In a medium, light slows down due to interaction with the material's electric and magnetic properties.
  • The speed of propagation \( v \) in a material is calculated with \( v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} \).
  • This reduction plays a significant role in technologies like fiber optics and lenses where precise control over light speed and paths is needed.
For the exercise, substituting the given values of dielectric constant and permeability helps us find the reduced speed of light in the medium.
Wavelength
Wavelength, denoted as \( \lambda \), is the distance between consecutive peaks of a wave, significant in understanding wave behavior. For electromagnetic waves, this property helps categorize and measure the waves.
In our specific exercise, it is determined with the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed, and \( f \) is frequency.
Important aspects of wavelength include:
  • It defines the color of light in the visible spectrum or the type of electromagnetic wave (radio, microwave, etc.).
  • Changes based on the wave's speed and the medium it travels through.
  • Critical in applications like antennas where resonance with specific wavelengths is needed.
For the problem, the wavelength is crucial to understand how the wave propagates in the material with the given frequency.
Magnetic Field Amplitude
The amplitude of the magnetic field, often represented as \( B_0 \), is a measure of the strength of the magnetic field at its peak in an electromagnetic wave.
This concept informs us about the intensity of the wave's magnetic component as it spreads through a medium.
To find it, the relation \( B_0 = \frac{E_0}{v} \) is used, where \( E_0 \) is the electric field's amplitude and \( v \) is the wave's speed of propagation.
Consider these aspects of magnetic field amplitude:
  • Higher magnetic field amplitudes imply stronger interactions and more potent magnetic effects.
  • It is directly influenced by the electric field amplitude and inversely related to the wave's speed.
  • Magnetic field characteristics are crucial in technologies using electromagnetic waves, such as MRI machines.
In our exercise, this amplitude translates the given electric field's information into the magnetic component, allowing a complete wave analysis.

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Most popular questions from this chapter

He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.500 mW spread over a cylindrical beam 1.00 mm in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

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 nm and delivers 250 mW of power spread over a circular spot 510 \(\mu\)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 would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed? (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?

A small helium-neon laser emits red visible light with a power of 5.80 mW in a beam of diameter 2.50 mm. (a) What are the amplitudes of the electric and magnetic fields of this light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a 1.00-m length of the beam?

Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm in air. What are the frequency, period, and wave number of such waves?

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 \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), 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 \(i^2R\) 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 \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); 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.
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