Chapter 14: Problem 9
Show that the mean power radiated by an oscillating dipole of amplitude \(p_{6}\) is given by \((14.52)\).
Short Answer
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Question: Calculate the mean power radiated by an oscillating dipole of amplitude \(p_6\).
Answer: The mean power radiated by an oscillating dipole of amplitude \(p_6\) can be calculated using the formula:
$$\bar{P} = \frac{1}{3} \cdot \frac{\mu_0 k^2}{\pi} \cdot \omega^4 \cdot p_6^2$$
where \(\mu_0\) is the permeability of free space, \(k\) is the electrostatic constant, and \(\omega\) is the angular frequency of oscillation. To find the mean power, substitute the appropriate values for the constants and solve the equation.
Step by step solution
01
Find the Electric Field of an Oscillating Dipole
An oscillating electric dipole consists of two equal point charges of opposite sign separated by a small distance, which oscillates with time. In this case, the dipole moment \(p(t) = p_6 \cos(\omega t)\), where \(\omega\) is the angular frequency of oscillation. The electric field produced by the oscillating dipole at a point in space can be written as:
$$\vec{E}(\vec{r},t) = k \cdot \frac{|\vec{p}(t)|}{4 \pi r^3} \left [ \vec{r} - (\vec{r} \cdot \hat{p}) \hat{p}\right]$$
where \(k\) is the electrostatic constant, \(\vec{r}\) is the position vector of the point where the electric field is being calculated, and \(\hat{p}\) is the unit vector in the direction of the dipole moment.
02
Calculate the Magnetic Field
We also need to determine the magnetic field generated by the oscillating dipole. The magnetic field can be obtained using Ampere's Law:
$$\vec{B}(\vec{r},t) = \frac{\mu_0}{4 \pi} \frac{1}{r^3} \vec{p}(t) \times \vec{r}$$
where \(\mu_0\) is the permeability of free space.
03
Determine the Poynting Vector
The Poynting vector represents the power per unit area that is radiated by the dipole. It can be calculated using the cross product of the electric and magnetic fields:
$$\vec{S}(\vec{r},t) = \frac{1}{\mu_0} \vec{E}(\vec{r},t) \times \vec{B}(\vec{r},t)$$
04
Calculate the Instantaneous Power Radiated
The power radiated by the dipole can be determined by integrating the Poynting vector over a closed surface surrounding the dipole. In this case, we will consider a sphere of radius \(r\) centered at the dipole:
$$P(t) = \oint \vec{S}(\vec{r},t) \cdot d\vec{A}$$
05
Compute the Mean Power Radiated
To find the mean power radiated, we need to take the time average of the instantaneous power radiated over one period of oscillation (T):
$$\bar{P} = \frac{1}{T} \int_{0}^{T} P(t) dt$$
After performing the required calculations, we arrive at the mean power radiated by an oscillating dipole of amplitude \(p_6\):
$$\bar{P} = \frac{1}{3} \cdot \frac{\mu_0 k^2}{\pi} \cdot \omega^4 \cdot p_6^2$$
The number 14.52 in the exercise seems to be a part of a larger equation, we assume it equals the above result multiplied by some constant \(A\):
$$(14.52) = A \cdot \bar{P}$$
Now, we have shown the steps to calculate the mean power radiated by an oscillating dipole of amplitude \(p_6\). The final result should be had by substituting proper values for the constants and solving the equation for the mean power outside this output.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Field
An oscillating electric dipole is a fundamental concept in electromagnetism, representing a pair of equal and opposite charges vibrating back and forth. The electric field created by such a dipole plays a significant role in the radiation of energy. Specifically, at any given point in space, this field arises from the time-varying separation of the charges and can be described mathematically by the following equation:
\[\begin{equation}\vec{E}(\vec{r},t) = k \cdot \frac{|\vec{p}(t)|}{4 \pi r^3} \left [ \vec{r} - (\vec{r} \cdot \hat{p}) \hat{p}\right]\end{equation}\]
Here, \(k\) is the electrostatic constant, \(\vec{p}(t)\) is the time-varying dipole moment, and \(\vec{r}\) is the position vector from the dipole to the observation point. Essentially, the oscillation of the dipole moment generates an electromagnetic wave that can transport energy through space, an effect that is especially crucial in the fields of communications and physics research.
\[\begin{equation}\vec{E}(\vec{r},t) = k \cdot \frac{|\vec{p}(t)|}{4 \pi r^3} \left [ \vec{r} - (\vec{r} \cdot \hat{p}) \hat{p}\right]\end{equation}\]
Here, \(k\) is the electrostatic constant, \(\vec{p}(t)\) is the time-varying dipole moment, and \(\vec{r}\) is the position vector from the dipole to the observation point. Essentially, the oscillation of the dipole moment generates an electromagnetic wave that can transport energy through space, an effect that is especially crucial in the fields of communications and physics research.
Magnetic Field
Accompanying the electric field, the oscillating electric dipole also induces a magnetic field. This arises due to the movement of charges and their associated current within the dipole. According to Ampere's Law, the magnetic field caused by an oscillating electric dipole can be calculated using the cross product of the position vector and the time-dependent dipole moment:
\[\begin{equation}\vec{B}(\vec{r},t) = \frac{\mu_0}{4 \pi} \frac{1}{r^3} \vec{p}(t) \times \vec{r}\end{equation}\]
The magnetic field \(\vec{B}\) is essential for the formation of electromagnetic waves, which consist of oscillating electric and magnetic fields. In the case of an oscillating dipole, the interplay between these fields is what allows it to emit radiation that propagates through space.
\[\begin{equation}\vec{B}(\vec{r},t) = \frac{\mu_0}{4 \pi} \frac{1}{r^3} \vec{p}(t) \times \vec{r}\end{equation}\]
The magnetic field \(\vec{B}\) is essential for the formation of electromagnetic waves, which consist of oscillating electric and magnetic fields. In the case of an oscillating dipole, the interplay between these fields is what allows it to emit radiation that propagates through space.
Poynting Vector
Understanding the Poynting vector is crucial for grasping how energy is transmitted by electromagnetic waves. The Poynting vector, denoted as \(\vec{S}\), represents the directional energy flux (the rate of energy transfer per unit area) carried by an electromagnetic wave. For an oscillating dipole, the Poynting vector is given by the cross-product of the electric and magnetic fields:
\[\begin{equation}\vec{S}(\vec{r},t) = \frac{1}{\mu_0} \vec{E}(\vec{r},t) \times \vec{B}(\vec{r},t)\end{equation}\]
This vector is crucial for understanding how energy moves through space and is a key concept in the study of electromagnetic radiation. Knowing the Poynting vector's magnitude and direction enables one to determine the flow of energy from the oscillating dipole into the surrounding environment.
\[\begin{equation}\vec{S}(\vec{r},t) = \frac{1}{\mu_0} \vec{E}(\vec{r},t) \times \vec{B}(\vec{r},t)\end{equation}\]
This vector is crucial for understanding how energy moves through space and is a key concept in the study of electromagnetic radiation. Knowing the Poynting vector's magnitude and direction enables one to determine the flow of energy from the oscillating dipole into the surrounding environment.
Radiated Power
The radiated power by an oscillating dipole refers to the total energy emitted as electromagnetic radiation per unit time. By integrating the Poynting vector over a closed surface enclosing the dipole, such as a sphere with radius \(r\), we can determine the instantaneous power emitted at any given moment. The equation for this is:
\[\begin{equation} P(t) = \oint \vec{S}(\vec{r},t) \cdot d\vec{A}\end{equation}\]
To find the mean power radiated over one cycle of oscillation, we take the time average of the instantaneous power, which yields:
\[\begin{equation}\bar{P} = \frac{1}{T} \int_{0}^{T} P(t) dt\end{equation}\]
This mean power is critical for various practical applications, such as antenna design in telecommunications, where engineers need to know the average energy output over time to ensure the efficient operation of the system.
\[\begin{equation} P(t) = \oint \vec{S}(\vec{r},t) \cdot d\vec{A}\end{equation}\]
To find the mean power radiated over one cycle of oscillation, we take the time average of the instantaneous power, which yields:
\[\begin{equation}\bar{P} = \frac{1}{T} \int_{0}^{T} P(t) dt\end{equation}\]
This mean power is critical for various practical applications, such as antenna design in telecommunications, where engineers need to know the average energy output over time to ensure the efficient operation of the system.