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What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically with an average power of 250 \(\mathrm{kW}\) ?

Short Answer

Expert verified
The average magnitude of the Poynting vector 5.00 miles from a 250 kW isotropically broadcasting radio transmitter is \(5.3219 \times 10^{-7} \frac{\text{W}}{\text{m}^2}\).

Step by step solution

01

Understand the Poynting Vector and Power Relationship

The Poynting vector, represented by \textbf{S}, describes the power per unit area (power flux) of an electromagnetic wave. For isotropic radiation, the power emanates equally in all directions from a point source. The magnitude of the Poynting vector at a distance r from the source is the power P divided by the surface area of a sphere with radius r: \(\mathbf{S} = \frac{P}{4\pi r^2}\).
02

Convert Miles to Meters

Since the Poynting vector is typically given in SI units, convert the distance from miles to meters: \(1 \text{ mile} = 1609.34 \text{ meters}\).
03

Calculate the Surface Area of the Sphere

The surface area A of a sphere with radius r is given by \(A = 4\pi r^2\). Fill in the distance in meters to find the sphere's surface area at the given distance.
04

Compute the Magnitude of the Poynting Vector

Use the formula for the Poynting vector magnitude, \(\mathbf{S}\), the power of the transmitter, and the surface area calculated in step 3 to find the average magnitude of the Poynting vector.

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Key Concepts

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

Understanding Power Flux Through the Poynting Vector
The Poynting vector is a fundamental concept in electromagnetism that characterizes the power flux, which is the power per unit area, of an electromagnetic wave. It's the directional energy transfer rate per unit area. Imagine sunlight passing through a window; the Poynting vector would describe the flow of light energy through that window.

In the exercise, it's essential to comprehend that the Poynting vector, denoted as \(\textbf{S}\), provides a description of how energy is distributed through space as electromagnetic waves travel. The power mentioned in the exercise, broadcast by the radio transmitter, indeed spreads out in the form of electromagnetic waves, and the Poynting vector helps quantify this distribution.

The expression given in the solution, \(\textbf{S} = \frac{P}{4\text{\textpi} r^2}\), encapsulates the power flux by relating the total power P radiated by the source to the area of a sphere that expands as the waves move away from the source. By doing so, we can calculate the average magnitude of the Poynting vector at any distance from the transmitter, understanding that the power flux decreases with the square of the distance.
Isotropic Radiation from a Point Source
The term 'isotropic radiation' refers to a point source emitting energy uniformly in all directions. This concept is vital in understanding how the radio transmitter in the exercise broadcasts power.

The key to isotropic radiation is symmetry. Imagine a sparkling orb, swelling in size, where every point on its surface glows equally; that's how an isotropic radiator behaves. In reality, perfect isotropic radiators don't exist, but they serve as a useful model for basic calculations and concepts.

For isotropic radiation, the surface over which the power spreads is a sphere, and as the wave travels outward, the sphere's surface area increases, causing the power flux to diminish. This can be visualized by picturing the sphere's radius stretching outward, with the same amount of power now spread across a more extensive area, thus reducing the density of the power - the Poynting vector's magnitude - with distance.
Electromagnetic Waves and Their Energy Distribution
An electromagnetic wave, like the radio waves from the transmitter in our exercise, consists of oscillating electric and magnetic fields. These waves are responsible for carrying energy across space which enables radios and other devices to pick up signals.

Electromagnetic waves travel at the speed of light in a vacuum, and the energy they carry is expressed by the Poynting vector. This is why the Poynting vector is significant; it tells us the energy flow rate at a given point in space, per unit area, due to the passage of an electromagnetic wave.

The solution's step-by-step approach elegantly demonstrates the direct relationship between electromagnetic waves and the Poynting vector as it pertains to power distribution. By analyzing this relationship, students can gain insight into the energy dynamics of electromagnetic radiation emitted by various sources, from simple antennas to celestial bodies.

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Most popular questions from this chapter

A plane electromagnetic wave varies sinusoidally at 90.0 \(\mathrm{MHz}\) as it travels along the \(+x\) direction. The peak value of the electric field is 2.00 \(\mathrm{mV} / \mathrm{m}\) , and it is directed along the \(\pm y\) direction. (a) Find the wavelength, the period, and the maximum value of the magnetic field. (b) Write expressions in SI units for the space and time variations of the electric field and of the magnetic field. Include numerical values and include subscripts to indicate coordinate directions. (c) Find the average power per unit area that this wave carries through space. (d) Find the average energy density in the radiation (in joules per cubic meter). (e) What radiation pressure would this wave exert upon a perfectly reflecting surface at normal incidence?

A plane electromagnetic wave has an intensity of 750 \(\mathrm{W} / \mathrm{m}^{2}\) . A flat, rectangular surface of dimensions \(50 \mathrm{cm} \times 100 \mathrm{cm}\) is placed perpendicular to the direction of the wave. The surface absorbs half of the energy and reflects half. Calculate (a) the total energy absorbed by the surface in 1.00 \(\mathrm{min}\) and \((\mathrm{b})\) the momentum absorbed in this time.

A radio wave transmits 25.0 \(\mathrm{W} / \mathrm{m}^{2}\) of power per unit area. A flat surface of area \(A\) is perpendicular to the direction of propagation of the wave. Calculate the radiation pressure on it, assuming the surface is a perfect absorber.

A 10.0 -mW laser has a beam diameter of 1.60 \(\mathrm{mm}\) . (a) What is the intensity of the light, assuming it is uniform across the circular beam? (b) What is the average energy density of the beam?

The United States Navy has long proposed the construction of extremely low- frequency (EIF) communication systems. Such waves could penetrate the oceans to reach distant submarines. Calculate the length of a quarterwavelength antenna for a transmitter generating ELF waves of frequency 75.0 Hz. How practical is this?

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