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An airplane flying at a distance of \(10 \mathrm{~km}\) from a radio transmitter receives a signal of intensity \(10 \mu \mathrm{W} / \mathrm{m}^{2} .\) What is the amplitude of the (a) electric and (b) magnetic component of the signal at the airplane? (c) If the transmitter radiates uniformly over a hemisphere, what is the transmission power?

Short Answer

Expert verified
(a) Electric field amplitude \( \approx 27.4 \ \mathrm{V/m} \), (b) Magnetic field amplitude \( \approx 9.13 \times 10^{-8} \ \mathrm{T} \), (c) Transmission power \( \approx 6.28 \ \mathrm{W} \).

Step by step solution

01

Formula for Electric Field Intensity

The intensity \( I \) of an electromagnetic wave is related to the amplitude \( E \) of the electric component by the formula \( I = \frac{1}{2} \epsilon_0 c E^2 \), where \( \epsilon_0 \) is the permittivity of free space \( 8.85 \times 10^{-12} \ \mathrm{F/m} \) and \( c \) is the speed of light \( 3 \times 10^8 \ \mathrm{m/s} \).
02

Solve for Electric Field Amplitude

Rearrange the formula to solve for \( E \): \[ E = \sqrt{\frac{2I}{\epsilon_0 c}} \]Plug in the given intensity, \( I = 10 \times 10^{-6} \ \mathrm{W/m^2} \), and solve for \( E \).
03

Calculate Electric Field Amplitude

Substitute the values into the equation:\[ E = \sqrt{\frac{2 imes 10 \times 10^{-6}}{8.85 \times 10^{-12} \times 3 \times 10^8}} \approx 27.4 \ \mathrm{V/m} \]
04

Formula for Magnetic Field Amplitude

The relationship between the electric field \( E \) and the magnetic field \( B \) in an electromagnetic wave is given by \( E = cB \).
05

Solve for Magnetic Field Amplitude

Using the equation \( B = \frac{E}{c} \) where \( c = 3 \times 10^8 \ \mathrm{m/s} \): \[ B = \frac{27.4}{3 \times 10^8} \ \mathrm{T} \]
06

Calculate Magnetic Field Amplitude

Compute \( B \): \[ B \approx 9.13 \times 10^{-8} \ \mathrm{T} \]
07

Transmission Power over Hemisphere

The intensity \( I \) is the power per unit area. If power is spread uniformly over a hemisphere, the total area is \( 2 \pi r^2 \) where \( r = 10 \times 10^3 \ \mathrm{m} \).
08

Calculate Transmission Power

The total power \( P \) is given by \( P = I \times 2 \pi r^2 \). Substituting \( I = 10 \times 10^{-6} \ \mathrm{W/m^2} \) and \( r = 10,000 \ \mathrm{m} \), we find:\[ P = 10 \times 10^{-6} \times 2 \pi \times (10^4)^2 \approx 6.28 \ \mathrm{W} \]

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Key Concepts

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

Electric Field Amplitude
Electromagnetic waves consist of oscillating electric and magnetic fields. The amplitude of the electric field in these waves is a crucial factor in determining how much energy the wave carries. Intensity, which is the power per unit area, provides a way to calculate the electric field amplitude. The equation involved is:\[I = \frac{1}{2} \epsilon_0 c E^2\] Here,
  • \(I\) is the intensity,
  • \(\epsilon_0\) is the permittivity of free space,
  • \(c\) is the speed of light, and
  • \(E\) is the electric field amplitude.
Rearranging the formula, you can solve for the electric field amplitude: \[E = \sqrt{\frac{2I}{\epsilon_0 c}}\] By substituting the known values, you can determine the desired amplitude, giving you insight into how powerful the electric component of the wave is.
Magnetic Field Amplitude
The magnetic field amplitude is inherently linked to the electric field amplitude in an electromagnetic wave. This relationship is defined by the speed of light, \(c\). The fundamental equation connecting the electric field \(E\) and the magnetic field \(B\) is:\[ E = cB \]Using this equation, you can easily derive the magnetic field amplitude from the electric field amplitude by:\[ B = \frac{E}{c} \]Here, having calculated \(E\) already gives you a straightforward path to finding \(B\). This interconnectedness between electric and magnetic fields underlines the compelling symmetry and elegance of electromagnetic theory. The values you calculate provide a measurable insight into the wave's profile at any given point in space.
Transmission Power
Transmission power relates to how much energy a transmitter can send out into its environment. When dealing with a radio transmitter, you must consider how the power distributes itself across space. In this case, the wave energy radiates uniformly over a hemisphere. The intensity \(I\) (power per unit area) combined with the geometry of the hemisphere allows us to find the total transmission power.In a hemisphere, the surface area is given by:\[ 2 \pi r^2 \]Where \(r\) is the radius from the transmitter. By multiplying this area by the intensity \(I\), the total transmission power \(P\) is:\[ P = I \times 2 \pi r^2 \]Plug in the numbers for the radius and the intensity, and you find the transmitter's power, providing a sense of how much energy it broadcasts across the air.
Intensity of Electromagnetic Waves
Intensity is a key concept to grasp when understanding electromagnetic waves. It represents the power per unit area carried by a wave and plays a central role in calculating other parameters like field amplitudes. The expression for the intensity of an electromagnetic wave is:\[ I = \frac{P}{A} \]Where,
  • \(I\) is the intensity,
  • \(P\) is the power,
  • \(A\) is the area.
This equation implies that the greater the power applied, the greater the intensity if the area remains constant. It's important when considering the signal strength received by objects such as airplanes, as it governs how strong or weak a signal may appear at a given distance from the radio transmitter.

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Most popular questions from this chapter

Sunlight just outside Earth's atmosphere has an intensity of \(1.40 \mathrm{~kW} / \mathrm{m}^{2} .\) Calculate (a) \(E_{m}\) and (b) \(B_{m}\) for sunlight there, assuming it to be a plane wave.

Someone plans to float a small, totally absorbing sphere \(0.500\) \(\mathrm{m}\) above an isotropic point source of light, so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere's density is \(19.0 \mathrm{~g} / \mathrm{cm}^{3}\), and its radius is \(2.00 \mathrm{~mm}\). (a) What power would be required of the light source? (b) Even if such a source were made, why would the support of the sphere be unstable?

An isotropic point source emits light at wavelength \(500 \mathrm{~nm}\), at the rate of \(200 \mathrm{~W}\). A light detector is positioned \(400 \mathrm{~m}\) from the source. What is the maximum rate \(\partial B / \partial t\) at which the magnetic component of the light changes with time at the detector's location?

In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity \(6.0 \mathrm{~mW} / \mathrm{m}^{2}\). The sphere is totally absorbing and has a radius of \(2.0 \mu \mathrm{m}\) and a uniform density of \(5.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} .\) What is the mag= nitude of the sphere's acceleration due to the light?

High-power lasers are used to compress a plasma (a gas of charged particles) by radiation pressure. A laser generating radiation pulses with peak power \(1.5 \times 10^{3} \mathrm{MW}\) is focused onto \(1.0\) \(\mathrm{mm}^{2}\) of high-electron-density plasma. Find the pressure exerted on the plasma if the plasma reflects all the light beams directly back along their paths.

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