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Chapter 32: Problem 38

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of \(10.0 \mathrm{~m}\) from this source, the amplitude of the electric field is measured to be \(3.50 \mathrm{~N} / \mathrm{C}\). What is the electric-field amplitude \(20.0 \mathrm{~cm}\) from the source?

Short Answer

Expert verified
The amplitude of the electric field 20 cm from the source is \( E = 87.5 \) N/C.

Step by step solution

01

Identify applicable formulas

The intensity of the wave is given by the formula I = P / (4 * 蟺 * r虏), where P is the power, I is the intensity, and r is the distance from the source. Also, the intensity is proportional to the square of the electric field (E), that is, I 鈭� E虏. Hence, if we get the ratio of intensities at different distances, on solving, we will get E = E鈧� * (r鈧� / r) where r is the distance from the source, r鈧� is the reference distance, E is the amplitude of the electric field and E鈧� is the amplitude of the electric field at the reference distance.
02

Substitute given values

The problem provides us the radius at two different locations (10m or 1000cm and 20cm) and the amplitude of the electric field at 10m. We can substitute these values into the formula E = E鈧� * (r鈧� / r) to find E.
03

Solve for the electric field E

Substitute E鈧� = 3.50 N/C (amplitude of the electric field at 10m), r鈧� = 1000 cm (reference distance 10m converted into cm to match the unit of the other distance given), and r = 20 cm into the formula E = E鈧� * (r鈧� / r). Then solve for E.
04

Verify the solution

Once we solve and find the answer for E, we check the units and verify if the magnitude of the answer makes sense in the context of the given problem.

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Key Concepts

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

Electric Field Amplitude
In electromagnetic waves, the electric field amplitude refers to the maximum strength of the electric field component. This is crucial because it directly influences how much energy the wave can transfer. The electric field amplitude is usually denoted as \( E \).When dealing with problems like the original exercise, it is essential to understand that this amplitude decreases with distance from the wave source. This decrease is tied to the fundamental nature of wave propagation in space.In the example given, the measured electric field at a reference distance provides a basis for calculations. The amplitude measured at 10 meters (or 1000 cm) from the source serves as the initial value, helping predict the electric field at a new distance. The equation \( E = E_0 \times \frac{r_0}{r} \) captures this relationship clearly, highlighting how the electric field weakens as you move further from the source.
Wave Intensity Formula
Wave intensity is a measure of how much power a wave carries through a given area. For electromagnetic waves, intensity is a crucial factor because it determines how much energy can be transferred over a specific period.The intensity \( I \) can be mathematically expressed using the formula: \[ I = \frac{P}{4\pi r^2} \] where \( P \) is the power output of the source and \( r \) is the distance from the source.Intensity is also related to the electric field amplitude by the relationship \( I \propto E^2 \). This relationship explains why changes in electric field amplitude can significantly alter wave intensity. This means if the amplitude is doubled, the intensity becomes four times greater. This quadratic relationship demonstrates the significant impact the electric field has on intensity. This understanding helps in solving exercises where you know either the amplitude or the intensity and need to find the other.
Distance and Field Relationship
The relationship between distance from the wave source and the electric field is a key concept in understanding electromagnetic wave behavior. As electromagnetic waves move away from the source, they spread out over a larger area. This spreading causes the electric field amplitude to decrease.This concept can be understood through the inverse relationship formula \( E = E_0 \times \frac{r_0}{r} \). Here:
  • \( E \) is the electric field amplitude at a new distance \( r \).
  • \( E_0 \) represents the electric field amplitude at an initial distance \( r_0 \).
As distances increase, \( r_0/r \) decreases, illustrating how field amplitude reduces with distance. This formula allows you to predict or calculate the change in electric field amplitude when distance changes, providing valuable insight into the wave's behavior at different points in space.

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Most popular questions from this chapter

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). How large a sail is necessary to propel a \(10,000 \mathrm{~kg}\) spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a \(50.0 \mathrm{~W}\) (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is \(1575.42 \mathrm{MHz}\). Assume that a satellite transmits half of its power at each frequency and that the waves travel uniformly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton's laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel \(1.50 \mathrm{~cm}\) on a side that absorbs all of the beam, what average pressure does the signal exert on it? (d) What wavelength must the receiver be tuned to?

A totally reflecting disk has radius \(8.00 \mu \mathrm{m}\) and average density \(600 \mathrm{~kg} / \mathrm{m}^{3}\). A laser has an average power output \(P_{\mathrm{av}}\) spread uniformly over a cylindrical beam of radius \(2.00 \mathrm{~mm}\). When the laser beam shines upward on the disk in a direction perpendicular to its flat surface, the radiation pressure produces a force equal to the weight of the disk. (a) What value of \(P_{\text {av }}\) is required? (b) What average laser power is required if the radius of the disk is doubled?

We can reasonably model a \(75 \mathrm{~W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\left.\mathrm{W} / \mathrm{m}^{2}\right)\) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

A circular wire loop has a radius of \(7.50 \mathrm{~cm}\). A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is \(0.0275 \mathrm{~W} / \mathrm{m}^{2},\) and the wavelength of the wave is \(6.90 \mathrm{~m} .\) What is the maximum emf induced in the loop?
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