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Chapter 32: Problem 46

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At 10.0\(\mathrm { m }\) from this source, the amplitude of the electric field is measured to be 1.50\(\mathrm { N } / \mathrm { C }\) . What is the electric-field amplitude at a distance of 20.0\(\mathrm { cm }\) from the source?

Short Answer

Expert verified
The electric field amplitude at 0.20 m is 3750 N/C.

Step by step solution

01

Understanding the Problem

We need to determine the electric field amplitude at a closer distance given the amplitude at a farther distance from a wave source that radiates uniformly. Since it is given that the waves radiate uniformly in all directions, we will use the inverse square law for this scenario.
02

Applying the Inverse Square Law

For waves radiating uniformly from a point source, the intensity (and thereby the amplitude) of the wave is inversely proportional to the square of the distance from the source. Mathematically, if the electric field amplitude at a distance \( d_1 \) is \( E_1 \), and at distance \( d_2 \) is \( E_2 \), then \( \frac{E_1}{E_2} = \left(\frac{d_2}{d_1}\right)^2 \).
03

Identifying Known Values

We are given that \( E_1 = 1.50\, \text{N/C} \), \( d_1 = 10.0\, \text{m} \), and \( d_2 = 0.20\, \text{m} \).
04

Calculating the Ratio of Distances

Calculate \( \left(\frac{d_1}{d_2}\right)^2 \) to find the ratio of Intensities. Here, \( d_1 = 10.0 \, \text{m} \) and \( d_2 = 0.20 \, \text{m} \).\[\left(\frac{d_1}{d_2}\right)^2 = \left(\frac{10.0}{0.20}\right)^2 = (50)^2 = 2500\]
05

Determining the Electric Field Amplitude at 0.20 m

Using the inverse square law relationship \( \frac{E_1}{E_2} = \left(\frac{d_2}{d_1}\right)^2 \), rearrange for \( E_2 \):\[E_2 = E_1 \cdot \left(\frac{d_1}{d_2}\right)^2 = 1.50 \, \text{N/C} \times 2500\]Thus, \( E_2 = 3750 \, \text{N/C} \).

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude represents the strength of the electric field at a particular point in space. It is a crucial parameter in understanding electromagnetic waves and is measured in newtons per coulomb (N/C). The amplitude determines the maximum extent of the field's oscillation and directly influences how much force would act on an electric charge placed within the field.
When considering sources of electromagnetic waves, such as sinusoidal ones, the electric field amplitude can change with distance from the source.
  • The further from the source, the lower the amplitude becomes.
  • This is due to the spreading of the wave energy over a larger area.
By using known values at one point, we can predict the field's amplitude at another location using mathematical relationships like the inverse square law.
Sinusoidal Electromagnetic Waves
Sinusoidal electromagnetic waves are waves that oscillate in a smooth, sinusoidal pattern. These waves consist of oscillating electric and magnetic fields that propagate through space. Such waves are fundamental in electromagnetic theory and display crucial behaviors:
  • They move at the speed of light.
  • Their electric and magnetic fields are perpendicular to each other and the direction of wave propagation.
Importantly, these waves can be generated by sources emitting uniformly in all directions, which simplifies analysis using laws like the inverse square law. As they radiate outward, the amplitude of these waves, including the electric field component, varies with distance from the source. This variation is often quantitatively analyzed to predict wave behavior in different scenarios.
Distance and Intensity Relationship
The relationship between distance and intensity for electromagnetic waves is often understood through the inverse square law. Essentially, this law states that the intensity of wave phenomena, like light or sound, decreases with the square of the distance from the source.
The intensity is directly related to the amplitude of the electric field because:
  • Higher amplitudes mean higher energy outputs.
  • As distance increases, the energy is spread over a larger area, reducing intensity.
In mathematical terms, if you double the distance from a source, the intensity of the wave becomes one-fourth of its original value. This relation allows us to calculate changes in amplitude as shown in the provided solution. By knowing the initial field amplitude and the distances involved, one can determine the new amplitude at a closer or farther point.

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Most popular questions from this chapter

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm { km }\) ? (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance from earth to Sirius in kilometers?

A satellite 575\(\mathrm { km }\) above the earth's surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0\(\mathrm { kW }\) . (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring 15.0\(\mathrm { cm }\) by 40.0\(\mathrm { cm }\) oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

Fields from a Light Bulb. We can reasomably 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 \(\left( \mathrm { in } \mathrm { W } / \mathrm { m } ^ { 2 } \right)\) 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 small helium-neon laser emits red visible light with a power of 4.60\(\mathrm { mW }\) in a beam that has a diameter of 2.50\(\mathrm { mm }\) . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field are the magnetic field? (c) What is the total energy contained in a \(1.00 - \mathrm { m }\) length of the beam?

A plane sinusoidal electromagnetic wave in air has a wavelength of 3.84\(\mathrm { cm }\) and an \(\vec { \boldsymbol { E } }\) -field amplitude of 1.35\(\mathrm { V } / \mathrm { m }\) . (a) What is the frequency? (b) What is the \(\vec { \boldsymbol { B } }\) -field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area 0.240\(\mathrm { m } ^ { 2 }\) perpendicular to the direction of propagation?
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