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Chapter 22: Problem 74

A sinusoidal EM wave has an electric field amplitude $E_{\mathrm{m}}=32.0 \mathrm{mV} / \mathrm{m} .$ What are the intensity and average energy density? [Hint: Recall the relationship between amplitude and rms value for a quantity that varies sinusoidally.]

Short Answer

Expert verified
Answer: The intensity of the sinusoidal EM wave is approximately \(2.85 \times 10^{-7}\, \text{W/m}^2\) and the average energy density is approximately \(9.5 \times 10^{-16}\, \text{J/m}^3\).

Step by step solution

01

Calculate the RMS value of electric field amplitude

To find the intensity and average energy density of the sinusoidal wave, we first need to find the root mean square (RMS) value of the electric field amplitude. The RMS value of a sinusoidal quantity is given by the amplitude divided by the square root of 2: \(E_{\text{rms}} = \frac{E_m}{\sqrt{2}}\) We are given the electric field amplitude \(E_{\text{m}}=32.0\, \text{mV/m}\). So we can calculate the RMS value: \(E_{\text{rms}} = \frac{32.0\, \text{mV/m}}{\sqrt{2}}\)
02

Use the relationship between electric field and intensity

Now that we have the RMS value of the electric field amplitude, we can use the relationship between the electric field and the intensity of the wave, which is given by the following formula: \(I = \frac{1}{2} \epsilon_0 c E_{\text{rms}}^2\) Where \(I\) is the intensity, \(\epsilon_0 = 8.854 \times 10^{-12}\, \text{F/m}\) is the vacuum permittivity, and \(c = 3.0 \times 10^8\, \text{m/s}\) is the speed of light in a vacuum. Substituting the values, we get: \(I = \frac{1}{2} \times 8.854 \times 10^{-12}\, \text{F/m} \times 3.0 \times 10^8\, \text{m/s} \times \left(\frac{32.0 \times 10^{-3}\, \text{V/m}}{\sqrt{2}}\right)^2\)
03

Calculate the intensity

Now we can calculate the intensity: \(I = \frac{1}{2} \times 8.854 \times 10^{-12}\, \text{F/m} \times 3.0 \times 10^8\, \text{m/s} \times \left(\frac{32.0 \times 10^{-3}\, \text{V/m}}{\sqrt{2}}\right)^2 \approx 2.85 \times 10^{-7}\, \text{W/m}^2\) So, the intensity of the sinusoidal EM wave is approximately \(2.85 \times 10^{-7}\, \text{W/m}^2\).
04

Use the relationship between intensity and average energy density

Next, we can use the relationship between intensity and average energy density to find the average energy density of the wave. The average energy density is given by the following formula: \(u=\frac{I}{c}\) Substituting the values, we get: \(u=\frac{2.85 \times 10^{-7}\, \text{W/m}^2}{3.0 \times 10^8\, \text{m/s}}\)
05

Calculate the average energy density

Now we can calculate the average energy density: \(u = \frac{2.85 \times 10^{-7}\, \text{W/m}^2}{3.0 \times 10^8\, \text{m/s}} \approx 9.5 \times 10^{-16}\, \text{J/m}^3\) So, the average energy density of the sinusoidal EM wave is approximately \(9.5 \times 10^{-16}\, \text{J/m}^3\). In conclusion, the intensity of the sinusoidal EM wave is approximately \(2.85 \times 10^{-7}\, \text{W/m}^2\), and the average energy density is approximately \(9.5 \times 10^{-16}\, \text{J/m}^3\).

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Most popular questions from this chapter

An unpolarized beam of light (intensity \(I_{0}\) ) is moving in the \(x\) -direction. The light passes through three ideal polarizers whose transmission axes are (in order) at angles \(0.0^{\circ}, 45.0^{\circ},\) and \(30.0^{\circ}\) counterclockwise from the \(y\) -axis in the \(y z\) -plane. (a) What is the intensity and polarization of the light that is transmitted by the last polarizer? (b) If the polarizer in the middle is removed, what is the intensity and polarization of the light transmitted by the last polarizer?
An electric dipole antenna used to transmit radio waves is oriented horizontally north-south.At a point due east of the transmitter, how should a magnetic dipole antenna be oriented to serve as a receiver?
A \(1.0-\mathrm{m}^{2}\) solar panel on a satellite that keeps the panel oriented perpendicular to radiation arriving from the Sun absorbs $1.4 \mathrm{kJ}\( of energy every second. The satellite is located at \)1.00 \mathrm{AU}\( from the Sun. (The Earth Sun distance is defined to be \)1.00 \mathrm{AU} .$. How long would it take an identical panel that is also oriented perpendicular to the incoming radiation to absorb the same amount of energy, if it were on an interplanetary exploration vehicle 1.55 AU from the Sun? ( W) tutorial: sunlight on Io).
Vertically polarized light with intensity \(I_{0}\) is normally incident on an ideal polarizer. As the polarizer is rotated about a horizontal axis, the intensity \(I\) of light transmitted through the polarizer varies with the orientation of the polarizer \((\theta),\) where \(\theta=0\) corresponds to a vertical transmission axis. Sketch a graph of \(I\) as a function of \(\theta\) for one complete rotation of the polarizer $\left(0 \leq \theta \leq 360^{\circ}\right)$.(W) tutorial: polarized light).
Verify that the equation \(I=\langle u\rangle c\) is dimensionally consistent (i.e., check the units).
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