Question

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

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

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: \(E_{\text{rms}} = \frac{32.0\, \text{mV/m}}{\sqrt{2}}\)

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: \(I = \frac{1}{2} \times 8.854 \times 10^{-12}\, \text{F/m} \times 3.0 \times 10^8\, \text{m/s} \times \left(\frac{32.0 \times 10^{-3}\, \text{V/m}}{\sqrt{2}}\right)^2\)

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 \(2.85 \times 10^{-7}\, \text{W/m}^2\).

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: \(u=\frac{2.85 \times 10^{-7}\, \text{W/m}^2}{3.0 \times 10^8\, \text{m/s}}\)

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 \(9.5 \times 10^{-16}\, \text{J/m}^3\).