Chapter 22, Problem 47GP

(a)

To determine

The total energy received by the antenna during 6.0 hours of viewing station's programs.

Answer to Problem 47GP

  $1.85\times {10}^{-10}\text{J}$

Explanation of Solution

Given:

Intensity, $\mathrm{I}=1.0\times {10}^{-13}{\text{W/m}}^2$

Diameter of satellite, $\mathrm{d}=33\text{ cm}=0.33\text{ m}$

Time interval, $\mathrm{t}=6\text{ h}=6\times 3600\text{ s}=2.16\times {10}^4\text{s}$

Formula used:

The total energy can be given as:

  $\mathrm{E}=\mathrm{P}\mathrm{t}$

Here, power is given as:

  $\mathrm{P}=\mathrm{I}\mathrm{A}$

So, the total energy is given by:

  $\mathrm{E}=\mathrm{I}\mathrm{A}\mathrm{t}$

Here, $\mathrm{I}$ is the intensity, $\mathrm{A}$ is the area and $\mathrm{t}$ is the time.

Calculation:

The area can be calculated as follows:

  $\begin{array}{l}\mathrm{A}=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2\\ \mathrm{A}=\left(3.14\right){\left(\frac{0.33\text{ m}}{2}\right)}^2\\ \mathrm{A}=0.0855{\text{ m}}^2\end{array}$

Putting all the values in the above expression, we get:

  $\begin{array}{l}\mathrm{E}=\left(1.0\times {10}^{-13}{\text{W/m}}^2\right)\left(0.0855{\text{ m}}^2\right)\left(2.16\times {10}^4\text{s}\right)\\ \mathrm{E}=1.85\times {10}^{-10}\text{J}\end{array}$

Conclusion:

Hence, the total energy is $1.85\times {10}^{-10}\text{J}$ .

(b)

To determine

To Find: The amplitudes of E and B field of the EM wave.

Answer to Problem 47GP

The amplitudes of E and B field of the EM wave are: ${\mathrm{E}}_0=8.7\times {10}^{-6}\text{V/m}$ and ${\mathrm{B}}_0=2.9\times {10}^{-14}\text{T}$ .

Explanation of Solution

Given:

Intensity, $\mathrm{I}=1.0\times {10}^{-13}{\text{W/m}}^2$

Diameter of satellite, $\mathrm{d}=33\text{ cm}=0.33\text{ m}$

Time interval, $\mathrm{t}=6\text{ h}=6\times 3600\text{ s}=2.16\times {10}^4\text{s}$

Formula Used:

The average intensity of an EM wave is given by:

  $\overline{\mathrm{I}}=\frac{1}{2}{\mathrm{\epsilon }}_0\mathrm{c}{\mathrm{E}}_0^2$

Calculation:

Putting all the values to get the value of ${\mathrm{E}}_0$ :

  $\begin{array}{l}{\mathrm{E}}_0=\sqrt{\frac{2\overline{\mathrm{I}}}{{\mathrm{\epsilon }}_0\mathrm{c}}}\\ {\mathrm{E}}_0=\sqrt{\frac{2\left(1.0\times {10}^{-13}{\text{W/m}}^2\right)}{\left(8.854\times {10}^{-12}{\text{C}}^2\text{N}{\text{.m}}^2\right)\left(3\times {10}^8\text{m/s}\right)}}\\ {\mathrm{E}}_0=8.7\times {10}^{-6}\text{V/m}\end{array}$

Similarly, using relation:

  $\begin{array}{l}{\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}\\ {\mathrm{B}}_0=\frac{8.7\times {10}^{-6}\text{V/m}}{3\times {10}^8\text{m/s}}\\ {\mathrm{B}}_0=2.9\times {10}^{-14}\text{T}\end{array}$

Conclusion:

The amplitudes of E and B field of the EM wave are: ${\mathrm{E}}_0=8.7\times {10}^{-6}\text{V/m}$ and ${\mathrm{B}}_0=2.9\times {10}^{-14}\text{T}$ .