Chapter 22, Problem 53GP

(a)

To determine

The amount of energy that crosses a transmitting antenna per second.

Answer to Problem 53GP

  $0.40\mu W$

Explanation of Solution

Given data:

Power, ${\text{P}}_{\text{0}}=50\times {10}^3\text{W}$

Distance (radius), $r=100\times {10}^3\text{m}$

Formula used:

  $S=\frac{P_0}{4\pi r^2}$

Where,

  $S$ is the power flux.

  $P_0$ is power.

  $\text{r}$ is the radius.

Calculation:

Substitute the given data into the formula:

  $\begin{array}{c}S=\frac{P_0}{4\pi r^2}\\ =\frac{50\times {10}^3\text{W}}{4\pi {\left(100\text{m}\times {\text{10}}^3\right)}^2}\\ =3.98\times {10}^{-7}W/m^2\end{array}$

Therefore, the power output is:

  $\begin{array}{c}P=SA\\ =\left(3.98\times {10}^{-7}W/m^2\right)\times {\left(1.0m\right)}^2\\ =3.98\times {10}^{-7}W\\ =0.40\mu W\end{array}$

(b)

To determine

The root mean square value of the electric field

Answer to Problem 53GP

  $0.012\text{V/m}$

Explanation of Solution

Given data:

Power over unit length, $S=3.98\times {10}^{-7}W/m^2$

Speed of the light, $\text{c}{\text{=3×10}}^{\text{8}}\text{m/s}$

Permittivity of vacuum, ${\text{ε}}_{\text{0}}\text{=8}{\text{.85×10}}^{\text{-12}}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$

Formula used:

  $E_{rms}=\sqrt{\frac{S}{c{\epsilon }_0}}$

Where,

  $\text{S}$ is the power flux.

  $c$ is speed of the light.

  ${\text{E}}_{\text{rms}}$ is root square mean value of the electric field.

  ${\text{ε}}_{\text{0}}$ is the permittivity of vacuum.

Calculation:

Substitute the given data into the formula

  $E_{rms}=\sqrt{\frac{S}{c{\epsilon }_0}}$ ,

  ${\text{E}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3}{\text{.98×10}}^{\text{-7}}{\text{W/m}}^{\text{2}}}{\left({\text{3×10}}^{\text{8}}\text{m/s×8}{\text{.85×10}}^{\text{-12}}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)}}\text{=0}\text{.012}\text{V/m}$

(c)

To determine

The voltage induced in antenna.

Answer to Problem 53GP

  $0.012\text{V}$

Explanation of Solution

Given data:

Distance, $\text{d=1}\text{.0}\text{m}$

Root square mean value of the electric field, ${\text{E}}_{\text{rms}}\text{=0}\text{.012}\text{V/m}$

Formula used:

  $E_{rms}=V_{rms}=E_{rms}d$

Where,

  ${\text{E}}_{\text{rms}}$ is root square mean value of the electric field.

Calculation:

Substitute the given data into the formula $V_{rms}=E_{rms}d$ ,

  ${\text{V}}_{\text{rms}}={\text{E}}_{\text{rms}}\text{d}=\left(0.012\text{V/m}\right)\times \left(\text{1}\text{.0}\text{m}\right)=0.012\text{V}$