Chapter 22, Problem 37P

(a)

To determine

The average power incident on the telescope due to a wave at normal incidence with intensity $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$.

Answer to Problem 37P

The average power incident on the telescope due to a wave at normal incidence with intensity $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$ is $\underset{\_}{7.3\times {10}^{-22}\text{W}}$

Explanation of Solution

Given that the diameter of the telescope is $305\text{m}$, and intensity of the radio waves is ${10}^{-26}\text{W}/{\text{m}}^2$.

Write the expression for the average power incident on a surface of area $A$.

    $\langle P\rangle =IA\cos \theta$                                                                                                           (I)

Here, $\langle P\rangle$ is the average power incident on the surface, $I$ is the intensity of the wave, $A$ is the area of the surface, and $\theta$ is the angle.

Write the expression for the area of the circular surface.

    $A=\pi {\left(\frac{d}{2}\right)}^2$                                                                                                             (II)

Here, $A$ is the area of the surface, $d$ is the diameter of the surface.

Use equation (II) in equation (I),

    $\langle P\rangle =I\pi {\left(\frac{d}{2}\right)}^2\cos \theta$                                                                                               (III)

Conclusion:

Substitute $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$ for $I$, $305\text{m}$ for $d$, $0°$ for $\theta$ in equation (III) to find $\langle P\rangle$.

    $\begin{array}{c}\langle P\rangle =\left(1.0\times {10}^{-26}\text{W}/{\text{m}}^2\right)\pi {\left(\frac{305\text{m}}{2}\right)}^2\cos 0°\\ =7.3\times {10}^{-22}\text{W}\end{array}$

Therefore, the average power incident on the telescope due to a wave at normal incidence with intensity $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$ is $\underset{\_}{7.3\times {10}^{-22}\text{W}}$

(b)

To determine

Average power incident on Earth’s surface.

Answer to Problem 37P

Average power incident on Earth’s surface is $\underset{\_}{1.3\times {10}^{-12}\text{W}}$.

Explanation of Solution

The diameter of the Earth is $1.274\times {10}^7\text{m}$.

The average power incident on the Earth’s surface can be found out by equation (III),

    $\langle P\rangle =I\pi {\left(\frac{d}{2}\right)}^2\cos \theta$

Conclusion:

Substitute $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$ for $I$, $1.274\times {10}^6\text{m}$ for $d$, $0°$ for $\theta$ in equation (III) to find $\langle P\rangle$.

  $\begin{array}{c}\langle P\rangle =\left(1.0\times {10}^{-26}\text{W}/{\text{m}}^2\right)\pi {\left(\frac{1.274\times {10}^7\text{m}}{2}\right)}^2\cos 0°\\ =1.3\times {10}^{-12}\text{W}\end{array}$

Therefore, the average power incident on Earth’s surface is $\underset{\_}{1.3\times {10}^{-12}\text{W}}$.

(c)

To determine

The rms electric and magnetic fields.

Answer to Problem 37P

The rms electric and magnetic fields are $\underset{\_}{1.9\times {10}^{-12}\text{V}/\text{m}\text{ and 6}\text{.5}\times {\text{10}}^{-21}\text{T}}$.

Explanation of Solution

Write the expression for the energy density.

    $\langle u\rangle ={\epsilon }_0{E}_{\text{rms}}^2$                                                                                                            (IV)

Here, $\langle u\rangle$ is the energy density, ${\epsilon }_0$ permittivity of free space, $E_{\text{rms}}$ is the rms electric field.

Write the expression for the energy density in terms of intensity.

    $\langle u\rangle =\frac{I}{c}$                                                                                                                    (V)

Here, $c$ is the speed of light.

Use equation (V) in equation (IV),and solve for $E_{\text{rms}}$,

$E_{\text{rms}}=\sqrt{\frac{I}{{\epsilon }_{0}c}}$                                                                                                       (VI)

Write the relation connecting rms electric field and rms magnetic field.

    $E_{\text{rms}}=c{B}_{\text{rms}}$                                                                                                           (VII)

Here, $B_{\text{rms}}$ is the rms magnetic field.

Solve equation (VII) for $B_{\text{rms}}$.

    $B_{\text{rms}}=\frac{B_{ms}}{c}$                                                                                                           (VIII)

Conclusion:

Substitute $1.0\times {10}^{-26}\text{W}/{\text{m}}^2$ for $I$ , $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VI) to find $E_{\text{rms}}$.

    $\begin{array}{c}{E}_{\text{rms}}=\sqrt{\frac{1.0\times {10}^{-26}\text{W}/{\text{m}}^2}{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}}\\ =1.9\times {10}^{-12}\text{V}/\text{m}\end{array}$

Substitute $1.94\times {10}^{-12}\text{V}/\text{m}$ for $E_{\text{rms}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VIII) to find $B_{\text{rms}}$.

    $\begin{array}{c}{B}_{\text{rms}}=\frac{1.94\times {10}^{-12}\text{V}/\text{m}}{3.00\times {10}^8\text{m}/\text{s}}\\ =6.5\times {10}^{-21}\text{T}\end{array}$

Therefore, the rms electric and magnetic fields are $\underset{\_}{1.9\times {10}^{-12}\text{V}/\text{m}\text{ and 6}\text{.5}\times {\text{10}}^{-21}\text{T}}$.