Chapter 34, Problem 60AP

(a)

To determine

The wavelength of the microwaves.

Answer to Problem 60AP

The wavelength of the microwaves is $1.5\times {10}^{-2}\text{m}$.

Explanation of Solution

Write the expression for the wavelength of an electromagnetic wave.

    $\lambda =\frac{c}{f}$                                                                                                                     (I)

Here, $\lambda$ is the wavelength, $c$ is the speed of light and $f$ is the frequency of the wave.

Conclusion:

Substitute $3.0\times {10}^8\text{m}/\text{s}$ for $c$ and $20.0\text{GHz}$ for $f$ in equation (I) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{3.0\times {10}^8\text{m}/\text{s}}{\left(20.0\text{GHz}\times \frac{{10}^9\text{Hz}}{1\text{GHz}}\right)}\\ =1.5\times {10}^{-2}\text{m}\end{array}$

Therefore, the wavelength of the microwaves is $1.5\times {10}^{-2}\text{m}$.

(b)

To determine

The total energy contained in each pulse.

Answer to Problem 60AP

The total energy contained in each pulse is $25\times {10}^{-6}\text{J}$.

Explanation of Solution

Write the expression for the total energy contained in each pulse.

    $U=P\left(\Delta t\right)$                                                                                                               (II)

Here, $U$ is the total energy contained in each pulse, $P$ is power during each pulse and $\Delta t$ is the duration of each pulse.

Conclusion:

Substitute $25.0\text{kW}$ for $P$ and $1.00\text{ns}$ $\Delta t$ in the equation (II) to find $U$.

    $\begin{array}{c}U=\left(25.0\text{kW}\times \frac{{10}^3\text{W}}{1\text{kW}}\right)\left(1.00\text{ns}\times \frac{{10}^{-9}\text{s}}{1\text{ns}}\right)\\ =25\times {10}^{-6}\text{J}\end{array}$

Therefore, the total energy contained in each pulse is $25\times {10}^{-6}\text{J}$.

(c)

To determine

The average energy density in each pulse.

Answer to Problem 60AP

The average energy density in each pulse is $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Write the expression for the energy density of electromagnetic wave.

    $U_{\text{avg}}=\frac{U}{V}$                                                                                                                 (III)

Here, $U_{\text{avg}}$ is the energy density, $U$ is the total energy associated with the wave and $V$ is the volume.

Write the expression for the volume.

    $V=\pi r^{2}l$                                                                                                                 (IV)

Here, $r$ is the radius of the dish and $l$ is the length travelled by wave.

Write the expression for the length travelled by wave.

    $l=c\left(\Delta t\right)$                                                                                                                  (V)

Here, $c$ is the speed of light.

Substitute $c\left(\Delta t\right)$ for $l$ in equation (IV) to find $V$.

    $V=\pi r^{2}c\left(\Delta t\right)$                                                                                                         (VI)

Conclusion:

Substitute, $3.0\times {10}^8\text{m}/\text{s}$ for $c$, $6.00\text{cm}$ for $r$ and $1.00\text{ns}$ for $\Delta t$ in equation (VI) to find $V$.

    $\begin{array}{c}V=\pi {\left(6.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\left(3.0\times {10}^8\text{m}/\text{s}\right)\left(1.00\text{ns}\times \frac{{10}^{-9}\text{s}}{1\text{ns}}\right)\\ =339.12\times {10}^{-5}{\text{m}}^{\text{3}}\end{array}$

Substitute, $339.12\times {10}^{-5}{\text{m}}^{\text{3}}$ for $V$ and $25\times {10}^{-6}\text{J}$ for $U$ in equation (III) to find $U_{\text{avg}}$.

    $\begin{array}{c}{U}_{\text{avg}}=\frac{25\times {10}^{-6}\text{J}}{339.12\times {10}^{-5}{\text{m}}^{\text{3}}}\\ =7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\end{array}$

Therefore, the average energy density in each pulse is $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$.

(d)

To determine

The amplitude of electric and magnetic fields associated with the wave.

Answer to Problem 60AP

The amplitude of the electric field is $40.8\times {10}^3\text{V}/\text{m}$ and the amplitude of the magnetic field is $1.36\times {10}^{-4}\text{T}$.

Explanation of Solution

Write the expression for the amplitude of the electric field associated with electromagnetic wave.

    $E_{\max }=\sqrt{\frac{2U_{\text{avg}}}{{\epsilon }_0}}$                                                                                                    (VII)

Here, $E_{\max }$ is the amplitude of the electric field, $U_{\text{avg}}$ is the average energy density in each pulse and ${\epsilon }_0$ is the permittivity of free space.

Write the expression for the amplitude of the magnetic field associated with electromagnetic field.

    $B_{\max }=\frac{E_{\max }}{c}$                                                                                                      (VIII)

Conclusion:

Substitute $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$ for $U_{\text{avg}}$ and $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ in equation (VII) to find $E_{\max }$.

    $\begin{array}{c}{E}_{\max }=\sqrt{\frac{2\left(7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\right)}{8.85\times {10}^{-12}\text{F}/\text{m}}}\\ =40.8\times {10}^3\text{V}/\text{m}\end{array}$

Substitute $40.8\times {10}^3\text{V}/\text{m}$ for $E_{\max }$ and $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VIII) to find $B_{\max }$.

    $\begin{array}{c}{B}_{\max }=\frac{40.8\times {10}^3\text{V}/\text{m}}{3.0\times {10}^8\text{m}/\text{s}}\\ =1.36\times {10}^{-4}\text{m}/\text{s}\end{array}$

Therefore, The amplitude of the electric field is $40.8\times {10}^3\text{V}/\text{m}$ and the amplitude of the magnetic field is $1.36\times {10}^{-4}\text{T}$.

(e)

To determine

The force exerted on the surface during the duration of the pulse.

Answer to Problem 60AP

The force exerted on the surface during the duration of the pulse is $8.33\times {10}^{-5}\text{N}$.

Explanation of Solution

Write the expression for the force exerted on a surface.

    $F=U_{\text{avg}}A$                                                                                                             (IX)

Here, $F$ is the force exerted on the surface and $A$ is the area.

Write the expression for the area.

    $A=\pi r^2$                                                                                                                 (X)

Here, $A$ is the area of surface and $r$ is the radius.

Conclusion:

Substitute $6.00\text{cm}$ for $r$ in equation (X) to find $A$.

    $\begin{array}{c}A=3.14\times {\left(6.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\\ =113.04\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$                                                                            (XI)

Substitute $113.04\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ and $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$ for $U_{\text{avg}}$ in equation (IX) to find $F$.

    $\begin{array}{c}F=7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\times \left(113.04\times {10}^{-4}{\text{m}}^{\text{2}}\right)\\ =8.33\times {10}^{-5}\text{N}\end{array}$

Therefore, the force exerted on the surface during the duration of the pulse is $8.33\times {10}^{-5}\text{N}$.