Chapter 34, Problem 34.60AP

A microwave source produces pulses of 20.0–GHz radiation, with each pulse lasting 1.00 ns. A parabolic reflector with a face area of radius 6.00 cm is used to focus the microwaves into a parallel beam of radiation as shown in Figure P34.60. The average power during each pulse is 25.0 kW. (a) What is the wavelength of these microwaves? (b) What is the total energy contained in each pulse? (c) Compute the average energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves. (e) Assuming that this pulsed beam strikes an absorbing surface, compute the force exerted on the surface during the 1.00-ns duration of each pulse.

Figure P34.60

To determine

The wavelength of microwaves.

Answer to Problem 34.60AP

The wavelength of microwaves is $0.015\text{m}$.

Explanation of Solution

Given Info: The frequency of pulse is $20.0\text{GHz}$, time duration of each pulse is $1.00\text{ns}$, radius of parabolic reflector is $6.00\text{cm}$ and the average power during each pulse is $25.0\text{kW}$.

Formula to calculate the wavelength is,

$\lambda =\frac{c}{f}$

Here,

$f$ is the frequency of pulse.

$c$ is the speed of waves.

Substitute $20.0\text{GHz}$ for $f$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in above equation.

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}/\text{s}}{20.0\text{GHz}}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{20.0\times {10}^9\text{Hz}}\\ =0.015\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of microwaves is $0.015\text{m}$.

To determine

The total energy contained in each pulse.

Answer to Problem 34.60AP

The total energy contained in each pulse is $\text{2}\text{.5}\times {\text{10}}^{-5}\text{J}$.

Explanation of Solution

Given Info: The frequency of pulse is $20.0\text{GHz}$, time duration of each pulse is $1.00\text{ns}$, radius of parabolic reflector is $6.00\text{cm}$ and the average power during each pulse is $25.0\text{kW}$.

Formula to calculate the total energy is,

$T_{\text{E}}=P_{\text{avg}}\times t$

Here,

$P_{\text{avg}}$ is the average power.

$t$ is the time duration of each pulse.

Substitute $25.0\text{kW}$ for $P_{\text{avg}}$ and $1.00\text{ns}$ for $t$ in the above equation.

$\begin{array}{c}{T}_{\text{E}}=25.0\text{kW}\times 1.00\text{ns}\\ =25.0\times {10}^3\text{W}\times 1.00\times {10}^{-9}\text{s}\\ =25\times {10}^{-6}\text{J}\\ \text{=2}\text{.5}\times {\text{10}}^{-5}\text{J}\end{array}$

Conclusion:

Therefore, the total energy contained in each pulse is $\text{2}\text{.5}\times {\text{10}}^{-5}\text{J}$.

To determine

The average energy density inside each pulse.

Answer to Problem 34.60AP

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

Explanation of Solution

Given Info: The frequency of pulse is $20.0\text{GHz}$, time duration of each pulse is $1.00\text{ns}$, radius of parabolic reflector is $6.00\text{cm}$ and the average power during each pulse is $25.0\text{kW}$.

Formula to calculate the average energy density for each pulse is,

$\begin{array}{c}{u}_{\text{g}}=\frac{I}{c}\\ =\frac{P_{\text{avg}}}{Ac}\\ =\frac{P_{\text{avg}}}{\pi r^{2}c}\end{array}$

Here,

$P_{\text{avg}}$ is the average power.

$r$ is the radius.

$c$ is the velocity of pulse.

$A$ is the cross sectional area.

Substitute $25.0\text{kW}$ for $P_{\text{avg}}$, 3.14 for $\pi$, $6.00\text{cm}$ for $r$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

$\begin{array}{c}{u}_{\text{g}}=\frac{25.0\text{kW}}{3.14\times {\left(6.00\text{cm}\right)}^2\times \left(3\times {10}^8\text{m}/\text{s}\right)}\\ =\frac{25.0\times {10}^3\text{W}}{3.14\times {\left(6.00\times {10}^{-2}\text{m}\right)}^2\times \left(3\times {10}^8\text{m}/\text{s}\right)}\\ =7.37\times {10}^{-3}\text{J}/{\text{m}}^3\end{array}$

Conclusion:

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

To determine

The amplitude of electric and magnetic fields in the microwaves.

Answer to Problem 34.60AP

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

Explanation of Solution

Given Info: The frequency of pulse is $20.0\text{GHz}$, time duration of each pulse is $1.00\text{ns}$, radius of parabolic reflector is $6.00\text{cm}$ and the average power during each pulse is $25.0\text{kW}$.

Formula to calculate the magnitude of electric field is,

$E_{\text{max}}=\sqrt{\frac{2u_{\text{g}}}{{\epsilon }_{\text{o}}}}$

Here,

$u_{\text{g}}$ is the average energy density for each pulse.

Substitute $7.37\times {10}^{-3}\text{J}/{\text{m}}^3$ for $u_{\text{g}}$ and $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_{\text{o}}$ in the above equation.

$\begin{array}{c}{E}_{\text{max}}=\sqrt{\frac{2\times 7.37\times {10}^{-3}\text{J}/{\text{m}}^3}{8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}}}\\ =40810.98\text{V}/\text{m}\\ =40.8\times {10}^3\text{V}/\text{m}\end{array}$

Formula to calculate the magnitude of magnetic field is,

$B_{\text{max}}=\frac{E_{\text{max}}}{c}$

Here,

$E_{\text{max}}$ is the electric field magnitude.

$c$ is the velocity of microwave.

Substitute for $40.8\times {10}^3\text{V}/\text{m}$ $E_{\text{max}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation.

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

Conclusion:

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

To determine

The force exerted on the surface on each pulse.

Answer to Problem 34.60AP

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

Explanation of Solution

Given Info: The frequency of pulse is $20.0\text{GHz}$, time duration of each pulse is $1.00\text{ns}$, radius of parabolic reflector is $6.00\text{cm}$ and the average power during each pulse is $25.0\text{kW}$.

Formula to calculate the force exerted is,

$\begin{array}{c}F=u_{\text{g}}A\\ =u_{\text{g}}\pi r^2\end{array}$

Here,

$u_{\text{g}}$ is the average energy density of each pulse.

$A$ is the cross sectional area.

$r$ is the radius of pulse.

Substitute $7.37\times {10}^{-3}\text{J}/{\text{m}}^3$ for $u_{\text{g}}$, 3.14 for $\pi$ and $6.00\text{cm}$ for $r$ in the above equation.

$\begin{array}{c}F=\left(7.37\times {10}^{-3}\text{J}/{\text{m}}^3\right)\times 3.14\times {\left(6.00\text{cm}\right)}^2\\ =\left(7.37\times {10}^{-3}\text{J}/{\text{m}}^3\right)\times 3.14\times {\left(6.00\times {10}^{-2}\text{m}\right)}^2\\ =8.33\times {10}^{-5}\text{N}\end{array}$

Conclusion:

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