Chapter 24, Problem 72P

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 micro-waves into a parallel beam of radiation as shown in Figure P24.72. 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.

To determine

The wavelength of the microwave.

Answer to Problem 72P

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

Explanation of Solution

Write the expression to calculate the wavelength.

  $\lambda =\frac{c}{f}$

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

Conclusion:

Substitute $20.0\text{GHz}$ for f and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $\lambda$.

  $\begin{array}{c}\lambda =\frac{3.00\times {10}^8\text{m}/\text{s}}{20.0\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)}\\ =0.015\text{m}\end{array}$

Thus, the wavelength is $0.015\text{m}$.

To determine

The total energy of each pulse.

Answer to Problem 72P

The total energy is $25.0\times {10}^{-6}\text{J}$.

Explanation of Solution

Write the expression to calculate the total energy.

  $U=Pt$

Here, U is the total energy, P is the power and t is the time.

Conclusion:

Substitute $25.0\text{kW}$ for P and $1.00\text{ns}$ for t in the above equation to calculate U.

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

Thus, the total energy is $25.0\times {10}^{-6}\text{J}$.

To determine

The average energy density inside the pulse.

Answer to Problem 72P

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

Explanation of Solution

Write the expression to calculate the average energy density.

  $u=\frac{U}{\pi r^{2}l}$

Here, u is the average energy density, r is the radius and l is the length of the pulse.

Write the expression to calculate the length of the pulse.

  $l=ct$

Use the expression $l=ct$ in the equation for u to rewrite.

  $u=\frac{U}{\pi r^2\left(ct\right)}$

Conclusion:

Substitute $25.0\times {10}^{-6}\text{J}$ for U, $1.00\text{ns}$ for t, $6.00\text{cm}$ for r and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate u.

  $\begin{array}{c}u=\frac{25.0\times {10}^{-6}\text{J}}{\pi {\left(6.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(3.00\times {10}^8\text{m}/\text{s}\right)\left(1.00\text{ns}\right)\left(\frac{{10}^{-9}\text{s}}{1\text{ns}}\right)}\\ =7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\end{array}$

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

To determine

The amplitude of electric and magnetic field.

Answer to Problem 72P

The amplitude of electric and magnetic field respectively $4.08\times {10}^4\text{V}/\text{m}$ and $1.36\times {10}^{-4}\text{T}$.

Explanation of Solution

Write the expression to calculate the magnitude of electric field.

  $E_{\max }=\sqrt{\frac{2u}{{\epsilon }_0}}$        (I)

Here, ${\epsilon }_0$ is the permittivity of free space and $E_{\max }$ is the magnitude of electric field.

Write the expression to calculate the magnitude of magnetic field.

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

Here, $B_{\max }$ is the magnitude of magnetic field.

Conclusion:

Substitute $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$ for u and $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ in the equation (I) to calculate $E_{\max }$.

  $\begin{array}{c}{E}_{\max }=\sqrt{\frac{2\left(7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\right)}{8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}}}\\ =4.08\times {10}^4\text{V}/\text{m}\end{array}$

Substitute $4.08\times {10}^4\text{V}/\text{m}$ for $E_{\max }$ and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation (II) to calculate $B_{\max }$.

  $\begin{array}{c}{B}_{\max }=\frac{4.08\times {10}^4\text{V}/\text{m}}{3.00\times {10}^8\text{m}/\text{s}}\\ =1.36\times {10}^{-4}\text{T}\end{array}$

Thus, the amplitude of electric and magnetic field respectively $4.08\times {10}^4\text{V}/\text{m}$ and $1.36\times {10}^{-4}\text{T}$.

To determine

The force exerted on the surface.

Answer to Problem 72P

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

Explanation of Solution

Write the expression to calculate the force exerted on the surface.

  $F=uA$

Here, F is the force and A is the area.

Write the expression to calculate the area.

  $A=\pi r^2$

Use the above equation to rewrite the equation for F.

  $F=u\left(\pi r^2\right)$

Conclusion:

Substitute $7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}$ for u and $6.00\text{cm}$ for r in the above equation to calculate F.

  $\begin{array}{c}F=7.37\times {10}^{-3}\text{J}/{\text{m}}^{\text{3}}\left(\pi {\left(6.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\right)\\ =8.33\times {10}^{-5}\text{N}\end{array}$

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