Chapter 34, Problem 83PQ

(a)

To determine

The value of $E_{\max }$.

Answer to Problem 83PQ

The value of $E_{\max }$ is $2.61\times {10}^3\text{V}/\text{m}$.

Explanation of Solution

Write the expression for the amplitude of electric field.

    $E_{\max }=c{B}_{\max }$                                                                                                      (I)

Here, $E_{\max }$ is the amplitude of electric field, $c$ is the speed of light and $B_{\max }$ is the amplitude of magnetic field.

Conclusion:

Substitute $8.70\text{μT}$ for $B_{\max }$ and $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find $E_{\max }$.

    $\begin{array}{c}{E}_{\max }=\left(3.0\times {10}^8\text{m}/\text{s}\right)\left(8.70\text{μT}\times \frac{{10}^{-9}\text{T}}{1\text{μT}}\right)\\ =2.61\times {10}^3\text{V}/\text{m}\end{array}$

Thus, the value of $E_{\max }$ is $2.61\times {10}^3\text{V}/\text{m}$.

(b)

To determine

The wave number of the wave.

Answer to Problem 83PQ

The wave number of the wave is $1.08\times {10}^3{\text{m}}^{-1}$.

Explanation of Solution

Write the expression for the wave number of the wave.

    $k=\frac{2\pi }{\lambda }$                                                                                                                  (II)

Here, $k$ is the wave number and $\lambda$ is the wavelength of the wave.

Conclusion:

Substitute $5.8\text{mm}$ for $\lambda$ in equation (II) to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{\left(\left(5.8\text{mm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{mm}}\right)\right)}\\ =1.08\times {10}^3\text{rad}/\text{m}\end{array}$

Therefore, the wave number of the wave is $1.08\times {10}^3{\text{m}}^{-1}$.

(c)

To determine

The angular frequency of the wave.

Answer to Problem 83PQ

The angular frequency of the wave is $3.25\times {10}^{11}\text{rad}/\text{s}$.

Explanation of Solution

Write the expression for the angular frequency of the wave.

    $\omega =\frac{2\pi c}{\lambda }$                                                                                                    (III)

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

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $5.8\text{mm}$ for $\lambda$ in equation (III) to find $\omega$.

    $\begin{array}{c}\omega =\frac{2\pi \left(3\times {10}^8\text{m}/\text{s}\right)}{\left(\left(5.8\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\\ =3.25\times {10}^{11}\text{rad}/\text{s}\end{array}$

Thus, the angular frequency of the wave is $3.25\times {10}^{11}\text{rad}/\text{s}$.

(d)

To determine

The plane in which electric field oscillates.

Answer to Problem 83PQ

The electric field oscillates in the $x\text{-}y$-plane.

Explanation of Solution

Since, the electric field is directed in the $z$-direction, the plane of oscillation of the electric field is the $x\text{-}y$-plane.

(e)

To determine

The average value of the Poynting vector.

Answer to Problem 83PQ

The average value of Poynting vector is $9.03\times {10}^3\stackrel{\wedge }{i}\text{W}/{\text{m}}^{\text{2}}$

Explanation of Solution

Write the expression for the average value of Poynting vector.

    $S_{\text{avg}\text{.}}=\frac{E_{\max }{B}_{\max }}{2{\mu }_0}$                                                                                            (IV)

Here, ${\mu }_0$ is the permeability of free space, $E_{\max }$ is the amplitude of electric field and $B_{\max }$ is the amplitude of magnetic field.

Conclusion:

Substitute $4\pi \times {10}^3\text{H}/\text{m}$ for ${\mu }_0$, $8.70\text{μT}$ for $B_{\max }$ and $2.61\times {10}^3\text{V}/\text{m}$ for $B_{\max }$ in equation (IV) to find $S_{\text{avg}\text{.}}$.

    $\begin{array}{c}{S}_{\text{avg}\text{.}}=\frac{\left(2.61\times {10}^3\text{V}/\text{m}\right)\left(8.70\times {10}^{-6}\text{T}\right)}{2\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)}\\ =9.03\times {10}^3\text{W}/{\text{m}}^{\text{2}}\end{array}$

Thus, the average value of Poynting vector is $9.03\times {10}^3\stackrel{\wedge }{i}\text{W}/{\text{m}}^{\text{2}}$.

(f)

To determine

The pressure exerted by the wave on a lightweight solar sail.

Answer to Problem 83PQ

The pressure exerted by the wave on the solar sail is $6.02\times {10}^{-5}\text{Pa}$.

Explanation of Solution

Write the expression for the pressure exerted by the wave on lightweight solar sail.

    $P=\frac{2I}{c}$                                                                                                                    (V)

Here, $P$ is the pressure, $I$ is the intensity, and $c$ is the speed of light.

Conclusion:

Substitute $9.03\times {10}^3\text{W}/{\text{m}}^{\text{2}}$ for $I$ and $3.0\times {10}^8\text{m}/\text{s}$ for $c$. in equation (V) to find $P$.

    $\begin{array}{c}P=\frac{\left(2\right)\left(9.03\times {10}^3\text{W}/{\text{m}}^{\text{2}}\right)}{3.0\times {10}^8\text{m}/\text{s}}\\ =6.02\times {10}^{-5}\text{Pa}\end{array}$

Thus, the pressure exerted by the wave on the solar sail is $6.02\times {10}^{-5}\text{Pa}$.

(g)

To determine

The acceleration of the solar sail.

Answer to Problem 83PQ

The acceleration of the solar sail is $6.98\times {10}^{-2}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression for the acceleration of the solar sail.

    $a=\frac{IA}{mc}$                                                  (VI)

Here, $a$ is the acceleration of the sail, $I$ is the intensity, $A$ is the area of the sail, $m$ is the mass of the sail, $c$ is the speed of light.

Write the expression for the area of the sail.

    $\begin{array}{c}A=\left(5.00\text{m}\times 8.00\text{m}\right)\\ =40{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Substitute $9.03\times {10}^3\text{W}/{\text{m}}^{\text{2}}$ for $I$, $40{\text{m}}^{\text{2}}$ for $A$, $3.0\times {10}^8\text{m}/\text{s}$ for $c$ and $34.5\text{g}$ for $\text{m}$ in equation (VI) to find $a$.

    $\begin{array}{c}a=\frac{\left(2\right)\left(9.03\times {10}^3\text{W}/{\text{m}}^{\text{2}}\right)\left(40{\text{m}}^{\text{2}}\right)}{\left(34.5\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(3.0\times {10}^8\text{m}/\text{s}\right)}\\ =6.98\times {10}^{-2}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration of the solar sail is $6.98\times {10}^{-2}\text{m}/{\text{s}}^{\text{2}}$