Chapter 34, Problem 73PQ

(a)

To determine

The direction in which the wave is travelling.

Answer to Problem 73PQ

The wave is travelling along the positive Z-direction.

Explanation of Solution

Since the given equation of the electric field of the wave contains the position variable ‘z’, therefore. The wave travels in the positive Z-direction.

Conclusion:

Therefore, the wave is travelling along the positive Z-direction.

(b)

To determine

The frequency, angular frequency and wave number of the wave.

Answer to Problem 73PQ

The frequency of the wave is $6.60\times {10}^{14}\text{Hz}$, the angular frequency of the wave is $4.15\times {10}^{15}\text{rad}/\text{s}$ and the wave number of the wave is $1.40\times {10}^7{\text{m}}^{-1}$.

Explanation of Solution

Write the expression for the frequency of an electromagnetic wave.

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

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

Write the expression for the angular frequency of the wave.

  $\omega =2\pi f$                                          (II)

Here, $\omega$ is the angular frequency and $f$ is the frequency of the wave.

Write the expression for the wave number of the wave.

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

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

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $454\text{nm}$ for $\lambda$ in equation (I) to find $f$.

    $f=\frac{3\times {10}^8\text{m}/\text{s}}{\left(454\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{m}}\right)}$

  $=6.60\times {10}^{14}\text{Hz}$

Substitute equation (II) in the above equation to find $\omega$.

    $\begin{array}{c}\omega =2\pi \left(6.60\times {10}^{14}\text{Hz}\right)\\ =4.15\times {10}^{15}\text{rad}/\text{s}\end{array}$

Substitute $454\text{nm}$ for  $\lambda$ in (III) to find $k$.

    $\begin{array}{c}k=\frac{2\pi }{\left(454\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{m}}\right)}\\ =1.40\times {10}^7{\text{m}}^{-1}\end{array}$

Therefore, the frequency of the wave is $6.60\times {10}^{14}\text{Hz}$, the angular frequency of the wave is $4.15\times {10}^{15}\text{rad}/\text{s}$ and the wave number of the wave is $1.40\times {10}^7{\text{m}}^{-1}$.

(c)

To determine

The direction in which the magnetic field of the wave oscillates.

Answer to Problem 73PQ

The magnetic field of the wave oscillates in the negative X-direction.

Explanation of Solution

Since, the wave is travelling along the Z direction, and the electric field oscillates along the Y-direction. Therefore, the magnetic field oscillates along the X-direction.

Conclusion:

Therefore, the magnetic field of the wave oscillates in the negative X-direction.

(d)

To determine

The equation of the magnetic field of the wave.

Answer to Problem 73PQ

The equation of the magnetic field of the wave is $B\left(z,t\right)=\left(1.67\times {10}^{-11}\text{T}\right)\sin \left[\left(1.40\times {10}^7{\text{m}}^{-1}\right)z-\left(4.15\times {10}^{15}\text{rad}/\text{s}\right)t\right]$

Explanation of Solution

Write the expression for the maximum magnetic field.

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

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

Write the equation for the magnetic field.

    $B\left(z,t\right)=B_{\max }\sin \left(kz-\omega t\right)$                                                           (II)

Conclusion:

Substitute $5.00\times {10}^{-3}\text{V}/\text{m}$ for $E_{\max }$ and $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in (I) equation to find $B_{\max }$.

    $\begin{array}{c}{B}_{\max }=\frac{5.00\times {10}^{-3}\text{V}/\text{m}}{3.0\times {10}^8\text{m}/\text{s}}\\ =1.67\times {10}^{-11}\text{T}\end{array}$

Substitute $1.67\times {10}^{-11}\text{T}$ for $B_{\max }$, $1.40\times {10}^7{\text{m}}^{-1}$ for $k$, and $4.15\times {10}^{15}\text{rad}/\text{s}$ for $\omega$ in (II) equation to find $B\left(z,t\right)$.

    $B\left(z,t\right)=\left(1.67\times {10}^{-11}\text{T}\right)\sin \left[\left(1.40\times {10}^7{\text{m}}^{-1}\right)z-\left(4.15\times {10}^{15}\text{rad}/\text{s}\right)t\right]$

Therefore, the equation for the magnetic field is $B\left(z,t\right)=\left(1.67\times {10}^{-11}\text{T}\right)\sin \left[\left(1.40\times {10}^7{\text{m}}^{-1}\right)z-\left(4.15\times {10}^{15}\text{rad}/\text{s}\right)t\right]$