Chapter 34, Problem 67PQ

(a)

To determine

The direction in which the wave is travelling.

Answer to Problem 67PQ

The direction in which the wave is travelling is $-\widehat{i}$.

Explanation of Solution

Write the expression for magnetic field of electromagnetic wave.

    $B\left(x,t\right)=B_{\text{max}}\sin \left(kx+\omega t\right)$                                                                              (I)

Here, $k$ is the wave number, $B_{\text{max}}$ is the maximum magnetic field, $\omega$ is the angular frequency, and $t$ is the time.

Compare equation (I) with given equation of magnetic field.

    $B_{\text{max}}\sin \left(kx+\omega t\right)=\left(4.0\times {10}^{-8}\right)\sin \left[\left(1.4\times {10}^4\text{rad}/\text{m}\right)x+\omega t\right]$

Conclusion:

Therefore, the direction in which the wave is travelling is $-\widehat{i}$.

(b)

To determine

The wave number, wave-length and frequency of the electro-magnetic wave.

Answer to Problem 67PQ

The wave number is $1.4\times {10}^4{\text{m}}^{-1}$, the wave-length of the electro-magnetic wave is $4.5\times {10}^{-4}\text{m}$ and the frequency of radiation is $6.7\times {10}^{11}\text{Hz}$.

Explanation of Solution

Write the expression for wave-length of the electro-magnetic wave.

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

Here, $k$ is the wave number.

Write the expression for frequency of radiation.

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

Here, $c$ is the velocity of light.

Conclusion:

Compare equation (I) with given equation of magnetic field.

    $B_{\text{max}}\sin \left(kx+\omega t\right)=\left(4.0\times {10}^{-8}\right)\sin \left[\left(1.4\times {10}^4{\text{m}}^{-1}\right)x+\omega t\right]$

The wave number is $1.4\times {10}^4{\text{m}}^{-1}$.

Substitute $1.4\times {10}^4\text{rad}/\text{m}$ for $k$ in equation (II) to calculate $\lambda$.

    $\begin{array}{c}\lambda =\frac{\left(2\right)\left(3.14\right)}{1.4\times {10}^4\text{rad}/\text{m}}\\ =\left(4.49\times {10}^{-4}\text{m}\right)\left(\frac{1\mu \text{m}}{{10}^{-6}\text{m}}\right)\\ =4.5\times {10}^{-4}\text{m}\end{array}$

Substitute $4.5\times {10}^{-4}\text{m}$ for $\lambda$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (III) to calculate $f$.

    $\begin{array}{c}f=\frac{3\times {10}^8\text{m}/\text{s}}{4.5\times {10}^{-4}\text{m}}\\ =6.67\times {10}^{11}\text{Hz}\\ \approx 6.7\times {10}^{11}\text{Hz}\end{array}$

Therefore, the wave number is $1.4\times {10}^4{\text{m}}^{-1}$, the wave-length of the electro-magnetic wave is $4.5\times {10}^{-4}\text{m}$ and the frequency of radiation is $6.7\times {10}^{11}\text{Hz}$.

(c)

To determine

The expression for electric field of electromagnetic wave

Answer to Problem 67PQ

The expression for electric field of electromagnetic wave is given below.

    $E\left(x,t\right)=\left(12\text{V}/\text{m}\right)\sin \left[\left(1.4\times {10}^4\text{rad}/\text{m}\right)x+\left(4.2\times {10}^{12}\text{rad}/\text{s}\right)t\right]$

Explanation of Solution

Write the expression for maximum value of electric field of

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

Write the expression for electric field of electromagnetic wave.

    $E\left(x,t\right)=E_{\text{max}}\sin \left(kx+\omega t\right)$                                                                               (V)

Write the expression for angular frequency.

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

Conclusion:

Substitute $4.0\times {10}^{-8}\text{ T}$ for $B_{\text{max}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (IV) to calculate $E_{\text{max}}$.

    $\begin{array}{c}{E}_{\text{max}}=3\times {10}^8\text{m}/\text{s}\times 4.0\times {10}^{-8}\text{ T}\\ =12\text{V}/\text{m}\end{array}$

Substitute $6.7\times {10}^{11}\text{Hz}$ for $f$ in equation (VI) to calculate $\omega$.

    $\begin{array}{c}\omega =2\pi \left(6.7\times {10}^{11}\text{Hz}\right)\\ =4.2\times {10}^{12}\text{rad}/\text{s}\end{array}$

Substitute $12.0\text{V}/\text{m}$ for $E_{\text{max}}$ , $4.2\times {10}^{12}\text{rad}/\text{s}$ for $\omega$ , and $1.4\times {10}^4\text{ rad/m}$ for $k$ in equation (V) to find $E\left(x,t\right)$.

    $E\left(x,t\right)=\left(12\text{V}/\text{m}\right)\sin \left[\left(1.4\times {10}^4\text{rad}/\text{m}\right)x+\left(4.2\times {10}^{12}\text{rad}/\text{s}\right)t\right]$

Therefore, the expression for electric field of electromagnetic wave is given below.

  $E\left(x,t\right)=\left(12\text{V}/\text{m}\right)\sin \left[\left(1.4\times {10}^4\text{rad}/\text{m}\right)x+\left(4.2\times {10}^{12}\text{rad}/\text{s}\right)t\right]$