Chapter 30, Problem 46P

(a)

To determine

The wavelength and direction of the propagating wave.

Answer to Problem 46P

The wavelength of the wave is $3.0\text{m}$ and direction of propagation is perpendicular to direction of magnetic field.

Explanation of Solution

Given:

The frequency is $f=100\text{MHz}$ .

The magnetic is $\overrightarrow{B}\left(z,t\right)=\left(1.00\times {10}^{-8}\text{T}\right)\cos \left(kz-\omega t\right)\widehat{i}$

Formula used:

The expression for wavelength is given by,

  $\lambda =\frac{c}{f}$

Calculation:

The wavelength of the wave is calculated as,

  $\begin{array}{c}\lambda =\frac{c}{f}\\ =\frac{3.0\times {10}^8\text{m}/\text{s}}{\left(\left(100\text{MHz}\right)\left(\frac{{10}^6\text{Hz}}{\text{1}\text{MHz}}\right)\right)}\\ =3.0\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the wave is $3.0\text{m}$ and direction of propagation is perpendicular to direction of magnetic field.

(b)

To determine

The electric field vector.

Answer to Problem 46P

The electric field vector is $-\left(3.0\text{V}/\text{m}\cos \left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\widehat{j}\right)$ .

Explanation of Solution

Formula used:

The expression for electric field vector is given by,

  $\overrightarrow{E}\left(z,t\right)=-\left(E\cos \left[k\widehat{z}-\omega t\right]\widehat{j}\right)$

The expression for amplitude of electric field is given by,

  $E=cB$

The expression for angular frequency is given by,

  $\omega =2\pi f$

The expression for wave number is given by,

  $k=\frac{2\pi }{\lambda }$

Calculation:

The amplitude of electric field is calculated as,

  $\begin{array}{c}E=cB\\ =\left(3.0\times {10}^8\text{m}/\text{s}\right)\left(\left(1.00\times {10}^{-8}\text{T}\right)\right)\\ =3.0\text{V}/\text{m}\end{array}$

The angular frequency is calculated as,

  $\begin{array}{c}\omega =2\pi f\\ =2\pi \left(100\text{MHz}\right)\left(\frac{{10}^6\text{Hz}}{\text{1}\text{MHz}}\right)\\ =6.28\times {10}^8/\sec \end{array}$

The wave number is calculated as,

  $\begin{array}{c}k=\frac{2\pi }{\lambda }\\ =\frac{2\pi }{3.0\text{m}}\\ =2.09/\text{m}\end{array}$

The electric field vector is calculated as,

  $\begin{array}{c}\overrightarrow{E}\left(z,t\right)=-\left(E\cos \left[k\widehat{z}-\omega t\right]\widehat{j}\right)\\ =-\left(3.0\text{V}/\text{m}\cos \left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\widehat{j}\right)\end{array}$

Conclusion:

Therefore, the electric field vector is $-\left(3.0\text{V}/\text{m}\cos \left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\widehat{j}\right)$ .

(c)

To determine

The Poynting vector and intensity of the wave.

Answer to Problem 46P

The Poynting vector is $\left(\left(23.9\text{mW}/{\text{m}}^{\text{2}}\right)\right){\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{k}\right)$ and intensity of the wave is $11.9\text{mW}/{\text{m}}^{\text{2}}$ .

Explanation of Solution

Formula used:

The expression for Poynting vector is given by,

  $\overrightarrow{S}\left(z,t\right)=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$

The expression for intensity is given by,

  $I=\left|\overrightarrow{S}\right|$

Calculation:

The Poynting vector is calculated as,

  $\begin{array}{c}\overrightarrow{S}\left(z,t\right)=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}\\ =\frac{1}{4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}}\left[\begin{array}{l}\left(-\left(3.0\text{V}/\text{m}\cos \left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\widehat{j}\right)\right)\\ \times \left(\left(1.00\times {10}^{-8}\text{T}\right)\cos \left(kz-\omega t\right)\widehat{i}\right)\end{array}\right]\\ =\frac{\left(-3.0\text{V}/\text{m}\right)\left({10}^{-8}\text{T}\right)}{4\pi \times {10}^{-7}\text{N}/{\text{A}}^{\text{2}}}{\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{j}\times \widehat{i}\right)\\ =\left(\left(23.9\times {10}^{-3}\text{W}/{\text{m}}^{\text{2}}\right)\left(\frac{{10}^3\text{mW}}{1\text{W}}\right)\right){\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{k}\right)\end{array}$

Further simplify the above,

  $\overrightarrow{S}\left(z,t\right)=\left(\left(23.9\text{mW}/{\text{m}}^{\text{2}}\right)\right){\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{k}\right)$

The intensity of the wave is calculated as,

  $\begin{array}{c}I=\left|\overrightarrow{S}\right|\\ =\left|\left(\left(23.9\text{mW}/{\text{m}}^{\text{2}}\right)\right){\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{k}\right)\right|\\ =0.5\left(23.9\text{mW}/{\text{m}}^{\text{2}}\right)\\ =11.9\text{mW}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the Poynting vector is $\left(\left(23.9\text{mW}/{\text{m}}^{\text{2}}\right)\right){\cos }^2\left[\left(2.09/\sec \right)\widehat{z}-\left(6.28\times {10}^8/\sec \right)t\right]\left(\widehat{k}\right)$ and intensity of the wave is $11.9\text{mW}/{\text{m}}^{\text{2}}$ .