Chapter 30, Problem 51P

(a)

To determine

The instantaneous Poynting vector.

Answer to Problem 51P

The instantaneous Poynting vector is $\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x-{\omega }_{2}t-\delta \right)\right)\right]\widehat{i}$ .

Explanation of Solution

Given:

The electric field of first wave is ${\overrightarrow{E}}_1=E_{10}\cos \left(k_{1}x-{\omega }_{1}t\right)\widehat{j}$ .

The electric field of second wave is ${\overrightarrow{E}}_2=E_{20}\cos \left(k_{2}x-{\omega }_{2}t+\delta \right)\widehat{j}$ .

Formula used:

The expression for instantaneous Poynting vector is given by,

  $\overrightarrow{S}=\frac{1}{{\mu }_0}\left[{\overrightarrow{E}}_1+{\overrightarrow{E}}_2\right]\times \frac{1}{c}\left[\left[{\overrightarrow{E}}_1-{\overrightarrow{E}}_2\right]\widehat{i}\right]$

Calculation:

The instantaneous Poynting vector is calculated as,

  $\begin{array}{c}\overrightarrow{S}=\frac{1}{{\mu }_0}\left[{\overrightarrow{E}}_1+{\overrightarrow{E}}_2\right]\times \frac{1}{c}\left[\left[{\overrightarrow{E}}_1-{\overrightarrow{E}}_2\right]\widehat{i}\right]\\ =\frac{1}{{\mu }_0}\left[\left(E_{10}\cos \left(k_{1}x-{\omega }_{1}t\right)\widehat{j}\right)+\left(E_{20}\cos \left(k_{2}x-{\omega }_{2}t-\delta \right)\widehat{j}\right)\right]\times \frac{1}{c}\left[\left(E_{10}\cos \left(k_{1}x-{\omega }_{1}t\right)\widehat{j}\right)-\left(E_{20}\cos \left(k_{2}x-{\omega }_{2}t-\delta \right)\widehat{j}\right)\right]\widehat{i}\\ =\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x-{\omega }_{2}t-\delta \right)\right)\right]\widehat{i}\end{array}$

Conclusion:

Therefore, the instantaneous Poynting vector is $\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x-{\omega }_{2}t-\delta \right)\right)\right]\widehat{i}$ .

(b)

To determine

The time averaged Poynting vector.

Answer to Problem 51P

The time averaged Poynting vector is $\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$ .

Explanation of Solution

Calculation:

Consider the relation,

  $\overrightarrow{S}=\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x-{\omega }_{2}t+\delta \right)\right)\right]\widehat{i}$

The time average of the square of the cosine term is $0.5$ so, the time averaged Poynting vector is,

  ${\overrightarrow{S}}_{\text{av}}=\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$

Conclusion:

Therefore, the time averaged Poynting vector is $\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$ .

(c)

To determine

The instantaneous and time average Poynting vector.

Answer to Problem 51P

The instantaneous and time averaged Poynting vector are $\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\right)\right]\widehat{i}$ and $\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$ respectively.

Explanation of Solution

Given:

The direction of propagation of second wave is ${\overrightarrow{E}}_2=E_{20}\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\widehat{j}$ .

Calculation:

The instantaneous Poynting vector is calculated as,

  $\begin{array}{c}\overrightarrow{S}=\frac{1}{{\mu }_0}\left[{\overrightarrow{E}}_1+{\overrightarrow{E}}_2\right]\times \frac{1}{c}\left[\left[{\overrightarrow{E}}_1-{\overrightarrow{E}}_2\right]\widehat{i}\right]\\ =\frac{1}{{\mu }_0}\left[\left(E_{10}\cos \left(k_{1}x-{\omega }_{1}t\right)\widehat{j}\right)+\left(E_{20}\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\widehat{j}\right)\right]\times \frac{1}{c}\left[\left(E_{10}\cos \left(k_{1}x-{\omega }_{1}t\right)\widehat{j}\right)-\left(E_{20}\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\widehat{j}\right)\right]\widehat{i}\\ =\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\right)\right]\widehat{i}\end{array}$

The time averaged Poynting vector is,

  ${\overrightarrow{S}}_{\text{av}}=\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$

Conclusion:

Therefore, the instantaneous and time averaged Poynting vector are $\frac{1}{{\mu }_{0}c}\left[E_{10}^2\left(\cos \left(k_{1}x-{\omega }_{1}t\right)\right)-E_{20}^2\left(\cos \left(k_{2}x+{\omega }_{2}t+\delta \right)\right)\right]\widehat{i}$ and $\frac{1}{2{\mu }_{0}c}\left[E_{10}^2-E_{20}^2\right]\widehat{i}$ respectively.