MCQ
An electromagnetic wave is represented by the electric field $\vec E = {E_0}\hat n\,\sin \,\left[ {\omega t + \left( {6y - 8z} \right)} \right]$. Taking unit vectors in $x, y$ and $z$ directions to be $\hat i,\hat j,\hat k$ ,the direction of propogation $\hat s$, is
  • $\hat S = \left( {\frac{{ - 3\hat j + 4\hat k}}{5}} \right)$
  • B
    $\hat S = \left( {\frac{{ 4\hat j - 3\hat k}}{5}} \right)$
  • C
    $\hat S = \left( {\frac{{ - 4\hat k + 3\hat j}}{5}} \right)$
  • D
    $\hat S = \left( {\frac{{ - 3\hat i - 4\hat j}}{5}} \right)$

Answer

Correct option: A.
$\hat S = \left( {\frac{{ - 3\hat j + 4\hat k}}{5}} \right)$
a
$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{n}} \sin \left(\omega \mathrm{t}+(6 \mathrm{y}-8 \mathrm{z})=\mathrm{E}_{0} \hat{\mathrm{n}} \sin (\omega \mathrm{t}+\overrightarrow{\mathrm{k}} \cdot \overrightarrow{\mathrm{r}})\right.$

where $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ and $\vec{k} \cdot \vec{r}=6 y-8 z$

$\Rightarrow \overrightarrow{\mathrm{k}}=6 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}$

$\text { direction of propagation } \hat{\mathrm{s}} =-\hat{\mathrm{k}}$

$ =\left(\frac{-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{5}\right) $

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