Question

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 :

• A. $\hat{s} = \frac{4 \hat{j} - 3 \hat{k}}{5}$
• B. $\hat{s} = \frac{3 \hat{j} - 4 \hat{k}}{5}$
• C. $\hat{s} = \left( \frac{-3 \hat{j} + 4 \hat{k}}{5} \right)$
• D. $\hat{s} = \frac{-4 \hat{j} + 3 \hat{k}}{5}$

Answer 1

Answer: (C)

$\vec{E} = E_0 \; \hat{n} \sin (\omega t + (6y - 8z))$
$ = E_0 \hat{n}\sin (\omega + \vec{k} . \vec{r})$
where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$
and $\vec{k} .\vec{r} = 6 y - 8z$
$\Rightarrow \; \vec{k} = 6 \hat{j} - 8 \hat{h}$
direction of propagation
$\hat{s} = - \hat{k}$
$ = \left( \frac{ -3 \hat{j} + 4 \hat{k}}{5} \right)$