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(A) The general form of an electromagnetic wave is $\vec{E} = E_0 \hat{n} \sin(\omega t + \vec{k} \cdot \vec{r})$.Comparing this with the given equati

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The general form of an electromagnetic wave is $\vec{E} = E_0 \hat{n} \sin(\omega t + \vec{k} \cdot \vec{r})$.Comparing this with the given equation $\vec{E} = E_0 \hat{n} \sin[\omega t + (6y - 8z)]$,we have the wave vector $\vec{k} = 6\hat{j} - 8\hat{k}$.The direction of propagation $\hat{s}$ is given by the unit vector in the direction of $-\vec{k}$.Thus,$\hat{s} = -\frac{\vec{k}}{|\vec{k}|} = -\frac{6\hat{j} - 8\hat{k}}{\sqrt{6^2 + (-8)^2}} = -\frac{6\hat{j} - 8\hat{k}}{\sqrt{36 + 64}} = -\frac{6\hat{j} - 8\hat{k}}{10} = \frac{-6\hat{j} + 8\hat{k}}{10} = \frac{-3\hat{j} + 4\hat{k}}{5}$.

An electromagnetic wave is represented by the electric field $\vec{E} = E_0 \hat{n} \sin [\omega t + (6y - 8z)]$. Taking unit vectors in $x, y$ and $z$ directions to be $\hat{i}, \hat{j}$ and $\hat{k}$,the direction of propagation $\hat{s}$ is:

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