An electromagnetic wave is represented by the electric field E = E₀nsin [ω t + (6y - 8z)]. Taking unit vectors in x, y and z directions to be i, j, k, the direction of propagation s, is: (a) s = (3i-4j)/(5) (b) s = (-4k+3j)/(5) (c) s = ((-3j+4k)/(5)) (d) s = (4j-3k)/(5)

Answer: (c). Phase ω t + k with k = 6j - 8k; a +ω t term means the wave moves along -k. -k = -6j + 8k, magnitude √(36+64) = 10. s = (-6j+8k)/(10) = (-3j+4k)/(5). So the answer is (c).

An electromagnetic wave is represented by the electric field E =… | JEE Main 2019 (12 Apr, Shift 1)

Q61JEE Main 2019 · 12 APR · SHIFT 1Easy

Nature of EM Waves & Wave Equation

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

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Answer: (c)

Phase $\omega t + \vec{k}\cdot\vec{r}$ with $\vec{k} = 6\hat{j} - 8\hat{k}$ ; a $+\omega t$ term means the wave moves along $-\vec{k}$ .
$-\vec{k} = -6\hat{j} + 8\hat{k}$ , magnitude $\sqrt{36+64} = 10$ .
$\hat{s} = \frac{-6\hat{j}+8\hat{k}}{10} = \frac{-3\hat{j}+4\hat{k}}{5}$ .
So the answer is (c).

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Nature of EM Waves & Wave Equation question from JEE Main 2019 (12 Apr, Shift 1)

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