An electromagnetic wave is represented by the electric field $\vec{E}=E_{0} \hat{n} \sin [\omega t+(6 y-8 z)]$. 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) $\hat{S}=\left(\frac{-3 \hat{j}+4 \hat{k}}{5}\right)$

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

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

(D) $\hat{S}=\left(\frac{-3 \hat{i}-4 \hat{j}}{5}\right)$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · EM Waves

An electromagnetic wave is represented by the electric field

E = E_{0} \overset{n}{^} sin [ω t + (6 y - 8 z)]

. Taking unit vectors in

x

y

and

z

directions to be

\hat{i}, \hat{j}, \hat{k}

, the direction of propogation

\overset{s}{^}

, is

Answer: (A)

Compare the given field

E = E_{0} \overset{n}{^} sin [ω t + (6 y - 8 z)]

with the standard travelling-wave form

E = E_{0} \overset{n}{^} sin (ω t + k \cdot r)

. Reading off,

k \cdot r = 6 y - 8 z

, so the wave vector is

k = 6 \hat{j} - 8 \hat{k}

with magnitude

6^{2} + 8^{2} = 10

. Because the spatial part appears as

+ k \cdot r

alongside

+ ω t

, the wave advances opposite to

k

, so the propagation direction is

\overset{s}{^} = - \frac{k}{∣ k ∣} = - \frac{6 j ^ - 8 k ^}{10} = \frac{- 3 j ^ + 4 k ^}{5}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option