Q 12-08-196JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
The electric field of a plane polarized electromagnetic wave in free space at time $t = 0$ is given by the expression $\vec E(x, y) = 10\,\hat j\cos(6x + 8z)$. The magnetic field $\vec B(x, z, t)$ is given by ($c$ is the velocity of light)
Answer: (D) $\dfrac1c\left(6\hat k - 8\hat i\right)\cos(6x + 8z - 10ct)$
The wave vector is $\vec k = 6\hat i + 8\hat k$, $|\vec k| = 10$, so $\omega = 10c$ and the wave is $\cos(6x + 8z - 10ct)$ travelling along $\hat k_0 = 0.6\hat i + 0.8\hat k$.
$\vec B = \dfrac1c\,\hat k_0\times\vec E$:
$$\hat k_0\times\hat j = 0.6\,\hat k - 0.8\,\hat i$$
$$\vec B = \frac{10}{c}(0.6\hat k - 0.8\hat i)\cos(\ldots) = \frac1c\left(6\hat k - 8\hat i\right)\cos(6x + 8z - 10ct)$$
Solution by Sreeraj P, M.Sc Physics