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(C) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec E 	imes \vec B$.Given that the wave travels

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$+z$-axis

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(C) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec E 	imes \vec B$.Given that the wave travels along the $y$-direction (unit vector $\hat j$) and the electric field $\vec E$ is along the $x$-axis (unit vector $\hat i$).Let the direction of the magnetic field $\vec B$ be $\hat n$.Then,$\hat i 	imes \hat n = \hat j$.We know that $\hat i 	imes \hat k = -\hat j$ and $\hat i 	imes (-\hat k) = \hat j$.Therefore,the direction of the $\vec B$ vector must be along the $+z$-axis. Note: The original option $(d)$ was $-z$-axis,but mathematically $\hat i 	imes \hat k = \hat j$ is incorrect; $\hat i 	imes \hat k = -\hat j$. Thus,for $\hat i 	imes \hat n = \hat j$,$\hat n$ must be $-\hat k$. Let us re-evaluate: $\hat i 	imes \hat k = \hat j$ is false,$\hat i 	imes \hat j = \hat k$,$\hat j 	imes \hat k = \hat i$,$\hat k 	imes \hat i = \hat j$. So,$\hat k 	imes \hat i = \hat j$. If $\vec E$ is along $x$ $(\hat i)$ and propagation is along $y$ $(\hat j)$,then $\vec E 	imes \vec B = \hat i 	imes \vec B = \hat j$. Since $\hat k 	imes \hat i = \hat j$,$\vec B$ must be along the $+z$-axis.

$A$ light wave is travelling along the $y$-direction. If the corresponding $\vec E$ vector at any time is along the $x$-axis,the direction of the $\vec B$ vector at that time is along:

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