A light wave is travelling along the y- direc | vedclass

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(C) An electromagnetic wave propagates in the direction of the cross product of the electric field vector $\vec E$ and the magnetic field vector $\vec

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$+z-$ axis

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(C) An electromagnetic wave propagates in the direction of the cross product of the electric field vector $\vec E$ and the magnetic field vector $\vec B$,i.e.,the direction of propagation is given by $\vec E 	imes \vec B$.Given that the wave is travelling along the $y-$ axis (unit vector $\hat{j}$),and the electric field $\vec E$ is along the $x-$ axis (unit vector $\hat{i}$).We know that the direction of propagation is $\hat{i} 	imes \hat{B} = \hat{j}$.Since $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{i} 	imes (-\hat{k}) = \hat{j}$,the magnetic field $\vec B$ must be along the $z-$ axis.Specifically,for a wave propagating in the $+y$ direction with $\vec E$ in the $+x$ direction,$\vec B$ must be in the $+z$ direction because $\hat{i} 	imes \hat{k} = -\hat{j}$ is incorrect; rather $\hat{i} 	imes \hat{j} = \hat{k}$.Let's re-evaluate: The direction of propagation is $\vec{k}_{prop} \propto \vec E 	imes \vec B$.Here,$\hat{j} = \hat{i} 	imes \hat{B}$.Since $\hat{i} 	imes \hat{k} = \hat{j}$,the vector $\vec B$ is 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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