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(A) The direction of propagation of an $EM$ wave is given by the direction of the Poynting vector $\overrightarrow{S}$,which is in the direction of $\

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Towards $+x$ direction

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(A) The direction of propagation of an $EM$ wave is given by the direction of the Poynting vector $\overrightarrow{S}$,which is in the direction of $\overrightarrow{E} 	imes \overrightarrow{B}$.Given,the direction of propagation is $-z$ direction,so $\overrightarrow{E} 	imes \overrightarrow{B} = -\widehat{k}$.From the figure,the electric field $\overrightarrow{E}$ is in the $+y$ direction,so $\overrightarrow{E} = E_0 \widehat{j}$.Substituting these into the relation: $(E_0 \widehat{j}) 	imes \overrightarrow{B} = -\widehat{k}$.Let $\overrightarrow{B} = B_x \widehat{i} + B_y \widehat{j} + B_z \widehat{k}$.Since $\overrightarrow{E} \perp \overrightarrow{B}$ and $\overrightarrow{B} \perp 	ext{propagation direction}$,$\overrightarrow{B}$ must be along the $x$-axis.So,$\widehat{j} 	imes (B_x \widehat{i}) = -B_x \widehat{k}$.Comparing this with $-\widehat{k}$,we get $-B_x = -1$,which means $B_x = 1$.Therefore,the magnetic field $\overrightarrow{B}$ is in the $+x$ direction.

The figure shows the electric field $\overrightarrow{E}$ of an $EM$ wave at a certain point and a certain instant. The wave is transporting energy in the negative $z$ direction ($-z$ direction). What is the direction of the magnetic field $\overrightarrow{B}$ of the wave at that point and instant?

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