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Question

(A) For plane electromagnetic waves,the electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular,i.e.,$\vec{E} \cdot \vec{B} =

Vedclass · Accepted Answer

$(-2 \hat{i}-3 \hat{j})$ and $(3 \hat{i}-2 \hat{j})$

Vedclass · Answer

(A) For plane electromagnetic waves,the electric field $\vec{E}$ and magnetic field $\vec{B}$ are mutually perpendicular,i.e.,$\vec{E} \cdot \vec{B} = 0$. Also,the direction of propagation is given by $\vec{E} 	imes \vec{B}$.Given the propagation is in the $+Z$-direction $(\hat{k})$,we must have $\vec{E} 	imes \vec{B} \propto \hat{k}$.Checking option $A$: $\vec{E} = (-2 \hat{i} - 3 \hat{j})$ and $\vec{B} = (3 \hat{i} - 2 \hat{j})$.Dot product: $\vec{E} \cdot \vec{B} = (-2)(3) + (-3)(-2) = -6 + 6 = 0$. This satisfies the perpendicularity condition.Cross product: $\vec{E} 	imes \vec{B} = (-2 \hat{i} - 3 \hat{j}) 	imes (3 \hat{i} - 2 \hat{j}) = 4(\hat{i} 	imes \hat{j}) - 9(\hat{j} 	imes \hat{i}) = 4\hat{k} - 9(-\hat{k}) = 13\hat{k}$.Since the result is in the $+Z$-direction,option $A$ is correct.

For plane electromagnetic waves propagating in the positive $Z$-direction,the combination which gives the correct possible direction for $\vec{E}$ and $\vec{B}$ fields respectively is:

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