An electromagnetic wave has its electric and | vedclass

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Question

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is the direction of $\vec{E} 	imes

Vedclass · Accepted Answer

$-\hat{k}$

Vedclass · Answer

(A) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is the direction of $\vec{E} 	imes \vec{B}$.First,we find the unit vectors for the electric and magnetic fields:$\hat{E} = \frac{\hat{i} + \hat{j}}{\sqrt{1^2 + 1^2}} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$$\hat{B} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$The direction of propagation $\hat{n}$ is given by $\hat{E} 	imes \hat{B}$:$\hat{n} = \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} ight) 	imes \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} ight)$$\hat{n} = \frac{1}{2} [(\hat{i} 	imes \hat{i}) - (\hat{i} 	imes \hat{j}) + (\hat{j} 	imes \hat{i}) - (\hat{j} 	imes \hat{j})]$Using the cross product rules $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{j} = 0$,$\hat{i} 	imes \hat{j} = \hat{k}$,and $\hat{j} 	imes \hat{i} = -\hat{k}$:$\hat{n} = \frac{1}{2} [0 - \hat{k} - \hat{k} - 0] = \frac{1}{2} [-2\hat{k}] = -\hat{k}$.

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