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(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.The electric field polarization is given by $\hat{E} = \hat{

Vedclass · Accepted Answer

$B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - k \frac{\hat{i}+\hat{j}}{\sqrt{2}} \cdot \vec{r}\right)$

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(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.The electric field polarization is given by $\hat{E} = \hat{k}$.In an electromagnetic wave,the direction of propagation $\hat{n}$ is given by the cross product of the electric field direction $\hat{E}$ and the magnetic field direction $\hat{B}$,i.e.,$\hat{n} = \hat{E} 	imes \hat{B}$.Substituting the known values: $\frac{\hat{i}+\hat{j}}{\sqrt{2}} = \hat{k} 	imes \hat{B}$.We know that $\hat{k} 	imes \hat{i} = \hat{j}$ and $\hat{k} 	imes \hat{j} = -\hat{i}$.Therefore,$\hat{k} 	imes \left( \frac{\hat{i}-\hat{j}}{\sqrt{2}} ight) = \frac{\hat{j} - (-\hat{i})}{\sqrt{2}} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.Thus,the direction of the magnetic field is $\hat{B} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$.Hence,the magnetic field vector is $\vec{B} = B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - \vec{k} \cdot \vec{r}ight)$.

$A$ plane electromagnetic wave is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}},$ with its polarization along the direction $\hat{k}$. The correct form of the magnetic field of the wave would be (here $B_{0}$ is an appropriate constant)

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