The electric field of a plane electromagnetic | vedclass

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Question

(B) The given electric field is $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$.The direction of propagation of the wave is along

Vedclass · Accepted Answer

$\frac{E_{0}}{c}(-\hat{x} + \hat{y}) \sin(kz - \omega t)$

Vedclass · Answer

(B) The given electric field is $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$.The direction of propagation of the wave is along the $+z$-axis,so $\hat{k} = \hat{z}$.The relation between the direction of propagation,electric field,and magnetic field is given by $\hat{k} = \hat{E} 	imes \hat{B}$.The unit vector for the electric field is $\hat{E} = \frac{\hat{x} + \hat{y}}{\sqrt{2}}$.Substituting the values: $\hat{z} = \left(\frac{\hat{x} + \hat{y}}{\sqrt{2}}ight) 	imes \hat{B}$.Solving for $\hat{B}$,we find $\hat{B} = \frac{-\hat{x} + \hat{y}}{\sqrt{2}}$.Since the magnitude of the magnetic field is $B_{0} = \frac{E_{0}}{c}$,the magnetic field is $\overrightarrow{B} = \frac{E_{0}}{c}(-\hat{x} + \hat{y}) \sin(kz - \omega t)$.

The electric field of a plane electromagnetic wave is given by $\overrightarrow{E} = E_{0}(\hat{x} + \hat{y}) \sin(kz - \omega t)$. Its magnetic field will be given by

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