A plane electromagnetic wave of frequency 25; | vedclass

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(D) For an electromagnetic wave propagating in vacuum,the relationship between the electric field $\overrightarrow{E}$,the magnetic field $\overrighta

Vedclass · Accepted Answer

$-15 \hat{i}\; V/m$

Vedclass · Answer

(D) For an electromagnetic wave propagating in vacuum,the relationship between the electric field $\overrightarrow{E}$,the magnetic field $\overrightarrow{B}$,and the velocity vector $\overrightarrow{c}$ is given by $\overrightarrow{E} = \overrightarrow{c} 	imes \overrightarrow{B}$.The wave propagates along the $z$-direction,so the velocity vector is $\overrightarrow{c} = c \hat{k} = (3 	imes 10^{8}) \hat{k}\; m/s$.The magnetic field is given as $\overrightarrow{B} = 5 	imes 10^{-8} \hat{j}\; T$.Substituting these values into the formula:$\overrightarrow{E} = (3 	imes 10^{8} \hat{k}) 	imes (5 	imes 10^{-8} \hat{j})$Using the cross product rules for unit vectors $(\hat{k} 	imes \hat{j} = -\hat{i})$:$\overrightarrow{E} = (3 	imes 5) 	imes (10^{8} 	imes 10^{-8}) 	imes (\hat{k} 	imes \hat{j})$$\overrightarrow{E} = 15 	imes 1 	imes (-\hat{i})$$\overrightarrow{E} = -15 \hat{i}\; V/m$.

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