The electric field in an electromagnetic wave | vedclass

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Question

(D) The given electric field is $\overrightarrow{E} = \hat{i} 40 \cos \omega(t - \frac{z}{c})$.Here,the electric field vector $\overrightarrow{E}$ is

Vedclass · Accepted Answer

$\overrightarrow{B} = \hat{j} \frac{40}{c} \cos \omega(t - \frac{z}{c}) 	ext{ T}$

Vedclass · Answer

(D) The given electric field is $\overrightarrow{E} = \hat{i} 40 \cos \omega(t - \frac{z}{c})$.Here,the electric field vector $\overrightarrow{E}$ is directed along the $+x$ axis.The wave propagates in the $+z$ direction (as indicated by the term $(t - z/c)$).In an electromagnetic wave,the direction of propagation is given by the direction of $\overrightarrow{E} 	imes \overrightarrow{B}$.Since $\hat{i} 	imes \hat{j} = \hat{k}$,the magnetic field $\overrightarrow{B}$ must be directed along the $+y$ axis $(\hat{j})$.The magnitude of the magnetic field is related to the electric field by $B_0 = \frac{E_0}{c}$.Given $E_0 = 40 	ext{ N/C}$,we have $B_0 = \frac{40}{c} 	ext{ T}$.Thus,the magnetic field is $\overrightarrow{B} = \hat{j} \frac{40}{c} \cos \omega(t - \frac{z}{c}) 	ext{ T}$.

The electric field in an electromagnetic wave is given by $\overrightarrow{E} = \hat{i} 40 \cos \omega(t - \frac{z}{c}) \text{ N/C}$. The magnetic field induction of this wave is (in $SI$ units):

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