The magnetic field of an text{E.M.} wave is g | vedclass

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Question

(A) The given magnetic field is $\overrightarrow{ B } = B_0 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right] \hat{n}$, where $\hat{n} = \frac{

Vedclass · Accepted Answer

$\overrightarrow{ E }=\left(\frac{1}{2} \hat{ i }-\frac{\sqrt{3}}{2} \hat{ j }\right) 30 c \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right]$

Vedclass · Answer

(A) The given magnetic field is $\overrightarrow{ B } = B_0 \sin \left[\omega\left( t -\frac{ z }{ c }ight)ight] \hat{n}$, where $\hat{n} = \frac{\sqrt{3}}{2} \hat{ i } + \frac{1}{2} \hat{ j }$ and $B_0 = 30 	ext{ T}$.Since the wave propagates in the $+z$ direction $(\hat{k})$, the electric field $\overrightarrow{ E }$ is related to the magnetic field $\overrightarrow{ B }$ by the relation $\overrightarrow{ E } = c (\overrightarrow{ B } 	imes \hat{k})$.Substituting the values: $\overrightarrow{ E } = c \left[ \left( \frac{\sqrt{3}}{2} \hat{ i } + \frac{1}{2} \hat{ j } ight) 30 \sin \left[\omega\left( t -\frac{ z }{ c }ight)ight] 	imes \hat{k} ight]$.Using the cross products $\hat{ i } 	imes \hat{k} = -\hat{ j }$ and $\hat{ j } 	imes \hat{k} = \hat{ i }$, we get:$\overrightarrow{ E } = 30 c \sin \left[\omega\left( t -\frac{ z }{ c }ight)ight] \left( \frac{\sqrt{3}}{2} (-\hat{ j }) + \frac{1}{2} \hat{ i } ight)$.Rearranging the terms, we get $\overrightarrow{ E } = \left( \frac{1}{2} \hat{ i } - \frac{\sqrt{3}}{2} \hat{ j } ight) 30 c \sin \left[\omega\left( t -\frac{ z }{ c }ight)ight]$.

The magnetic field of an $\text{E.M.}$ wave is given by $\overrightarrow{ B }=\left(\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }\right) 30 \sin \left[\omega\left( t -\frac{ z }{ c }\right)\right]$ ($\text{S.I.}$ units). The corresponding electric field in $\text{S.I.}$ units is:

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