The magnetic field of an E.M. wave is given by B→=32i^+12j^30sin⁡ωt−zc (S.I.Units)The corresponding electric field in S.I. units is :

B→=32i^+12j^30sin⁡ωt−zcE→=B→×c→ and E=B0c Here E→32(−j^)+12i^E0=30cE→=12i^−32j^30csin⁡ωt−zc

The magnetic field of an E.M. wave is given by B→=32i^+12j^30sin⁡ωt−zc (S.I.Units)The corresponding electric field in S.I. units is :

B→=32i^+12j^30sin⁡ωt−zcE→=B→×c→ and E=B0c Here E→32(−j^)+12i^E0=30cE→=12i^−32j^30csin⁡ωt−zc

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

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$\vec{E} = (\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) 30 csin [ω (t + \frac{z}{c})]$

$\vec{E} = (\frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j}) 30 csin [ω (t + \frac{z}{c})]$

$\vec{E} = (\frac{3}{4} \hat{i} + \frac{1}{4} \hat{j}) 30 ccos [ω (t - \frac{z}{c})]$

$\vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 csin [ω (t - \frac{z}{c})]$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{B} = (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 \sin [ω (t - \frac{z}{c})] \\ \vec{E} = \vec{B} \times \vec{c} and E = B_{0} c \\ Here \vec{E} (\frac{\sqrt{3}}{2} (- \hat{j}) + \frac{1}{2} \hat{i}) \\ E_{0} = 30 c \\ \vec{E} = (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}) 30 csin [ω (t - \frac{z}{c})] \end{array}$