JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 6)
The magnetic field of an E.M. wave is given by $\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]$ (S.I. Units).
The corresponding electric field in S.I. units is:
$\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]$
$\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]$
$\overrightarrow{\mathrm{E}}=\left(\frac{\sqrt{3}}{2} \hat{i}-\frac{1}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]$
$\overrightarrow{\mathrm{E}}=\left(\frac{3}{4} \hat{i}+\frac{1}{4} \hat{j}\right) 30 \mathrm{c} \cos \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]$
Explanation
$$\begin{aligned}
& \vec{B}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin \left[\omega\left(t-\frac{z}{c}\right)\right] \\
& \vec{E}=\vec{B} \times \vec{c} \text { and } E=B_0 c \\
& \text { Here } \vec{E}\left(\frac{\sqrt{3}}{2}(-\hat{j})+\frac{1}{2} \hat{i}\right) \\
& E_0=30 c \\
& \vec{E}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 c \sin \left[\omega\left(t-\frac{z}{c}\right)\right]
\end{aligned}$$
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