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

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right) 30 \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right] \\ & \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{c}} \text { and } \mathrm{E}=\mathrm{B}_0 \mathrm{c}
\end{aligned}$
Here $\overrightarrow{\mathrm{E}}\left(\frac{\sqrt{3}}{2}(-\hat{\mathrm{j}})+\frac{1}{2} \hat{\mathrm{i}}\right)$
$\begin{aligned}
& \mathrm{E}_0=30 \mathrm{c} \\ & \overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]
\end{aligned}$

The magnetic field of an E.M. wave is given by $\overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2}…

Solution