JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2025
- Easy
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right) 30 \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right] \\
& \overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{c}} \text { and } \mathrm{E}=\mathrm{B}_0 \mathrm{c}
\end{aligned}\)
Here \(\overrightarrow{\mathrm{E}}\left(\frac{\sqrt{3}}{2}(-\hat{\mathrm{j}})+\frac{1}{2} \hat{\mathrm{i}}\right)\)
\(\begin{aligned}
& \mathrm{E}_0=30 \mathrm{c} \\
& \overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]
\end{aligned}\)
& \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 \(\overrightarrow{\mathrm{E}}\left(\frac{\sqrt{3}}{2}(-\hat{\mathrm{j}})+\frac{1}{2} \hat{\mathrm{i}}\right)\)
\(\begin{aligned}
& \mathrm{E}_0=30 \mathrm{c} \\
& \overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]
\end{aligned}\)
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