Electromagnetic Waves JEE Main 2025 Solution

JEE MAIN 2025

28-01-2025 SHIFT-2

Question

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 :

Select the correct option:

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

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

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

$\overrightarrow{\mathrm{E}}=\left(\frac{3}{4} \hat{i}+\frac{1}{4} \hat{j}\right) 30 \mathrm{c} \cos \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\vec{B}=\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right) 30 \sin \left[\omega\left(t-\frac{z}{c}\right)\right]$
$\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{c}}$ and $\mathrm{E}=\mathrm{B}_{0} \mathrm{c}$
Here $\vec{E}\left(\frac{\sqrt{3}}{2}(-\hat{j})+\frac{1}{2} \hat{i}\right)$ $\mathrm{E}_{0}=30 \mathrm{c}$
$\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{csin}\left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$

Question Tags

JEE Main

Physics

Hard

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