Physics · Electromagnetic Waves

JEE Main 2026 — 22 January, Evening Shift — Question 39

A laser beam has intensity of $4.0 \times 10^{14} \mathrm{~W} / \mathrm{m}^{2}$ . The amplitude of magnetic field associated with beam is $\_\_\_\_$ T. ( Take $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ )

Option A:
2
Option B:
18.3
Option C:
5.5
Option D:
1.83
Correct

Answer: D

Step-by-step solution

$\mathrm{I}=\frac{1}{2} \in_{0} \mathrm{E}_{0}^{2} \cdot \mathrm{C}$ $\therefore \mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\epsilon_{0} \mathrm{C}}}$ $\& \frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}=\mathrm{C}$ $\therefore \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}=\frac{1}{\mathrm{C}} \sqrt{\frac{2 \mathrm{I}}{\epsilon_{0} \mathrm{C}}}$ $\therefore \mathrm{B}_{0}=\frac{1}{3 \times 10^{8}} \sqrt{\frac{2 \times 4 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^{8}}}$ $\mathrm{B}_{0}=\frac{10}{3} \sqrt{\frac{8}{8.85 \times 3}}$ $\mathrm{B}_{0}=1.83 \mathrm{~T}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Power , Energy and Intensity of EM Waves