Physics · Electromagnetic Waves

JEE Main 2025 — 28 January, Morning Shift — Question 62

Due to presence of an em-wave whose electric component is given by $\mathrm{E}=100 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$ , a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

Option A:
$25 \sin (\omega t-k x) \mathrm{NC}^{-1}$
Option B:
$200 \sin (\omega t-k x) \mathrm{NC}^{-1}$
Correct
Option C:
$400 \sin (\omega t-k x) \mathrm{NC}^{-1}$
Option D:
$50 \sin (\omega t-k x) \mathrm{NC}^{-1}$

Answer: B

Step-by-step solution

Energy density $=\frac{1}{2} \epsilon_{0} E^{2} \times c$

Energy $=\frac{1}{2} \epsilon_{0} E^{2} \times c \times$ volume

$(\text { Energy })_{1}=(\text { Energy })_{2} \quad($ Given $)$

$\frac{1}{2} \in_{0} E_{1}^{2} c \pi R_{1}^{2} \times L_{1}=\frac{1}{2} \in_{0} E_{2}^{2} c \pi R_{2}^{2} \times L_{2}$

$\mathrm{E}_{1}^{2} \mathrm{R}_{1}^{2}=\mathrm{E}_{2}^{2} \mathrm{R}_{2}^{2}$

$\mathrm{E}_{1} \mathrm{R}_{1}=\mathrm{E}_{2} \mathrm{R}_{2}$

$100 \times \mathrm{R}_{1}=\mathrm{E}_{2} \times \frac{\mathrm{R}_{1}}{2}$

$\mathrm{E}_{2}=200 \mathrm{~N} / \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Power , Energy and Intensity of EM Waves