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Physics · Electromagnetic Waves

JEE Main 2025 — 23 January, Evening Shift — Question 64

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3\mathrm{E}_{\mathrm{y}}=9.3 Vm−1\mathrm{Vm}^{-1}. Then, the magnetic field vector of the wave at that point is-

  1. Option A:

    Bz=9.3×10−8 T\mathrm{B}_{\mathrm{z}}=9.3 \times 10^{-8} \mathrm{~T}

  2. Option B:

    Bz=1.55×10−8 T\mathrm{B}_{\mathrm{z}}=1.55 \times 10^{-8} \mathrm{~T}

  3. Option C:

    Bz=6.2×10−8 T\mathrm{B}_{\mathrm{z}}=6.2 \times 10^{-8} \mathrm{~T}

  4. Option D:

    Bz=3.1×10−8 T\mathrm{B}_{\mathrm{z}}=3.1 \times 10^{-8} \mathrm{~T}

    Correct

Answer: D

Step-by-step solution

E=BC\mathrm{E}=\mathrm{BC}

9.3=B×3×1089.3=\mathrm{B} \times 3 \times 10^{8}

B=9.33×108=3.1×10−8 TB=\frac{9.3}{3 \times 10^{8}}=3.1 \times 10^{-8} \mathrm{~T}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves
Topic
Power , Energy and Intensity of EM Waves
A plane electromagnetic wave of frequency 20 MHz travels in free… | JEE Main 2025 PYQ with Solution · DhiX AI