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Physics · Electromagnetic Waves

JEE Main 2024 — 29 January, Shift 2 — Question 44

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E→=9.6j^V/m\overrightarrow{\mathrm{E}}=9.6 \hat{\mathrm{j}} \mathrm{V} / \mathrm{m}. The value of magnetic field at this point is :

  1. Option A:

    3.2×10−8k^T3.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}

    Correct
  2. Option B:

    3.2×10−8i^T3.2 \times 10^{-8} \hat{\mathrm{i}} \mathrm{T}

  3. Option C:

    9.6j^9.6 \hat{\mathrm{j}}

  4. Option D:

    9.6×10−8kT9.6 \times 10^{-8} \mathrm{k} \mathrm{T}

Answer: A

Step-by-step solution

EB=C\frac{E}{B}=C

EB=3×108\frac{E}{B}=3 \times 10^{8}

B=E3×108=9.63×108B=\frac{E}{3 \times 10^{8}}=\frac{9.6}{3 \times 10^{8}}

B=3.2×10−8 T\mathrm{B}=3.2 \times 10^{-8} \mathrm{~T}

B^=v^×E^\hat{B}=\hat{v} \times \hat{E}

=i^×j^=k^=\hat{\mathrm{i}} \times \hat{\mathrm{j}}=\hat{\mathrm{k}}

So, B→=3.2×10−8k^T\overrightarrow{\mathrm{B}}=3.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Physics
Chapter
Electromagnetic Waves
Topic
Displacement Current and Equation of EM Waves
A plane electromagnetic wave of frequency 35 MHz travels in free… | JEE Main 2024 PYQ with Solution · DhiX AI