Verified Solution • 2025–26 Session moderate

QJEEM24Ph3918: JEE Main 2024 Physics - Class 12 JEEM24Ph3918 Solution: A plane electromagnetic wave of frequency travels in free space …

A plane electromagnetic wave of frequency 35 MHz35 \mathrm{~MHz} travels in free space along the XX -direction. At a particular point (in space and time) E⃗=9.6j^ V/m\vec{E}=9.6 \hat{j} \mathrm{~V} / \mathrm{m} . The value of magnetic field at this point is :

Multiple Choice Options

  • A 9.6j^T9.6 \hat{j} T
  • B 3.2×10−8i^T3.2 \times 10^{-8} \hat{i} T
  • C 9.6×10−8k^T9.6 \times 10^{-8} \hat{k} T
  • D 3.2×10−8k^T3.2 \times 10^{-8} \hat{k} T

Step-by-Step Verified Solution

EBamp;=CEBamp;=3×108Bamp;=E3×108=9.63×108Bamp;=3.2×10−8TB^amp;=v^×E^amp;=i^×j^=k^\begin{aligned}\frac{E}{B} & =C \\\frac{E}{B} & =3 \times 10^8 \\B & =\frac{E}{3 \times 10^8}=\frac{9.6}{3 \times 10^8} \\B & =3.2 \times 10^{-8} T \\\hat{B} & =\hat{v} \times \hat{E} \\& =\hat{i} \times \hat{j}=\hat{k}\end{aligned}

So,

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

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