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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2024 · 31 January · Morning Shift · Physics · Electromagnetic Waves

In a plane EM wave, the electric field oscillates sinusoidally at a...

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010Hz5 \times {10}^{10}Hz and an amplitude of 50Vm–150V{m}^{– 1}. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×10–12C2N−1m−2{ε}_{0} = 8 . 85 \times {10}^{– 12}{C}^{2}{N}^{- 1}{m}^{- 2}]

Options

  1. A1.106×10–8Jm–31 . 106 \times {10}^{– 8}J{m}^{– 3}
  2. B4.425×10–8Jm–34 . 425 \times {10}^{– 8}J{m}^{– 3}
  3. C2.212×10–8Jm–32 . 212 \times {10}^{– 8}J{m}^{– 3}
  4. D2.212×10–10Jm–32 . 212 \times {10}^{– 10}J{m}^{– 3}

Correct answer

A. 1.106×10–8Jm–31 . 106 \times {10}^{– 8}J{m}^{– 3}

Step-by-step solution

Total average energy density of the electromagnetic field is given by, UE=12ε0(E0)2{U}_{E} = \dfrac{1}{2}{ε}_{0}{\left({E}_{0}\right)}^{2}.

⇒UE=12×8.85×10−12×(50)2⇒ {U}_{E} = \dfrac{1}{2} \times 8 . 85 \times {10}^{- 12} \times {\left(50\right)}^{2}

=1.106×10–8Jm–3= 1 . 106 \times {10}^{– 8}J{m}^{– 3}

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