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Friday, 9 October

JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 31 January 2024, Shift 1, Physics Q44

Question 44 of 90 in this shift, Physics question 14 of 30, Section A.

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×10105\times10^{10} Hz and an amplitude of 50 Vm−1^{-1}. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×10−12\varepsilon_0=8.85\times10^{-12} C2^2/Nm2^2]
  1. (1)1.106×10−81.106\times10^{-8} Jm−3^{-3}Official answer
  2. (2)2.212×10−102.212\times10^{-10} Jm−3^{-3}
  3. (3)2.212×10−82.212\times10^{-8} Jm−3^{-3}
  4. (4)4.425×10−84.425\times10^{-8} Jm−3^{-3}

Official answer

Option 1

NTA final key.

Chapter
Electromagnetic Waves
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.