JEE Main 2024 · 31 January · morning · Single correct · +4 / −1
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ Hz and an amplitude of 50 Vm⁻¹. The total average energy density of the electromagnetic field of the wave is : [Use ε₀=8.85 × 10⁻¹² C² / Nm² ]
- (a) 4.425 × 10⁻⁸ Jm⁻³
- (b) 2.212 × 10⁻¹⁰ Jm⁻³
- (c) 2.212 × 10⁻⁸ Jm⁻³
- (d) 1.106 × 10⁻⁸ Jm⁻³
Answer and solution
Answer (d) 1.106 × 10⁻⁸ Jm⁻³
Solution
Physics sir · 4 steps
Electric and magnetic parts carry equal energy, and averaging sin² gives a factor of half.
- Equal shares. In an EM wave B = E/c, so the magnetic energy density B²/2μ₀ always equals the electric one, 1/2ε₀ E². The instantaneous total is ε₀ E².
- Time average. E = E₀ sin(ω t - kx), and the average of sin² over a cycle is 1/2. The frequency does not affect the average, so it is not needed.
u_(avg) = 1/2ε₀ E₀²
- Put in numbers. u_(avg) = 1/2 × 8.85 × 10⁻¹² × (50)² = 1/2 × 8.85 × 10⁻¹² × 2500 = 1.106 × 10⁻⁸ J m⁻³.
- So. The total average energy density is 1.106 × 10⁻⁸ J m⁻³, option (d).
The idea to keep. Total average energy density of an EM wave is 1/2ε₀ E₀²: each field contributes 1/4ε₀ E₀². Use Eᵣₘₛ instead and it becomes ε₀ Eᵣₘₛ².
Answer from the NTA final answer key. Solution by PrepWiser.