JEE Main 2023 · 11 April · morning · Single correct · +4 / −1
The electric field in an electromagnetic wave is given as
E⃗=20 sin ω(t-x/c) j⃗ NC⁻¹
where ω and c are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5 × 10⁻⁴ m³ will be
(Given ε₀=8.85 × 10⁻¹² C² / Nm² )
- (a) 17 · 7 × 10⁻¹³ J
- (b) 28 · 5 × 10⁻¹³ J
- (c) 8 · 85 × 10⁻¹³ J
- (d) 88 · 5 × 10⁻¹³ J
Answer and solution
Answer (c) 8 · 85 × 10⁻¹³ J
Solution
Physics sir · 4 steps
Average energy density is 1/2ε₀ E₀², shared equally by E and B. Multiply by the volume.
- Read the amplitude. The field is 20 sin ω(t - x/c), so the peak value is E₀ = 20 N C⁻¹.
- Average energy density. The electric field gives an average of 1/4ε₀ E₀². The magnetic field gives exactly the same amount, so the total average energy density is 1/2ε₀ E₀².
U = 1/2ε₀ E₀² V = 1/2 × 8.85 × 10⁻¹² × 400 × 5 × 10⁻⁴ J
- Put in numbers. The energy density is 1/2 × 8.85 × 10⁻¹² × 400 = 1.77 × 10⁻⁹ J m⁻³. Multiply by 5 × 10⁻⁴ m³ to get 8.85 × 10⁻¹³ J.
- So. The energy in the volume is 8.85 × 10⁻¹³ J, which is option (c).
The idea to keep. For an EM wave, the time-averaged total energy density is 1/2ε₀ E₀². Half comes from the electric field and half from the magnetic field.
Answer from the NTA final answer key. Solution by PrepWiser.