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MCQEasyJEE 2023Characteristics of EM Waves

JEE Physics 2023 Question with Solution

The electric field in an electromagnetic wave is given as E⃗=20sin⁡(ωt−xc) j^ N/C\vec{E} = 20 \sin \left( \omega t - \frac{x}{c} \right) \, \hat{j} \, N/C where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×104 m35 \times 10^4 \, \text{m}^3 will be (Given ϵ0=8.85×10−12 C2/Nm2\epsilon_0 = 8.85 \times 10^{-12} \, C^2 / Nm^2):

  • A

    8.85×10−13 J8.85 \times 10^{-13} \, J

  • B

    17.7×10−13 J17.7 \times 10^{-13} \, J

  • C

    8.85×10−10 J8.85 \times 10^{-10} \, J

  • D

    28.5×10−13 J28.5 \times 10^{-13} \, J

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Peak electric field is E0=20 N/CE_0 = 20 \, \text{N/C}, volume is V=5×104 m3V = 5 \times 10^4 \, \text{m}^3, and ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12}.

Find: Total energy contained in the given volume.

For electromagnetic waves, the energy density is

u=12ϵ0E02u = \frac{1}{2} \epsilon_0 E_0^2

Substituting the given values,

ν=12×8.85×10−12×(20)2=8.85×10−13 J/m3\nu = \frac{1}{2} \times 8.85 \times 10^{-12} \times (20)^2 = 8.85 \times 10^{-13} \, \text{J/m}^3

The total energy in volume VV is

Energy=ν×V=8.85×10−13×5×104=8.85×10−13 J\text{Energy} = \nu \times V = 8.85 \times 10^{-13} \times 5 \times 10^4 = 8.85 \times 10^{-13} \, \text{J}

Therefore, the correct option is A.

Common mistakes

  • Using the instantaneous electric field instead of the peak value E0E_0. The formula given in the solution uses the amplitude, so identify 2020 as the peak field from the wave equation.

  • Forgetting to multiply energy density by the given volume. Energy density is per unit volume, so total energy must be found using U=uVU = uV.

  • Confusing average energy density with another electromagnetic-wave relation without checking what the solution uses. Follow the stated formula u=12ϵ0E02u = \frac{1}{2} \epsilon_0 E_0^2 for this question.

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