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MCQEasyJEE 2024Characteristics of EM Waves

JEE Physics 2024 Question with Solution

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 5×1010 Hz5 \times 10^{10} \, \text{Hz} and an amplitude of 50 V/m50 \, \text{V/m}. The total average energy density of the electromagnetic field of the wave is:

  • A

    1.106×10−8 J/m31.106 \times 10^{-8} \, \text{J/m}^3

  • B

    4.425×10−8 J/m34.425 \times 10^{-8} \, \text{J/m}^3

  • C

    2.212×10−8 J/m32.212 \times 10^{-8} \, \text{J/m}^3

  • D

    2.212×10−10 J/m32.212 \times 10^{-10} \, \text{J/m}^3

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Electric field amplitude is E=50 V/mE = 50 \, \text{V/m}.

Find: The total average energy density of the electromagnetic wave.

For a plane electromagnetic wave, the average energy density is written as

u=12ϵ0E2+12μ0H2u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \mu_0 H^2

In vacuum, the electric and magnetic field contributions are equal, so the total average energy density can be evaluated as

u=ϵ0E2u = \epsilon_0 E^2

Using the average-value substitution shown in the working,

UE=12ϵ0E2U_E = \frac{1}{2} \epsilon_0 E^2

Substitute the given values:

UE=12×8.85×10−12×(50)2U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2 UE=12×8.85×10−12×2500U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500 UE=12×2.2125×10−8U_E = \frac{1}{2} \times 2.2125 \times 10^{-8} UE=1.106×10−8 J/m3U_E = 1.106 \times 10^{-8} \, \text{J/m}^3

Therefore, the total average energy density is 1.106×10−8 J/m31.106 \times 10^{-8} \, \text{J/m}^3, so the correct option is A.

Direct Substitution

Given: E=50 V/mE = 50 \, \text{V/m}, ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12}.

Find: Average energy density.

The average energy density of the electric field is given by

UE=12ϵ0E2U_E = \frac{1}{2} \epsilon_0 E^2

Substituting the given values,

UE=12×8.85×10−12×(50)2U_E = \frac{1}{2} \times 8.85 \times 10^{-12} \times (50)^2

Calculating,

UE=1.106×10−8 J/m3U_E = 1.106 \times 10^{-8} \, \text{J/m}^3

Hence, the correct option is A.

Common mistakes

  • Using u=ϵ0E2u = \epsilon_0 E^2 directly with the field amplitude when the question asks for average energy density. This overestimates the result by a factor of 22. Use the time-averaged form uavg=12ϵ0E02u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 when amplitude is given.

  • Choosing the option corresponding to the electric and magnetic contributions added incorrectly. In an electromagnetic wave, the average electric and magnetic energy densities are equal, but the average total is already accounted for in 12ϵ0E02\frac{1}{2} \epsilon_0 E_0^2 when expressed using the electric-field amplitude. Do not double-count.

  • Treating the frequency 5×1010 Hz5 \times 10^{10} \, \text{Hz} as necessary for the calculation. The average energy density here depends on the field amplitude and ϵ0\epsilon_0, not on frequency. Identify the relevant formula before substituting values.

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