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MCQEasyJEE 2025Characteristics of EM Waves

JEE Physics 2025 Question with Solution

Due to presence of an em-wave whose electric component is given by E=100sin⁡(ωt−kx) NC−1E = 100 \sin(\omega t - kx) \, \text{NC}^{-1}, a cylinder of length 200 cm200 \, \text{cm} holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:

  • A

    200sin⁡(ωt−kx) NC−1200 \sin(\omega t - kx) \, \text{NC}^{-1}

  • B

    25sin⁡(ωt−kx) NC−125 \sin(\omega t - kx) \, \text{NC}^{-1}

  • C

    50sin⁡(ωt−kx) NC−150 \sin(\omega t - kx) \, \text{NC}^{-1}

  • D

    400sin⁡(ωt−kx) NC−1400 \sin(\omega t - kx) \, \text{NC}^{-1}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The electric field is E=100sin⁡(ωt−kx) NC−1E = 100 \sin(\omega t - kx) \, \text{NC}^{-1}. The cylinder length remains the same, while the diameter of the second cylinder is half of the first.

Find: The modified electric field expression when both cylinders contain the same electromagnetic energy.

For an electromagnetic wave, energy density is proportional to the square of the electric field amplitude.

u∝E02u \propto E_0^2

The total energy stored in the cylinder is proportional to energy density times volume.

U∝E02VU \propto E_0^2 V

For a cylinder of fixed length, volume is proportional to cross-sectional area, and hence to the square of the diameter.

V∝D2V \propto D^2

Therefore,

U∝E02D2U \propto E_0^2 D^2

Since the total energy remains the same,

E12D12=E22D22E_1^2 D_1^2 = E_2^2 D_2^2

Given that

D2=D12D_2 = \frac{D_1}{2}

Substituting,

E12D12=E22(D12)2E_1^2 D_1^2 = E_2^2 \left(\frac{D_1}{2}\right)^2 E12D12=E22D124E_1^2 D_1^2 = E_2^2 \frac{D_1^2}{4} E22=4E12E_2^2 = 4E_1^2 E2=2E1=2×100=200 NC−1E_2 = 2E_1 = 2 \times 100 = 200 \, \text{NC}^{-1}

Hence the modified electric field is

E=200sin⁡(ωt−kx) NC−1E = 200 \sin(\omega t - kx) \, \text{NC}^{-1}

Therefore, the correct option is A.

Discrepancy Note

The solution concludes that the new amplitude becomes 200 NC−1200 \, \text{NC}^{-1}, and both extracted approaches show the same working. However, the answer key marks option (3). Using the solution, the answer must be taken as A, not C.

Common mistakes

  • Assuming the electric field should decrease when the cylinder diameter decreases. This is wrong because the total energy is kept the same while the volume becomes smaller. Use U∝E2VU \propto E^2 V and increase the field amplitude accordingly.

  • Using diameter directly as proportional to volume. This is incorrect because for a cylinder of fixed length, volume is proportional to cross-sectional area, so V∝D2V \propto D^2, not DD.

  • Forgetting that the energy density of an electromagnetic wave varies as the square of the field amplitude. If you use U∝EU \propto E instead of U∝E2U \propto E^2, the final amplitude will be incorrect.

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