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Electromagnetic Waves · Question 25

JEE Main 2025 (Online) 28th January Morning ShiftMCQEasy📈 Topic in 65% papers+4−1
1

Due to presence of an em-wave whose electric component is given by E = 100 ( t - kx) NC-1 a cylinder of length 200 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

  1. 50 ( t - kx) NC-1
  2. 400 ( t - kx) NC-1
  3. 200 ( t - kx) NC-1
  4. 25 ( t - kx) NC-1

Answer

(3) 200 ( t - kx) NC-1

Solution

Average energy density: u = 12 0 E02

Energy in the cylinder: U = u d24l E02 d2 (same length l)

Same energy E0 d = constant

E0' = E0 dd' = 100 2 = 200 NC-1

E = 200 ( t - kx) NC-1

Answer: (3)

JEE Main 2025 (Online) 28th January Morning Shift

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2

The magnetic field of an E.M. wave is given by B = ( 32 i + 12 j )30 [ (t - zc ) ] (S.I. Units). The corresponding electric field in S.I. units is:

  1. E = ( 12 i + 32 j )30c [ (t + zc ) ]
  2. E = ( 12 i - 32 j )30c [ (t - zc ) ]
  3. E = ( 32 i - 12 j )30c [ (t + zc ) ]
  4. E = ( 34 i + 14 j )30c [ (t - zc ) ]
JEE Main 2025 (Online) 28th January Evening Shift

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