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JEE Advanced Physics Electromagnetic Waves 2025 JEE Advanced 2025 (Paper 1)

JEE Advanced Physics Question (2025) — Solution

Question

A cube of unit volume contains 35 10^7 photons of frequency 10^ 15 ~Hz . If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is 10^ -9 ~T . Taking permeability of free space _0=4 10^ -7 Tm / A , Planck's constant h=6 10^ -34 Js and = 22 7 , the value of is_______

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Total energy in cube =35 10^7 hf aligned & =35 10^7 6 10^ -34 10^ 15 \\ & =2.1 10^ -10 ~J aligned Total energy of EM waves = B_0^2 2 _0 volume aligned & B _0^2= 2.1 10^ -10 8 10^ -7 1^3 \\ & ~B _0=22.98 10^ -9 ~T aligned Ans. 22.98

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