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Electromagnetic Waves question

2025 · Shift 1 · Q41
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Electromagnetic Waves question

2025 · Shift 1 · Q41

JEE AdvancedPhysicsElectromagnetic WavesNumerical+4 / −1
A cube of unit volume contains 35×10735 \times 10^735×107 photons of frequency 1015 Hz10^{15} \mathrm{~Hz}1015 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 α×10−9 T\alpha \times 10^{-9} \mathrm{~T}α×10−9 T. Taking permeability of free space μ0=4π×10−7Tm/A\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}μ0​=4π×10−7Tm/A, Planck's constant h=6×10−34Jsh=6 \times 10^{-34} \mathrm{Js}h=6×10−34Js and π=227\pi=\frac{22}{7}π=722​, the value of α\alphaα is ‾\underline{\hspace{2cm}}​.
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