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JEE Advanced 2025 · Paper 1, Physics question 9

A cube of unit volume contains 35×10735\times10^7 photons of frequency 101510^{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 α×10−9\alpha\times10^{-9} T. Taking permeability of free space μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} Tm/A, Planck's constant h=6×10−34h = 6\times10^{-34} Js and π=227\pi = \dfrac{22}{7}, the value of α\alpha is _____.

Answer and worked solution

Official answer: 21

Energy density.
u=Nhν=35×107×6×10−34×1015=2.1×10−10 J/m3.u=Nh\nu=35\times10^7\times6\times10^{-34}\times10^{15}=2.1\times10^{-10}\ \text{J/m}^3.

Field amplitude. The average energy density of an EM wave is B022μ0\frac{B_0^2}{2\mu_0}, so
B02=2μ0u=2×4π×10−7×2.1×10−10=8⋅227⋅2.1×10−17=5.28×10−16.B_0^2=2\mu_0u=2\times4\pi\times10^{-7}\times2.1\times10^{-10}=8\cdot\tfrac{22}{7}\cdot2.1\times10^{-17}=5.28\times10^{-16}.

B0≈2.3×10−8B_0\approx2.3\times10^{-8} T =23×10−9=23\times10^{-9} T, so α≈23\alpha\approx23.

Topics:Energy and momentum of EM wavesPhotoelectric effect
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