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Electromagnetic Waves

Section: Physics

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Question : 3 of 8

Marks: +1, -0

A cube of unit volume contains

35 \times 10^{7}

photons of frequency

10^{15} \;\mathrm{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

\alpha \times 10^{-9} \;\mathrm{T}

. Taking permeability of free space

\mu_0 = 4 \pi \times 10^{-7} \;\mathrm{Tm}/\mathrm{A}

, Planck's constant

h = 6 \times 10^{-34} \;\mathrm{Js}

\pi = \;\frac{22}{7}

\alpha

[JEE Adv 2025 P1]

Your Answer:

Solution:

\;E = N \cdot h f

\;E = 35 \times 10^{7} \times 6 \times 10^{-34} \times 10^{15}

\;\;\frac{B_0^2}{2 \mu_0} = E

\;\;B_0^2 = 2 \times 4 \times \;\frac{22}{7} \times 10^{-7} \times 35 \times 10^{7} \times 6 \times 10^{-34} \times 10^{15}

\;\;\;= 5280 \times 10^{-19}

\;\;B_0^2 = 528 \times 10^{-18}

\;\;B_0 = 22.978 \times 10^{-9}

\;\;\alpha = 22.98

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