Physics · Electromagnetic Waves

JEE Advanced 2025 — Paper 1 — Question 7

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 $h=6 \times 10^{-34} \mathrm{Js}$ and $\pi=\frac{22}{7}$ , the value of $\alpha$ is $\qquad$

Answer: 22.98

Numerical answer — enter this value.

Step-by-step solution

Total energy in cube $=35 \times 10^{7} \times \mathrm{hf}$

\begin{aligned} & =35 \times 10^{7} \times 6 \times 10^{-34} \times 10^{15} \\ & =2.1 \times 10^{-10} \mathrm{~J} \end{aligned}

Total energy of EM waves $=\frac{\mathrm{B}_{0}^{2}}{2 \mu_{0}} \times$

volume $\begin{aligned} & \mathrm{B}_{0}^{2}=\frac{2.1 \times 10^{-10} \times 8 \pi \times 10^{-7}}{1^{3}} \\ & \Rightarrow \mathrm{~B}_{0}=22.98 \times 10^{-9} \mathrm{~T} \end{aligned}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 1
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Power , Energy and Intensity of EM Waves