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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Electromagnetic Waves

JEE Advanced 2025 · Paper 1 · Physics · Electromagnetic Waves

A cube of unit volume contains 35 × 10^7 photons of frequency 10^...

A cube of unit volume contains 35×10735 \times 10^7 photons of frequency 1015 Hz10^{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 α×10−9 T\alpha \times 10^{-9} \mathrm{~T}. Taking permeability of free space μ0=4π×10−7Tm/A\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}, Planck's constant h=6×10−34Jsh=6 \times 10^{-34} \mathrm{Js} and π=227\pi=\frac{22}{7}, the value of α\alpha is_______

Correct answer

0

Step-by-step solution

Total energy in cube =35×107×hf=35 \times 10^7 \times \mathrm{hf} =35×107×6×10−34×1015=2.1×10−10 J\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 =B022μ0×=\frac{B_0^2}{2 \mu_0} \times volume B02=2.1×10−10×8π×10−713⇒ B0=22.98×10−9 T\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} Ans. 22.98
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