Q41 / 48 · JEE Advanced 2025 (Paper 1) · Physics
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_______
Correct answer: 22.98
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 $=\frac{B_0^2}{2 \mu_0} \times$ volume
$\begin{aligned} & \mathrm{B}_0^2=\frac{2.1 \times 10^{-10} \times 8 \pi \times 10^{-7}}{1^3} \\ & \Rightarrow \mathrm{~B}_0=22.98 \times 10^{-9} \mathrm{~T}\end{aligned}$
Ans. 22.98
Hence, the correct answer is 22.98.