ExamBro
ExamBro
JEE Advanced · Physics · 23. EM Waves

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_______

  1. A 22.98
  2. B 23.98
  3. C 24.98
  4. D 25.98
Verified Solution

Answer & Solution

Correct Answer

(A) 22.98

Step-by-step Solution

Detailed explanation

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
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app