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_______
- A 22.98
- B 23.98
- C 24.98
- D 25.98
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
\(\begin{aligned} & =35 \times 10^7 \times 6 \times 10^{-34} \times 10^{15} \\ & =2.1 \times 10^{-10} \mathrm{~J}\end{aligned}\)
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
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