Question

A cube of unit volume contains $ 35 \times 10^7 $ photons of frequency $ 10^{15} \, \text{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} \, \text{T} $. Taking permeability of free space $ \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} $, Planck’s constant $ h = 6 \times 10^{-34} \, \text{Js} $ and $ \pi = \frac{22}{7} $, the value of $ \alpha $ is ____

Hint:

Total electromagnetic energy density in a volume is given by \( u = \frac{B^2}{2\mu_0} \). For problems involving photons, use \( E = h \nu \) and sum over total number of photons to connect quantum and classical results.

Answer 1

Correct Answer: 22 - 23

Step 1: Total energy contained in the cube Each photon has energy: \[ E_{\text{photon}} = h \nu = 6 \times 10^{-34} \times 10^{15} = 6 \times 10^{-19} \, \text{J} \] Total number of photons: \( N = 35 \times 10^7 \) So total energy in the cube is: \[ E = N \cdot E_{\text{photon}} = 35 \times 10^7 \times 6 \times 10^{-19} = 210 \times 10^{-12} = 2.1 \times 10^{-10} \, \text{J} \] Step 2: Energy density in terms of magnetic field amplitude Electromagnetic energy per unit volume: \[ u = \frac{B^2}{2\mu_0} \] Since volume is unity, total energy = energy density: \[ E = \frac{B^2}{2\mu_0} \quad \Rightarrow \quad B^2 = 2\mu_0 E \] Substitute values: \[ \mu_0 = 4\pi \times 10^{-7}, \quad E = 2.1 \times 10^{-10} \] \[ B^2 = 2 \times 4\pi \times 10^{-7} \times 2.1 \times 10^{-10} = 8\pi \cdot 2.1 \cdot 10^{-17} \] Use \( \pi = \dfrac{22}{7} \): \[ B^2 = 8 \cdot \frac{22}{7} \cdot 2.1 \cdot 10^{-17} = \frac{176}{7} \cdot 2.1 \cdot 10^{-17} \] Calculate numerically: \[ \frac{176}{7} \approx 25.14, \quad 25.14 \cdot 2.1 = 52.794 \Rightarrow B^2 = 52.794 \times 10^{-17} \] Step 3: Solve for \( B \) \[ B = \sqrt{52.794 \times 10^{-17}} = \sqrt{52.794} \cdot 10^{-8.5} \] Now calculate: \[ \sqrt{52.794} \approx 7.27, \quad 10^{-8.5} = \sqrt{10^{-17}} = 3.16 \times 10^{-9} \Rightarrow B \approx 7.27 \times 3.16 \times 10^{-9} \approx 22.98 \times 10^{-9} \, \text{T} \] Final Answer: \( \alpha = \boxed{22.98} \)