A cube of unit volume contains photons of frequency . 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 . [2025]

Question

A cube of unit volume contains $35 \times 10^{7}$ photons of frequency $10^{15} Hz$ . 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 $α \times 10^{- 9} T$ . Taking permeability of free space $μ_{0} = 4 π \times 10^{- 7} Tm / A,$ Planck's constant $h = 6 \times 10^{- 34} Js$ and $π = \frac{22}{7},$ the value of $α$ is ______. [2025]

Smart Achievers · Accepted Answer

(22.98)

$Total energy in cube$ $E = n h f = 35 \times 10^{7} \times h f$

$= 35 \times 10^{7} \times 6 \times 10^{- 34} \times 10^{15} = 2.1 \times 10^{- 10} J$

$Total energy of EM waves = \frac{B_{0}^{2}}{2 μ_{0}} \times volume$

$B_{0}^{2} = \frac{2.1 \times 10^{- 10} \times 8 π \times 10^{- 7}}{1^{3}} = 22.98 \times 10^{- 9} T$

$∴$ $α = 22.98$

Solution

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Solution

Ans. (22.98) Total energy in cube E=nhf=35×107×hf =35×107×6×10-34×1015=2.1×10-10 J Total energy of EM waves=B022μ0×volume B02=2.1×10-10×8π×10-713=22.98×10-9 T ∴ α=22.98

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