JEE Advanced 2025 — Paper 1 — PHY-III solution

JEE Advanced2025Paper 1PHY-III

A cube of unit volume contains $35 \times 1 0^{7}$ photons of frequency $1 0^{15}$ 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 $α \times 1 0^{- 9}$ T. Taking permeability of free space $μ_{0} = 4 π \times 1 0^{- 7}$ Tm/A, Planck's constant $h = 6 \times 1 0^{- 34}$ Js and $π = \frac{22}{7}$ , the value of $α$ is ____

Your answer

Solution

Total photon energy in the cube:

$U = N h f = (35 \times 1 0^{7}) (6 \times 1 0^{- 34}) (1 0^{15}) = 2.1 \times 1 0^{- 10}$ J

For an EM wave, the time-averaged energy density is $u = \frac{B _{0}^{2}}{2 μ _{0}}$ . Equating $U = u V$ with $V = 1$ m $^{3}$ :

$B_{0}^{2} = \frac{2 μ _{0} U}{V} = \frac{2 \times 4 π \times 1 0 ^{- 7} \times 2.1 \times 1 0 ^{- 10}}{1}$

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

Using $π = 22/7$ , $B_{0}^{2} \approx 5.28 \times 1 0^{- 16}$ T $^{2}$ , so $B_{0} \approx 22.98 \times 1 0^{- 9}$ T.

$∴ α = 22.98$ .

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