EM Waves — JEE Advanced · 2025 · 18 May 2025 · Paper 1 Question with Solution

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 ____.

Answer: 22.98

Solution

Total energy of photons:

E_{t o t a l} = N \times h f = 35 \times 1 0^{7} \times 6 \times 1 0^{- 34} \times 1 0^{15} = 2.1 \times 1 0^{- 10} J

For electromagnetic waves, energy density:

u = \frac{B _{0}^{2}}{2 μ _{0}}

Total energy in unit volume:

E_{t o t a l} = u \times V = \frac{B _{0}^{2}}{2 μ _{0}} \times 1

Solving for

B_{0}

B_{0}^{2} = 2.1 \times 1 0^{- 10} \times 2 \times 4 π \times 1 0^{- 7}

B_{0}^{2} = 2.1 \times 8 π \times 1 0^{- 17}

B_{0} = 16.8 π \times 1 0^{- 17} = 52.8 \times 1 0^{- 17} = 22.98 \times 1 0^{- 9} T

Therefore,

α = 22.98

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · EM Waves

Solution

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