A cube of unit volume contains 35 × 10⁷ photons of frequency 10¹⁵ Hz. Treating their total energy as the average electromagnetic-wave energy in the same volume, the magnetic-field amplitude is α × 10⁻⁹ T. Given μ₀ = 4π × 10⁻⁷ T m A⁻¹, h = 6 × 10⁻³⁴ J s and π = 22/7, find α.
Physics · Circular Motion
JEE Advanced 2025 Paper 1

Question

A cube of unit volume contains 35 × 10⁷ photons of frequency 10¹⁵ Hz. Treating their total energy as the average electromagnetic-wave energy in the same volume, the magnetic-field amplitude is α × 10⁻⁹ T. Given μ₀ = 4π × 10⁻⁷ T m A⁻¹, h = 6 × 10⁻³⁴ J s and π = 22/7, find α.

Physics · Circular Motion

JEE Advanced 2025 Paper 1

G1 Physics · Accepted Answer

Total photon energy: U = Nhν = (35 × 10⁷)(6 × 10⁻³⁴)(10¹⁵) = 2.1 × 10⁻¹⁰ J. For unit volume, average energy density u = 2.1 × 10⁻¹⁰ J m⁻³. For an electromagnetic wave, u = B₀²/(2μ₀). Thus B₀ = √(2μ₀u) = √[2(4π × 10⁻⁷)(2.1 × 10⁻¹⁰)] ≈ 2.3 × 10⁻⁸ T = 23 × 10⁻⁹ T. Therefore α ≈ 23.