JEE Advanced 2025 Paper 1 · Q09 · EM Waves
A cube of unit volume contains $35 \times 10^{7}$ photons of frequency $10^{15}$ Hz. If the total photon energy is interpreted as the average EM-wave energy in the same volume, find the magnetic-field amplitude $\alpha \times 10^{-9}$ T (i.e., find $\alpha$). Use $\mu_0 = 4\pi \times 10^{-7}$ T·m/A, $h = 6 \times 10^{-34}$ J·s, $\pi = 22/7$.
Reveal answer + step-by-step solution
Correct answer:23
Solution
Total photon energy in volume $1$ m³: $U/V = N h \nu = 35 \times 10^{7} \times 6 \times 10^{-34} \times 10^{15} = 35 \times 6 \times 10^{(7-34+15)} = 210 \times 10^{-12}$ J/m³. Average EM energy density $= B_0^2/(2\mu_0)$ (since average over a cycle is half the peak; for the field amplitude $B_0$, total energy density $u = B_0^2/(2\mu_0)$ when summed over $E$ and $B$). So $B_0^2 = 2\mu_0 u = 2 \times 4\pi \times 10^{-7} \times 210 \times 10^{-12} = 1680\pi \times 10^{-19}$. With $\pi = 22/7$: $1680 \times (22/7) = 240 \times 22 = 5280$. So $B_0^2 = 5280 \times 10^{-19} = 528 \times 10^{-18}$. $B_0 = \sqrt{528} \times 10^{-9} \approx 22.98 \times 10^{-9}$ T, so $\alpha \approx 23$. Within official acceptance band $[21, 25]$.
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