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 \alpha \times 10^-9 ~T . Taking permeability of free space \mu_0=4 \pi \times 10^-7 Tm / A , Planck's constant h=6 \times 10^-34 Js and \pi=\frac227 , the value of \alpha is \underline\hspace2cm .

Electromagnetic Waves PYQ — JEE Advanced 2025 · Shift 1 · Q41 with Solution

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Correct answer: 21TO25

Given frequency,

u = 10^{15}\,\text{Hz}$$ Energy of one photon is $$E_1 = h\nu = 6\times 10^{-34}\times 10^{15} = 6\times 10^{-19}\,\text{J}$$ 2. **Total energy of all photons in unit volume** Number of photons: $$N = 35\times 10^7$$ So total energy is $$U = N E_1 = (35\times 10^7)(6\times 10^{-19})$$ $$U = 210\times 10^{-12} = 2.1\times 10^{-10}\,\text{J}$$ Since the cube has **unit volume**, this is also the **energy density**: $$u = 2.1\times 10^{-10}\,\text{J/m}^3$$ 3. **Energy density of electromagnetic wave** For an electromagnetic wave, the average total energy density is $$u = \frac{B_0^2}{2\mu_0}$$ where $B_0$ is the amplitude of magnetic field. Thus, $$\frac{B_0^2}{2\mu_0} = 2.1\times 10^{-10}$$ So, $$B_0^2 = 2\mu_0 u$$ Substitute $$\mu_0 = 4\pi\times 10^{-7}$$ Hence, $$B_0^2 = 2(4\pi\times 10^{-7})(2.1\times 10^{-10})$$ $$B_0^2 = 8\pi\times 2.1\times 10^{-17}$$ $$B_0^2 = 16.8\pi\times 10^{-17}$$ Using $\pi = \frac{22}{7}$, $$B_0^2 = 16.8\times \frac{22}{7}\times 10^{-17}$$ $$B_0^2 = 52.8\times 10^{-17} = 5.28\times 10^{-16}$$ Therefore, $$B_0 = \sqrt{5.28\times 10^{-16}}$$ $$B_0 = \sqrt{5.28}\times 10^{-8}$$ Now, $$\sqrt{5.28} \approx 2.3$$ So, $$B_0 \approx 2.3\times 10^{-8}\,\text{T} = 23\times 10^{-9}\,\text{T}$$ Thus, $$\alpha = 23$$ 4. **Comparison with stored answer** Stored correct answer: `21TO25` Our derived answer is $23$, which lies in the given range. So the answer agrees.

View written solutionFree

Correct answer: 21TO25

Given frequency,

u = 10^{15}\,\text{Hz}$$ Energy of one photon is $$E_1 = h\nu = 6\times 10^{-34}\times 10^{15} = 6\times 10^{-19}\,\text{J}$$ 2. **Total energy of all photons in unit volume** Number of photons: $$N = 35\times 10^7$$ So total energy is $$U = N E_1 = (35\times 10^7)(6\times 10^{-19})$$ $$U = 210\times 10^{-12} = 2.1\times 10^{-10}\,\text{J}$$ Since the cube has **unit volume**, this is also the **energy density**: $$u = 2.1\times 10^{-10}\,\text{J/m}^3$$ 3. **Energy density of electromagnetic wave** For an electromagnetic wave, the average total energy density is $$u = \frac{B_0^2}{2\mu_0}$$ where $B_0$ is the amplitude of magnetic field. Thus, $$\frac{B_0^2}{2\mu_0} = 2.1\times 10^{-10}$$ So, $$B_0^2 = 2\mu_0 u$$ Substitute $$\mu_0 = 4\pi\times 10^{-7}$$ Hence, $$B_0^2 = 2(4\pi\times 10^{-7})(2.1\times 10^{-10})$$ $$B_0^2 = 8\pi\times 2.1\times 10^{-17}$$ $$B_0^2 = 16.8\pi\times 10^{-17}$$ Using $\pi = \frac{22}{7}$, $$B_0^2 = 16.8\times \frac{22}{7}\times 10^{-17}$$ $$B_0^2 = 52.8\times 10^{-17} = 5.28\times 10^{-16}$$ Therefore, $$B_0 = \sqrt{5.28\times 10^{-16}}$$ $$B_0 = \sqrt{5.28}\times 10^{-8}$$ Now, $$\sqrt{5.28} \approx 2.3$$ So, $$B_0 \approx 2.3\times 10^{-8}\,\text{T} = 23\times 10^{-9}\,\text{T}$$ Thus, $$\alpha = 23$$ 4. **Comparison with stored answer** Stored correct answer: `21TO25` Our derived answer is $23$, which lies in the given range. So the answer agrees.