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PhyMedAdv 2025 • Paper 1

Magnetic Field Amplitude of Electromagnetic Wave in Cube

A cube of unit volume contains 35×10735 \times 10^7 photons of frequency 1015 Hz10^{15}\text{ 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×10−9 T\alpha \times 10^{-9}\text{ T}. Taking permeability of free space μ0=4π×10−7 Tm/A\mu_0 = 4\pi \times 10^{-7}\text{ Tm/A}, Planck's constant h=6×10−34Jsh = 6 \times 10^{-34}\text{Js} and π=227\pi = \frac{22}{7}, the value of α\alpha is_____

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Step-by-Step Solution

To find the amplitude of the magnetic field of the electromagnetic wave, we calculate the total energy of the photons and equate it to the average energy stored in the electromagnetic wave within the unit volume.

Step 1: Calculate the total energy of the photons The energy of a single photon is given by: Ep=hνE_p = h\nu

For N=35×107N = 35 \times 10^7 photons of frequency ν=1015 Hz\nu = 10^{15}\text{ Hz}: Etotal=NhνE_{\text{total}} = N h \nu

Substituting the given values (h=6×10−34 J⋅sh = 6 \times 10^{-34}\text{ J}\cdot\text{s}): Etotal=(35×107)×(6×10−34)×(1015)=210×10−12 J=2.1×10−10 JE_{\text{total}} = (35 \times 10^7) \times (6 \times 10^{-34}) \times (10^{15}) = 210 \times 10^{-12}\text{ J} = 2.1 \times 10^{-10}\text{ J}

Step 2: Relate to the average energy density of the electromagnetic wave Since the volume of the cube is V=1 m3V = 1\text{ m}^3, the average energy density uavgu_{\text{avg}} is: uavg=EtotalV=2.1×10−10 J/m3u_{\text{avg}} = \frac{E_{\text{total}}}{V} = 2.1 \times 10^{-10}\text{ J/m}^3

The average energy density of an electromagnetic wave in vacuum in terms of the magnetic field amplitude B0B_0 is: uavg=B022μ0u_{\text{avg}} = \frac{B_0^2}{2\mu_0}

Step 3: Calculate the amplitude of the magnetic field B0B_0 Rearranging the equation for B0B_0: B02=2μ0uavgB_0^2 = 2\mu_0 u_{\text{avg}}

Using μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A} with π=227\pi = \frac{22}{7}: μ0=4×227×10−7 T⋅m/A\mu_0 = 4 \times \frac{22}{7} \times 10^{-7}\text{ T}\cdot\text{m/A}

Now, substituting the values: B02=2×(4×227×10−7)×(2.1×10−10)B_0^2 = 2 \times \left(4 \times \frac{22}{7} \times 10^{-7}\right) \times (2.1 \times 10^{-10}) B02=2×4×22×(210×10−127)×10−7B_0^2 = 2 \times 4 \times 22 \times \left(\frac{210 \times 10^{-12}}{7}\right) \times 10^{-7} B02=8×22×(30×10−12)×10−7B_0^2 = 8 \times 22 \times (30 \times 10^{-12}) \times 10^{-7} B02=5280×10−19=528×10−18 T2B_0^2 = 5280 \times 10^{-19} = 528 \times 10^{-18}\text{ T}^2

Taking the square root: B0=528×10−9 T≈22.98×10−9 TB_0 = \sqrt{528} \times 10^{-9}\text{ T} \approx 22.98\times 10^{-9}\text{ T}

Comparing with B0=α×10−9 TB_0 = \alpha \times 10^{-9}\text{ T}: α≈22.98≈23\alpha \approx 22.98 \approx 23

Magnetic Field Amplitude of Electromagnetic Wave in Cube | Physics PYQ Solution - JEE Challenger