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JEE Main 2021 (26 Feb, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2021 (26 Feb, Shift 1)

Q23JEE Main 2021 · 26 FEB · SHIFT 1Medium

Energy Density & Intensity

A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x ×\times 10−-1 V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take ε0=8.85×10−12\varepsilon_0 = 8.85 \times 10^{-12} C2N−1{}^{-1} m−2{}^{-2}, c = 3×1083 \times 10^{8} ms−1{}^{-1}]

Numerical answer type. Work it out, then open the answer.

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Answer: 137

  1. Radiated power P=0.0125×1000=12.5 WP = 0.0125 \times 1000 = 12.5\ \text{W}
  2. I=12.54π×4≈0.249 W/m2I = \dfrac{12.5}{4\pi \times 4} \approx 0.249\ \text{W/m}^2
  3. E0=2Iε0c=0.4978.85×10−12×3×108=187E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{0.497}{8.85 \times 10^{-12} \times 3 \times 10^8}} = \sqrt{187}
  4. E0≈13.7 V/m=137×10−1 V/mE_0 \approx 13.7\ \text{V/m} = 137 \times 10^{-1}\ \text{V/m}, so x=137x = 137.
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