Question:
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 \(x\times10^{-1}\) V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take \(\varepsilon_0 = 8.85\times10^{-12}\) C2N-1m-2, \(c = 3\times10^{8}\) ms-1]
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 \(x\times10^{-1}\) V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take \(\varepsilon_0 = 8.85\times10^{-12}\) C2N-1m-2, \(c = 3\times10^{8}\) ms-1]
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Intensity at distance r is \(\frac{\eta P}{4\pi r^2}\), and \(I = \frac{1}{2}\varepsilon_0 c E_0^2\).
Updated On: Nov 4, 2026
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Correct Answer: 137
Solution and Explanation
Radiated power \(= 1000\times0.0125 = 12.5\) W.
Intensity at 2 m: \(I = \frac{12.5}{4\pi(2)^2} = \frac{12.5}{16\pi} \approx 0.2487\) W/m2.
\(I = \frac{1}{2}\varepsilon_0 c E_0^2\), so \(E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2\times0.2487}{8.85\times10^{-12}\times3\times10^{8}}} = \sqrt{\frac{0.4974}{2.655\times10^{-3}}} = \sqrt{187.3}\approx 13.69\) V/m.
\(13.69\) V/m \(= 136.9\times10^{-1}\) V/m, so \(x \approx 137\).
Answer: 137
Intensity at 2 m: \(I = \frac{12.5}{4\pi(2)^2} = \frac{12.5}{16\pi} \approx 0.2487\) W/m2.
\(I = \frac{1}{2}\varepsilon_0 c E_0^2\), so \(E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2\times0.2487}{8.85\times10^{-12}\times3\times10^{8}}} = \sqrt{\frac{0.4974}{2.655\times10^{-3}}} = \sqrt{187.3}\approx 13.69\) V/m.
\(13.69\) V/m \(= 136.9\times10^{-1}\) V/m, so \(x \approx 137\).
Answer: 137
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