Electromagnetic Waves

JEE Main level, numerical answer, medium

87 of 165

Question 87 of 165

A radiation is emitted by 1000W1000W bulb and it generates an electric field and magnetic field at PP, placed at a distance of 2m2m. The efficiency of the bulb is 1.251.25%. The value of peak electric field at PP is x×10−1Vm−1x\times {10}^{-1}V{m}^{-1}. Value of xx is (Rounded-off to the nearest integer) [Take ϵ0=8.85×10−12C2N−1m−2,c=3×106ms−1{\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2},c=3\times {10}^{6}m{s}^{-1}]

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Answer: 137

Iavg=12ϵ0E02c{I}_{avg}=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}c
1.25100×10004π(2)2=12×8.85×10−12×3×108×E02\dfrac{1.25}{100}\times \dfrac{1000}{4\pi {\left(2\right)}^{2}}=\dfrac{1}{2}\times 8.85\times {10}^{-12}\times 3\times {10}^{8}\times {E}_{0}^{2}
E02=187.4{E}_{0}^{2}=187.4
∴E0=13.689Vm−1∴{E}_{0}=13.689V{m}^{-1}
=136.89×10−1Vm−1=136.89\times {10}^{-1}V{m}^{-1}
∴x=136.89∴x=136.89
Rounding off to nearest integer x=137x=137

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