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QJEEM21Phd3d5: JEE Main 2021 Physics - Class 12 JEEM21Phd3d5 Solution: A radiation is emitted by 1000W bulb and it generates an…

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 m −- 2, c = 3×1083 \times {10^8} ms −- 1]

Step-by-Step Verified Solution

Intensity of electro magnetic wave is,

I=12cε0E02=P4πr2I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}

124πε0×c×E02=Pr2{1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}

12×3×105×E029×109=1000×1.25(2)2×1100{1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}

E02=60×1000×1.254×100=125×32E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}

E02=3752=187.5E_0^2 = {{375} \over 2} = 187.5

E0=13.69{E_0} = 13.69

E0≈137×10−1{E_0} \approx 137 \times {10^{ - 1}} v/m

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