Electromagnetic Waves

JEE Main level, single correct, medium

69 of 165

Question 69 of 165

An electric bulb is rated as 200W200W. What will be the peak magnetic field at 4m4m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.53.5% efficiency.

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Answer: (B) 1.71×10−8T1.71\times {10}^{-8}T

Intensity of light is I=PowerArea=200×3.51004π(4)2=0.034Wm−2I=\dfrac{Power}{Area}=\dfrac{200\times \dfrac{3.5}{100}}{4\pi {\left(4\right)}^{2}}=0.034W{m}^{-2}
Using the relation between magnetic field and intensity, I=B02c2μ0I=\dfrac{{B}_{0}^{2}c}{2{\mu }_{0}}
We have, magnetic field as B0=I×2μ0c=0.034×2×4π×10−73×108=1.69×10−8T≈1.71×10−8T{B}_{0}=\sqrt{\dfrac{I\times 2{\mu }_{0}}{c}}=\sqrt{\dfrac{0.034\times 2\times 4\pi \times {10}^{-7}}{3\times {10}^{8}}}=1.69\times {10}^{-8}T\approx 1.71\times {10}^{-8}T

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