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  • What is the electric field produced by radiation coming from a  bub at a distance of <i

What is the electric field produced by radiation coming from a 100 \mathrm{W} bub at a distance of 3 \mathrm{~m}. (Assume that the efficiency of bulb is 2.5 \% ).

Option: 1

1.4 \ \mathrm{v} / \mathrm{m}


Option: 2

1.9 \ \mathrm{v} / \mathrm{m}


Option: 3

2.4 \ \mathrm{v} / \mathrm{m}


Option: 4

2.9 \ \mathrm{~v} / \mathrm{m}


Answers (1)

best_answer

Intensity at a distance r from a point source is I=\frac{p}{4 \pi r^2}.
Since efficiency at 3 \mathrm{~m} is 2.5 \%

I=\frac{100}{4 \pi \times 3^2} \times \frac{2.5}{100}=0.0220 \mathrm{~w} / \mathrm{m}^2
Intensity is equally divided between the electric field & magnetic field

\begin{aligned} & \Rightarrow I_E=\frac{I}{2}=0.011 \mathrm{~W} / \mathrm{m}^2 .=\frac{1}{2} \epsilon_0 E^2 C . \\ & \Rightarrow E=\sqrt{\frac{2 I_E}{\epsilon_0 C}}=\sqrt{\frac{2 \times 0.011}{8.85 \times 10^{-12} \times 3 \times 10^8}}=2.9 \mathrm{~v} / \mathrm{m} . \end{aligned}

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Rakesh

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