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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Waves

Calculate the electric field produced by the radiation coming from a $100\: W$ bulb at a distance of $3\: m$. Assume that the efficiency of the bulb is $2.5\ %$ and it is a point source.

$\begin {array} {1 1} (a)\;29 V/m & \quad (b)\;4.07 V/m \\ (c)\;2.9 V/m & \quad (d)\;40.7 V/m \end {array}$


1 Answer

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The bulb, as a point source, radiates light in all directions uniformly. At a distance of $3\: m$, the surface area of the surrounding sphere is
$A = 4 \pi^2 = 4\pi(3)^2 = 113 m^2$
The intensity at this distance is
$I = \large\frac{power}{area}$$ = \large\frac{100 \times 2.5\%}{113}$$ = 0.022 W/m^2$
Half of this intensity is provided by the electric field and half by the magnetic field
$\large\frac{1}{2}$$ I =\large\frac{1}{2}$$ (\in_oE^2_{rms}c)$
$\Rightarrow \large\frac{1}{2}$$ (0.022 W/m^2 ) = \large\frac{1}{2}$$ (\in_oE^2_{rms}c)$
$\Rightarrow E_{rms} = \sqrt{ \bigg[ \large\frac{(0.022)}{(8.85 \times 10^{-12} \times 3 \times 10^8 )} \bigg]}$$ V/m = 2.9 V/m$
Since the electric field is sinusoidal, the peak electric field $E_o$ is
$E_o = \sqrt{2} E_{rms}= \sqrt{2} \times 2.9 V/m = 4.07 V/m $
Ans : (b)
answered Mar 4, 2014 by thanvigandhi_1

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