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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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)

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...