Question

A radiation is emitted by a $1000W$ 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 the peak electric field at $P$ is $x\times {10}^{-1}V/m$. Value of $x$ is __________. (Rounded-off to the nearest integer)

Answer 1

Step 1: Given data

Power, $P=1000W$

Distance, $r=2m$

Speed of light, $c=3\times {10}^{8}m{s}^{-1}$

The permittivity of free space, ${\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$

Step 2: Find peak electric field

Let $E_0$ be the peak electric field.

Let the Intensity of electromagnetic wave be $I$

The intensity of electromagnetic wave is given by,

$I=\frac{1}{2}c{\epsilon }_0{E}_0^2$

It is also given by

$I=\frac{P}{4{\mathrm{\pi r}}^2}\times$efficiency

On equating both,

$\frac{1}{2}c{\epsilon }_0{E}_0^2=\frac{P}{4{\mathrm{\pi r}}^2}\times$efficiency

$\frac{1}{2}\times 4\pi {\epsilon }_0\times c\times E_0^2=\frac{P}{r^2}\times$efficiency

The value of $4\pi {\epsilon }_0$is,

$$\begin{gathered} 4{\mathrm{\pi \epsilon }}_0=4\times 3.14\times 8.85\times {10}^{-12} \\ 4{\mathrm{\pi \epsilon }}_0=\frac{1}{9\times {10}^9} \end{gathered}$$

Putting values in above equation, we get

$\begin{array}{rcl}\frac{1}{2}\times 3\times {10}^8\times \frac{E_0^2}{9\times {10}^9}& =& \frac{1000\times 1.25}{{\left(2\right)}^2\times 100}\\ E_0^2& =& \frac{60\times 1000\times 1.25}{400}\\ E_0^2& =& 125\times \frac{3}{2}\\ E_0^2& =& \frac{375}{2}\\ E_0& =& 13.69\\ E_0& \approx & 137\times {10}^{-1}V{m}^{-1}\\ x& =& 137\end{array}$

[ $c=3\times {10}^8$ is speed of the light, ${\epsilon }_0$ is permittivity]

Hence, the value of $x$ is $\mathbf{137}$.