Question

A point source of electromagnetic radiation has an average power output of $800\;W$ . The maximum value of the electric field at a distance $4.0\;m$ from the source is
(A) $\dfrac{54.77V}{m}$
(B) $\dfrac{44.7V}{m}$
(C) $\dfrac{55V}{2m}$
(D) None of these

Answer 1

Hint : The electromagnetic radiation from a point source at some distance creates a spherical surface. The maximum value of the electric field can be found using the formula for the average intensity ${{I}_{avg}}=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}c$ . Also, the average intensity can be defined as the average power output per unit area.

Complete Step By Step Answer:
Let us note down the given data;
The average power output of the wave $P=800W$
The distance at which the maximum electric field is to be found $d=4m$
Now, we know that an electromagnetic wave originating from a point source moves forward in a spherical pattern.
Hence, the surface of the electric field at the given distance is spherical. Hence the given distance from the point source can be taken as the radius of the spherical surface.
The radius of the spherical surface $r=4m$
$\therefore$ The surface area of the spherical surface $A=4\pi {{r}^{2}}$
Substituting the value of the radius,
$\therefore A=4\pi {{(4)}^{2}}$
$\therefore A=64\pi$
Now, we know that the average intensity is defined as the average output power per unit area which can be expressed as,
${{I}_{avg}}=\dfrac{P}{A}$
Substituting the given values,
$\therefore {{I}_{avg}}=\dfrac{800}{64\pi }$ …… $(1)$
The average intensity in terms of the electric field can be expressed as,
${{I}_{avg}}=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}_{0}}^{2}c$
Where, ${{\varepsilon }_{0}}$ = Permittivity of free space = $8.85\times {{10}^{-12}}C{{N}^{-1}}{{m}^{-2}}$
$c$ = Velocity of the electromagnetic wave in vacuum = $3\times {{10}^{8}}$
Substituting the values from the equation $(1)$ and the given values,
$\therefore \dfrac{800}{64\pi }=\dfrac{1}{2}\left( 8.85\times {{10}^{-12}} \right){{E}_{0}}^{2}\left( 3\times {{10}^{8}} \right)$
To find the value of the maximum electric field,
${{E}_{0}}=\dfrac{2\times 800}{64\pi \times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}}$
$\therefore {{E}_{0}}=54.77V{{m}^{-1}}$
Hence, the correct answer is Option $(A)$ .

Note :
The electromagnetic wave, as its name suggests has two components namely an electric field component and a magnetic field component which are perpendicular to each other. The intensity of the EM wave is distributed equally between the electric field and magnetic field. Hence, the maximum value of the magnetic field can be found by the similar method by ${{I}_{avg}}=\dfrac{1}{2}{{\varepsilon }_{0}}{{B}_{0}}^{2}c$ .