JEE Main 2021 — Electromagnetic Waves Question with Solution

Firstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\pi r^2$ where r is the distance from the source. So,

$$I=\frac{P}{4\pi r^2}\text{\hspace{1em}(1)}$$ ........(1)

This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :

$$I=\frac{1}{2}c{\epsilon }_0{E}^2\text{\hspace{1em}(2)}$$ .........(2)

where $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=$ speed of light in vacuum and ${\epsilon }_0$ is the permittivity of free space.

From equations (1) and (2), we can solve for E and square root it to find the peak value of the electric field $E_{\text{peak}}$ :

$$E_{\text{peak}}=\sqrt{\frac{2P}{c{\epsilon }_{0}4\pi r^2}}=\sqrt{\frac{2P{\mu }_{0}c}{4\pi r^2}}$$

Since $${\epsilon }_0=\frac{1}{{\mu }_0{c}^2}$$

Given the efficiency of the bulb is 10%, the actual power radiated is $0.10\times 8\text{W}=0.8\text{W}$.

So, substituting $P=0.8\text{W}$, $r=10\text{m}$, $c=3\times 10^8\text{m/s}$, and ${\mu }_0=4\pi \times 10^{-7}\text{T m/A}$, we have :

$$E_{\text{peak}}=\sqrt{\frac{2\times 0.8\times 4\pi \times 10^{-7}\times 3\times 10^8}{4\pi \times 100}}=\frac{x}{10}\sqrt{\frac{{\mu }_{0}c}{\pi }}$$

Comparing with the expression in the question, we find that $x=2$.

Therefore, the answer is $x=2$.