Question 181: EM waves

Question

The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is $${x \over {10}}\sqrt {{{{\mu _0}c} \over \pi }} {V \over m}$$. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.

Accepted Answer

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,
$$I = \frac{P}{4\pi r^2} 	ag{1}$$ ........(1)
This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :
$$I = \frac{1}{2} c \varepsilon_0 E^2 	ag{2}$$ .........(2)
where $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=$ speed of light in vacuum and $\varepsilon_0$ 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 $E_{	ext{peak}}$ :
$$E_{	ext{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$$
Since $$
\varepsilon_0=\frac{1}{\mu_0 c^2}
$$
Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 	imes 8\, 	ext{W} = 0.8\, 	ext{W}$.
So, substituting $P = 0.8\, 	ext{W}$, $r = 10\, 	ext{m}$, $c = 3 	imes 10^8\, 	ext{m/s}$, and $\mu_0 = 4\pi 	imes 10^{-7}\, 	ext{T m/A}$, we have :
$$E_{	ext{peak}} = \sqrt{\frac{2 	imes 0.8 	imes 4\pi 	imes 10^{-7} 	imes 3 	imes 10^8}{4\pi 	imes 100}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}}$$
Comparing with the expression in the question, we find that $x = 2$.
Therefore, the answer is $x = 2$.

EM waves

The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is ${x \over {10}}\sqrt {{{{\mu _0}c} \over \pi }} {V \over m}$. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.

Integer / Numeric Answer Type