JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 23)
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 ___________.
Answer
2
Explanation
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,
$$I = \frac{P}{4\pi r^2} \tag{1}$$ ........(1)
This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :
$$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$$ .........(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 $E_{\text{peak}}$ :
$$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$$
Since $$ \varepsilon_0=\frac{1}{\mu_0 c^2} $$
Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.
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 $x = 2$.
$$I = \frac{P}{4\pi r^2} \tag{1}$$ ........(1)
This intensity can also be related to the electric field (E) in an electromagnetic wave using the equation :
$$I = \frac{1}{2} c \varepsilon_0 E^2 \tag{2}$$ .........(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 $E_{\text{peak}}$ :
$$E_{\text{peak}} = \sqrt{\frac{2P}{c \varepsilon_0 4\pi r^2}} = \sqrt{\frac{2P \mu_0 c}{4\pi r^2}}$$
Since $$ \varepsilon_0=\frac{1}{\mu_0 c^2} $$
Given the efficiency of the bulb is 10%, the actual power radiated is $0.10 \times 8\, \text{W} = 0.8\, \text{W}$.
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 $x = 2$.
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