The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is x/10 √((μ₀ c)/(π)) V/m. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 2. I = 1/2ε₀ c E₀² and (1)/(ε₀ c) = μ₀ c, so E₀² = 2μ₀ c I With 10% efficiency: P = 0.8 W, I = (0.8)/(4π × 100) = (1)/(500π), giving E₀ = √(1/250)√((μ₀ c)/(π)) ≈ 0.063√((μ₀ c)/(π)) The official answer uses the full 8 W: I = (8)/(400π) = (0.02)/(π) E₀² = 2μ₀ c × (0.02)/(π) = 0.04 (μ₀ c)/(π) ⇒ E₀ = 2/10√((μ₀ c)/(π)) So x = 2 (official key).

The peak electric field produced by the radiation coming from the 8W… | JEE Main 2021 (25 Feb, Shift 2)

Q24JEE Main 2021 · 25 FEB · SHIFT 2Medium

Energy Density & Intensity

The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is $\frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}} \frac{V}{m}$ . The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

$I = \tfrac{1}{2}\varepsilon_0 c E_0^2$ and $\dfrac{1}{\varepsilon_0 c} = \mu_0 c$ , so $E_0^2 = 2\mu_0 c\, I$
With 10% efficiency: $P = 0.8\ \text{W}$ , $I = \dfrac{0.8}{4\pi \times 100} = \dfrac{1}{500\pi}$ , giving $E_0 = \sqrt{\dfrac{1}{250}}\sqrt{\dfrac{\mu_0 c}{\pi}} \approx 0.063\sqrt{\dfrac{\mu_0 c}{\pi}}$
The official answer uses the full $8\ \text{W}$ : $I = \dfrac{8}{400\pi} = \dfrac{0.02}{\pi}$
$E_0^2 = 2\mu_0 c \times \dfrac{0.02}{\pi} = 0.04\,\dfrac{\mu_0 c}{\pi} \Rightarrow E_0 = \dfrac{2}{10}\sqrt{\dfrac{\mu_0 c}{\pi}}$
So $x = 2$ (official key).

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