The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is x/10√((μ ₀c)/π) V/m. The bulb is a point source. The value of x is ___________.

Solution

Physics sir · 4 steps

Spread the bulb's power over a sphere to get the intensity, then turn that intensity into the peak field.

1. Intensity at 10 m. A point source sends its power out equally over a sphere of area 4π r². So I = P/(4π r²) = 8/(4π × 100) = 1/50π W m⁻².
2. Link to peak field. The intensity of an EM wave is the average energy density times c. In terms of the peak field E₀, this gives:

   I = 1/2ε₀ E₀² c = E₀²/(2μ₀ c)

3. Solve for E0. E₀² = 2μ₀ c I = (2μ₀ c)/50π = 1/25·(μ₀ c)/π. Taking the square root, E₀ = 1/5√((μ₀ c)/π) = 2/10√((μ₀ c)/π) V m⁻¹.
4. So. Comparing with x/10√((μ₀ c)/π) gives x = 2.

The common slip: 4. Using the area π r² instead of the full sphere 4π r² makes the intensity four times too large and gives x = 4.

The idea to keep. For a point source, I = P/(4π r²). For the peak field use I = E₀²/(2μ₀ c). The factor 2 is there because E₀ is the peak value, not the rms value.

Answer from the NTA final answer key. Solution by PrepWiser.