An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

Solution

Physics sir · 4 steps

Only 3.5% of the 200 W is light. Spread it over a sphere, then use intensity to find the peak field.

1. Radiated power. Only the useful fraction comes out as electromagnetic radiation: P = 0.035 × 200 = 7 W.
2. Intensity at 4 m. A point source spreads this power evenly over a sphere of area 4π r². So I = 7/(4π × 4²) = 7/201.1 = 3.48 × 10⁻² W m⁻².

   I = (c B₀²)/2μ₀ ⇒ B₀ = √((2μ₀ I)/c)

3. Put in numbers. B₀² = (2 × 4π × 10⁻⁷ × 3.48 × 10⁻²)/(3 × 10⁸) = 2.92 × 10⁻¹⁶ T², so B₀ = 1.71 × 10⁻⁸ T.
4. So. The peak magnetic field at 4 m is 1.71 × 10⁻⁸ T, option (b).

The idea to keep. For any point source: take the radiated power, divide by 4π r² to get intensity, then use I = cB₀²/2μ₀. The factor 2 is there because B₀ is the peak value.

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