A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x × 10⁻¹ V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take ε ₀ = 8.85 × 10⁻¹² C²N⁻¹ m⁻², c = 3 × 10⁸ ms⁻¹]

Solution

Physics sir · 4 steps

Only 12.5 W leaves as light: spread it over a sphere, then the intensity gives the peak field.

1. Useful power. The bulb radiates only 1.25% of its input as light: P = 0.0125 × 1000 = 12.5 W.
2. Intensity at P. This power spreads evenly over a sphere of radius 2 m, so I = P/(4π r²) = 12.5/16π = 0.2487 W m⁻².

   I = 1/2ε₀ c E₀² ⇒ E₀ = √(2I/(ε₀ c))

3. Put in numbers. ε₀ c = 8.85 × 10⁻¹² × 3 × 10⁸ = 2.655 × 10⁻³. Then E₀² = (2 × 0.2487)/(2.655 × 10⁻³) = 187.3, so E₀ = 13.69 V m⁻¹.
4. So. E₀ = 136.9 × 10⁻¹ V m⁻¹, so x rounds to 137.

The common slip: 97. Using I = ε₀ c E² without the 1/2 gives the rms field, 9.68 V m⁻¹, so x comes out as 97, not 137.

The idea to keep. For a point source, find the radiated power first, divide by 4π r² for intensity, then use I = 1/2ε₀ c E₀². The factor 1/2 is what makes it the peak field.

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