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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.

Asked in JEE Main 24th June 2nd Shift 2022 · Point source, efficiency and power on an area

Answer: (2) 1.71 × 10⁻⁸ T

Step-by-step solution

Radiated power P = 0.035 × 200 = 7 W

I = P/4π r² = 7/(4π × 16) = 3.48 × 10⁻² W m⁻²

I = 1/2ε₀cE₀² ⇒ E₀ = √2I/ε₀c = (2 × 3.48 × 10⁻²/2.655 × 10⁻³) = 5.12 V m⁻¹

B₀ = E₀/c = 5.12/(3 × 10⁸) = 1.71 × 10⁻⁸ T

Why the other options are wrong

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