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A 27 mW laser beam has a circular cross-sectional area of 10 mm². Taking the permittivity of free space as epsilon₀ = 9 * 10⁻¹² F/m and the speed of light as c = 3 * 10⁸ m/s, find the amplitude (peak value) of the electric field in this electromagnetic wave.

1. 1 kV/m
2. 2 kV/m
3. 1.4 kV/m
4. 0.7 kV/m

Correct answer: 1.4 kV/m

Solution

The intensity gives the time-averaged energy flux. Using I = (1/2) * epsilon₀ * c * E₀², the peak electric field comes out to approximately 1414 V/m ≈ 1.4 kV/m.

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