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Rms Fields And Intensity

Hardphysics

A plane electromagnetic wave in vacuum has a peak electric field of 30 V/m, and an analyst computes its average intensity using free-space constants.

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About This Question

Subject
physics
Chapter
electromagnetic waves
Topic
rms fields and intensity
Difficulty
Hard
Year
2025
Tags
average intensityhalf epsilon c E squaredtime averagingpeak electric fieldplane wave power

Solution

Correct Answer:

Extending NCERT Class 12, Chapter 8 (Electromagnetic Waves), the average intensity of a plane electromagnetic wave is I = ½ ε₀ c E₀², where the factor of one half comes from time-averaging the squared sinusoidal field over a cycle, since the average of sin² over a full period is exactly one half. The intensity represents the mean power carried per unit area by the combined electric and magnetic fields, and it can equally be written as the product of the average total energy density and the speed of light. Substituting ε₀ = 8.85 × 10⁻¹² C²/N·m², c = 3 × 10⁸ m/s, and E₀ = 30 V/m gives I = ½ × 8.85 × 10⁻¹² × 3 × 10⁸ × 900 = 1.19 W/m². The value 2.39 W/m² is double the correct result because it omits the one-half averaging factor. The value 0.60 W/m² is roughly half of the true value, arising from an extra erroneous factor of one half. The value 4.78 W/m² overcounts by a factor of four, likely from mishandling both the averaging factor and the square of the field. A magnitude check confirms that a modest 30 V/m field yields an intensity of about one watt per square metre, consistent with the correct answer.

This hard difficulty physics question is from the chapter electromagnetic waves, covering the topic of rms fields and intensity. It appeared in the 2025 exam.

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