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Energy transfer rate of em wave

Short answer

The Poynting vector S = (E×B)/μ₀ gives instantaneous energy flux; for a plane wave S = E²/(μ₀c). Time-averaging yields intensity I = E₀²/(2μ₀c) = E_rms²/(μ₀c), so I ∝ E_rms².

Obtain an expression for energy transfer rate by electromagnetic wave. From your result show that I ∝ E²rms, where I is the intensity of the em wave and Erms is the root mean square value of electric field.

Energy transfer rate (Poynting vector)

Energy flows with the wave. The rate of energy transfer per unit area is the Poynting vector

S = (E × B)/μ₀

For a plane wave E and B are perpendicular, so the magnitude is S = EB/μ₀.

  1. In an EM wave E/B = c, so B = E/c.

  2. Substitute: S = E²/(μ₀ c) (instantaneous energy flow per unit area, W/m²).

  3. For E = E₀ sin(kx − ωt), the instantaneous S = (E₀²/μ₀c) sin²(kx − ωt).

  4. The time average of sin² over a cycle is ½. So the intensity (average S) is

    I = S_avg = E₀²/(2μ₀c)

  5. The rms value of E is E_rms = E₀/√2, so E₀² = 2E_rms². Therefore

I = E_rms² / (μ₀ c)

Since μ₀ and c are constants, I ∝ E_rms².

Equivalent forms: I = ½ε₀cE₀² = ε₀ c E_rms², and I = c × (average energy density), since u = ε₀E² for the wave with u_E = u_B.

Example: for I = 1 kW/m², E_rms = √(μ₀cI) = √(4π×10⁻⁷ × 3×10⁸ × 1000) ≈ 614 V/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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