Home Solver Electromagnetic wave — intensity and radiation pressure Physics 2 · Exercise 9 of 13
EM Waves Medium
Electromagnetic wave — intensity and radiation pressure A laser emits an EM wave with electric field amplitude E₀ = 500 V/m .\nCalculate: (a) average intensity, (b) amplitude B₀, (c) radiation pressure on an absorbing surface, (d) force on a mirror of area A = 1 cm² (total reflection).
Given data
E₀ = 500 V/m c = 3×10⁸ m/s ε₀ = 8.85×10⁻¹² F/m A = 10⁻⁴ m² Review the theory: Onde EM 1 Calculate the average intensity I = cε₀E₀²/2 (in W/m²).
2 Calculate the magnetic field amplitude B₀ (in μT).
3 Calculate the radiation pressure P_rad on an absorbing surface (in μPa).
4 Calculate the force F on the mirror of A = 1 cm² with total reflection (in pN).
Calculate the average intensity I = cε₀E₀²/2 (in W/m²).
I = 1 2 c ε 0 E 0 2 = 1 2 × 3 × 10 8 × 8.85 × 10 − 12 × 500 2 = 1 2 × 3 × 10 8 × 8.85 × 10 − 12 × 2.5 × 10 5 = 1 2 × 663.75 ≈ 331.9 W / m 2 I = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2} \times 3\times10^8 \times 8.85\times10^{-12} \times 500^2 = \frac{1}{2} \times 3\times10^8 \times 8.85\times10^{-12} \times 2.5\times10^5 = \frac{1}{2} \times 663.75 \approx \mathbf{331.9\,W/m^2} I = 2 1 c ε 0 E 0 2 = 2 1 × 3 × 1 0 8 × 8.85 × 1 0 − 12 × 50 0 2 = 2 1 × 3 × 1 0 8 × 8.85 × 1 0 − 12 × 2.5 × 1 0 5 = 2 1 × 663.75 ≈ 331.9 W/ m 2 .
Calculate the magnetic field amplitude B₀ (in μT).
B 0 = E 0 c = 500 3 × 10 8 ≈ 1.667 × 10 − 6 T = 1.667 μ T B_0 = \frac{E_0}{c} = \frac{500}{3\times10^8} \approx \mathbf{1.667\times10^{-6}\,T} = \mathbf{1.667\,\mu T} B 0 = c E 0 = 3 × 1 0 8 500 ≈ 1.667 × 1 0 − 6 T = 1.667 μ T . In an electromagnetic wave, the electric and magnetic field amplitudes are related by
E 0 / B 0 = c E_0/B_0 = c E 0 / B 0 = c .
Calculate the radiation pressure P_rad on an absorbing surface (in μPa).
P r a d = I c = 331.9 3 × 10 8 ≈ 1.106 × 10 − 6 P a = 1.11 μ P a P_{\mathrm{rad}} = \frac{I}{c} = \frac{331.9}{3\times10^8} \approx \mathbf{1.106\times10^{-6}\,Pa} = \mathbf{1.11\,\mu Pa} P rad = c I = 3 × 1 0 8 331.9 ≈ 1.106 × 1 0 − 6 Pa = 1.11 μ Pa . The radiation pressure on an absorbing surface is the intensity divided by the speed of light.
Calculate the force F on the mirror of A = 1 cm² with total reflection (in pN).
F = 2 I c × A = 2 × 331.9 3 × 10 8 × 10 − 4 = 663.8 3 × 10 8 × 10 − 4 = 2.213 × 10 − 6 × 10 − 4 ≈ 2.213 × 10 − 10 N = 221 p N F = \frac{2I}{c} \times A = \frac{2 \times 331.9}{3\times10^8} \times 10^{-4} = \frac{663.8}{3\times10^8} \times 10^{-4} = 2.213\times10^{-6} \times 10^{-4} \approx \mathbf{2.213\times10^{-10}\,N} = \mathbf{221\,pN} F = c 2 I × A = 3 × 1 0 8 2 × 331.9 × 1 0 − 4 = 3 × 1 0 8 663.8 × 1 0 − 4 = 2.213 × 1 0 − 6 × 1 0 − 4 ≈ 2.213 × 1 0 − 10 N = 221 pN . For total reflection, the momentum transfer is doubled.
Result: I = 331.9 W/m² — B₀ = 1.67 μT — P_rad = 1.11 μPa — F_mirror = 221 pN .
