A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 m2 . At the window, the electric field of the wave has rms value 0.0200 N/C. How much energy does this wave carry through the window during a 30.0-s commercial?

31.8 × 10⁻⁴ J = 3.18 mJ

Explanation:

We know the intensity I of a wave is I = P/A where P = power and A = area = 0.500 m²

The intensity of an electromagnetic wave is also equal to I = E₀²/μ₀c

where E₀ = maximum electric field strength = √2E where E = rms value of electric field = 0.0200 N/C, μ₀ = 4π × 10⁻⁷ H/m ,c = 3 × 10⁸ m/s

P/A = E₀²/μ₀c = 2E²/μ₀c

P = 2E²A/μ₀c = 2 × (0.02 N/C)² × 0.5 m²/(4π × 10⁻⁷ H/m × 3 × 10⁸ m/s)

  = 1.06 × 10⁻⁴ W = 0.106 mW

Since P = E/t where E = Energy and t = time

E = Pt with t = 30 s

E = 1.06 × 10⁻⁴ W × 30 s =  31.8 × 10⁻⁴ J = 3.18 mJ

So the wave carries 3.18 mJ of energy through the window in 30 s

the components of a vector are the parts of the vectors that are to each other.

b perpendicular