The electric field in an electromagnetic wave is given by E = 56.5 sin ω(t - x/c) NC⁻¹. Find the intensity of the wave if it is propagating along x-axis in the free space.
(Given : ε₀ = 8.85 × 10⁻¹²C²N⁻¹m⁻²)

Solution

Physics sir · 4 steps

Pick out the amplitude E₀ and put it into I = 1/2ε₀ c E₀². That is the whole job.

1. Read the amplitude. The number in front of the sine is the peak field, so E₀ = 56.5 N C⁻¹. The wave travels in free space, so its speed is c = 3 × 10⁸ m s⁻¹.
2. Intensity formula. Intensity is the average energy density times the speed of the wave. The average energy density is 1/2ε₀ E₀², with electric and magnetic parts sharing it equally.

   I = 1/2ε₀ c E₀²

3. Put in numbers. E₀² = 3192.25 and 1/2 × 8.85 × 10⁻¹² × 3 × 10⁸ = 1.3275 × 10⁻³. Multiplying gives I = 1.3275 × 10⁻³ × 3192.25 ≈ 4.24 W m⁻².
4. So. The intensity of the wave is 4.24 W m⁻², which is option (b).

The idea to keep. For any EM wave, intensity is 1/2ε₀ c E₀², using the peak field E₀. The 1/2 is there because the sine is averaged over time.

Answer from the NTA final answer key. Solution by PrepWiser.