A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10⁻² Am⁻¹ at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given : Permeability of free space μ₀ = 4π × 10⁻⁷ NA⁻², speed of light in vacuum c = 3 × 10⁸ ms⁻¹)

Solution

Physics sir · 4 steps

The medium's impedance links E to H directly: E = ZH, and here the medium's factor is exactly one half.

1. E links to H. In any medium, E = vB and B = μ H. So E = μ v H, and the factor μ v is the wave impedance Z of the medium.

   Z = √(μ/ε) = √(μ₀/ε₀)√(μᵣ/εᵣ) = μ₀ c √(μᵣ/εᵣ)

2. Find Z. μ₀ c = 4π × 10⁻⁷ × 3 × 10⁸ = 120π ≈ 377 Ω. Also 1.61/6.44 = 1/4, so its square root is 1/2. That gives Z = 60π ≈ 188.5 Ω.
3. Put in numbers. E = ZH = 60π × 4.5 × 10⁻² = 2.7π ≈ 8.48 V m⁻¹.
4. So. The electric field is about 8.48 V m⁻¹, which is option (c).

The idea to keep. For E and H in a medium, use E = ZH with Z = 377√(μᵣ/εᵣ) Ω. Look out for a ratio μᵣ/εᵣ that comes out as a neat number, as it does here.

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