If Electric field intensity of a uniform plane electromagnetic wave is given as E = - 301.6 sin (kz - ω t)aₓ + 452.4 sin (kz - ω t)a_yV/m. Then magnetic intensity 'H' of this wave in Am⁻¹ will be :
[Given : Speed of light in vacuum c = 3 × 10⁸ ms⁻¹, Permeability of vacuum μ ₀ = 4π × 10⁻⁷ NA⁻²]

Solution

Physics sir · 4 steps

Divide E by the impedance of free space, then let k̂ × E⃗ fix the direction of each part.

1. Size of H. In vacuum, |E|/|H| equals the impedance of free space, η₀ = μ₀ c = 4π × 10⁻⁷ × 3 × 10⁸ = 120π ≈ 377 Ω. So 301.6/377 = 0.8 and 452.4/377 = 1.2, both in A m⁻¹.
2. Direction of travel. The phase (kz - ω t) means the wave moves along +z, so k̂ = a_z. Then H⃗ follows from the rule below.

   H⃗ = 1/η₀ a_z × E⃗, a_z × aₓ = a_y, a_z × a_y = -aₓ

3. Each component. The -301.6 aₓ part of E⃗ gives -0.8 a_y in H⃗. The +452.4 a_y part gives -1.2 aₓ.
4. So. H⃗ = -0.8 sin(kz - ω t) a_y - 1.2 sin(kz - ω t) aₓ A m⁻¹, which is option (c).

The idea to keep. For H, divide E by μ₀ c ≈ 377 Ω. For B, divide E by c. Then E⃗, H⃗ and the travel direction must form a right-handed set.

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