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Question:

The electric field component of a monochromatic radiation is given by ⃗E=2E₀^icos(kz)cos(ωt). Its magnetic field ⃗B is then given by:

2E₀c^jsin(kz)sin(ωt)

2E₀c^jsin(kz)cos(ωt)

−E₀c^jsin(kz)sin(ωt)

2E₀c^jcos(kz)cos(ωt)

Solution:

⃗E=2E₀^icos(kz)cos(ωt)
⃗E is perpendicular to ⃗B
We know, |B|=E/c
∴B=2E₀/c^jcos(kz)cos(ωt)