The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E = E₀jcos (ω t - kx). The magnetic field B, at the moment t = 0 is: (a) B = (E₀)/(√(μ₀ε₀))cos (kx)j (b) B = (E₀)/(√(μ₀ε₀))cos (kx)k (c) B = E₀√(μ₀ε₀)cos (kx)k (d) B = E₀√(μ₀ε₀)cos (kx)j

Answer: (c). B = E₀/ccos (ω t - kx) k (since i× j = k), and 1c = √(μ₀ε₀). At t = 0: cos (-kx) = cos kx. So the answer is (c) B = E₀√(μ₀ε₀)cos (kx) k.

The electric field of a plane electromagnetic wave propagating along… | JEE Main 2020 (3 Sep, Shift 2)

Q53JEE Main 2020 · 3 SEP · SHIFT 2Medium

Nature of EM Waves & Wave Equation

The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is $\vec{E} = E_0\hat{j}\cos(\omega t - kx)$ .

The magnetic field $\vec{B}$ , at the moment t = 0 is:

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Answer: (c)

$\vec B = \dfrac{E_0}{c}\cos(\omega t - kx)\hat k$ (since $\hat i\times\hat j = \hat k$ ), and $\dfrac1c = \sqrt{\mu_0\varepsilon_0}$ .
At $t = 0$ : $\cos(-kx) = \cos kx$ .
So the answer is (c) $\vec B = E_0\sqrt{\mu_0\varepsilon_0}\cos(kx)\hat k$ .