The electric field of a plane electromagnetic wave is given by E = E₀ i cos (kz) cos (ω t) The corresponding magnetic field B is then given by (a) B = E₀/C j sin (kz) sin (ω t) (b) B = E₀/C j sin (kz) cos (ω t) (c) B = E₀/C j cos (kz) sin (ω t) (d) B = E₀/C k sin (kz) cos (ω t)

Answer: (a). This is a standing wave; use × E = -(∂ B)/(∂ t). ×(Ex i) = (∂ Ex)/(∂ z) j = -kE₀sin kzcos ω t j (∂ By)/(∂ t) = kE₀sin kzcos ω t ⇒ By = (kE₀)/(ω)sin kzsin ω t = E₀/csin kzsin ω t So the answer is (a) B = E₀/c jsin (kz)sin (ω t).

The electric field of a plane electromagnetic wave is given by E = E₀… | JEE Main 2019 (10 Apr, Shift 1)

Q66JEE Main 2019 · 10 APR · SHIFT 1Medium

Nature of EM Waves & Wave Equation

The electric field of a plane electromagnetic wave is given by

$\vec{E} = E_0 \hat{i} \cos(kz) \cos(\omega t)$

The corresponding magnetic field $\vec{B}$ is then given by

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Answer: (a)

This is a standing wave; use $\nabla\times\vec E = -\dfrac{\partial\vec B}{\partial t}$ .
$\nabla\times(E_x\hat i) = \dfrac{\partial E_x}{\partial z}\hat j = -kE_0\sin kz\cos\omega t\,\hat j$
$\dfrac{\partial B_y}{\partial t} = kE_0\sin kz\cos\omega t \Rightarrow B_y = \dfrac{kE_0}{\omega}\sin kz\sin\omega t = \dfrac{E_0}{c}\sin kz\sin\omega t$
So the answer is (a) $\vec B = \dfrac{E_0}{c}\hat j\sin(kz)\sin(\omega t)$ .

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Nature of EM Waves & Wave Equation question from JEE Main 2019 (10 Apr, Shift 1)

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