The electric field component of a monochromatic radiation is given by E = 2 E0 i cos kz cos ω t Its magnetic field B is then given by: (a) 2E₀/c j sin kz cos ω t (b) -2E₀/c j sin kz sin ω t (c) 2E₀/c j sin kz sin ω t (d) 2E₀/c j cos kz cos ω t

Answer: (c). Faraday's law: × E = -(∂ B)/(∂ t); for E = Ex(z,t)i, ( × E)y = (∂ Ex)/(∂ z). (∂ Ex)/(∂ z) = -2E₀ k sin kz cos ω t, so (∂ By)/(∂ t) = 2E₀ k sin kz cos ω t. Integrate: By = (2E₀ k)/(ω) sin kz sin ω t = 2E₀/c sin kz sin ω t. So the answer is (c) 2E₀/cj sin kz sin ω t.

The electric field component of a monochromatic radiation is given by… | JEE Main 2017 (9 Apr, Shift 1)

Q76JEE Main 2017 · 9 APR · SHIFT 1Difficult

Nature of EM Waves & Wave Equation

The electric field component of a monochromatic radiation is given by

$\vec{E}$ = 2 E0 $\hat{i}$ cos kz cos $\omega$ t

Its magnetic field $\vec{B}$ is then given by:

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Answer: (c)

Faraday's law: $\nabla \times \vec{E} = -\dfrac{\partial \vec{B}}{\partial t}$ ; for $\vec{E} = E_x(z,t)\hat{i}$ , $(\nabla \times \vec{E})_y = \dfrac{\partial E_x}{\partial z}$ .
$\dfrac{\partial E_x}{\partial z} = -2E_0 k \sin kz \cos \omega t$ , so $\dfrac{\partial B_y}{\partial t} = 2E_0 k \sin kz \cos \omega t$ .
Integrate: $B_y = \dfrac{2E_0 k}{\omega} \sin kz \sin \omega t = \dfrac{2E_0}{c} \sin kz \sin \omega t$ .
So the answer is (c) $\dfrac{2E_0}{c}\hat{j} \sin kz \sin \omega t$ .

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Nature of EM Waves & Wave Equation question from JEE Main 2017 (9 Apr, Shift 1)

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