• # question_answer 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: [JEE Main 10-4-2019 Morning] A) $\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)cos(\omega t)$B) $\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)sin(\omega t)$C) $\vec{B}=\frac{{{E}_{0}}}{C}\hat{k}\sin (kz)cos(\omega t)$D) $\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)sin(\omega t)$

Correct Answer: B

Solution :

$\because$$\vec{E}\times \vec{B}||\vec{v}$ Given that wave is propagating along positive z-axis and $\vec{E}$along positive x-axis. Hence $\vec{B}$along y-axis. From Maxwell equation$\vec{\nabla }\times \vec{E}=-\frac{\partial B}{\partial t}$ i.e.$\frac{\partial E}{\partial Z}=-\frac{\partial B}{dt}$and${{B}_{0}}=\frac{{{E}_{0}}}{C}$

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