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\[{{B}_{0}}=\frac{{{E}_{0}}}{C}\]You need to login to perform this action.
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