Question

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 :

• A. $\vec{B} = \frac{E_{0}}{C} \hat{j} \sin\left(kz\right)\cos\left(\omega t\right) $
• B. $ \vec{B} = \frac{E_{0}}{C} \hat{j} \sin \left(kz\right)\sin\left(\omega t\right)$
• C. $ \vec{B} = \frac{E_{0}}{C} \hat{k} \sin \left(kz\right)\cos \left(\omega t\right)$
• D. $\vec{B} = \frac{E_{0}}{C} \hat{j} \cos \left(kz\right)\sin\left(\omega t\right)$

Answer 1

Answer: (A)

$\because \overrightarrow{ E } \times \overrightarrow{ B }|| \overrightarrow{ v }$
Given that wave is propagating along positive $z$ -axis and $\overrightarrow{ E }$ along positive $x$ -axis. Hence $\vec{B}$ along $y$ -axis. From Maxwel equation $\overrightarrow{ V } \times \overrightarrow{ E }=-\frac{\partial B }{\partial t }$
i.e. $\frac{\partial E }{\partial Z }=-\frac{\partial B }{\partial t } \operatorname{andB}_{0}=\frac{ E _{0}}{ C }$
so, $\overrightarrow{ B }=\frac{\overrightarrow{ E _{0}}}{ c } \hat{ j } \sin ( k z ) \cos (\omega t )$