An electromagnetic wave is travelling in x-di | vedclass

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(C) The electric field is given by $\vec{E} = E_{0} \sin(kx - \omega t) \hat{j}$.The wave travels in the $+x$-direction,so the direction of propagatio

Vedclass · Accepted Answer

$\vec{B}_{z} = \frac{E_{0}}{C} \sin(kx - \omega t) \hat{k}$

Vedclass · Answer

(C) The electric field is given by $\vec{E} = E_{0} \sin(kx - \omega t) \hat{j}$.The wave travels in the $+x$-direction,so the direction of propagation is $\hat{i}$.The relation between the magnitudes of the electric and magnetic fields is $E_{0} = C B_{0}$,which implies $B_{0} = \frac{E_{0}}{C}$.The direction of the magnetic field $\vec{B}$ is given by the direction of $\vec{c} 	imes \vec{E}$,where $\vec{c}$ is the direction of wave propagation.Here,$\hat{i} 	imes \hat{j} = \hat{k}$.Therefore,the magnetic field vector is $\vec{B}_{z} = \frac{E_{0}}{C} \sin(kx - \omega t) \hat{k}$.

An electromagnetic wave is travelling in $x$-direction with electric field vector given by $\vec{E}_{y} = E_{0} \sin(kx - \omega t) \hat{j}$. The correct expression for the magnetic field vector is:

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