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Magnetic field in a plane of electromagnetic wave is given by$\overrightarrow{B}={{B}_{\circ }}\sin (kx+\omega t)\widehat{\text{j}}\text{ T}$ Expression for corresponding electric field will be: where c is the speed of lighta)$\overrightarrow{E}={{B}_{\circ }}c\sin (kx+\omega t)\widehat{k}\text{ V/m}$ b)$\overrightarrow{E}=\dfrac{{{B}_{\circ }}}{c}\sin (kx+\omega t)\widehat{k}\text{ V/m}$ c)$\overrightarrow{E}=-{{B}_{\circ }}c\sin (kx+\omega t)\widehat{k}\text{ V/m}$ d)$\overrightarrow{E}={{B}_{\circ }}c\sin (kx-\omega t)\widehat{k}\text{ V/m}$

Last updated date: 18th Sep 2024
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Hint: An electromagnetic wave consists of varying electric and magnetic fields vibrating perpendicular to each other. It is a transverse wave as it oscillates perpendicular to the direction of motion. Because of varying electric and magnetic field, it does not require medium for its propagation
Formula used:
$\dfrac{{{\text{E}}_{\circ }}}{{{\text{B}}_{\circ }}}=c$

Maxwell showed that any variation in electric field would generate magnetic field and vice versa both serving as a source for each other. More precisely hence they do not require any medium to travel. Let us say there is an electromagnetic wave propagating along the x-axis and whose electric field vibrates along z-axis and magnetic field along y-axis. Therefore the electric field vector ($\overrightarrow{\text{E}}$)and the magnetic field vector ($\overrightarrow{\text{B}}$) is given by,
\begin{align} & \overrightarrow{\text{E}}={{E}_{\circ }}\sin \left[ 2\pi \left( \dfrac{x}{\lambda }-\dfrac{t}{T} \right) \right]\widehat{\text{k}}\ \text{ }\And \\ & \overrightarrow{\text{B}}={{B}_{\circ }}\sin \left[ 2\pi \left( \dfrac{x}{\lambda }-\dfrac{t}{T} \right) \right]\widehat{\text{j}} \\ \end{align}
Where ${{E}_{\circ }}$and ${{B}_{\circ }}$ are the amplitudes of the electric and magnetic field,
$\lambda$ is the wavelength of the wave, ‘x’ is a point on the wave at any time ‘t’ and T is the time period. The relation between the amplitudes of the above waves is given by
$\dfrac{{{\text{E}}_{\circ }}}{{{\text{B}}_{\circ }}}=c$
Where ‘c’ is the speed of light. Hence the corresponding value of electric field is, $\overrightarrow{E}={{B}_{\circ }}c\sin (kx+\omega t)\widehat{k}\text{ V/m}$