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# The electric field of a plane electromagnetic wave is given by $\vec y = {E_0}\hat i\cos (kz)(\omega t)$. The corresponding magnetic field $\vec B$ is then given by:A.)$\vec B = \dfrac{{{E_0}}}{c}\hat i\sin (kz)\cos (\omega t)$B.)$\vec B = \dfrac{{{E_0}}}{c}\hat j\sin (kz)\cos (\omega t)$C.)$\vec B = \dfrac{{{E_0}}}{c}\hat k\sin (kz)\cos (\omega t)$D.)$\vec B = \dfrac{{{E_0}}}{c}\hat j\cos (kz)\sin (\omega t)$

Last updated date: 17th Sep 2024
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Hint – You can start by defining what an electromagnetic wave is and what its properties are. Then use the Maxwell equation for electromagnetic wave, i.e. $\vec \nabla \times \vec E = - \dfrac{{\partial B}}{{\partial t}}$ to calculate the magnetic field ($\vec B$).

In the problem given we are given the following equation
$\vec y = {E_0}\hat i\cos (kz)(\omega t)$
As you can see that the wave is travelling in the z-axis and the vibrations of the electric field ($\vec E$) are along the positive x-axis. Hence the vibrations in the magnetic field ($\vec B$) will be along the y-axis.
Using Maxwell’s equation, we know

$\vec \nabla \times \vec E = - \dfrac{{\partial B}}{{\partial t}}$
$\dfrac{{\partial E}}{{\partial Z}} = - \dfrac{{\partial B}}{{\partial t}}$
And ${B_0} = \dfrac{{{E_0}}}{c}$
So, $\vec B = \dfrac{{{{\vec E}_0}}}{c}\hat j\sin (kz)\cos (wt)$