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(C) Given,the electric field component of the monochromatic radiation is $\vec E = 2E_0 \hat i \cos kz \cos \omega t$.From Maxwell's equations,the rel

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$\frac{2E_0}{c} \hat j \sin kz \sin \omega t$

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(C) Given,the electric field component of the monochromatic radiation is $\vec E = 2E_0 \hat i \cos kz \cos \omega t$.From Maxwell's equations,the relationship between the electric field $\vec E$ and the magnetic field $\vec B$ is given by $
abla 	imes \vec E = -\frac{\partial \vec B}{\partial t}$.For a wave propagating in the $z$-direction with the electric field in the $x$-direction,the curl simplifies to $\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t}$.Calculating the partial derivative of $E_x$ with respect to $z$:$\frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z} (2E_0 \cos kz \cos \omega t) = -2E_0 k \sin kz \cos \omega t$.Substituting this into the relation: $-2E_0 k \sin kz \cos \omega t = -\frac{\partial B_y}{\partial t}$.Therefore,$\frac{\partial B_y}{\partial t} = 2E_0 k \sin kz \cos \omega t$.Integrating with respect to $t$:$B_y = \int 2E_0 k \sin kz \cos \omega t \, dt = 2E_0 k \sin kz \left( \frac{\sin \omega t}{\omega} ight) = \frac{2E_0 k}{\omega} \sin kz \sin \omega t$.Since $c = \frac{\omega}{k}$,we have $\frac{k}{\omega} = \frac{1}{c}$.Thus,$B_y = \frac{2E_0}{c} \sin kz \sin \omega t$.Since the direction of propagation is along the $z$-axis and the electric field is along the $x$-axis,the magnetic field must be along the $y$-axis $(\hat j)$.Hence,$\vec B = \frac{2E_0}{c} \hat j \sin kz \sin \omega t$.

The electric field component of a monochromatic radiation is given by
$\vec E = 2E_0 \hat i \cos kz \cos \omega t$
Its magnetic field $\vec B$ is then given by

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The electric field component of a monochromatic radiation is given by$\vec E = 2E_0 \hat i \cos kz \cos \omega t$Its magnetic field $\vec B$ is then given by