The relation between electric field E and magnetic field H in electromagnetic wave is:
A. \[E = H\]
B. \[E = \dfrac{{{\mu _0}}}{{{\varepsilon _0}}}H\]
C. \[E = \sqrt {\dfrac{{{\mu _0}}}{{{\varepsilon _0}}}} H\]
D. \[E = \sqrt {\dfrac{{{\varepsilon _0}}}{{{\mu _0}}}} H\]

Hint:Recall Maxwell's equation and calculate the curl of the electric field.Express the solutions of differential equations of electric field and magnetic field and substitute it into the Maxwell’s equation you obtained after taking the curl. We know that speed of light is expressed as, \[c = \dfrac{\omega }{k} = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}\].

Complete step by step answer:
To answer this question, we can derive the relation between electric field and magnetic field of the electromagnetic wave using Maxwell’s equation. We will start with one of Maxwell’s equation,
\[\nabla \times E = - {\mu _0}\dfrac{{\partial B}}{{\partial t}}\] …… (1)
We assume the electric field is along the y-axis and magnetic field is along the z-axis. Since electric field and magnetic field are only function of distance x and time x, we can write the equation for electric field and magnetic field as follows,
\[\vec E\left( {x,t} \right) = E\left( {x,t} \right)\hat j\] and, \[\vec B\left( {x,t} \right) = B\left( {x,t} \right)\hat k\] …… (2)

We take the curl of electric field as follows,
\[\nabla \times \vec E\left( {x,t} \right) = \left[ {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
{\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\
0&{E\left( {x,t} \right)}&0
\end{array}} \right]\]
\[ \Rightarrow \nabla \times \vec E\left( {x,t} \right) = - \dfrac{{\partial E}}{{\partial x}}\hat k\] …… (3)
From equation (1) and (2), we can write,
\[\dfrac{{\partial E}}{{\partial x}} = - {\mu _0}\dfrac{{\partial B}}{{\partial t}}\] …… (4)

We have the solutions of differential equations of electric field and magnetic field is,
\[E\left( {x,t} \right) = {E_{\max }}\cos \left( {kx - \omega t} \right)\] …… (5)
And,
\[B\left( {x,t} \right) = {B_{\max }}\cos \left( {kx - \omega t} \right)\] …… (6)
Substituting equation (5) and (6) in equation (4), we have,
\[\dfrac{\partial }{{\partial x}}\left( {{E_{\max }}\cos \left( {kx - \omega t} \right)} \right) = - {\mu _0}\dfrac{\partial }{{\partial t}}\left( {{B_{\max }}\cos \left( {kx - \omega t} \right)} \right)\]
\[ \Rightarrow {E_{\max }}\cos \left( {kx - \omega t} \right)\left( k \right) = - {\mu _0}{B_{\max }}\cos \left( {kx - \omega t} \right)\left( { - \omega } \right)\]
\[ \Rightarrow k{E_{\max }} = \omega {\mu _0}{B_{\max }}\]
\[ \Rightarrow {E_{\max }} = \dfrac{\omega }{k}{\mu _0}{B_{\max }}\]
We know that, the speed of light is expressed as,
\[c = \dfrac{\omega }{k} = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}\]
Therefore, the above equation becomes,
\[E = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}{\mu _0}B\]
\[ \therefore E = \sqrt {\dfrac{{{\mu _0}}}{{{\varepsilon _0}}}} B\]

So, the correct answer is option C.

Note:We can also answer this question by referring to the Poynting vector or energy transferred by the electromagnetic wave. The energy density of the electromagnetic wave is,
\[S = \dfrac{{{\mu _0}B_{\max }^2}}{2} = \dfrac{{{\varepsilon _0}E_{\max }^2}}{2}\]
\[ \Rightarrow \dfrac{{{E_{\max }}}}{{{B_{\max }}}} = \sqrt {\dfrac{{{\mu _0}}}{{{\varepsilon _0}}}} \]
Electric fields and magnetic fields are perpendicular to each other only for electromagnetic waves.