The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?

Given,
                $\mathrm{Amplitude} \mathrm{of} \mathrm{magnetic} \mathrm{field}, {\mathrm{B}}_0 = 510 \mathrm{nT} = 510 \times {10}^{-9}\mathrm{T}$
Speed of EM wave, $\mathrm{C} = 3 \times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}$ 

For electromagnetic waves,
                     $\boxed{\mathrm{C} = \frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}}$ 

i.e.,               ${\mathrm{E}}_0 = \mathrm{C} {\mathrm{B}}_0$ 

                        $$\begin{gathered} = 3 \times {10}^{8 }\times 510 \times {10}^{-9} \\ = 153 {\mathrm{NC}}^{-1} \end{gathered}$$ 

which is the required amplitude of electric field.        

Here,          $$\begin{gathered} \mathrm{Freque}n\mathrm{cy} \mathrm{of} \mathrm{the} \mathrm{plane} \mathrm{EM} \mathrm{wave}, \mathrm{v} = 2 \times {10}^{10}\mathrm{Hz} \\ \mathrm{Amplitude} \mathrm{of} \mathrm{electric} \mathrm{field}, {\mathrm{E}}_0 = 48 {\mathrm{Vm}}^{-1} \end{gathered}$$  

Therefore, 

Wavelength of the wave, 

$\mathrm{\lambda } = \frac{\mathrm{c}}{\mathrm{v}} =\frac{3 \times {10}^8}{2 \times {10}^{10}}\mathrm{m} = 1.5 \times {10}^{-2}\mathrm{m}$ 

Amplitude of oscillating magnetic field,

${\mathrm{B}}_0 = \frac{{\mathrm{E}}_0}{\mathrm{c}} = \frac{48}{3 \times {10}^8}\mathrm{T} = 1.6 \times {10}^{-7}\mathrm{T}$ 

Energy density in electric field, 

         $\boxed{{\mathrm{\mu }}_{\mathrm{E}} = \frac{1}{2} {\mathrm{\epsilon }}_0 {\mathrm{E}}^2}$

Energy density in magnetic field,
         $\boxed{{\mathrm{\mu }}_{\mathrm{B}} = \frac{1}{2{\mathrm{\mu }}_0}{\mathrm{B}}^2}$ 

Using the relation, we have 

               $$\begin{gathered} \boxed{\mathrm{E} = \mathrm{cB}}, \\ {\mathrm{u}}_{\mathrm{E}} = \frac{1}{2} {\mathrm{\epsilon }}_0 (\mathrm{cB}{)}^2 \\ = {\mathrm{c}}^{2 }\left(\frac{1}{2} {\mathrm{\epsilon }}_{0 }{\mathrm{B}}^2\right) \end{gathered}$$ 

But,           $\mathrm{c} = \frac{1}{\sqrt{{\mathrm{\mu }}_0 {\mathrm{\epsilon }}_0}}$ 

$\therefore$ ${\mathrm{\mu }}_{\mathrm{E}} = \frac{1}{{\mathrm{\mu }}_0 {\mathrm{\epsilon }}_0}\left(\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{B}}^2\right) = \frac{1}{2{\mathrm{\mu }}_0}{\mathrm{B}}^2 = {\mathrm{\mu }}_{\mathrm{B}}$ 
Hence, the average energy density of electric field equals the average energy density of magnetic field.

(a) (i)        

or              

                      

(ii)    

(iii)    

                
(iv)     
(b) Let the electromagnetic wave travel along +x-axis, and  are along y-axis and z-axis respectivelty. Then,