91Ó°ÊÓ

The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude E amplitude of magnetic field B and it is given in the form $\mathbf{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{cE}}_{\mathbf{max}}^{\mathbf{2}}$Also, the intensity I is proportional to the incident power P by area A.$\mathbf{l}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{VI}}{\mathbf{4}{\mathbf{Ï„Ï„°ù}}^{\mathbf{2}}}$

The intensity I is proportional to the incident power P by area A and given by$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$ Where P= VI, the voltage times the current. Both equations (1) and (2) have the same left side, so we can get the

next expression of ${\mathrm{E}}_{\max }$as

$$\begin{gathered} \frac{\mathrm{VI}}{\mathrm{A}}=\frac{1}{2}{\mathrm{Î\mu }}_0{\mathrm{cE}}_{\max }^2 \\ {\mathrm{E}}_{\max }=\sqrt{\frac{2\mathrm{VI}}{{\mathrm{´¡Î\mu }}_0\mathrm{c}}} \\ =\sqrt{\frac{2\left(5×{10}^5\right)\left(1000\right)}{\left(100\right)\left(8.85×{10}^{-12}\right)\left(3×{10}^8\right)}} \\ =6.14×{10}^4\mathrm{V}/\mathrm{m} \end{gathered}$$

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by ${\mathrm{B}}_{\max }=\frac{{\mathrm{E}}_{\max }}{\mathrm{c}}$ Substitute the values we have,

${\mathrm{B}}_{\max }=\frac{6.14×{10}^4}{3×{10}^8}=2.05×{10}^{-4}\mathrm{T}$

Therefore,${\mathrm{B}}_{\max }=2.05×{10}^{-4}\mathrm{T}$