91Ó°ÊÓ

The intensity of light is expressed by the equation$\mathbf{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$where A is the cross-sectional area of the beam. The energy densities of electric and magnetic field is given by the equation${\mathbf{μ}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{E}}_{\mathbf{0}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{μ}}_{\mathbf{0}}}$

Given that the light has a power P =5.80 mW and diameter D= 2.50 mm .Using the equation $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$were $\mathrm{A}=4{\mathrm{Ï€°ù}}^2=4\mathrm{Ï€}(575×{10}^3\mathrm{m})$,we have .

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{5.80×{10}^{-3}\mathrm{W}}{\mathrm{π}{\left(1.25×{10}^{-3}\mathrm{m}\right)}^2}\mathrm{I}=1182\mathrm{W}/{\mathrm{m}}^2$

Substituting the intensity from above in the formula

$$\begin{gathered} 1182\mathrm{W}/{\mathrm{m}}^2=\frac{{\mathrm{B}}_0^2\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}{2\left(4\mathrm{Ï€}×{10}^{-7}\right)} \\ {\mathrm{B}}_0^2=\frac{1182\mathrm{W}/{\mathrm{m}}^2\left(8×{10}^{-7}\right)}{3×{10}^8\mathrm{m}/\mathrm{s}} \\ =9.89×{10}^{-12}\mathrm{T} \\ {\mathrm{B}}_0=3.14\mathrm{μ°Õ} \end{gathered}$$

By using the magnetic field, we find the electric field using the equation E =CB as${\mathrm{E}}_0=\mathrm{cB}$ we get,

$$\begin{gathered} {\mathrm{E}}_0={\mathrm{cB}}_0 \\ =3×{10}^8\mathrm{m}/\mathrm{s}\left(3.14\mathrm{μ°Õ}\right) \\ =943\mathrm{V}/\mathrm{m} \end{gathered}$$

Therefore, ${\mathrm{E}}_0=943\mathrm{V}/\mathrm{m}$and${\mathrm{B}}_0=3.14\mathrm{μ°Õ}$

The energy densities of electric and magnetic field is given by the equation${\mathrm{μ}}_0=\frac{{\mathrm{Î\mu }}_0{\mathrm{E}}_0^2}{4}=\frac{{\mathrm{B}}_0^2}{4{\mathrm{μ}}_0}$ So, by using one of the equations above we can find out the energy densities of both the electric and magnetic field we have,

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{Eav}}=\frac{{\mathrm{Î\mu }}_0{\mathrm{E}}_0^2}{4} \\ =\frac{8.85×{10}^{-12}\left(943\mathrm{W}/{\mathrm{m}}^2\right)}{4} \\ =1.96×{10}^{-6}\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

Similarly,

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{Bav}}=\frac{{\mathrm{B}}_0^2}{4{\mathrm{μ}}_0} \\ =\frac{{\left(3.14×{10}^{-6}\right)}^2}{4\left(4\mathrm{π}×{10}^{-7}\right)} \\ =1.961×{10}^{-6}\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

Therefore, the average energy densities are$1.961×{10}^{-6}\mathrm{J}/{\mathrm{m}}^3$

The expression to find average energy per unit volume is,${\mathrm{μ}}_{\mathrm{av}}={\mathrm{μ}}_{\mathrm{Eav}}+{\mathrm{μ}}_{\mathrm{Bav}}$.Substitute the values we have,

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{av}}=1.96×{10}^{-6}+1.96×{10}^{-6} \\ =3.92×{10}^{-6}\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

The expression to find the total energy contained in length I is

$$\begin{gathered} \mathrm{U}={\mathrm{μ}}_{\mathrm{av}}\mathrm{IA} \\ =3.92×{10}^{-6}\left(1\right)\left({\mathrm{Ï€°ù}}^2\right) \\ =3.92×{10}^{-6}\left(1\right)\mathrm{Ï€}{\left(1.25×{10}^{-3}\mathrm{m}\right)}^2 \\ =1.92×{10}^{-11}\mathrm{J} \end{gathered}$$

Therefore, the total energy contained in a 1.00-m length of the beam is$\mathrm{U}=1.92×{10}^{-11}\mathrm{J}$