A small helium–neon laser emits red visible light with a power of 5.80 mW in a beam of diameter 2.50 mm. (a) What are the amplitudes of the electric and magnetic fields of this light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a 1.00-m length of the beam?

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${\mathbf{\mu }}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{E}}_{\mathbf{0}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\mu }}_{\mathbf{0}}}$

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 $\mathrm{A}=4{\mathrm{\pi r}}^2=4\mathrm{\pi }(575\times {10}^3\mathrm{m})$,we have .

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

Substituting the intensity from above in the formula

$$\begin{gathered} 1182\mathrm{W}/{\mathrm{m}}^2=\frac{{\mathrm{B}}_0^2\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{2\left(4\mathrm{\pi }\times {10}^{-7}\right)} \\ {\mathrm{B}}_0^2=\frac{1182\mathrm{W}/{\mathrm{m}}^2\left(8\times {10}^{-7}\right)}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =9.89\times {10}^{-12}\mathrm{T} \\ {\mathrm{B}}_0=3.14\mathrm{\mu T} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{E}}_0={\mathrm{cB}}_0 \\ =3\times {10}^8\mathrm{m}/\mathrm{s}\left(3.14\mathrm{\mu T}\right) \\ =943\mathrm{V}/\mathrm{m} \end{gathered}$$

Therefore, ${\mathrm{E}}_0=943\mathrm{V}/\mathrm{m}$and${\mathrm{B}}_0=3.14\mathrm{\mu T}$

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

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

Similarly,

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

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

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

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

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

$$\begin{gathered} \mathrm{U}={\mathrm{\mu }}_{\mathrm{av}}\mathrm{IA} \\ =3.92\times {10}^{-6}\left(1\right)\left({\mathrm{\pi r}}^2\right) \\ =3.92\times {10}^{-6}\left(1\right)\mathrm{\pi }{\left(1.25\times {10}^{-3}\mathrm{m}\right)}^2 \\ =1.92\times {10}^{-11}\mathrm{J} \end{gathered}$$

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