Question

Lasers can be constructed that produce an extremely high-intensity electromagnetic wave for a brief time called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{V}\mathbf{/}\mathit{m}$for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}{\mathbf{mm}}^{\mathbf{2}}$area?

Short answer

(a) The magnetic field strength is $3.33\times {10}^2\mathrm{T}$.

(b) The intensity is $1.33\times {10}^{19}\mathrm{W}/{\mathrm{m}}^2$.

(c) The energy is $1.33\mathrm{kJ}$.

Step-by-step solution

Define electric field strength

The intensity of an electric field at a certain area is quantified as electric field strength. The volt per meter is the standard unit.

Evaluating the magnetic field strength

(a)

The magnetic field, electric field, and light speed are all connected by the formula

$$\begin{gathered} E_0=c{B}_0 \\ {B}_0=\frac{E_0}{c} \end{gathered}$$

Where${\mathrm{B}}_0$ is the magnetic field,c is the speed of light, and${\mathrm{E}}_0$ is the electric field.

Substitute the values in the above equation, and we get,

$$\begin{gathered} {\mathrm{B}}_0=\frac{1\times {10}^{11}\mathrm{V}/\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ =334\mathrm{T} \\ \mathrm{Therefore},\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{strength}\mathrm{is}334\mathrm{T}. \end{gathered}$$

Evaluating the intensity

(b)

In terms of magnetic and electric fields, the average intensity is given by

${\mathrm{I}}_{\mathrm{avg}}=\frac{{\mathrm{E}}_0{\mathrm{B}}_0}{2{\mathrm{\mu }}_0}$

Where ${\mathrm{I}}_{\mathrm{avg}}$is average intensity, ${\mathrm{\mu }}_0$is the permeability of free space

Substitute the values in the above equation,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{avg}}=\frac{\left(1.00\times {10}^{11}\mathrm{V}/\mathrm{m}\right)\times \left(334\mathrm{T}\right)}{2\times 4\mathrm{\tau \tau }\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^2} \\ =1.33\times {10}^{19}\mathrm{W}/{\mathrm{m}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{intensity}\mathrm{is}1.33\times {10}^{19}\mathrm{W}/{\mathrm{m}}^2. \end{gathered}$$

Evaluating the energy

(c)

Intensity is measured in terms of power per unit area and may be calculated using the following formula

$$\begin{gathered} {\mathrm{I}}_0=\frac{\mathrm{P}}{\mathrm{A}} \\ \mathrm{P}={\mathrm{I}}_0\mathrm{A} \end{gathered}$$

Energy is also given by the product of power and time, which is given by

$$\begin{gathered} \mathrm{E}=\mathrm{Pt} \\ \mathrm{E}={\mathrm{I}}_0\mathrm{At} \end{gathered}$$

Substitute the values in the above equation,

$$\begin{gathered} \mathrm{E}=\left(1.33\times {10}^{19}\mathrm{W}/{\mathrm{m}}^2\right)\left({10}^{-6}{\mathrm{m}}^2\right)\left({10}^{-9}\mathrm{s}\right) \\ =1.33\times {10}^3\mathrm{J}\left(\frac{1\mathrm{kJ}}{{10}^3\mathrm{J}}\right) \\ =1.33\mathrm{kJ} \\ \mathrm{Therefore},\mathrm{the}\mathrm{energy}\mathrm{is}1.33\mathrm{kJ}. \end{gathered}$$