Question

A laser beam of a plane monochromatic radiation has an intensity of $2.5\times {10}^{14}W·m^{-2}$. The amplitude of the magnetic field in this beam is

Answer 1

Step 1: Given data

The given intensity of the beam is, $I=2.5\times {10}^{14}W·m^{-2}$.

Let the RMS value of the electric field be $E_{rms}$.

The speed of light in vacuum, $c=3\times {10}^{8}m{s}^{-1}$.

The permittivity of the vacuum, ${\mu }_0=4\pi \times {10}^{-7}F{m}^{-1}$.

Step 2: Calculate the electric field in the given beam

It is known that, ${\left(E_{rms}\right)}^2=Ic{\mu }_0$.

$$\begin{gathered} \Rightarrow {\left(E_{rms}\right)}^2=2.5\times {10}^{14}\times 3\times {10}^8\times 4\mathrm{\pi }\times {10}^{-7} \\ \Rightarrow {\mathrm{E}}_{\mathrm{rms}}=\sqrt{94.25\times {10}^{15}} \\ \Rightarrow {\mathrm{E}}_{\mathrm{rms}}=3.07\times {10}^8 \end{gathered}$$

Thus, the electric field of the beam can be given by, $E=\sqrt{2}{E}_{rms}$.

$$\begin{gathered} \Rightarrow E=\sqrt{2}\times 3.07\times {10}^{8}V{m}^{-1} \\ \Rightarrow E=4.34\times {10}^{8}V{m}^{-1} \end{gathered}$$

Step 3: Calculate the magnetic field in the given beam

The magnetic field of the beam can be given by, $B=\frac{E}{c}$.

$$\begin{gathered} \Rightarrow B=\frac{4.34\times {10}^{8}V{m}^{-1}}{3\times {10}^{8}m{s}^{-1}} \\ \Rightarrow B=1.45T \end{gathered}$$

Hence, the magnetic field in the given beam is $1.45T$.