Question

A 15.0-mW laser puts out a narrow beam 2.00 mm indiameter.

Answer 1

Explanation:

It is known that formula for energy per unit area per unit time is as follows.

             S = $\frac{P}{A}$

                = $\frac{P}{\pi \times r^{2}} = \frac{4P}{\pi d^{2}}$

      $c \epsilon_{o} E^{2} = \frac{4P}{\pi d^{2}}$

              E = $\sqrt{\frac{4P}{\pi d^{2} c \epsilon_{o}}}$        

Now, putting the values we will calculate the electric field as follows.

            E = $\sqrt{\frac{4P}{\pi d^{2} c \epsilon_{o}}}$

               = $\sqrt{\frac{4 \times 15 \times 10^{-3} W}{3.14 \times (2mm \times \frac{1m}{1000 mm})^{2} \times 3 \times 10^{8} \times 8.85 \times 10^{-12}}}$  

               = 1341.03 V/m

Now, we will calculate the average magnetic field as follows.

                   B = $\frac{E}{c}$

                      = $\frac{1341.03 V/m}{3 \times 10^{8} m/s}$

                      = $4.47 \times 10^{-6}$ T

                      = $4.47 \mu T$

Thus, we can conclude that the average (rms) value of the magnetic field is $4.47 \mu T$.