A laser beam is of 9 mw, diameter = 2 mm. then what is the amplitude of magnetic field associated with it?
Answers
Answered by
69
We have the following equation and calculations:
[tex]S = \frac{P}{A} = c{\varepsilon _0}{E_{rms}}^2 {E_{rms}} = \sqrt {\frac{P}{{Ac{\varepsilon _0}}}} = \sqrt {\frac{{0.009}}{{\pi \times {{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {3 \times {{10}^8}} \right) \times \left( {8.85 \times {{10}^{ - 12}}} \right)}}} = 1038.7\;V/m {B_{rms}} = \frac{{{E_{rms}}}}{c} = \frac{{1038.7}}{{3 \times {{10}^8}}} = 3.46\mu T B = \sqrt 2 {B_{rms}} = 4.9\mu T[/tex]
[tex]S = \frac{P}{A} = c{\varepsilon _0}{E_{rms}}^2 {E_{rms}} = \sqrt {\frac{P}{{Ac{\varepsilon _0}}}} = \sqrt {\frac{{0.009}}{{\pi \times {{\left( {{{10}^{ - 3}}} \right)}^2} \times \left( {3 \times {{10}^8}} \right) \times \left( {8.85 \times {{10}^{ - 12}}} \right)}}} = 1038.7\;V/m {B_{rms}} = \frac{{{E_{rms}}}}{c} = \frac{{1038.7}}{{3 \times {{10}^8}}} = 3.46\mu T B = \sqrt 2 {B_{rms}} = 4.9\mu T[/tex]
Similar questions