Answer:
[tex]E_0 = 2180.53N/C.[/tex]
[tex]B_0 = 7.27*10^{-6}T.[/tex]
[tex]U_{avg} = 4.2*10^{-5}J/m^3.[/tex]
Explanation:
The Intensity [tex]I[/tex] of the beam is
[tex]I = P /A[/tex]
The diameter of the beam is 0.900mm; therefore, the area is
[tex]A = \pi( \dfrac{0.900*10^{-3}}{2} )^2[/tex]
[tex]A = 6.36*10^{-7}m^2[/tex]
and since [tex]P = 4.00*10^{-3}W[/tex], the intensity of the beam is
[tex]I = \dfrac{4.00*10^{-3}W}{6.36*10^{-7}m^2}[/tex]
[tex]\boxed{I = 6289.3W/m^2.}[/tex]
Now, the intensity [tex]I[/tex] is related to [tex]E_0[/tex] by the relation
[tex]I = \dfrac{E_0^2}{2\mu_0 c}[/tex]
solving for [tex]E_0[/tex] we get
[tex]E_0 = \sqrt{2\mu_0 c I}[/tex]
putting in the numbers we get:
[tex]E_0 = \sqrt{2(1.26*10^{-6}) (3*10^8) (6289.3)}[/tex]
[tex]\boxed{E_0 = 2180.53N/C.}[/tex]
The amplitude of magnetic field [tex]B_0[/tex] is related to [tex]E_0[/tex] by
[tex]B_0 = \dfrac{E_0}{c}[/tex]
putting in numerical values we get:
[tex]B_0 = \dfrac{2180.53}{3*10^8}[/tex]
[tex]\boxed{B_0 = 7.27*10^{-6}T. }[/tex]
The average energy density of the laser light is
[tex]U_{avg} = \epsilon_0 E_0^2[/tex]
[tex]U_{avg} = (8.85*10^{-12}) (2180.53)^2[/tex]
[tex]\boxed{U_{avg} = 4.2*10^{-5}J/m^3.}[/tex]
