Question

An instructor’s laser pointer produces a beam of light with a circular cross section of diameter 0.900 mm and a total power output of 4.00 mW. The beam size stays nearly constant when the instructor uses the laser pointer in the classroom. What is the amplitude E 0 of the electric field of the laser’s light? E 0 = N/C What is the amplitude B 0 of the magnetic field of the laser’s light? B 0 = T What is the average energy density u ave of the laser’s light? u ave J/m 3

Answer 1

Answer:

$E_0 = 2180.53N/C.$

$B_0 = 7.27*10^{-6}T.$

$U_{avg} = 4.2*10^{-5}J/m^3.$

Explanation:

The Intensity $I$ of the beam is

$I = P /A$

The diameter of the beam is 0.900mm; therefore, the area is

$A = \pi( \dfrac{0.900*10^{-3}}{2} )^2$

$A = 6.36*10^{-7}m^2$

and since $P = 4.00*10^{-3}W$, the intensity of the beam is

$I = \dfrac{4.00*10^{-3}W}{6.36*10^{-7}m^2}$

$\boxed{I = 6289.3W/m^2.}$

Now, the intensity $I$ is related to $E_0$ by the relation

$I = \dfrac{E_0^2}{2\mu_0 c}$

solving for $E_0$ we get

$E_0 = \sqrt{2\mu_0 c I}$

putting in the numbers we get:

$E_0 = \sqrt{2(1.26*10^{-6}) (3*10^8) (6289.3)}$

$\boxed{E_0 = 2180.53N/C.}$

The amplitude of magnetic field $B_0$ is related to $E_0$ by

$B_0 = \dfrac{E_0}{c}$

putting in numerical values we get:

$B_0 = \dfrac{2180.53}{3*10^8}$

$\boxed{B_0 = 7.27*10^{-6}T. }$

The average energy density of the laser light is

$U_{avg} = \epsilon_0 E_0^2$

$U_{avg} = (8.85*10^{-12}) (2180.53)^2$

$\boxed{U_{avg} = 4.2*10^{-5}J/m^3.}$