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lidiya [134]
3 years ago
8

An instructor’s laser pointer produces a beam of light with a circular cross section of diameter 0.900 mm and a total power outp

ut 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
Physics
1 answer:
il63 [147K]3 years ago
8 0

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.}

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