Respuesta :
A) [tex]524.0 W/m^2[/tex]
B) 0.531 V/m
C) [tex]1.77\cdot 10^{-3} \mu T[/tex]
D) [tex]2.5\cdot 10^{-12} J/m^3[/tex]
Explanation:
A)
The intensity of an electromagnetic wave is given by
[tex]I=\frac{P}{A}[/tex]
where
I is the intensity
P is the power of the wave
A is the area over which the radiation is spread
In this problem:
[tex]P=0.750 mW = 0.750\cdot 10^{-3}W[/tex] is the power of the light emitted by the laser
This power spread over a cylindrical beam, so the area over which it spreads is the area of the base of the cylinder, which is the area of a circle:
[tex]A=\pi (\frac{d}{2})^2[/tex]
where
[tex]d=1.35 mm = 1.35\cdot 10^{-3} m[/tex] is the diameter of the beam
Therefore, the intensity of the laser beam is:
[tex]I=\frac{P}{\pi (\frac{d}{2})^2}=\frac{0.750\cdot 10^{-3}}{\pi (\frac{1.35\cdot 10^{-3}}{2})^2}=524.0 W/m^2[/tex]
B)
The relationship between power of an electromagnetic wave and maximum value of the electric field is the following:
[tex]P=\epsilon_0 E^2 c[/tex]
where
P is the power
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]c[/tex] is the speed of light
E is the maximum value of the electric field
In this problem we have:
[tex]P=0.750 mW = 0.750\cdot 10^{-3}W[/tex] is the power of the light emitted by the laser
[tex]\epsilon_0 =8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
Solving for E, we find the value of the electric field:
[tex]E=\sqrt{\frac{P}{\epsilon_0 c}}=\sqrt{\frac{0.750\cdot 10^{-3}}{(8.85\cdot 10^{-12})(3.0\cdot 10^8)}}=0.531 V/m[/tex]
C)
For an electromagnetic wave, the relationship between amplitude of the electric field and amplitude of the magnetic field is:
[tex]E=cB[/tex]
where
E is the maximum value of the electric field
c is the speed of light
B is the maximum value of the magnetic field
In this problem we have:
E = 0.531 V/m is the maximum value of the electric field
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
Solving the equation for B, we find the maximum value of the magnetic field:
[tex]B=\frac{E}{c}=\frac{0.531}{3.0\cdot 10^8}=1.77\cdot 10^{-9} T = 1.77\cdot 10^{-3} \mu T[/tex]
D)
The energy density of an electromagnetic wave is the amount of energy per unit volume of the wave.
The average energy density of an electromagnetic wave is related to the amplitude of the electric field by the equation:
[tex]w=\epsilon_0 E^2[/tex]
where
w is the energy density
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the maximum value of the electric field
Here we have for this laser beam:
[tex]\epsilon_0 =8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
E = 0.531 V/m is the maximum value of the electric field
Substituting, we find the average energy density of the laser beam:
[tex]w=(8.85\cdot 10^{-12})(0.531)^2=2.5\cdot 10^{-12} J/m^3[/tex]
