Answer :
A) 983.2 V/m
First of all, we need to find the intensity of the laser light:
[tex]I=\frac{P}{A}[/tex]
where
[tex]P=5.80 mW=5.8\cdot 10^{-3} W[/tex] is the power
A is the area
the diameter is 2.40 mm, so the radius is 1.20 mm ([tex]1.2\cdot 10^{-3} m[/tex]), therefore the area is
[tex]A=\pi r^2 = \pi (1.2\cdot 10^{-3} m)^2=4.52\cdot 10^{-6}m^2[/tex]
And the intensity is
[tex]I=\frac{5.80\cdot 10^{-3} W}{4.52\cdot 10^{-6} m^2}=1283.2 W/m^2[/tex]
The amplitude of the electric field is related to the intensity by
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Solving for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1283.2 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=983.2 V/m[/tex]
B) [tex]3.28\cdot 10^{-6} T[/tex]
The amplitude of the magnetic field is given by
[tex]B=\frac{E}{c}[/tex]
where
E = 983.2 V/m is the amplitude of the electric field
c is the speed of light
Substituting,
[tex]B=\frac{983.2 V/m}{3\cdot 10^8 m/s}=3.28\cdot 10^{-6} T[/tex]
C) [tex]4.28\cdot 10^{-6} J/m^3[/tex]
The average energy density associated with the electric field is
[tex]u_E = \frac{1}{2}\epsilon_0 E^2[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
E = 983.2 V/m is the amplitude of the electric field
Substituting,
[tex]u_E = \frac{1}{2}(8.85\cdot 10^{-12} F/m) (983.2 V/m)^2=4.28\cdot 10^{-6} J/m^3[/tex]
D) [tex]4.28\cdot 10^{-6} J/m^3[/tex]
The average energy density associated with the magnetic field is
[tex]u_B = \frac{1}{2 \mu_0}B^2[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
[tex]B=3.28\cdot 10^{-6} T[/tex] is the amplitude of the magnetic field
Substituting,
[tex]u_B = \frac{1}{2 (4\pi 10^{-7} H/m)} (3.28\cdot 10^{-6} T)^2=4.28\cdot 10^{-6} J/m^3[/tex]
In fact, the average energy density associated with the electric field in an electromagnetic wave is equal to the average energy density associated with the magnetic field.
E) [tex]3.87\cdot 10^{-11} J[/tex]
The area of the beam is (calculated at point A)
[tex]A=4.52\cdot 10^{-6}m^2[/tex]
The lenght of the beam is
L = 1.00 m
So the volume of the beam is
[tex]V=A\cdot L=(4.52\cdot 10^{-6}m^2)(1.00 m)=4.52\cdot 10^{-6}m^3[/tex]
The total energy density is the sum of the energy density of the electric and magnetic field:
[tex]u=u_E + u_B = 4.28\cdot 10^{-6} J/m^3+4.28\cdot 10^{-6} J/m^3=8.56\cdot 10^{-6} J/m^3[/tex]
so, the total energy in the beam will be:
[tex]U= u\cdot V=(8.56\cdot 10^{-6} J/m^3)(4.52\cdot 10^{-6}m^3)=3.87\cdot 10^{-11} J[/tex]