Let\'s apply the energy density and intensity equations to a laser. In particula
ID: 2035526 • Letter: L
Question
Let's apply the energy density and intensity equations to a laser. In particular, we will consider an industrial laser cutter that is used for cutting through thin sheets of soft materials. It emits a beam with electric-field amplitude Emax=2.76×105V/m over an area of 2.00 mm2. Find (a) the maximum magnetic field Bmax, (b) the maximum energy density umax, (c) the intensity Sav=I of the beam, and (d) the average power of the beam.
a) Find the maximum energy density for a 3000 W laser with an electric-field amplitude Emax=7.08×105N/C.
Express your answer in joules per cubic meter to three significant figures.
Explanation / Answer
a)
Bmax = Emax / C
Bmax = (2.76 x 10^5 N/C) / (3 x 10^8)
Bmax = 9.2 x 10^-4 T
b)
Umax = eo*Emax^2
Umax = 8.854 x 10^-12 x (2.76 x 10^5)^2
Umax = 0.674 J/m^3
Uavg = Umax / 2 = 0.337 J/m^3
c)
I = Sav = Uavg x C
I = 0.337 J/m^3 x 3 x 10^8 m/s
I = 1.01 x 10^8 w/m^2
d)
Pavg = (1.01 x 10^8 w/m^2 x 2 x 10^-6 m^2)
Pavg = 202 W
A)
E_max = 7.08 x 10^5 N/C
maximum energy density = eo*E_max^2
= 8.854 x 10^-12 x (7.08 x 10^5)^2
= 4.438 J/m^3
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