SOLUTION Let's apply the energy density and intensity equations to a laser. SET UP AND SOLVE Part (a): From E = cB, the maximum magnetic field Bmax is In particular, we will consider an industrial laser cutter that is used for cutting through thin sheets of soft materials. It emits a beam Emax 2.76x105 V/m Bmax = 9.20 x 10-4 T with electric-field amplitude Emax = 2.76 x 109 V/m over an C 3.00 x 108 m/s area of 2.00 mm . Find (a) the maximum magnetic field Bmax, (b) the maximum energy density Umax, (c) the intensity Sav = I Part (b): The maximum energy density Umax is of the beam, and (d) the average power of the beam. Umax = co Emax = [8.85 x 10-12 C2 /(N . m2 )](2.76 x 105 N/C)2 = 0.674 N/m2 = 0.674 J/m3 The average energy density is half of this: Wav =- 60 Emax? = 0.337 J/m3 Part (c): The intensity I = Sav is given by the following: I = Sav = WavC = (0.337 J/m3) (3.00 x 108 m/s) = 1.01 x 108 J/(m2 . s) = 1.01 x 108 W/m2 Alternatively, I = Sav = cocEmax2 = [8.85 x 10-12 C2 / (N . m2)](3.00 x 108 m/s) (2.76 x 105 N/C)2 = 1.01 x 108 W/m Part (d): The average total power Pav is the intensity Sav multiplied by the cross-sectional area A of the beam: Pav = (1.01 x 108 W/m2) (2.00 x 10-6 m2) 202 W REFLECT A power of 200 W is high enough to cut cardboard and thin wood. A typical laser pointer has an output power on the order of a few milliwatts.electric field amplitude - Emax = 865 X 102 N /C . Average energy density = 2 6 Emax 6 = 8 854 X 10 - 12 = 2X 8. 854 x 15 2 , (865 X 10 = 3 . 312 5 / m 3 Maximeemn energy density . Vwray = 6 Emax 2 . ( VArg ) . = 6: 624 J/me An