Peak values for a neodymium-glass laser are 600 J for 1.0 nanosecond.If the cross-section of the laser beam is 1.64 cm²,what are the maximum values of E and B? (µ₀ = 4π × 10−⁷ T⋅m/A,c = 3.00 × 10⁸ m/s) A) 1.7E+9 V/m,1.6 T B) 4.7E+8 V/m,5.5 T C) 1.7E+9 V/m,5.5 T D) 4.7E+8 V/m,1.6 T E) Not enough information.

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Quizplus · Accepted Answer

The intensity (I) of the laser can be calculated using I = Power/Area. Power is energy/time, so Power = 600 J / 1.0 ns = 6.0E+11 W. The area is 1.64 cm² = 1.64E-4 m². Thus, I = 6.0E+11 W / 1.64E-4 m² = 3.66E+15 W/m². The electric field (E) is given by E = sqrt(2I/ε₀c), where ε₀ = 8.85E-12 F/m. E = sqrt(2 * 3.66E+15 W/m² / (8.85E-12 F/m * 3.00E+8 m/s)) = 1.7E+9 V/m. The magnetic field (B) is given by B = E/c = 1.7E+9 V/m / 3.00E+8 m/s = 5.5 T.

Peak Values for a Neodymium-Glass Laser Are 600 J for 1.0