Peak values for a neodymium-glass laser are 1200 J for 1 nanosecond.If the cross section of the laser beam is 1 cm²,what are the maximum values of E and B? (µ₀ = 4π × 10−⁷ T⋅m/A,c = 3.00 × 10⁸ m/s) A) 3 × 10⁷ V/m,2 T B) 2 × 10⁸ V/m,10 T C) 1 × 10¹⁰ V/m,3 T D) 3 × 10⁹ V/m,10 T

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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 I = 1200 J / (1 × 10^-9 s) / (1 × 10^-4 m²) = 1.2 × 10^16 W/m². The maximum electric field (E) is given by E = √(2I/ε₀c), and the maximum magnetic field (B) is B = E/c. Calculating these gives E ≈ 3 × 10^9 V/m and B ≈ 10 T.

Peak Values for a Neodymium-Glass Laser Are 1200 J for 1