Peak values for a neodymium-glass laser are 600 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 = 4p x 10^-7 T*m/A,c = 3.00 x 10⁸ m/s) A)2 x 10⁹ V/m,2 T B)4 x 10⁸ V/m,7 T C)2 x 10⁹ V/m,7 T D)4 x 10⁸ V/m,2 T

Question

Quizplus · Accepted Answer

We know that the maximum electric field E is given by:

E = √(2× Energy density / (cross-section area × permittivity of free space × time duration))

Energy density = Energy / (cross-section area × time duration)
    = 600 J / (1 cm² × 10⁻⁹ s)
    = 6×10¹² J/m³

Using the given values for cross-section and time duration, we get:

E = √(2×6×10¹² / (1×10⁻⁴ × 8.85×10⁻¹² × 1×10⁻⁹))
  = 2×10¹⁹ V/m

The maximum magnetic field B is given by:

B = √(2× Energy density / (c² × permittivity of free space))
  = √(2×6×10¹² / (9×10¹⁶))
  = 7 T

Therefore, the correct choice is C) 2 x 10¹⁹ V/m, 7 T.

Peak Values for a Neodymium-Glass Laser Are 600 J for 1