To manufacture a solenoid of length ' ' and inductance ' L ', the length of the thin wire required is (Diameter of the solenoid is very less than length, ₀= permeability of free space)

Step-by-step solution

The self-inductance of a long solenoid is given by L = ₀ n^2 A , where n is turns per unit length and A is cross-sectional area. Expressing n = N/ and A = r^2 yields L = ₀ (N/ )^2 ( r^2) = ₀ N^2 r^2 / . The wire length for N turns of radius r is L_ wire = 2 r N , so N = L_ wire /(2 r) . Substituting gives L = ₀ ( L_ wire 2 r )^2 r^2 = ₀ L_ wire ^2 4 . Solving for L_ wire results in L_ wire = 4 L ₀ , matching option A . Final answer: A