A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having n turns. If the same current is passed in both the cases, the ratio of magnetic field produced at the centre for one turn to that of n turns is

Step-by-step solution

Let the length of the wire be L . For a circular coil of one turn, the radius R₁ is given by L = 2 R₁ R₁ = L 2 . The magnetic field at the centre of the coil of one turn is B₁ = ₀ I 2 R₁ . For a circular coil of n turns, the radius R_n is given by L = n(2 R_n) R_n = L 2 n = R₁ n . The magnetic field at the centre of the coil of n turns is B_n = ₀ n I 2 R_n . Substituting R_n = R₁ n , we get B_n = ₀ n I 2 ( R₁ n ) = n^2 ( ₀ I 2 R₁ ) = n^2 B₁ . The ratio of the magnetic field for one turn to that of n turns is B₁ B_n