In a particular case X , a certain length of insulated copper wire is bent to form double loops of equal radii. In another case Y the same copper wire is bent to form three loops of equal radii. If the same steady current is passed through the copper wire in both the cases, the ratio of magnetic field at the centre in the case X to that in case Y is:

Step-by-step solution

Let L be the total length of the wire. In case X , the wire is bent into n₁ = 2 loops of radius r₁ . The circumference of these loops is 2 (2 r₁) = L , so r₁ = L 4 . The magnetic field at the center of n₁ loops is B_X = n₁ ₀ I 2 r₁ = 2 ₀ I 2 (L / 4 ) = 4 ₀ I L . In case Y , the wire is bent into n₂ = 3 loops of radius r₂ . The circumference of these loops is 3 (2 r₂) = L , so r₂ = L 6 . The magnetic field at the center of n₂ loops is B_Y = n₂ ₀ I 2 r₂ = 3 ₀ I 2 (L / 6 ) = 9 ₀ I L . The ratio of the magnetic fields