The magnetic field at the center of circular coil carrying current ' I ' for a single turn of a given length of wire is ' B '. The same wire is bent in a circular coil having two turns. When the same current passes through it, the value of magnetic field becomes

Step-by-step solution

Let the length of the wire be L . For a single turn coil, the radius R₁ is given by 2 R₁ = L R₁ = L 2 . The magnetic field at the center of the single turn coil is B = ₀ I 2 R₁ . When the same wire is bent into a coil of N=2 turns, the new radius R₂ is given by 2 2 R₂ = L R₂ = L 4 = R₁ 2 . The magnetic field at the center of the two-turn coil is B' = ₀ N I 2 R₂ . Substituting N = 2 and R₂ = R₁ 2 , we get B' = ₀ (2) I 2 ( R₁ 2 ) = 4 ( ₀ I 2 R₁ ) = 4B . Answer: 4B