A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is B. If it is bent into a circular loop of radius 'r' having n turns, the magnetic field at the centre of the coil for same current is

Step-by-step solution

Let the length of the wire be L . For a single turn circular loop of radius R , the length is L = 2 R R = L 2 . The magnetic field at the centre is B = ₀ I 2R . When the same wire is bent into a circular loop of n turns of radius r , the length is L = n(2 r) r = L 2 n = R n . The new magnetic field at the centre is B' = ₀ n I 2r . Substituting r = R n , we get B' = ₀ n I 2(R/n) = n^2 ( ₀ I 2R ) = n^2 B . Answer: n ^2 B