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 ₁ ^ having ' n ' turns, the magnetic field at the centre of the coil for same current is

Step-by-step solution

Magnetic field comparison for wire configurations A wire of length L carrying current I produces different magnetic fields when configured as a single turn versus n turns. For a single circular turn of circumference L , the radius is R = L / (2 ) . The magnetic field at the center is B = ₀ I / (2R) . Substituting R gives B = ₀ I / L . For n turns, the total circumference constraint requires L = n 2 r₁ , so r₁ = L / (2 n) . The magnetic field becomes B' = ₀ n I / (2r₁) . Substituting r₁ yields B' = ₀ n^2 I / L . Com