A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is then bent to form a circular coil of three turns. The ratio of magnetic inductions at the centre of the coil in the two cases is

Step-by-step solution

Magnetic induction at the centre of the coil with N turns is given by B_ N = ( ₀ I 2 r ) N Where r is the radius of each coil is, I is the current flowing through. for single loop coil 2 r=LB₁= ₀ I(1) 2 ( L 2 ) = ₀ I L & For the three-loop coil (2 r) 3=LB₃= ₀ I(3) 2 ( L 6 ) = 9 ₀ I L Or B₁ B₂ = 1 9