The magnitude of the magnetic induction (B) at a point on the axis at a large distance 'r' from the centre of a circular coil of 'n' turns and area 'A' carrying current ' i ' is given by x 10⁻⁷ SI unit. The value of x is ( ₀ = 4 10⁻⁷ SI unit )

Step-by-step solution

The magnetic field on the axis of a circular coil of radius R at a distance r from its centre is given by: B = ₀ n i R^2 2(R^2 + r^2)^ 3/2 For a large distance r R , the expression simplifies to: B = ₀ n i R^2 2r^3 Multiplying the numerator and denominator by , we get: B = ₀ n i ( R^2) 2 r^3 Since the area of the coil is A = R^2 , we can write: B = ₀ n i A 2 r^3 Substituting ₀ = 4 10⁻⁷ SI units: B = 4 10⁻⁷ n i A 2 r^3 = 2 n i A r^3 10⁻⁷ Comparing this with B = x 10⁻⁷ , we find: x = 2 n i A r^3 Answer: 2niA r^3