A circular coil having 'n' turns, each of radius 'r' carries a current 'I'. The magnetic field at a distance x = 8 r from its centre along axis of the coil is ( ₀ = permeability of free space)

Step-by-step solution

The magnetic field at a distance x on the axis of a circular coil of n turns, radius r , carrying current I is given by B = ₀ n I r^2 2(r^2 + x^2)^ 3/2 Given x = 8 r , substituting this value in the formula: B = ₀ n I r^2 2(r^2 + ( 8 r)^2)^ 3/2 B = ₀ n I r^2 2(r^2 + 8r^2)^ 3/2 B = ₀ n I r^2 2(9r^2)^ 3/2 B = ₀ n I r^2 2(27r^3) B = ₀ n I 54 r Answer: ₀ nI 54 r