A circular coil having ' n ' turns, each of radius ' r ' carries a current ' I '. The magnetic field at a distance x = 3 ,r from its center, along the 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 radius r with n turns carrying current I is given by B = ₀ n I r^2 2(r^2 + x^2)^ 3/2 Given x = 3 r , substituting this value into the formula: B = ₀ n I r^2 2(r^2 + ( 3 r)^2)^ 3/2 B = ₀ n I r^2 2(r^2 + 3r^2)^ 3/2 B = ₀ n I r^2 2(4r^2)^ 3/2 B = ₀ n I r^2 2(8r^3) B = ₀ n I 16r Answer: ₀ nI 16r