A circular coil carrying current 'I' has a radius ' r ' and ' n ' turns. The magnetic field along the axis of a coil at a distance ' 2 2 r ' from its centre is ( ₀= permeability of free space, n is very small)

Step-by-step solution

The magnetic field along the axis of a circular coil at distance x from its center is given by B = ₀ n I r^2 2 (r^2 + x^2)^ 3/2 . Substituting x = 2 2 r , the expression becomes B = ₀ n I r^2 2 (r^2 + (2 2 r)^2)^ 3/2 . Since (2 2 r)^2 = 8 r^2 , the denominator simplifies to (r^2 + 8 r^2)^ 3/2 = (9 r^2)^ 3/2 . The term (9 r^2)^ 3/2 = 27 r^3 , giving B = ₀ n I r^2 2 27 r^3 = ₀ n I 54 r . This result corresponds to option C .