A circular coil of radius R carries a steady current. The magnetic field at a point on the axis of the coil is found to be 1 8 of the magnetic field at its centre. The distance of this point from the centre of the coil is:

Step-by-step solution

The magnetic field at the centre of a circular coil of radius R carrying current I is: B_ centre = ₀ I 2R The magnetic field at a distance x on its axis is: B_ axis = ₀ I R^2 2(R^2 + x^2)^ 3/2 Given that B_ axis = 1 8 B_ centre : ₀ I R^2 2(R^2 + x^2)^ 3/2 = 1 8 ( ₀ I 2R ) Simplifying the expression gives: R^3 (R^2 + x^2)^ 3/2 = 1 8 Taking the cube root on both sides: R R^2 + x^2 = 1 2 Squaring both sides: R^2 R^2 + x^2 = 1 4 R^2 + x^2 = 4R^2 x^2 = 3R^2 x = 3 R Answer: 3 R