A circular coil carrying current has radius ' R '. The distance from the centre of the coil on the axis where the magnetic induction will be 1 27 th to its value at the centre of the coil is

Step-by-step solution

The magnetic induction B at distance x along the axis of a circular coil of radius R carrying current I is given by B_x = ₀ I R^2 2(R^2 + x^2)^ 3/2 . At the coil’s center ( x = 0 ), the magnetic field simplifies to B_c = ₀ I 2R . Given B_x = 1 27 B_c , substitute the expressions: ₀ I R^2 2(R^2 + x^2)^ 3/2 = 1 27 ₀ I 2R . Cancel the common factors ₀ I / 2 from both sides: R^2 (R^2 + x^2)^ 3/2 = 1 27R . Cross-multiplying yields 27R^3 = (R^2 + x^2)^ 3/2 . Raising both sides to the power 2 3 gives (27R^3)^ 2/3 = (R^2 +