A circular coil consisting of 200 turns has a radius of 10 cm and carries a current of 2.0 A . At what distance from the centre along the axis of the coil will the magnetic field drop to half of its value at the centre? (Take [3] 4 = 1.5874 )

Step-by-step solution

The magnetic field at the centre of a circular coil is given by B_ c = ₀ N I 2 R The magnetic field at a distance x from the centre on its axis is given by B_ x = ₀ N I R² 2 (R² + x²)^ 3/2 Given that B_ x = B_ c 2 ₀ N I R² 2 (R² + x²)^ 3/2 = ₀ N I 4 R R² (R² + x²)^ 3/2 = 1 2 R (R² + x²)^ 3/2 = 2 R³ Taking the power of 2/3 on both sides: R² + x² = 2^ 2/3 R² x² = R² (4^ 1/3 - 1) x = R 4^ 1/3 - 1 Substituting R = 10 cm and 4^ 1/3 = 1.5874 : x = 10 1.5874 - 1 x = 10 0.5874 x 10 0.7664 = 7.664 cm