For a circular coil of radius a , at what distance x along its axis the magnetic field is exactly 1 27 of its value at the center?

Step-by-step solution

The magnetic field at the center of a circular coil of radius a is given by B_c = ₀ I 2a . The magnetic field at a distance x along the axis is given by B_x = ₀ I a^2 2(a^2 + x^2)^ 3/2 . Given that B_x = 1 27 B_c , we have: ₀ I a^2 2(a^2 + x^2)^ 3/2 = 1 27 ( ₀ I 2a ) a^3 (a^2 + x^2)^ 3/2 = 1 27 Taking the cube root on both sides: a a^2 + x^2 = 1 3 Squaring both sides: a^2 a^2 + x^2 = 1 9 9a^2 = a^2 + x^2 x^2 = 8a^2 x = 2 2 a Answer: 2 2 a