A 100 -turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively : (Take ₀ = 4 10⁻⁷ T m/A)

Step-by-step solution

The magnetic field at the centre of a circular coil is given by: B = ₀ N I 2R Substituting the given values ( B = 3.14 10⁻³ T, N = 100 , R = 0.05 m, ₀ = 4 10⁻⁷ T m/A): 3.14 10⁻³ = 4 10⁻⁷ 100 I 2 0.05 Taking = 3.14 : 3.14 10⁻³ = 4 3.14 10⁻⁵ I 0.1 10⁻³ = 4 10⁻⁴ I I = 10⁻³ 4 10⁻⁴ = 2.5 A The magnetic moment of the coil is given by: M = N I A = N I ( R^2) M = 100 2.5 3.14 (0.05)^2 M = 250 3.14 0.0025 M = 1.9625 A m ^2 2 A m ^2 Answer: 2.5 A, 2 A m ^2