A circular coil of 50 turns and radius 10 cm has a magnetic moment of 3.14 A m ^2 . What is the magnitude of the magnetic field at the centre of the coil? (Take = 3.14 and ₀ = 4 10⁻⁷ T m/A)

Step-by-step solution

The magnetic moment of a circular coil is given by: M = N I A = N I R^2 Given M = 3.14 A m ^2 and taking = 3.14 , we can write M = A m ^2 . Substituting the given values ( N = 50 , R = 0.1 m): = 50 I (0.1)^2 1 = 50 I 0.01 1 = 0.5 I I = 2 A The magnetic field at the centre of the coil is: B = ₀ N I 2R B = 4 10⁻⁷ 50 2 2 0.1 B = 400 10⁻⁷ 0.2 B = 2000 10⁻⁷ = 2 10⁻⁴ T Alternatively, substituting NI = M R^2 directly into the magnetic field formula yields: B = ₀ M 2 R^3 = 4 10⁻⁷ 2 (0.1)^3 = 2 10⁻⁴ T Answer: 2 10⁻⁴ T