A circular coil carrying current has 100 turns and a radius of 5.0 cm . It generates a magnetic field of 6.0 10⁻⁵ T at its centre. Determine the magnitude of the current.

Step-by-step solution

The magnetic field at the centre of a circular coil of N turns is given by: B = ₀ N I 2R Rearranging for current I : I = B 2R ₀ N Substituting the given values, B = 6.0 10⁻⁵ T , R = 5.0 cm = 0.05 m , N = 100 , and ₀ = 4 10⁻⁷ T m/A : I = 6.0 10⁻⁵ 2 0.05 4 10⁻⁷ 100 I = 6.0 10⁻⁶ 4 10⁻⁵ I = 0.15 A Using 3.14 : I 0.15 3.14 0.048 A