Determine the magnetic field B at the centre of curvature of a semicircular wire that has a radius of 10.0 cm and carries a current of 5.0 A .

Step-by-step solution

The magnetic field at the centre of a semicircular wire carrying current I is given by: B = ₀ I 4r Given: I = 5.0 A r = 10.0 cm = 0.1 m ₀ = 4 10⁻⁷ T m/A Substituting the values into the formula: B = 4 10⁻⁷ 5.0 4 0.1 B = 10⁻⁷ 5.0 0.1 B = 50 10⁻⁷ T B = 5 10⁻⁶ T Substituting = 3.14 : B = 5 3.14 10⁻⁶ T B = 15.7 10⁻⁶ T B = 1.57 10⁻⁵ T