Determine the maximum magnitude of the magnetic field B produced by a copper wire having a diameter of 1.6 mm and carrying a current of 20 A .

Step-by-step solution

Given: Diameter of the wire, d = 1.6 mm Radius of the wire, R = d 2 = 0.8 mm = 0.8 10⁻³ m Current, I = 20 A The magnetic field produced by a current-carrying wire is maximum at its surface. Using Ampere's law, the maximum magnetic field is given by: B_ max = ₀ I 2 R Substituting the given values: B_ max = 4 10⁻⁷ 20 2 0.8 10⁻³ B_ max = 2 10⁻⁷ 20 0.8 10⁻³ B_ max = 40 10⁻⁷ 8 10⁻⁴ B_ max = 5.0 10⁻³ T