A long straight wire carries a current of 5 A. At a distance of 3 cm, what is the magnetic field? Use μ0 = 4π×10^-7.

A long straight wire carries a current of 5 A. At a distance of 3 cm, what is the magnetic field? Use μ0 = 4π×10^-7.

• A. Approximately 1.06×10^-5 T
• B. Approximately 5.00×10^-5 T
• C. Approximately 3.33×10^-5 T
• D. Approximately 2.00×10^-5 T

Answer: (C)

The magnetic field around a long straight current-carrying wire forms circles around the wire, and its magnitude is B = μ0 I /(2π r). Here the current is 5 A and the distance is 3 cm, which is 0.03 m, so plug into the formula: B = (4π×10^-7 × 5) / (2π × 0.03). The π cancels, leaving B = (20×10^-7) / 0.06 = (2×10^-6) / 0.06 ≈ 3.33×10^-5 T. So the magnetic field at 3 cm from the wire is about 3.33×10^-5 tesla. The field direction would be tangential, determined by the right-hand rule.

The magnetic field around a long straight current-carrying wire forms circles around the wire, and its magnitude is B = μ0 I /(2π r). Here the current is 5 A and the distance is 3 cm, which is 0.03 m, so plug into the formula: B = (4π×10^-7 × 5) / (2π × 0.03). The π cancels, leaving B = (20×10^-7) / 0.06 = (2×10^-6) / 0.06 ≈ 3.33×10^-5 T. So the magnetic field at 3 cm from the wire is about 3.33×10^-5 tesla. The field direction would be tangential, determined by the right-hand rule.