A straight wire carrying of 14 A is bent into a semi-circular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre (O) of the arc is ____ × 10–4 T.
To determine the magnetic field at the center (O) of a semi-circular arc of wire, we use the Biot-Savart law. For a semi-circular arc, the magnetic field at the center is given by the formula: B = (μ₀I)/(4R) where:
μ₀ = 4π × 10-7 T·m/A (permeability of free space)
I = 14 A (current)
R = 0.022 m (radius)
Substituting the given values: B = (4π × 10-7 T·m/A × 14 A) / (4 × 0.022 m) Simplifying, B = (56π × 10-7) / 0.088 T B ≈ 2 × 10-4 T This magnetic field value lies within the expected range (2 to 2). Thus, the magnetic field at the center is 2 × 10-4 T.
