A straight wire carrying 14 A is bent into a semicircular arc of radius 2.2 cm as shown. The magnetic field at the centre O i $s ______$ × 10⁻⁴ T.

Question

A straight wire carrying 14 A is bent into a semicircular arc of radius 2.2 cm as shown. The magnetic field at the centre O i s ______ × 10⁻⁴ T.

G1 Physics · Accepted Answer

The straight portions are radial with respect to O, so for them dℓ is parallel to r and their magnetic-field contribution at O is zero. Field at the centre due to a semicircular arc is B = μ₀I/(4R) Given: I = 14 A R = 2.2 cm = 2.2×10⁻² m Therefore, B = (4π×10⁻⁷)(14) / [4(2.2×10⁻²)] ≈ 2.0×10⁻⁴ T Hence the required numerical value is 2. Final value: 2