A wire carrying current 12 A is bent into a semicircular arc of radius 2.0 cm. The magnetic field at the centre due to the arc is half that of a full loop. Its magnitude is:

1. Full-loop centre field: $B_{loop} = \dfrac{\mu_0 I}{2R}$. 2. A semicircle gives half this: $B = \dfrac{\mu_0 I}{4R}$. 3. Substitute $\mu_0 = 4\pi\times10^{-7}$, $I = 12$, $R = 0.02$ m: $B = \dfrac{4\pi\times10^{-7}\times12}{4\times0.02}$. 4. Numerator $= 1.508\times10^{-5}$; denominator $= 0.08$. 5. $B = 1.9\times10^{-4}$ T (the straight segments contribute nothing since dl ∥ r). _Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", p.147_