$A$ straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way does the contribution to $B$ from the semicircle differ from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$?

(N/A) For the straight segments,the current element $dl$ and the position vector $r$ are parallel to each other at every point. According to the Biot-Savart law,$dB = \frac{\mu_0}{4\pi} \frac{I(dl \times r)}{r^3}$. Since $dl \times r = 0$,the straight segments do not contribute to the magnetic field $B$ at the centre.
$(b)$ For the semicircular arc,the contributions from all current elements $dl$ are in the same direction (perpendicular to the plane of the paper,pointing inwards). Thus,they add up in magnitude. The magnitude of the magnetic field at the centre of a full circular loop is $B_{loop} = \frac{\mu_0 I}{2R}$. For a semicircle,the magnitude is exactly half of this,i.e.,$B = \frac{\mu_0 I}{4R}$.
Given $I = 12\; A$ and $R = 2.0 \times 10^{-2}\; m$,$B = \frac{4\pi \times 10^{-7} \times 12}{4 \times 2.0 \times 10^{-2}} = 1.884 \times 10^{-4}\; T \approx 1.9 \times 10^{-4}\; T$. The direction is normal to the plane of the paper,going into it.
$(c)$ Yes,the magnitude of $B$ remains the same $(1.9 \times 10^{-4}\; T)$,but the direction is reversed (out of the plane of the paper) because the current flows in the opposite sense.