In Example 4.5, a current-carrying wire has straight segments meeting at the centre of a semicircular arc. The contribution of the straight segments to the field at the centre is:

1. The Biot–Savart contribution depends on $d\mathbf{l}\times\mathbf{r}$. 2. For the straight segments the element dl and the position vector r to the centre are parallel. 3. A cross product of parallel vectors is zero, so each such element gives $d\mathbf{B} = 0$. 4. Hence the straight segments do not contribute to the field at the centre. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.147_