Find the magnetic field at the ccentre of a current carrying conductor bent in the form of an arc subtending angle `theta` at its centre. Radius of the arc is R
Strategy: Let the arc lie in x-y plane with its centre at the origin. Consider a small current element `i vec(dl)` as shown. The field due to this element at the centre can be obtained by Biot-Savart law

Explanation

`dB = (mu_(0))/(4pi) (idl sin 90^(@))/(R^(2))` (`:' i vec(dl) and R` are perpendicular)
Now `dl = R d phi`
`:. dB = (mu_(0))/(4pi) (iR dphi)/(R^(2)) rArr dB = (mu_(0))/(4pi) (i)/(R) d phi`
The direction of field is outward perpendicular to plane of paper.
Total magnetic field
`B = int dB :. B = (mu_(0)i)/(4piR) int_(0)^(0) d phi = (mu_(0)i)/(4pi R) [phi]_(0)^(theta)`
`:. B = (mu_(0)i)/(4pi R) theta`