1. I see three “parts” to the wire.
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . A´
I see three “parts” to the wire.
A’ to A A
A to C C´
C to C’ I  C R O
As usual, break the problem up into simpler parts.
Thanks to Dr. Waddill for the use of the diagram.

2. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . A´ I ds
For segment A’ to A: A C´  C R O

3. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . A´
For segment C to C’: A ds C´ I  C R O

4. Important technique, handy for homework and exams:
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle . A´
Important technique, handy for homework and exams: ds A
The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths. ds C´  C R O

5. Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .
Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . A´ A
For segment A to C: ds C´ I  C R O

6. A´ A ds C´ I  C
The integral of ds is just the arc length; just use that if you already know it. R O
We still need to provide the direction of the magnetic field.

7. Cross into . The direction is “into” the page, or .ds
If we use the standard xyz axes, the direction is C´ I  C x y z R O

8. Important technique, handy for exams:
Along path AC, ds is perpendicular to . A ds C´ I  C R O