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. http://autosystempro.com/tag/generators/

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

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

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

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

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.

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

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