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 . I  A´ C A R Thanks to Dr. Waddill for the use of the diagram. C´ 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 . I see three “parts” to the wire. As usual, break the problem up into simpler parts. A’ to A A to C C to C’

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 . I  A´ C A R C´ 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 . For segment A’ to A: ds

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´ C A R C´ 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 . For segment C to C’: ds I

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´ C A R C´ Important technique, handy for homework and exams: The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths. O ds

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 . For segment A to C:  A´ C A R C´ O I ds

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

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

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