rude man
Science Advisor
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TSny said:Yes, everything looks good. I was misinterpreting how you were using the word "equivalent". I thought you were implying that the semicircle current (alone) and the D loop produce the same result for the integral of Bnet around the circular path.
What if you replaced the semicircular current from A to B with a current of a different shape, such as shown in the attachment?
Are the points A and B along the axis of the integration circle?