Ampere's Law and Conductor: Finding ∫B.dl for a Semicircle in the x-z Plane

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The discussion centers on applying Ampere's Law to a semicircular conductor carrying current, specifically finding the line integral ∫B.dl for a semicircle in the x-z plane. Participants clarify that Ampere's Law requires a complete circuit, prompting the suggestion to complete the circuit by adding a straight path from point B to A. The importance of superposition is emphasized, where the magnetic field contributions from both the semicircular and straight sections must be considered. The correct approach involves calculating the magnetic field using the Biot-Savart Law and integrating appropriately, while also addressing potential sign issues in the calculations. Ultimately, the discussion highlights the nuances of applying Ampere's Law to non-ideal circuits and the necessity of understanding the underlying principles.
  • #31
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?
 
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  • #32
If A and B are along the axis I would say the OP's a is correct but not b or c.

(I might as well benefit from this too!) :biggrin:
 
  • #33
rude man said:
If A and B are along the axis I would say the OP's a is correct but not b or c.

I wish TSny is a bit lenient in his marking :wink:. I somehow have the feeling that getting one answer correct should be enough to pass the test :-p.
 
  • #34
rude man said:
Are the points A and B along the axis of the integration circle?

Yes, I should have made that clear. A and B are still in the same location relative to the path of integration.
 
  • #35
Tanya Sharma said:
I wish TSny is a bit lenient in his marking :wink:. I somehow have the feeling that getting one answer correct should be enough to pass the test :-p.

Well, I'm retired. So, I refuse to do any more grading! :-p
But, if you were to use Tanya's suggested grading scheme, then you both passed. Tanya gets A+ on this one. (That is, her answers agree with mine, anyway.)
 
  • #36
Wow ! I can't believe I topped the class :-p

Have to say the Prof. is really cool :biggrin:.
 
  • #37
TSny said:
Well, I'm retired. So, I refuse to do any more grading! :-p
But, if you were to use Tanya's suggested grading scheme, then you both passed. Tanya gets A+ on this one. (That is, her answers agree with mine, anyway.)

Tanya's answers for b and c seem to have the wrong sign, otherwise I agree.

A downward current gives a positive B circulation as I understand it.
 
  • #38
Tanya Sharma said:
Wow ! I can't believe I topped the class :-p

Have to say the Prof. is really cool :biggrin:.

Congrats! And yes, we owe Dr. T. one - or actually several by now!
 

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